Inter 1st Year

Andhra Pradesh BIEAP AP Inter 1st Year Economics Study Material 3rd Lesson Theory of Demand Textbook Questions and Answers.

AP Inter 1st Year Economics Study Material 3rd Lesson Theory of Demand

Essay Questions

Question 1.
Explain the Law of Demand and examine Exceptions to it.
Answer:
Demand means a desire which is backed up by ability to buy and willingness to pay the price is called demand in Economics. Thus demand will be always to a price and time. Demand has the following features.

1. Desire for the commodity
2. Ability to buy the commodity
3. Willing to pay the price of commodity
4. Demand is always at a price
5. Demand is per unit of time i.e, per day, week etc.

Therefore the price demand may be expressed in the form of small equation.
D_x = f(P_x)
Price demand explains the relation between price and quantity demanded of a commodity. Price demand states that there is an inverse relationship between price and demand.

Law of demand : Marshall defines the law of demand as, “The amount demanded increases with a fall in price and diminishes with arise in price when other things remain the same”. So, the law of demand explains the inverse relationship between the price and quantity demanded of a commodity.

Demand schedule : It means a list of the quantities demanded at various prices in a given period of time in a market. An imaginary example given below.

The table shows that as the price falls to ₹ 1/- the quantity demanded 50 units, when price ₹ 5/- he is buying 10 units. So, there is inverse relationship between price and demand. Price is low demand will be high and price is high demand will be low. We can illustrate the above schedule in a diagram.

In the above diagram on X-axis demand is shown and price is on Y-axis. DD is the ad curve. Demand curves slopes downward from left to right.
Assumptions :

1. No change in the income of consumer
2. The taste and preferences consumers remain same.
3. The prices of related goods remain the same.
4. New substitutes are not discovered.
5. No expectation of future price changes.

Exceptions : In certain situations, more will be demanded at higher price and less will be demanded at a lower price. In such cases the demand curve slopes upward from left to right which is called an exceptional demand curve. This can be shown in the following diagram.

In the diagram when price increases from OP to OP₁, demand also increases from OQ to OQ₁. This is opposite to law of demand.
1) Giffen’s Paradox: This was stated by Sir Robert Giffen. He observed that poor people will demand more of inferior goods, if their prices raise. Inferior goods are known as Giffen goods.
Ex : Ragee, Jowar etc. He pointed out that in case of the English workers, the law of demand does not apply to bread. Giffen noticed that workers spend a major portion of their income on bread and only small portion on meat.

2) Veblen Effect (Prestigious goods) : This exception was stated by Veblen. Costly goods like diamonds and precious stones are called prestige goods or veblen goods. Generally rich people purchase those goods for the sake of prestige. Hence rich people may buy more such goods when their prices rise.

3) Speculation : When the price of a commodity rises the group of speculators expect that it will rise still further. Therefore, they buy more of that commodity. If they expect that there is a fall in price, the demand may not expand.
Ex : Shares in the stock market.

4) Illusion : Some times, consumer develop to false idea that a high priced good will have a better quality instead of low priced good. If the price of such good falls, demand decreases, which is contrary to the law of demand.

Question 2.
What is Demand Function ? What are the factors that determine the demand for a good ? [March 16]
Answer:
The functional relationship between the demand for a commodity and its various determinants may be explained mathematically in terms of a demand function.
D_x = f(P_x,P₁, ………… P_n, Y, T)
Where D_x = Demand for good X;
P_x = price of X;
P₁ …. P_n = Prices of substitutes and complementaries
Y = Income,
T = Taste of the consumer.
Determinants of demand :

1. Price of commodity: The demand for any good depends on its price, more will be demanded at lower price and vice-versa.
2. Prices of substitutes and complementaries : Demand is influenced by changes in price of related goods either substitutes or complementary goods.
   Ex: Increase in the price of coffee leads an increase in the demand for tea in the case of substitutes positive relation and complementaries negative relationship between price and demand.
3. Income of the consumer : Demand always changes with a change in the incomes of the people. When income increases the demand for several commodities increases and vice-versa.
4. Population : A change in the size and composition of population will effect the demand for certain goods like food grains, clothes etc.
5. Taste and preferences: A change in the taste and the fashions bring about a change in the demand for a commodity.
6. Technological changes: Due to economic progress technological changes the quantity the quality of goods available to the consumers increase.
   Ex : Demand for cell phones reduced the demand for landline phones.
7. Change in the weather : Demand for commodity may change due to change in a climatic condition.
   Ex : During summer demand for cool drinks, in winter demand for wollen clothes.
8. State of business : During the period of prosperity demand for commodities will expand and during depression demand will contract.

Question 3.
Explain the Concept of Demand and various types of Demand.
Answer:
The concept of demand has immerse significance in economics. In general language demand means a desire but in economics the desire backed up by ability to buy and willingness to pay the price.
Types of demands :
The demand may be classified into 3 types.

1. Price demand
2. Income demand
3. Cross demand

1) Price demand : Price demand explains the relation between price and quantity demanded by a commodity it shows the inverse relationship between price and demand when the other things like consumers income, taste etc., remains constant. It means the price falls demand extends and the price raises demand contracts. The price demand can be expressed D_x = f(P_x).
whereD_x = Demand for X commodity
P_x = Price of X
F = Function.

2) Income demand: It explains the relationship between consumers income and various quantities of various levels of income assuming other factors like price of goods, related goods, taste etc; remain the same. It means if income increases quantity demand increases and vice versa. This can be shown in the following form.
D_x = F(Y)
Where D_x = demand of X good;
Y = income of consumer.

3) Cross demand: It refers to change in demand for a commodity as a result of change in prices of its related commodities when other factors remains constant. This can be shown
D_x = f(P_y)
D_x = demand of X good;
P_y = Price of Y commodity;
F = function
Related goods are 2 types,

1. Substitutes : Goods which satisfy same want called substitutes.
   Ex : Coffee and tea. Here there is a positive relation between price and demand.
2. Complementary goods : These goods which satisfy the same want jointly.
   Ex : Car and petrol, shoes and sockes etc; are complementary goods.
   Here there exists inverse relation between complementary goods.

Question 4.
Explain the three Forms of Demand with suitable diagrams.
Answer:
The concept of demand has great significance in economics. In general language demand means a desire but in economics the desire backed up by ability to buy and willingness to pay the price.
Types of demand : The demand may be classified into three types.

1. Price demand
2. Income demand
3. Cross demand

1) Price demand : Price demand explains the relationship between price and quantity demanded of a commodity it shows the inverse relationship between price and demand when the other things like consumers income, taste etc., remains constant. It means the price falls demand extends and price raises demand contracts. The price demand can be expressed D_x = f(P_x)
Price demand can be explained with the help of demand schedule.

As price falls to ₹ 1/- the quantity demand is 50 units, when price of apple is ₹ 5/- he is buying 10 units. So, the table shows inverse relationship between price and demand. Price demand can be explained with the help of the demand curve.

On OX axis shows demand, OY axis shows price. We can obtain the demand curve ‘DD’ by joining all the points A, B, C, D, E which represents various quantities of demand at various prices. ‘DD’ is demand curve. It slopes downwards from left to right. It shows the inverse relationship between price and demand.

2) Income demand: It explains the relationship between consumers income and various quantities of various levels of income assuming other factors like price of goods, related goods, taste etc; remain the same. It means if income increases quantity demand increases and vice versa. This can be shown in the following form.
D_x = f(Y)
The functional relationship between income and demand may be inverse or direct depending on the nature of the commodity. This can be shown in the following table.

Superior goods : In case of superior goods quantity demanded will increase when there is an increase in the income of consumers.
In the diagram ‘X’ axis represents demand, OY axis represents income, YD represents the income demand curve. It showing positive slope whenever income increased from OY to OY₁; the demand of superior or normal goods increases from OQ to OQ₁.
This may happens in case of Veblen goods.

Inferior goods : On the contrary quantity demanded of inferior goods decreases with the increase in incomes of consumers.
In the diagram on ‘OX’ axis measures demand and OY axis represents income of the consumer. When the consumer income increases from OY to OY₁ the demand for a commodity decreases from OQ to OQ₁ So the YD’ curve is negative sloping.

3) Cross demand: Cross demand refers to the relationship between any two goods which are either complementary to each other or substitute for each other. It explains the functional relationship between the price of one commodity and quantity demanded of another commodity is called cross demand.
D_x = f(Py)
Where D_x – demand for ‘X’ commodity
P_y Price of Y’ commodity
f = function
Substitutes : The goods which satisfy the same want are called substitutes.
Ex : Tea and coffee; pepsi and coca-cola etc. In the case of substitutes, the demand curve has a positive slope.

In the diagram ‘OX’ axis represents demand tea and OY axis represents price of coffee. Increase in the price of coffee from OY to OY₂ leads to increase in the demand of tea from OQ to OQ₂.

Complementaries : In case of complementary goods, with the increase in price of one commodity, the quantity demanded of another commodity falls.
Ex : Car and Petrol. Hence the demand curve of these goods slopes downward to the right.

In the diagram if price of car decreases from OP to OP₂ the quantity demand of petrol increases from OQ to OQ₂. So cross demand i.e., CD curve is downward slopes.

Question 5.
Define the Concept of Elasticity of Demand and explain the concepts of price, income and cross elasticity of demand.
Answer:
The concept of elasticity demand was first introduced by Cournot and Mill. Later it was developed in a scientific manner by Marshall. Elasticity of demand means the degree of sensitiveness or responsiveness of demand to a change in its price.

According to Marshall “The elasticity of demand in a market is great or small according as the amount demanded increases much or little for a given fall in price”.

The concept of elasticity of demand explains how much or to what extent a change in any one of the independent variables leads to change in the dependent variable.
There are three kinds of elasticity of demand.

1. Price elasticity of demand
2. Income elasticity of demand
3. Cross elasticity of demand

1) Price Elasticity of Demand : Alfred Marshall developed the concept of price elasticity of demand. Price elasticity of demand is generally defined as the degree of responsiveness or sensitiveness of demand for a commodity to the changes in its price. Thus price elasticity of demand is the ratio of percentage change in quantity demanded of a good and percentage change in its price. The following formula to measure price elasticity of demand.
E_p = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in price }}\)
= \(\frac{\text { Change in quantity demanded }}{\text { Original quantity demanded }} \times \frac{\text { Original price }}{\text { Change in price }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{p}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{p}} \times \frac{\mathrm{p}}{\mathrm{q}}\)
where q = quantity;
p = price;
∆q = change in quantity demanded;
∆p = change in price.
There are five kinds of price elasticity of demand. They are :

1. Perfectly Elastic demand (Ed = α)
2. Perfectly Inelastic demand (Ed = 0)
3. Unitary Elastic demand (Ed = 1)
4. Relatively Elastic demand (Ed > 1)
5. Relatively Inelastic demand (Ed < 1)

2) Income Elasticity of Demand : Income elasticity of demand shows the degree of responsiveness of quantity demanded of a commodity to a change in the income of the consumer, other things remain constant.
E_y = \(=\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change inconsumer’s income }}\)
= \(=\frac{\text { Change in quantity demanded }}{\text { Original quantity demanded }} \times \frac{\text { Original income }}{\text { Change in income }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{y}}{\Delta \mathrm{y}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{y}} \times \frac{\mathrm{y}}{\mathrm{q}}\)
where q = Quantity;
y = income;
∆Q = change in quantity demanded;
∆y = change in income.
Income elasticity of demand will be positive in case of superior goods like milk and meat and negative in case of inferior goods like porridge and broken rice.

3) Cross Elasticity of Demand : Cross elasticity of demand refers to change in the quantity demanded of one good in response to change in the price of related good, other things remaining constant. There are certain goods whose demand depend not only to their price but also on the prices of related goods.
Ec = \(\frac{\text { Percentage change in quantity demanded of X }}{\text { Percentage change in the price of } Y}\)
Ec = \(\frac{\Delta \mathrm{Qx}}{\mathrm{Qx}} \div \frac{\Delta \mathrm{Py}}{\mathrm{Py}}=\frac{\Delta \mathrm{Qx}}{\mathrm{Qx}} \times \frac{\mathrm{Py}}{\Delta \mathrm{Py}}=\frac{\Delta \mathrm{Qx}}{\Delta \mathrm{Px}} \times \frac{\mathrm{Py}}{\mathrm{Px}}\)
Where Q(x) = Quantity demanded for X; P(y) = price of commodity (Y), ∆Q(x) = change in quantity demanded of X commodity, ∆P(y) = change in price of commodity Y.
Substitute goods like tea and coffee have positive cross elasticity demand where as complementary goods like shoes and socks have negative cross elasticity of demand.

Question 6.
What is Price Elasticity of Demand ? Explain the various types of Price Elasticity of Demand.
Answer:
Alfred Marshall developed the concept of price elasticity of demand. Price elasticity measures, other things remaining constant, change in the demanded of a good in response to a change in its price. Thus price elasticity of demand is the ratio of percentage change in quantity demanded of a good and percentage change in its price. Price elasticity can be written as stated below.
Ep = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in price }}\)
= \(\frac{\text { Change in demand }}{\text { Original demand }} \times \frac{\text { Original price }}{\text { Change in price }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{p}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{p}} \times \frac{\mathrm{p}}{\mathrm{q}}\)
Where q = quantity; p = price; ∆q = change in demand; ∆p = change in price
Types of price elasticity of demand : Based on numerical value, price elasticity of demand can be five types.

1. Perfectly Elastic demand (Ed = ∞)
2. Perfectly Inelastic demand (Ed = 0)
3. Unitary Elastic demand (Ed = 1)
4. Relatively Elastic demand (Ed > 1)
5. Relatively Inelastic demand (Ed < 1)

1) Perfectly Elastic demand : It is also known as “infinite elastic demand”. A small change in price leads to an infinite change in demand is called perfectly elastic demand. It is horizontal straight line to ‘X’ axis. The numerical value of perfectly elastic demand is infinite (Ed = ∞). It can be shown in the diagram.

In the diagram, Ed = \(\frac{\mathrm{OQQ}_1}{\mathrm{OQ}} \div \frac{\mathrm{O}}{\mathrm{OP}}\)
= \(\frac{\mathrm{QQ}_1}{\mathrm{OQ}} \times \frac{\mathrm{OP}}{\mathrm{O}}\) = ∞

2) Perfectly Inelastic demand: It is also known as “zero elastic demand”. In this case even a great rise ‘or fall in price does not lead to any change in quantity demanded is known as perfectly inelastic demand. The ‘demand curve will be vertical to the Y axis. The numerical value is ‘OE. This can be shown in the following diagram.

In the diagram, Ed = \(\frac{\text { Zero }}{\mathrm{OQ}} \div \frac{\mathrm{PP}_1}{\mathrm{OP}}\) = 0
∴ Ed = 0

3) Unitary Elastic demand: The percentage change in price leads to same percentage change in demand is called unitary elastic demand. In this case the elasticity of demand is equal to one. The shape of demand curve is “Rectangular Hyperbola”. This can be shown in the following.

In the diagram, Ed = OP₁Q₁ = OPQ
(or) OQ₁ = PP₁
∴ Ed = 1

4) Relatively Elastic demand: When a percentage change in price leads to more than percentage change in quantity demand is called relatively elastic demand. In this case the numerical value of Ed is greater than one (Ed > 1)

In the diagram, Ed = OQ₁ > PP₁
∴ Ed > 1

5) Relatively Inelastic demand : When the percentage change in price leads to a less than percentage change in quantity demand is called relatively inelastic demand. Here the numerical value is less than one (Ed < 1) . This can be shown to following diagram.

In the diagram, Ed = QQ₁ < PP₁
∴ Ed < 1

Question 7.
Explain the three methods of Measuring Price Elasticity of Demand.
Answer:
The concept of elasticity of demand is one of the original contributions of Dr.Marshall. The concepts of elasticity of demand clearly explains ‘how much’ demand increases due to a certain fall in price and ‘how much’ demand decreases due to certain rise in price.

According to Mrs. Joan Robinson, “The elasticity of demand at any price or at any output is the proportional change of amount purchased in response to a small change in price, divided by the proportional change in price”.

Methods of measurements of Price Elasticity of demand :
The Elasticity of demand can be measured mainly in three ways.

1. Total outlay (or) Expenditure method
2. Point method and
3. Arc method.

1) Total outlay (or) Expenditure method : This method was introduced by Alfred Marshall. Price elasticity of demand can be measured on the basis of change in the total outlay due to a change in the price of a commodity. This method helps us to compare the total expenditure from a buyer or total revenue from the seller before and after the change in price.
Total outlay = Price × Quantity demanded
According to this method the price elasticity of demand is expressed in three forms, they are elastic demand, unitary elastic and inelastic demand. This can be explained with the help of table.

In this table shows that

1. If the total expenditure increases due to a fall in price is known as relatively elastic demand.
2. If total expenditure remains constant even the price falls is known as unitary elastic demand.
3. If the total expenditure decreases due to a fall in price is known as relatively inelastic demand.

2) Point method : This method is introduced by Marshall. In this method elasticity of demand is measured at a point on the demand curve. So, this method is also called as “geometrical method”. In this method to measure elasticity at a point on demand curve the following formula is applied.
Ed = \(\frac{\text { The distance from the point to the } \mathrm{X} \text {-axis }}{\text { The distance from the point to the } \mathrm{Y} \text {-axis }}\)
In the below diagram ‘AE’ is straight line demand curve. Which is 10 cm length. Applying the formula we get
Ed = 1 Ed < 1 Ed = 0

Elasticity at point A, Ed = \(\frac{\mathrm{AE}}{\mathrm{A}}=\frac{10}{0}\) = 0
Elasticity at point B, Ed = \(\frac{\mathrm{BE}}{\mathrm{BA}}=\frac{7.5}{2.5}\) = 3 > 1
Elasticity at point C, Ed = \(\frac{\mathrm{CE}}{\mathrm{CA}}=\frac{5}{5}\) = 1
Elasticity at point D, Ed = \(\frac{\mathrm{DE}}{\mathrm{DA}}=\frac{1}{3}\) = < 1
Elasticity at point E, Ed = \(\frac{\mathrm{E}}{\mathrm{DA}}=\frac{0}{10}\) = 0
if the demand curve is non-linear. It means if the demand curve is not straight line will be drawn at the point on the demand curve where to measure elasticity.

In the diagram at C point where Elasticity of demand will be equal to \(\frac{\mathrm{CB}}{\mathrm{CA}}\).

3) Arc method : The word ‘ArC means a portion or a segment of a demand curve. In this method mid points between the old and new price and quantities demanded are used. This method used to known small changes in price. This method is also known as ‘Average Elasticity of demand”. This method studies a segment of the demand curve between two points the formula for measuring elasticity is given below.

Suppose we take price of the commodity is ₹ 4/- demand is 300 units. If price falls ₹ 3/ – demand increases 400 units.
Then applying above formula Arc elasticity of demand is
\(\frac{100}{300+400} \div \frac{1}{4+3}=\frac{100}{700} \div \frac{1}{7}\)
= \(\frac{100}{700} \times \frac{7}{1}=\frac{700}{700}\) = 1
∴ Ed = 1

Question 8.
What are the factors that determine Price Elasticity of Demand ?
Answer:
The term elasticity refers to the measure of extent of relationship between two related variables. The elasticity of demand is the measure of responsiveness or sensitiveness of demand for a commodity to the change in it demand.
Determinants of Elasticity of Demand : The. elasticity of demand varies from commodity to commodity.
1) Nature of commodity: Commodities can be grouped as necessaries, comforts and luxuries. In case of necessaries, the elasticity of demand will be inelastic.
Ex : Rice, salt etc. On the other hand in case of luxuries the demand will be more elastic.
Ex : Diamonds & gold etc.

2) Availability of substitutes : Prices of substitutes influence the demand for a com-modity upto a certain extent. The closer the substitute, the greater the elasticity of demand for the commodity. For Ex : Cool drinks, soaps etc. but in case of non-availability of substitutes the elasticity of demand will be low.

3) Complementary goods: Price elasticity for a good is also depends on the nature of price elasticity of jointly demand goods. If the demand for car is elastic, then the demand for petrol will also be elastic.

4) Multiple uses of the commodity : The wider the range of alternative uses of a product, the higher the elasticity of demand and vice-versa.
Ex : Coal and electricity have multiple uses and will have elastic demand.

5) Proportion of income spent: If proportion of income spent on commodity is very small, its demand will be less elastic and vice-versa.

6) Period of time : In the long run, demand will be more elastic. Longer the time period considered, greater will be the possibility of substitution. For a cheaper good.
Ex : If the price of petrol increases in the short run, it may not be possible to replace the petrol engines with diesel engines but in the long run it can be possible.

7) Price level: Goods which are in very high range or in very low range have inelastic demand but it is high at moderate price.

8) Habit : The demand for a commodity to which the consumer is accustomed is generally inelastic.
Ex : Tobacco and alcohol.

9) Income group : The demand of higher income groups will be inelastic as they do not bother about price changes. On the other hand, the demand of middle and lower income groups will be elastic.

10) Postponement of purchase: The demand for a commodity, the consumption of which can be postponed is more eiastic than that of the use of the commodity cannot be postpone the purchases of such goods like life saving medicines.

Question 9.
Explain importance of Elasticity of Demand.
Answer:
The term elasticity refers to the measure of extent of relationship between two related variable. The elasticity of demand is the measure of responsiveness or sensitiveness of demand for a commodity to the change in it demand.
Importance :
1) Useful to monopolist: Monopolist should study the elasticity of demand for his commodity before fixing up the price. Monopolist will fix a higher price when the commodity inelastic demand, but he will fix a lower price when the commodity has elastic demand.

2) Useful to joint products : It is useful in the price fixation of joint goods like meat and fur. In such case the producer wall be guided by elasticity of demand to fix the prices of the joint goods.

3) Useful to the government: The concept of elasticity can be used in formulating government policies relating public utility service like Railways, drug industry etc.

4) Useful to international trade : In calculating the terms of trade both countries have to take into account the mutual elasticities of demand for the products.

5) Useful to finance minister : The concept of elasticity is useful to the Finance Minister in imposing taxes on goods. The finance minister studies the elasticity of commodities before he imposes new taxes or enhances old taxes.

6) Useful to Management: Before asking for higher wages trade union leaders must know the elasticity of demand of the product produced by them. Trade union leaders may demand for higher wages only when the goods produced by them have inelastic demand.

7) Useful to producers : Volume of goods must be produced in accordance with demand for the commodity. Whenever the demand for the commodity is inelastic, the producer will produce more commodities to take advantage of higher price. So, it helps in determining the volume of output.

Short Answer Questions

Question 1.
Explain the Law of Demand or Price Demand.
Answer:
Demand means a desire which is backed up by ability to buy and willingness to pay the price is called demand in Economics. Thus demand will be always to a price and time. Demand has the following features.

1. Desire for the commodity
2. Ability to buy the commodity
3. Willing to pay the price of commodity
4. Demand is always at a price ‘
5. Demand is per unit of time i.e, per day, week etc.

Therefore the price demand may be expressed in the form of small equation.
D_x = f(P_x)
Price demand explains the relation between price and quantity demanded of a commodity. Price demand states that there is an inverse relationship between price and demand.

Law of demand : Marshall defines the law of demand as, “The amount demanded increases with a fall in price and diminishes with arise in price when other things remain the same”. So, the law of demand explains the inverse relationship between the price and quantity demanded of a commodity.

Demand schedule : It means a list of the quantities demanded at various prices in a given period of time in a market. An imaginary example given below.

The table shows that as the price falls to ₹ 1/- the quantity demanded 50 units, when price ₹ 5/- he is buying 10 units. So, there is inverse relationship between price and demand. Price is low demand will be high and price is high demand will be low. We can illustrate the above schedule in a diagram.

In the above diagram on X-axis demand is shown and price is on Y-axis. DD is the demand curve. Demand curves slopes downward from left to right.
Assumptions :

1. No change in the income of consumer
2. The taste and preferences consumers remain same.
3. The prices of related goods remain the same.
4. New substitutes are not discovered.
5. No expectation of future price changes.

Exceptions : In certain situations, more will be demanded at higher price and less will be demanded at a lower price. In such cases the demand curve slopes upward from left to right which is called an exceptional demand curve. This can be shown in the following diagram.

In the diagram when price increases from OP to OP₁, demand also increases from OQ to OQ₁. This is opposite to law of demand.

Question 2.
Explain the Exceptions to the Law of Demand or Price Demand.
Answer:
In Economics demand means a desire which is backed up by ability to buy and willingness to pay the price. Thus demand will be always at a price and time.

According to Marshal “The amount demanded increases with a fall in price and diminishes with rise in price when other things remain the same”.

Exceptions : In certain situations, more will be demanded at higher price and less will be demanded at a lower price. In such cases the demand curve slopes upward from left to right which is called an exceptional demand curve. This can be shown in the following diagram.

In the diagram when price increases from OP to OP₁, demand also increases from OQ to OQ₁. This is opposite to law of demand.

Question 3.
Why a Demand Curve has a negative slope or why Demand Curve slopes downward ? [March 18]
Answer:
According to Marshall “The amount demanded increases with a fall in price and diminishes with a rise in price when other things remain the same”.

The law of demand explains inverse relationship between the price and quantity demanded of a commodity. Therefore the demand curve slopes downward from left to right.

There are some other reasons also responsible for downward sloping demand curve.
1) Old and new buyers : If the price of a good falls, the real income of the old buyers will increase. Hence the demand for the good will increase. In the same way, the fall in price attracts new buyers and will be able to built after a fall in its price. So the demand curve slopes downards from left to right.

2) Income effect: Fall in price of commodity the real income of its consumers increase. The increase in real income encourages demand for the commodity with reduced price. The increase in demand on account of increased in real income is known as income effect.

3) Substitution effect: When the price of commodity falls, it will become relatively cheaper than its substitutes. The increase in demand on account of increased in real income is known as income effect.

4) Law of diminishing marginal utility: According to this law, if consumer goes on consuming more units of the commodity, the additional utility goes on diminishing. Therefore, the consumer prefers to buy at a lower price. As a result the demand curve has a negative slope.

Question 4.
Explain the concept of Income Demand.
Answer:
Income demand : It explains the relationship between consumers income and various quantities of various levels of income assuming other factors like price of goods, related goods, taste etc; remain the same. It means if income increases quantity demand increases and vice versa. This can be shown in the following form.
D_x = f(Y)
The functional relationship between income and demand may be inverse or direct depending on the nature of the commodity. This can be shown in the following table.

Superior goods : In case of superior goods quantity demanded will increase when there is an increase in the income of consumers.

In the diagram ‘X’ axis represents demand, OY axis represents income, YD represents the income demand curve. It showing positive slope whenever income increased from OY to OY₁ the demand of superior or normal goods increases from OQ to OQ₁.
This may happens in case of Veblen goods.

Inferior goods : On the contrary quantity demanded of inferior goods decreases with the increase in incomes of consumers.

In the diagram on ‘OX’ axis measures demand and OY axis represents income of the consumer. When the consumer income increases from OY to OY₁ the demand for a commodity decreases from OQ to OQ₁ So the YD’ curve is negative sloping.

Question 5.
Explain the concept of Cross Demand.
Answer:
Cross demand: Cross demand refers to the relationship between any two goods which are either complementary to each other or substitute for each other. It explains the functional relationship between the price of one commodity and quantity demanded of another commodity is called cross demand.
D_x = f(P_y)
Where D_x = demand for ‘X’ commodity
P_x = Price of ‘y’ commodity
f = function
Substitutes : The goods which satisfy the same want are called substitutes.
Ex : Tea and coffee; pepsi and coca-cola etc. In the case of substitutes, the demand curve has a positive slope.

In the diagram ‘OX’ axis represents demand of tea and OY axis represents price of coffee. Increase in the price of coffee from OY to OY₂ leads to increase in the demand of tea from OQ to OQ₂.

Complementaries: In case of complementary goods, with the increase in price of one commodity, the quantity demanded of another commodity falls.
Ex : Car and Petrol. Hence the demand curve of these goods slopes downward to the right.

In the diagram if price of car decreases from OP to OP₂ the quantity demand of petrol increases from OQ to OQ₂. So cross demand i.e., CD curve is down ward slopes.

Question 6.
What are the factors that determine Demand ? [March 18, 17]
Answer:
The functional relationship between the demand for a commodity and its various determinants may be explained mathematically in terms of a demand function.
D_x = f(P_x,P₁, ………… P_n, Y, T)
Where D_x = Demand for good X;
P_x = price of X;
P₁ …. P_n = Prices of substitutes and complementaries
Y = Income,
T = Taste of the consumer.
Determinants of demand :
1) Price of commodity: The demand for any good depends on its price, more will be demanded at lower price and vice-versa.

2) Prices of substitutes and complementaries : Demand is influenced by changes in price of related goods either substitutes or complementary goods.
Ex: Increase in the price of coffee leads an increase in the demand for tea in the case of substitutes positive relation and complementaries negative relationship between price and demand.

3) Income of the consumer : Demand always changes with a change in the incomes of the people. When income increases the demand for several commodities increases and vice-versa.

4) Population : A change in the size and composition of population will effect the demand for certain goods like food grains, clothes etc.

5) Taste and preferences: A change in the taste and the fashions bring about a change in the demand for a commodity.

6) Technological changes: Due to economic progress technological changes the quantity the quality of goods available to the consumers increase.
Ex : Demand for cell phones reduced the demand for landline phones.

7) Change in the weather : Demand for commodity may change due to change in a climatic condition.
Ex : During summer demand for cool drinks, in winter demand for wollen clothes.

8) State of business : During the period of prosperity demand for commodities will expand and during depression demand will contract.

Question 7.
What is Elasticity of Demand ?
Answer:
In Economic theory, the concept of elasticity of demand has a significant role. Elasticity of demand means the percentage change in quantity demanded in response to the percentage change in one of the variables on which demand depends.
Elasticity of demand changes from person to person, place to place, time to time and one commodity to another.
Accoridng to Marshall “The elasticity of demand in a market is great or small according as the amount demanded increases much or little for a given fall in price”.
The concept of elasticity of demand explains how much or to what extent a change in any one of the independent variables leads to a change in the dependent variable.
There are three kinds of elasticity of demand.

1. Price Elasticity of demand
2. Income Elasticity of demand .
3. Cross Elasticity of demand

Question 8.
Explain the three types of Elasticity of Demand.
Answer:
The concept of elasticity demand was first introduced by Cournot and Mill. Later it was developed in a scientific manner by Marshall. Elasticity of demand means the degree of sensitiveness or responsiveness of demand to a change in its price.

According to Marshall “The elasticity of demand in a market is great or small according as the amount demanded increases much or little for a given fall in price”.

The concept of elasticity of demand explains how much or to what extent a change in any one of the independent variables leads to change in the dependent variable.
There are three kinds of elasticity of demand.

1. Price elasticity of demand
2. Income elasticity of demand
3. Cross elasticity of demand

1) Price Elasticity of Demand : Alfred Marshall developed the concept of price elasticity of demand. Price elasticity of demand is generally defined as the degree of responsiveness or sensitiveness of demand for a commodity to the changes in its price. Thus price elasticity of demand is the ratio of percentage change in quantity demanded of a good and percentage change in its price. The following formula to measure price elasticity of demand.
E_p = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in price }}\)
= \(\frac{\text { Change in quantity demanded }}{\text { Original quantity demanded }} \times \frac{\text { Original price }}{\text { Change in price }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{p}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{p}} \times \frac{\mathrm{p}}{\mathrm{q}}\)
where q = quantity;
p = price;
∆q = change in quantity demanded;
∆p = change in price.
There are five kinds of price elasticity of demand. They are :

1. Perfectly Elastic demand (Ed = α)
2. Perfectly Inelastic demand (Ed = 0)
3. Unitary Elastic demand (Ed = 1)
4. Relatively Elastic demand (Ed > 1)
5. Relatively Inelastic demand (Ed < 1)

2) Income Elasticity of Demand : Income elasticity of demand shows the degree of responsiveness of quantity demanded of a commodity to a change in the income of the consumer, other things remain constant.
E_y = \(=\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change inconsumer’s income }}\)
= \(=\frac{\text { Change in quantity demanded }}{\text { Original quantity demanded }} \times \frac{\text { Original income }}{\text { Change in income }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{y}}{\Delta \mathrm{y}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{y}} \times \frac{\mathrm{y}}{\mathrm{q}}\)
where q = Quantity;
y = income;
∆Q = change in quantity demanded;
∆y = change in income.
Income elasticity of demand will be positive in case of superior goods like milk and meat and negative in case of inferior goods like porridge and broken rice.

3) Cross Elasticity of Demand : Cross elasticity of demand refers to change in the quantity demanded of one good in response to change in the price of related good, other things remaining constant. There are certain goods whose demand depend not only to their price but also on the prices of related goods.
Ec = \(\frac{\text { Percentage change in quantity demanded of X }}{\text { Percentage change in the price of } Y}\)
Ec = \(\frac{\Delta \mathrm{Qx}}{\mathrm{Qx}} \div \frac{\Delta \mathrm{Py}}{\mathrm{Py}}=\frac{\Delta \mathrm{Qx}}{\mathrm{Qx}} \times \frac{\mathrm{Py}}{\Delta \mathrm{Py}}=\frac{\Delta \mathrm{Qx}}{\Delta \mathrm{Px}} \times \frac{\mathrm{Py}}{\mathrm{Px}}\)
Where Q(x) = Quantity demanded for X; P(y) = price of commodity (Y), ∆Q(x) = change in quantity demanded of X commodity, ∆P(y) = change in price of commodity Y.
Substitute goods like tea and coffee have positive cross elasticity demand where as complementary goods like shoes and socks have negative cross elasticity of demand.

Quantity 9.
Define Price Elasticity of Demand. Explain briefly various types of Price Elasticity of Demand.
Answer:
Alfred Marshall developed the concept of price elasticity of demand. Price elasticity measures, other things remaining constant, change in the demanded of a good in response to a change in its price. Thus price elasticity of demand is the ratio of percentage change in quantity demanded of a good and percentage change in its price. Price elasticity can be written as stated below.
Ep = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in price }}\)
= \(\frac{\text { Change in demand }}{\text { Original demand }} \times \frac{\text { Original price }}{\text { Change in price }}\)
= \(\frac{\Delta \mathrm{q}}{\mathrm{q}} \times \frac{\mathrm{p}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{p}} \times \frac{\mathrm{p}}{\mathrm{q}}\)
Where q = quantity; p = price; ∆q = change in demand; ∆p = change in price
Types of price elasticity of demand : Based on numerical value, price elasticity of demand can be five types.

1. Perfectly Elastic demand (Ed = ∞)
2. Perfectly Inelastic demand (Ed = 0)
3. Unitary Elastic demand (Ed = 1)
4. Relatively Elastic demand (Ed > 1)
5. Relatively Inelastic demand (Ed < 1)

1) Perfectly Elastic demand : It is also known as “infinite elastic demand”. A small change in price leads to an infinite change in demand is called perfectly elastic demand. It is horizontal straight line to ‘X’ axis. The numerical value of perfectly elastic demand is infinite (Ed = ∞). It can be shown in the diagram.

In the diagram, Ed = \(\frac{\mathrm{OQQ}_1}{\mathrm{OQ}} \div \frac{\mathrm{O}}{\mathrm{OP}}\)
= \(\frac{\mathrm{QQ}_1}{\mathrm{OQ}} \times \frac{\mathrm{OP}}{\mathrm{O}}\) = ∞

2) Perfectly Inelastic demand: It is also known as “zero elastic demand”. In this case even a great rise ‘or fall in price does not lead to any change in quantity demanded is known as perfectly inelastic demand. The ‘demand curve will be vertical to the Y axis. The numerical value is ‘OE. This can be shown in the following diagram.

In the diagram, Ed = \(\frac{\text { Zero }}{\mathrm{OQ}} \div \frac{\mathrm{PP}_1}{\mathrm{OP}}\) = 0
∴ Ed = 0

3) Unitary Elastic demand: The percentage change in price leads to same percentage change in demand is called unitary elastic demand. In this case the elasticity of demand is equal to one. The shape of demand curve is “Rectangular Hyperbola”. This can be shown in the following.

In the diagram, Ed = OP₁Q₁ = OPQ
(or) OQ₁ = PP₁
∴ Ed = 1

4) Relatively Elastic demand: When a percentage change in price leads to more than percentage change in quantity demand is called relatively elastic demand. In this case the numerical value of Ed is greater than one (Ed > 1)

In the diagram, Ed = OQ₁ > PP₁
∴ Ed > 1

5) Relatively Inelastic demand : When the percentage change in price leads to a less than percentage change in quantity demand is called relatively inelastic demand. Here the numerical value is less than one (Ed < 1) . This can be shown to following diagram.

In the diagram, Ed = QQ₁ < PP₁
∴ Ed < 1

Question 10.
Explain the Total Outaly method of Measuring Elasticity of Demand.
Answer:
Total Outlay (or) Expenditure method : This method was introduced by Alfred Marshall. Price elasticity of demand can be measured on the basis of change in the total outlay due to a change in the price of a commodity. This method helps us to compare the total expenditure from a buyer or total revenue from the seller before and after the change in price.
Total outlay = Price × Quantity demanded
According to this method the price elasticity of demand is expressed in three forms, they are elastic demand, unitary elastic and inelastic demand. This can be explained with the help of table.


In this table shows that

1. If the total expenditure increases due to a fall in price is known as relatively elastic demand.
2. If total expenditure remains constant even the price falls is known as unitary elastic demand.
3. If the total expenditure decreases due to a fall in price is known as relatively inelastic demand.

In the diagram on ‘OX’ axis measure total expenditure and ‘OY axis measures price. The total outlay curve AD is shown in three parts i.e., A to B; B to C and C to D.

Question 11.
Explain the Point method of Measuring Price Elasticity of Demand or How do you measure Elasticity of Demand on straight line Demand Curve ?
Answer:
Point method : This method is introduced by Marshall. In this method elasticity of demand is measured at a point on the demand curve. So, this method is also called as “geometrical method”. In this method to measure elasticity at a point on demand curve the following formula is applied.
Ed = \(\frac{\text { The distance from the point to the } \mathrm{X} \text {-axis }}{\text { The distance from the point to the } \mathrm{Y} \text {-axis }}\)
In the below diagram ‘AE’ is straight line demand curve. Which is 10 cm length. Applying the formula we get
Ed = 1 Ed < 1 Ed = 0

Elasticity at point A, Ed = \(\frac{\mathrm{AE}}{\mathrm{A}}=\frac{10}{0}\) = 0
Elasticity at point B, Ed = \(\frac{\mathrm{BE}}{\mathrm{BA}}=\frac{7.5}{2.5}\) = 3 > 1
Elasticity at point C, Ed = \(\frac{\mathrm{CE}}{\mathrm{CA}}=\frac{5}{5}\) = 1
Elasticity at point D, Ed = \(\frac{\mathrm{DE}}{\mathrm{DA}}=\frac{1}{3}\) = < 1
Elasticity at point E, Ed = \(\frac{\mathrm{E}}{\mathrm{DA}}=\frac{0}{10}\) = 0
If the demand curve is non-linear. It means if the demand curve is not straight line will be drawn at the point on the demand curve where to measure elasticity.

In the diagram at C point where Elasticity of demand will be equal to \(\frac{\mathrm{CB}}{\mathrm{CA}}\).

Question 12.
What are the basic determinants of Elasticity of Demand ?
Answer:
The term elasticity refers to the measure of extent of relationship between two related variables. The elasticity of demand is the measure of responsiveness or sensitiveness of demand for a commodity to the change in it demand.
Determinants of Elasticity of Demand : The. elasticity of demand varies from commodity to commodity.
1) Nature of commodity: Commodities can be grouped as necessaries, comforts and luxuries. In case of necessaries, the elasticity of demand will be inelastic.
Ex : Rice, salt etc. On the other hand in case of luxuries the demand will be more elastic.
Ex : Diamonds & gold etc.

2) Availability of substitutes : Prices of substitutes influence the demand for a com-modity upto a certain extent. The closer the substitute, the greater the elasticity of demand for the commodity. For Ex : Cool drinks, soaps etc. but in case of non-availability of substitutes the elasticity of demand will be low.

3) Complementary goods: Price elasticity for a good is also depends on the nature of price elasticity of jointly demand goods. If the demand for car is elastic, then the demand for petrol will also be elastic.

4) Multiple uses of the commodity : The wider the range of alternative uses of a product, the higher the elasticity of demand and vice-versa.
Ex : Coal and electricity have multiple uses and will have elastic demand.

5) Proportion of income spent: If proportion of income spent on commodity is very small, its demand will be less elastic and vice-versa.

6) Period of time : In the long run, demand will be more elastic. Longer the time period considered, greater will be the possibility of substitution. For a cheaper good.
Ex : If the price of petrol increases in the short run, it may not be possible to replace the petrol engines with diesel engines but in the long run it can be possible.

7) Price level: Goods which are in very high range or in very low range have inelastic demand but it is high at moderate price.

8) Habit : The demand for a commodity to which the consumer is accustomed is generally inelastic.
Ex : Tobacco and alcohol.

9) Income group : The demand of higher income groups will be inelastic as they do not bother about price changes. On the other hand, the demand of middle and lower income groups will be elastic.

10) Postponement of purchase: The demand for a commodity, the consumption of which can be postponed is more eiastic than that of the use of the commodity cannot be postpone the purchases of such goods like life saving medicines.

Question 13.
Explain the importance of the Concept of Elasticity of Demand.
Answer:
The term elasticity refers to the measure of extent of relationship between two related variable. The elasticity of demand is the measure of responsiveness or sensitiveness of demand for a commodity to the change in it demand.
Importance :
1) Useful to monopolist: Monopolist should study the elasticity of demand for his commodity before fixing up the price. Monopolist will fix a higher price when the commodity inelastic demand, but he will fix a lower price when the commodity has elastic demand.

2) Useful to joint products : It is useful in the price fixation of joint goods like meat and fur. In such case the producer wall be guided by elasticity of demand to fix the prices of the joint goods.

3) Useful to the government: The concept of elasticity can be used in formulating government policies relating public utility service like Railways, drug industry etc.

4) Useful to international trade : In calculating the terms of trade both countries have to take into account the mutual elasticities of demand for the products.

5) Useful to finance minister : The concept of elasticity is useful to the Finance Minister in imposing taxes on goods. The finance minister studies the elasticity of commodities before he imposes new taxes or enhances old taxes.

6) Useful to Management: Before asking for higher wages trade union leaders must know the elasticity of demand of the product produced by them. Trade union leaders may demand for higher wages only when the goods produced by them have inelastic demand.

7) Useful to producers : Volume of goods must be produced in accordance with demand for the commodity. Whenever the demand for the commodity is inelastic, the producer will produce more commodities to take advantage of higher price. So, it helps in determining the volume of output.

Very Short Answer Question

Question 1.
Demand
Answer:
The desire backed up by willingness and ability to pay a sum of money for some quantity of a good or service.

Question 2.
Demand Schedule [March 17]
Answer:
It shows the functional relationship between the quantity of commodity demanded and its price. The demand schedule may be two types.

1. Individual demand schedule
2. Market demand schedule

Question 3.
Individual Demand Schedule
Answer:
It explains the relationship between various quantities purchased at various prices by a single consumer in the market.

Question 4.
Market Demand Schedule
Answer:
It shows the total demand for a group at a particular time at different prices in the market.

Question 5.
Demand Function
Answer:
Demand function shows the functional relationship between quantity demanded at various factors that determine the demand for a commodity. It can be expressed as follows.
D_x = f(P_x, P₁, ………… P_n, Y, T)
Where
D_x = Demand for good X
P_x = price of X
P₁ …. P_n = Prices of substitutes and complementary
Y = Income of consumer
T = Tastes
f = Functional relationship

Question 6.
Giffen’s Paradox (or) Giffen Goods [March 18, 16]
Answer:
It means necessary goods Sir Robert Giffen in mid 19th century observed that the low paid workers in England purchased more bread when its price increase by decrease in the purchase of meat. The increase in demand for bread when price increased is an exception to the law of demand, it is known as Giffen’s paradox.

Question 7.
Veblen Goods (or) Prestigious Goods
Answer:
This is associated with the name of T.Veblen costly goods like diamonds and cars are called Veblen goods generally rich people purchase those goods. For the sake of prestage. Hence rich people may buy more such goods when their prises rise.

Question 8.
Speculation
Answer:
When the price of commodity rises the group of speculats except that be rise still further. Therefore, they buy more of the commodity. If they expect that there is a fallen price, the demand may not expand. Ex : shares.

Question 9.
Price Demand
Answer:
It explains the functional relationship between price of good and quantity of demanded when the remaining factors constant. It shows inverse relationship between price and demand.
D_x = f(P_x)
D_x = Demand for X commodity
P_x = Price of X

Question 10.
Income Demand [March 17]
Answer:
It shows the direct relationship between the income of the consumer and quantity demanded when the other factors remain constant. There is direct relationship between income and demand for superior goods. Inverse relationship between income and demand for inferior goods.
D_x = f(Y)

Question 11.
Cross Demand [March 18]
Answer:
Cross demand refers to the relationship between any two goods which are either complementary to each other or substitute of each other at different prices.
D_x = f(P_y)

Question 12.
Substitutes
Answer:
These are goods which satisfy the same want.
Ex : tea and coffee. In this case the relationship between demand for a product and the price of its substitute is positive in its nature.

Question 13.
Complementaries
Answer:
These are goods which satisfy the same wants jointly.
Ex : Shoes and socks, car and petrol. The relationship between complementary goods is inverse.

Question 14.
Inferior Goods
Answer:
The goods whose income elasticity of demand is negative for levels of income are termed as inferior goods. In case of inferior goods if income increases demand decreases and vice-versa. The income demand for inferior goods has a negative slope.

Question 15.
Elasticity of Demand
Answer:
It means the degree of responsiveness of demand or the sensitiveness of demand to change in price. This was developed by Marshall It explains how much demand increases due to fall in price and how much demand decreases due to rise in price.

Question 16.
Price Elasticity of Demand
Answer:
It is the percentage change in quantity demanded of a commodity as a result of percentage change in price of a commodity.
Ed = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in price }}\)

Question 17.
Income Elasticity of Demand
Answer:
It is the percentage change in quantity demanded of a commodity as a result of percentage change in the income of the consumer.
Ey = \(\frac{\text { Percentage change in quantity demanded }}{\text { Percentage change in consumer’s income }}\)

Question 18.
Cross Elasticity of Demand
Answer:
It is the percentage change in the quantity demanded of a commodity as a result of proportional change in the price of related commodity.
Ec = \(\frac{\text { Percentage change in quantity demanded of } \mathrm{X}}{\text { Percentage change in the price of } \mathrm{Y}}\)

Question 19.
Perfectly Elastic Demand
Answer:
If a negligible change in price leads to an infinite change in demand is called perfectly elastic demand. In this case the demand curve is horizontal to ‘X’ axis.

Question 20.
Perfectly Inelastic Demand [March 16]
Answer:
Even a great rise or fall in price does not lead and change in quantity demanded is known as perfectly inelastic demand. The demand curve is vertical to ‘Y’ axis.

Question 21.
Unitary Elastic Demand
Answer:
The proportionate change demand is equal to the proportionate change in price. In this case the demand curve will be a rectangular hyperbola.

Question 22.
Relatively Elastic Demand
Answer:
When a proportionate change in price leads to more than proportionate change in quantity demand is called relatively elastic demand.

Question 23.
Relatively Inelastic Demand
Answer:
When the proportionate change in price leads to a less than proportionate change in quantity demand is called relatively inelastic demand.

Question 24.
Arc method
Answer:
Arc method is the elasticity of the mid point of an arc of a demand curve. It studies a portion of the demand curve between two points. This is used when the change in price is not very large.

Question 25.
Importance of Price Elasticity of Demand.
Answer:

1. It is useful to finance minister in imposing taxes.
2. Useful to monopolist for fixing the price.
3. Useful in determination of wages.
4. Useful in determination of prices of factors of production.

Additional Questions

Question 26.
Terms of Trade
Answer:
It is the ratio of an index of a country’s export price to an index of its important price.

Question 27.
Tax
Answer:
Tax is a compulsory payment collected from individuals or firms by central, state and local governments.

Andhra Pradesh BIEAP AP Inter 1st Year Commerce Study Material 1st Lesson Concept of Business Textbook Questions and Answers.

AP Inter 1st Year Commerce Study Material 1st Lesson Concept of Business

Essay Answer Questions

Question 1.
Define Business. What are its characteristics? [A.P. Mar. 2019, 17]
Answer:
The term Business refers to “the state of being busy”. Every individual is engaged in some activities to fulfill his/her set of needs and wants. All these activities are intended to satisfy human needs. Business is one of the human economic activities.

Business – Definitions :
“A human activity directed towards producing or acquiring wealth through buying and selling of goods.” -L.H. Haney

“Business is an institution organized and operated to provide goods and services to society under the incentive of private gain.” -B.O. Wheeler

“Business is a sum of all activities involved in the production and distribution of goods and services for private profits.” – Keith and Carlo

Business – Characteristics :
Following are the essential characteristics of business.

1. Creation of utilities
2. Deals with goods and services
3. Continuity in dealings
4. Sale, transfer or exchange
5. Profit motive
6. Risk and uncertainty
7. Economic activity
8. Art as well as science

1) Creation of utilities :
Business makes goods more useful to satisfy human wants. It adds to production, the utilities of person, time, place, form, knowledge, etc. Businessman is able to satisfy the customer’s demands effectively and economically with the help of business transactions.

2) Deals with goods and services :
Business deals with goods and services. The goods may be consumer goods such as cloths, soaps, milk, shoes, furniture, etc. They may be industrial goods such as machinery, equipment, etc. which are used for further production. Business also deals with services such as transport, warehousing, banking, insurance, etc.

3) Continuity in dealings :
Dealings in goods and services become business only if undertaken on a regular basis. A single isolated transaction of purchase and sale does not constitute business. Recurring or repeated transactions of purchase and sale constitutes business. E.g.: If a person sells his old scooter or a car, it is not business though the seller gets money in exchange. But if he opens a shop and sells scooters or cars regularly, it will become business. Therefore, regularity of dealings in an essential feature of business.

4) Sale, transfer or exchange :
In a business activity there should be two parties i.e. a buyer and a seller. There should be exchange, sale, transfer of goods or services between these two parties for money. For instance, cooking food for personal consumption does not constitute business. But cooking food and selling it to others for a price becomes business. E.g.: Students’ mess.

5) Profit motive:
The primary objective of business is to earn profits. Profits are essential for the survival as well as growth of business. Profits must, however, be earned through legal and fair means. Business should never exploit society to make money.

6) Risk and uncertainty :
Profit is the reward for assuming risk. Risk implies uncertainty of profit or the possibility of loss. Risk is a part and parcel of business. Business enterprises function in uncertain and uncontrollable environment. E.g.: Changes in customers’ tastes and fashions, demand, competition, government policies, etc. create risk. Flood, fire, earthquake, strike by employees, theft, etc. also cause loss. A businessman can reduce risks through correct forecasting and insurance. But all risks cannot be eliminated.

7) Economic activity:
Business is primarily an economic function. It involves production and distribution of goods and services for the satisfaction of human wants. However, business is a part of society and it reflects on aspiration, values and beliefs of people. Therefore, business may be described as a socio-economic function.

8) Art as well as science :
Business is an art because it requires personal skills and experience. It is also a science because it is based on certain principles and laws.
The above mentioned features are common to all business enterprises irrespec¬tive of their nature, size and form of ownership.

Question 2.
Explain the objectives of a business.
Answer:
Objectives of business mean the purposes for which business is established and carried on. Proper selection of objectives is essential for the success of a business. Therefore, every businessman must select and define his business objectives carefully and clearly.

Objectives of business are classified as given below.

1) Economic Objectives:
Business is basically an economic activity. Therefore, its primary objectives are economic in nature. The main economic objectives of business are as follows.

i) Earning profits :
Every business enterprise’s main object is profit. It is the hope of earning profits that inspires people to start business. Profit is essential for the survival of every business unit. Profit also serves as the barometer of stability, efficiency and progress of a business enterprise.

ii) Creating customers :
Profits arise from the businessman’s efforts to satisfy the needs and wants of customers. A businessman can earn profits only when there are enough customers to buy and pay for his goods and services. The customer is the foundation of business and keeps it in existence. Business exists to satisfy the wants, tastes and preferences of customers.

iii) Innovation :
Innovation refers to “creation of new things resulting from the study and experimentation, research and development”. In these days of competition a business can be successful only when it creates new designs, better machines, improved techniques, new varieties, etc. Modern science and technology have created a great scope for innovation in the business world.

2) Social Objectives:
Business does not exist in a vaccum. It is a part of society. It cannot survive and grow without the support of society. So, business must have some social objectives. They are given below.

i) Supplying desired goods at reasonable prices :
Business is expected to supply the goods and services required by the society. Goods and services should be of good quality and these should be supplied at reasonable prices. It is also the social obligation of business toaKoid malpractices tike smuggling, black makreting and misleading advertising.

ii) Fair Remuneration to employees:
Employees must be given fair compensation for their work. In addition to wages and salary a reasonable part of profits should be distribuited among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees.

It is the obligation of business to provide healthy and safe work environment for employees. Employees work day and night to ensure smooth functioning of business. It is, therefore, the duty of employers to provide hygienic working and living conditions for workers.

iii) Employment generation :
Business should provide opportunities for gainful employment to members of the society. In a country like India unemployment has become a serious problem and no government can offer jobs to all. Therefore, provision of adequate and full employment opportunities is a significant service to society.

iv) Social welfare :
Business should provide support to social, cultural and reli¬gious organisations. Business enterprises can build schools, colleges, libraries, dharamshalas, hospitals, sports bodies and research institutions. They can help non-government organisations (NGOs) like CRY (Child Relief and You), Help Age, and others which render services to weaker sections of society.

v) Payment of government dues :
A business should not shut its eyes to its obligations towards the government. Therefore, business owes it to the government to pay its tax dues honestly and in time. It must also dutifully abide by the laws of the land.

3) Human Objectives:
i) Labour welfare :
Business must recognise the dignity of labour and human factor should be given due recognition. Adequate provisions should be made for their health, safety and social security.

ii) Developing human resources :
Employees must be provided with the opportunities for developing new skills and attitudes. This can be done by training the employees and conducting workshops on skill development and attitude. Human resources are the most valuable asset of business and their development will help in the growth of business.

iii) Participative management :
Employees should be allowed to take part in decision making process of business. This will help in the development of employees. Workers’ participation in management will usher in industrial democracy.

iv) Labour – Management cooperation :
Business should strive for creating and maintaining cordial employer- employee relations so as to ensure peace and progress, in industry.

4) National Objectives:
i) Optimum utilisation of resources :
Business should use the nation’s resources in the best possible manner. Judicious allocation and optimum utilisation of scarce resources is essential for rapid and balanced economic growth of the country. Business should produce goods in accordance with national priorities and interests. It should minimise the wastage of scarce natural resources.

ii) National self-reliance :
It is the duty of business to help the government in increasing exports and in reducing dependence on imports. This will help a country to achieve economic independence. .

iii) Development of small scale industries :
Big business firms are expected to encourage growth of small scale industries which are necessary for generating employment. Small scale firms can be developed as ancillaries which provide inputs to large scale industries.

iv) Development of backward areas :
Business is expected to give preference to the industrialisation of backward regions of the country. Balanced regional development is necessary for peace and progress in the country. It will also help to raise standard of living in backward areas. Government offers special incen¬tives to the businessmen who set up factories in notified backward areas.

Question 3.
Discuss the social responsibility of business.
Answer:
Business organisations are obliged to consider social impact of their decisions. The obligation of any business to protect and serve public interest is known as social responsibility of business. Any responsibility business has, particularly towards members of the society with whom they interact or towards the society in general called is social responsibility.

The Concept of Social Responsibility:
Every business operates within a society. It uses the resources of the society and depends on the society for its functioning. This creates an obligation on the part of business to look after the welfare of society. Therefore, all the activities of the business should be such that they will not harm, rather they will protect and contribute to the interests of the society.

Social responsibility of business refers to all such duties and obligations of business directed towards the welfare of society. So, every business must contribute in some way or the other for their benefit. E.g. : Every business must ensure a satisfactory rate of return to investors, provide good salary, security and proper working condition to its employees, make available quality products at reasonable price to its consumers, maintain the environment properly, etc. Social responsibility implies that a business should not do anything harmful to the society in course of business activities of a businessman.

Social Responsibility Towards Different Interest Groups :
The business generally interacts with owners, investors, employees, suppliers, customers, competitors, gov-ernment and society. They are called interest groups. Such interest groups are given below.

Social responsibility towards different interest groups

• Responsibility towards owners
• Responsibility towards employees
• Responsibility towards suppliers
• Responsibility towards customers
• Responsibility towards government
• Responsibility towards society

1) Responsibility towards owners :
Owners are the persons who own the business. They contribute capital and bear the business risks. The primary responsibilities of business towards its owners are to :

1. Run the business efficiently.
2. Proper utilisation of capital and other resources.
3. Growth and appreciation of capital.
4. Regular and fair return on capital invested by way of dividends.

2) Responsibility towards employees :
Business needs employees or workers to work for it. These employees put their best effort for the benefit of the business. The responsibility of business towards its employees include:

1. Timely and regular payment of wages and salaries.
2. Proper working conditions and welfare amenities.
3. Opportunity for better career prospects.
4. Job security as well as social security like facilities of provident fund, group insurance, pension, retirement benefits, etc.

3) Responsibility towards suppliers :
Suppliers are businessmen who supply raw materials and other items required by manufacturers and traders. Certain suppliers, called distributors, supply finished products to the customers. The responsibilities of business towards these suppliers are :

1. Giving qualitative goods at reasonable prices.
2. Dealing on fair terms and conditions.
3. Availing reasonable credit period.
4. Timely payment of dues.

4) Responsibility towards customers :
No business can survive without customers. As a part of the responsibility of business towards them the business should provide the following facilities.

1. Products and services must be qualitative
2. Giving delivery of goods within stipulated time
3. Reasonable price
4. There must be proper after-sales services.
5. Complaints and grievances of the customers, if any, must be settled quickly.
6. Unfair means like underweighing the product, adulteration, etc. must be avoided.

5) Responsibility towards government:
Business activities are governed by the rules and regulations framed by the government. The various social responsibilities of the government are:

1. Setting up units as per guidelines of the government.
2. Payment of fees, duties and taxes regularly as well as honestly.
3. Conforming to pollution control norms set up by government.
4. Not to indulge in corruption through bribing and other unlawful activities.

6) Responsibility towards society :
A society consists of individuals, groups, organizations, families, etc. They all are the members of the society. Thus, it has certain responsibilities towards society, which may be as follows :

1. to help the weaker and backward sections of the society
2. to preserve and promote social and cultural values
3. to generate employment
4. to protect the environment
5. to conserve natural resources and wildlife
6. to promote sports and culture

Question 4.
Classify and describe each type of Economic activities.
Answer:
All the activities in which the people participate from morning till night are called human activities. Every individual is engaged in some activities to fulfil his/her set of needs and wants. All these activities are intended to satisfy human needs.

All the activities of human beings can be classified into two types. They are :

Non – economic Activities :
Those human activities do not involve money or money’s worth, such activities are termed as non-economic activities. Human beings engage themselves in non-economic activities due to love, affection, patriotism, charity, sympathy, and other such sentiments.
E.g. : A mother looks after her children, young man helps a blind man to cross the road, etc.

Economic Activities:
Human beings undertake certain economic activities for earning money or livelihood. Working as a teacher in a school, a doctor in a hospital, a worker in a factory, a farmer in a field, an emloyee in an office, a merchant selling goods, etc.

In other words, an human being involves in any activity together with money or money’s worth, such activities are termed as human economic activities. Such human economic activities are classified into three types. They are given below ;

1. Profession
2. Employment
3. Business

1) Profession:
An activity which involves the rendering of personalized services of a specialized nature based on professional knowledge, education and training is called a profes¬sion. Services rendered by doctors, lawyers, chartered accountants, engineers, etc. come under this category.

2) Employment:
An employment is a contract of service. A person who works under the contract for a salary is called an employee and the person who has given the job to the employee is called employer. An employee works under an agreement as per the rules of service and performs tasks assigned to him by the employer. The relationship between the employer and the employee is that of a ‘Master’ and ‘Servant’.

3) Business:
Business is one of the human economic activities. Business is an economic activity involving production, exchange, distribution and sale of goods and services with an objective of making profits.

Short Answer Questions

Quelition 1.
Business objectives.
Answer:
Objectives of business mean the purposes for which business is established and carried on. Proper selection of objectives is essential for the success of a business. Objectives serve as the guidelines for the future direction and management of business. Therefore, every businessman must select and define his business objectives carefully and clearly.

Objectives of business may be classified into four broad categories. They are :

1. Economic objectives
2. Social objectives
3. Human objectives
4. National objectives

Every businessman seeks to earn profits by satisfying the wants of people. It is the hope of earning profits which induce people to enter into business. No business can survive without making adequate profits. Thus profit is the fundamental economic objective of business.

If profit maximization is regarded as the sole objective of business, it is likely to result in unfair practices such as hoarding, black marketing, etc. The profit making and social service objectives are not contradictory to each other.

Business must discharge social responsibilities in addition to earning profits. It should aim at servicing the community.

Quelition 2.
Social objectives.
Answer:
Business does not exist in a vacuum. It is a part of society. It cannot survive and grow without the support of society. Business must therefore discharge social responsibilities in addition to earning profits.

Social objectives – Definition :
“The primary aim of business should be service and subsidiary aim should be earning of profit.” – Henry Ford

Some important social objectives are given below :

1. Business is expected to supply the goods and services required by the society. Goods and services should be of good quality and these should be supplied at reasonable prices.
2. Employees must be given fair compensation for their work. In addition to wages and salary a reasonable part of profits should be distributed among employees by way of bonus. Such sharing of profits will help to increase the motivation and efficiency of employees. So, fair remuneration to employees is an important social objective.
3. Business should provide job opportunities to the members of the society. In a country like India unemployment has become a serious problem and no government can offer jobs to all. So, employment generation is also one of the social objectives.
4. Business should provide support to social, cultural and religious organisations. Business enterprises can build schools, colleges, libraries, hospitals, etc.
5. Every business enterprise should pay tax dues to the government honestly and at the right time. These direct and indirect taxes provide revenue to the government for spending on public welfare. So, payment of government dues is also one of the important social objectives.

Quelition 3.
Role of profit in business.
Answer:
A business enterprise is established for earning some income. It is the hope of earning profits that inspires people to start business. Profit is essential for the survival of every business unit. Just as a person cannot live without food, a business firm cannot survive without profit. Profits enable a businessman to stay in business by maintaining intact the wealth producing capacity of its resources.

Profit is also necessary for the expansion and growth of business. Profits ensure continuous flow of capital for the modernisation and extension of business operations in future. Profit also serves as the barometer of stability, efficiency and progress of a business enterprise.

Quelition 4.
Brief explaination of economic activities.
Answer:
The term business refers to “the state of being busy”. Every business individual is engaged in some activities to fulfil his/her set of needs and wants. All these activities are intended to satisfy human needs. E.g.: A farmer engages himself in agricultural activities and an employee works in the office, a teacher teaches in the classroom, etc. for satisfying his needs, comforts and luxuries.

All the activities of human being can be divided into two types. They are :

1. Economic Activities
2. Non-economic Activities

Economic activities :
Those human activities that are involved in money, such activities are termed as economic activities. Human beings undertake certain economic activities for earning money or livelihood. Working as a teacher in a school, a doctor in a hospital, a worker in a factory, a merchant selling goods or an industrialist manufacturing goods, all these are economic activities. These economic activities are concerned with production, exchange and distribution of goods and services.

Very Short Answer Questions

Quelition 1.
Define Business. [Mar. 2018, 17 ; May 17 – A.P.]
Answer:
A business is an economic institution. It is concerned with production and distribution of goods and rendering of service in order to earn profits and acquire wealth. Business may be defined as “a human activity directed towards producing or acquiring wealth through buying and selling of goods”. – L.H. Haney

Profits are consideration of Business.

Quelition 2.
What is a Profession?
Answer:
Profession is one of the human economic activities. An activity which involves the rendering of personalised services of a specialized nature based on professional knowledge, education and training is called a profession. E.g. : Doctors, Lawyers, Chartered Accountants, Engineers, etc.

Remuneration is consideration of Profession.

Quelition 3.
What is Employment?
Answer:
Employment is also one of human economic activities. Any activity assigned to a person by the employer under an agreement or rules of services comes under the category of employment.

A person who undertakes such activity is called employee.

Salary is consideration of Employment.

Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 9th Lesson Bank Reconciliation Statement Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 9th Lesson Bank Reconciliation Statement

Short Answer Questions

Question 1.
What is meant by Bank Reconciliation Statement? (May ’17 – T.S.)
Answer:
Bank Reconciliation Statement is a statement prepared to reconcile the difference between the balances as per the bank column of the cash book and the pass book on any given date.

Question 2.
What do you mean by Favourable Balance ? (Mar. ’17 – T.S.)
Answer:
It means cash book shows debit balance at the same time pass book shows credit balance. Favourable balance means our money is in the Bank Account. Hence cash book debit balance and pass book credit balance is called favourable balance.

Question 3.
What do you mean by Unfavourable Balance ?
Answer:
Sometimes a businessman withdraws/excess amount from the bank account and the closing bank balance of a month is a debit balance. This balance amount is called unfavourable balance or overdraft balance.

Question 4.
Describe the Overdraft. (Mar. 2019, ’18 – T.S.; Mar ’15 – A.P.)
Answer:
When withdrawals exceed deposits, cash book bank column shows credit balance and the pass book shows debit balance. It is called unfavourable or overdraft balance. Sometimes business¬man is allowed by the bank to withdraw the amount, in excess of what he has in his bank account after moving prior agreement with banker. This facility is called overdraft.

Question 5.
Cheques deposited with a bank for collection, what is the impact on Cash Book?
Answer:
When the cheques are deposited with a bank for collection, that the cheques amount was debited in the cash book, before sending them for collection. Hence the impact on cash book is the cash balance increased.

Essay Type Questions

Question 1.
Explain the nature and importance of the Bank Reconciliation Statement.
Answer:
A statement prepared to reconcile the balance of cash book and pass book is called the ‘Bank Reconciliation Statement’.
Generally the business concerns would like to maintain an account with a bank and prepare a Bank Reconciliation Statement.

Importance:

1. Locating the mistakes or errors either side of both cash book and pass book.
2. Preventing any fraud and misappropriations.
3. Enabling the business concern to get up-to-date record of transactions from the bank.
4. Ensuring a proper evidence of payment.
5. Help to know exact cash balance at bank.

Question 2.
Enumerate the reasons for differences between the balance shown in the Cash Book and Pass Book.
Answer:
I. Items entered in the bank columns of the Cash Book, but not entered in Pass Book or bank statement.

1. Cheques sent for collection, but not collected by the bank. This will appear only on the debit side of the cash book.
2. Cheques issued but not yet presented for payment. This appears only on credit side of the cash book.

II. Items entered in the bank statement or bank Pass Book, but not in Cash Book:

1. Direct payment into the bank by a customer. This appears only on the credit side of the bank statement.
2. Bank charges : Charges paid by the business for using some of the bank services. This will appear on the debit side of the pass book.
3. When the business instructs the bank to make regular payments of fixed amounts, rent, insurance premium, etc. These will appear on the debit side of the bank statement.
4. Interest on overdrafts or loans appears on the debit side of the bank statement.
5. Interest on deposits appears on the credit side of the bank statement.
6. Dishonour of cheques and bills – first appears on the debit side of the bank statement. But the firm records the same when it receives the information from the bank. As a result, the balance as per cash book and that of pass book will differ.

III. Difference caused by Errors:

1. Errors committed in recording transactions by the firm in cash book : Omission of transaction, wrong recording, wrong totaling, over / under casting, etc.
2. Errors committed in recording transactions by the bank : Sometimes bank may also commit errors. E.g. Omission or wrong recording of transaction, wrong totaling, over/ under casting, etc.

Question 3.
Explain the procedure of preparing the Bank Reconciliation Statement by taking imaginary items and figures.
Answer:
Bank reconciliation statement is prepared to reconcile the two balances of Cash Book and Pass Book. The preparation of Bank Reconciliation Statement starts with banking adjustments to one balance to reach the other balance, which ensures agreement between both the balances.

The BRS is prepared usually at the end of the period, i.e. a month, a quarter, half a year or a year whichever is convenient to the firm. When both the books, cash book and pass book are given in problem, then see whether the two books are related to the same period or different periods. If the books are for different periods, then common items should be considered and if it is for same period, then items not appearing in both the books should be taken into consideration.

The way how to prepare BRS may be illustrated as follows.
Bank Reconciliation Statement of …… as on …….

Problems

Question 1.
Pass Book of a trader shows a balance of Rs. 12,600. On comparing the Pass Book with the cash Book, the following discrepancies were noted. (Mar. 2018 – A.P.)
a) Cheques deposited in bank but not collected Rs. 2,100
b) Cheques issued but not presented for payment Rs. 1,800
c) Bank Charges Rs. 175
d) Bank paid insurance premium Rs. 1,500
e) The Debtor paid directly into bank account Rs. 1200
Answer:
Bank Reconciliation Statement

Question 2.
Murthy and Son’s Pass Book showed a balance of Rs. 21,700 as on 30^th September, 2013. On comparing the Cash Book the following discrepancies were noted.
a) Cheques issued but not yet presented for payment Rs. 2,500
b) Directly deposited by a customer Rs. 3,000
c) Interest credited by bank is found in Pass Book only Rs. 575
d) Cheques deposited in bank but not credited Rs. 3,500
e) Bank Charges Rs. 150
Prepare a Bank Reconciliation Statement showing balance as per Cash Book.
Answer:
Bank Reconciliation Statement of Murthy and Son as on 30th September 2013

Question 3.
Giri Ind Ltd’s bank balance as per Pass Book is Rs. $,900. There is disagreement between Cash Book and Pass Book balances as, on 31.3.2014. Prepare Bank Reconciliation statement by considering following transactions.
a) Cheque issued but not yet presented for payment Rs. 2100
b) Cheque deposited for collection, but not yet realized Rs. 900
c) A wrong debit given by bank in Pass Book Rs. 500
d) Bank charges debited only in Pass Book Rs. 210
e) Direct payment of insurance premium as per standing instructions Rs. 600
Answer:
Bank Reconciliation Statement of Giri lnd Ltd’s as on 31.03.2014

Question 4.
On comparing the bank Pass Book of BBR Ltd., with its Cash Book (bank column), the following differences were noticed. Prepare BRS with the help of Cash Book Balance Rs. 15,000.
a) Cheque sent for collection, notyet realized Rs. 5,600
b) Cheques issued but not yet presented for payment Rs. 4,200
c) The receipts side of Cash Book has been overcast by Rs. 300
d) A cheque drawn on firm’s current a/c, wrongly debited in its savings a/c, Rs. 2,100
e) A cheque ofRs. 900 deposited into bank, bit forgot to enter in Cash Book
Answer:
Bank Reconciliation Statement of BBR Ltd

Question 5.
Reddy’s Cash Book shows a favourable balance ofRs. 25,500 as on 31^st December, 2013. On comparing the same with his Pass Book following differences were noticed. Calculate bank balance as per Pass Book.
a) A cheque for Rs. 2,450 received from Saritha & Co was entered twice in the Cash Book.
b) The receipts column of the Cash Book has been over added by Rs. 1940.
c) Several cheques, totaling Rs. 6,000 were issued to different suppliers. Of these, cheques worth Rs. 1,500 were debited in Pass Book on 2nd January, 2014 and Rs. 2,500 on 4th January. The balance being debited before 31st December, 2013.
d) Bills discounted, got dishonored Rs. 750.
e) A cheque ofRs. 400 was credited in Pass Book, but was not recorded in Cash Book.
f) Uncredited cheque Rs. 1,000
Answer:
Bank Reconciliation Statement of Reddy’s as on 31^st December 2013

Question 6.
On 31^st December, 2013 the Cash Book showed an unfavourable balance Rs. 29,000. Prepare a Reconciliation Statement with the following information.
a) Cheques had been deposited into the bank but were not collected Rs, 4,530
b) A cheque issued to Karthik Reddy, the supplier, has not been encashed Rs. 5,040.
c) There was a debit entry in the Pass Book ofRs. 600 for bank charges.
d) Bills worth Rs. 2000 were discounted but dishonored.
Answer:
Bank Reconciliation Statement as on 31^st December 2013

Question 7.
From the following particulars prepare a Bank Reconciliation Statement.
a) Bank Overdraft as per Cash Book Rs. 16,100
b) Debit side of the bank column of Cash Book cast short Rs. 200
c) Bills collected directly by bank Rs. 3,500
d) Bank charges recorded twice in the Cash Book Rs. 240
e) A cheque deposited as per bank statement but not recorded in the Cash Book Rs. 1100
f) The cheques of 6,000 deposited but collections as per statement Rs. 2,600
g) Interest on investment collected by the banker, same was shown only in Pass Book Rs. 2,000
Answer:
Bank Reconciliation Statement

Question 8.
From the following particulars prepare Bank Reconciliation Statement as on 31^st March, 2014.
a) Overdraft balance as on 31-3-2014 as per Bank statement Rs. 22,470.
b) As per standing instructions given to bank, Chamber of Commerce fee Rs. 2,530 was paid by the bank but was not recorded in the Cash Book.
c) On 23-3-2014, the credit side of the bank column of the Cash Book was cast Rs. 1,900 short.
d) Cheque deposited into the bank but not recorded in Cash Book Rs. 2,500
e) In the Cash Book, a Bank charge ofRs. 290 was recorded twice while another bank charge of Rs. 120 was not recorded at all.
f) Divided on shares Rs. 3,200 was collected by bank directly the trader has no information.
g) Two cheques ofRs. 1850 and Rs. 1,500 were issued but out of them only one cheque ofRs. 1850 was presented for payment up to reconcile day.
Answer:
Bank Reconciliation Statement as on 31.03.2014

Question 9.
Prepare Bank Reconciliation statement ofKarthik as on 31 – 03 – 2014.
a) Bank overdraft as per Pass Book Rs. 6,500
b) Cheques deposited into bank Rs. 5,000, but only Rs. 2,000 was collected.
c) Cheques issued but not presented for payment Rs. 1500
d) A customer directly deposited in our bank Rs. 1200
e) Bank charges Rs. 200; Insurance premium Rs. 300 has debited in the Pass Book only
f) Divided Rs. 300 collected by the bank has credited in the Pass Book only
Answer:
Bank Reconciliation Statement of Karthik as on 31.03.2014

Question 10.
Prepare Bank Reconciliation Statement of P.R.G. Rao & Sons as on 31.03.2014
a) Bank overdraft as per Cash Book Rs. 14,500
b) Cheques issued but not yet presented for payment Rs. 4,500
c) Directly deposited by a Customer in our bank account Rs. 3,500.
d) Cheques deposited in bank but not credited Rs. 7,500
e) Bank charges debited in pass book only Rs. 200
f) Interest debited in the pass book only Rs. 500
Answer:
Bank Reconciliation Statement of P.R.G. Rao as on 31.03.2014

Student Activity

Visit any business careers and enquire about what discrepancies generally they notice in the items. Make a list of the discrepancies and show the effect on the bank balance.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

I.

Question 1.
Verify Rolle’s theorem for the following functions.
i) x² – 1 on [-1, 1]
Solution:
Let f(x) = x² – 1
f is continuous on [-1,1]
since f(-1) = f(1) = 0 and
f is differentiable on [-1, 1]
∴ By Rolle’s theorem ∃ c ∈ (-1, 1) such that f'(c) = 0
f'(x) = 2x = 0
∴ = f'(c) = 0
2c = 0
c = 0
The point c = 0 ∈ (-1, 1)
Then Rolle’s theorem is verified.

ii) sin x – sin 2x on [0, π].
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π]
since f(0) = f(π)= 0 and
f is differentiable on [0, π]
By Rolle’s theorem ∃ c ∈ (0, π)
such that f'(c) = 0
f'(x) = cos x – 2 cos 2x
f'(c) = 0 ⇒ cosc – 2 cos 2c = 0
⇒ cosc – 2(2cos²c – 1):
cosc – 4 cos²c + 2=0 2
4 cos² c – cosc – 2 = 0

iii) log (x² + 2) – log 3 on [-1, 1]
Solution:
Let f(x) = log (x² + 2) – log 3
f is continuous on [-1, 1]
Since f(-1) = f(1) = 0 and f is
Differentable on [-1, 1]
By Rolle’s theorem ∃ c ∈ (-1, 1)
Such that f'(c) = 0
f'(x) = \(\frac{1}{x^{2}+2}\) = (2x)
f'(x) = \(\frac{2c}{c^{2}+2}\) = 0
2c = 0
c = 0
c = 0 ∈ (-1, 1).

Question 2.
It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2t + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Solution:
Given f(x) = x³ + bx² + ax
f'(x) = 3x² + 2bx + a
∴ f'(c) = 0 6 ⇔ 3c² + 2bc + a = 0

⇔ b = 6 and b² – 3a = 3
36 – 3 = 3a
33 = 3a
a = 11
Hence a = 11 and b = -6.

Question 3.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].
Solution:
Clearly f(0) = f(c)
0 – 0 + k = 1 – 3 + k
0 = -2
Which is not possible
∴ There is no real number K.

Question 4.
Find a point on the graph of the curve y = (x – 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).
Solution:
Given points (3, 0) and (4, 1)
The slope of chord = \(\frac{1-0}{4-3}\) = 1
Given y = (x- 3)²
\(\frac{dy}{dx}\) = 2(x – 3)
⇒ Slope = 2(x – 3)
1 = 2(x – 3)
\(\frac{1}{2}\) = x – 3
x = \(\frac{1}{2}\) + 3 = \(\frac{7}{2}\)
y = (x – 3)² = (\(\frac{7}{2}\) – 3)² = \(\frac{1}{4}\)
∴ The point on the curve is (\(\frac{7}{2}\), \(\frac{1}{4}\))

Question 5.
Find a point on the graph of the curve y = x³, where the tangent is parallel to the chord joining (1, 1) and (3, 27).
Solution:
Given points (1, 1) and (3, 27)
Slope of chord = \(\frac{27-1}{3-1}\) = 13
Given y = x³
\(\frac{dy}{dx}\) = 3x²
⇒ Slope = 3x²
13 = 3x²

∴ The point on the curve is (\(\frac{\sqrt{39}}{3}\), \(\frac{13\sqrt{39}}{9}\))

Question 6.
Find ‘c’, so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases.
i) f(x) = x² – 3x – 1, a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\).
Solution:

f'(x) = 2x – 3
f'(c) = 2c – 3
Given f'(c)= \(\frac{f(b)-f(a)}{b-a}\)

ii) f(x) = e^x ; a = 0, b = 1
Solution:
f(b) = f(1) = e’ = e
f(a) = f(0) = e° = 1
Given f(x) = e^x
f'(x) = e^x
Given condition f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

Question 7.
Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [-1, 2]. Find the point in the interval where the derivate vanishes.
Solution:
Let f(x) = (x² – 1) (x- 2) = x³ – 2x² – x + 2
f is continous on [-1, 2]
since f(-1) = f(2) = 0 and f is
Differentiable on [-1, 2]
By Rolle’s theorem ∃ C ∈ (-1, 2)
Let f'(c) = 0
f'(x) = 3x² – 4x – 1
3c² – 4c – 1 = 0

Question 8.
Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem
i) x² -1 on [2, 3]
Solution:
Solution:
Let f(x) = x² – 1
f is continous on [2, 3]
and f is differentiable
Given f(x) = x² – 1
f'(x) = 2x
By Lagrange’s mean value theorem ∃ C ∈ (2, 3) such there

ii) sin x – sin 2x on [0, π]
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π] and f is differentiable
Given f(x) = sin x – sin 2x
f'(x) = cos x – 2 cos 2x
By Lagrange’s mean value than ∃ C ∈ (0, π) such there
f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)
cosc – 2 cos 2c = 0
cosc 2(2cos² – 1) = 0
cosc – 4 cos²c + 2 = 0
4 cos² c – cos c – 2 = 0

iii) log x on [1, 2].
Solution:
Let f(x) = log x
f is continuous on [1, 2] and f is differentiable
Given f(x) = log x
f'(x) = \(\frac{1}{x}\)
By Lagrange’s mean value theorem ∃ c ∈ (1, 2) such that

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions

Question 1.
Find the values of
(i) sin \(\frac{5 \pi}{3}\)
(ii) tan (855°)
(iii) sec \(\left(\frac{13 \pi}{3}\right)\)
Solution:
i) sin \(\frac{5 \pi}{3}\) = sin \(\left(2 \pi-\frac{\pi}{3}\right)\) = -sin \(\frac{\pi}{3}\) = \(-\frac{\sqrt{3}}{2}\)

ii) tan (855°) = tan (2 × 360° + 135°)
= tan (135°)
= tan (180° – 45°)
= -tan 45° = -1

iii) sec \(\left(13 \frac{\pi}{3}\right)\) = sec \(\left(4 \pi+\frac{\pi}{3}\right)\)
= sec \(\frac{\pi}{3}\) = 2

Question 2.
Simplify.
i) Cot (θ – \(\frac{13 \pi}{2}\))
ii) tan \(\left(-23 \frac{\pi}{3}\right)\)
Solution:

Question 3.
Find the value of sin² \(\frac{\pi}{10}\) + sin² \(\frac{4 \pi}{10}\) + sin² \(\frac{6 \pi}{10}\) + sin² \(\frac{9 \pi}{10}\)
Solution:

Question 4.
If sin θ = \(\frac{4}{5}\) and θ is not in the first qua-drant, find the value of cos θ.
Solution:
∵ sin θ = \(\frac{4}{5}\) and θ is not in the first quadrant
⇒ θ lies in 2^nd quadrant, ∵ sin θ is +ve

Question 5.
If sec θ + tan θ = \(\frac{2}{3}\), find the value of sin θ and determine the quadrant in which θ lies.
Solution:
∵ sec θ + tan θ = \(\frac{2}{3}\)

∵ tan θ is negative, sec θ is positive
⇒ θ lies in fourth quadrant.

Question 6.
Prove that
cot\(\frac{\pi}{16}\).cot\(\frac{2 \pi}{16}\).cot\(\frac{3 \pi}{16}\)…..cot\(\frac{7 \pi}{16}\) = 1
Solution:

Question 7.
If 3 sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3 cos θ.
Solution:
Given 3 sin θ + 4 cos θ = 5 and let 4 sin θ – 3 cos θ = x
By squaring and adding, we get
(3 sin θ + 4 cos θ)² + (4 sin θ – 3 cos θ)² = 5² + x²
⇒ 9 sin² θ + 16 cos² θ + 24 sin θ cos θ + 16 sin² θ + 9 cos² θ – 24 sin θ cos θ = 25 + x²
⇒ 9 + 16 = 25 + x²
⇒ x² = 0 ⇒ x = 0
∴ 4 sin θ – 3 cos θ = 0.

Question 8.
If cos θ + sin θ = \(\sqrt{2}\) cos θ, prove that cos θ – sin θ = \(\sqrt{2}\) sin θ. (May ’11)
Solution:
Given cos θ + sin θ = \(\sqrt{2}\) cos θ
⇒ sin θ = (\(\sqrt{2}\) – 1) cos θ
Now multiplying both sides by (\(\sqrt{2}\) + 1)

Question 9.
Find the value of 2(sin² θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)
Solution:
2 (sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)
= 2[(sin² θ)³ + (cos² θ)³] – 3[(sin² θ)² + (cos² θ)²]
= 2[(sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin² θ + cos² θ)] – 3[(sin² θ + cos² θ)² – 2 sin² θ cos² θ]
= 2[1 – 3 sin² θ cos² θ] – 3[1 – 2 sin² θ cos² θ]
= 2 – 6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ
= -1

Question 10.
Prove that (tan θ + cot θ)² = sec² θ + cosec² θ = sec² θ cosec² θ.
Solution:
(tan θ + cot θ)²
=tan² θ + cot ² θ + 2 tan θ cot θ
= tan² θ + cot² θ + 2
= (1 + tan² θ) + (1 + cot² θ)
= sec² θ + cosec² θ

Question 11.
If cos θ > 0, tan θ + sin θ = m and tan θ – sin θ = n, then show that m² – n² = n, then show that m² – n² = 4\(\sqrt{m n}\)
Solution:
Given that m = tan θ + sin θ
n = tan θ – sin θ
⇒ m + n = 2 tan θ and m – n = 2 sin θ
Now(m + n)(m – n) = 4 tan θ sin θ

Question 12.
If tan 20° = λ, then show that \(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \cdot \tan 110^{\circ}}\) = \(\frac{1-\lambda^{2}}{2 \lambda}\) (A.P.) Mar. ’16
Solution:
Given tan 20° = λ

Question 13.
Find the values of sin 75°, cos 75°, tan 75° and cot 75°.
Solution:
i) sin 75° = sin (45° + 30°)
= sin 45°. cos 30° + cos 45°. sin 30°

ii) cos (75°) = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

Question 14.
If 0 < A, B < 90°.cos A = \(\frac{5}{13}\) and sin B = \(\frac{4}{5}\)-then find sin(A + B).
Solution:
0 < A < 90°and cos A = \(\frac{5}{13}\) ⇒ sin A = \(\frac{12}{13}\)
0 < B < 90°and sin B = \(\frac{4}{5}\) ⇒ cos B = \(\frac{3}{5}\)
∴ sin (A + B) = sin A cos B + cos A sin B
= \(\frac{12}{13}\) • \(\frac{3}{5}\) + \(\frac{5}{13}\) • \(\frac{4}{5}\) = \(\frac{56}{65}\)

Question 15.
Prove that
sin² \(\left(52 \frac{1}{2}\right)^{\circ}\) – sin² \(\left(22 \frac{1}{2}\right)^{\circ}\) = \(\frac{\sqrt{3}+1}{4 \sqrt{2}}\)
Solution:

Question 16.
Prove that tan 70° – tan 20° = 2 tan 50°.
Solution:
tan 50° = tan (70° – 20°)
= \(\frac{\tan 70^{\circ}-\tan 20^{\circ}}{1+\tan 20^{\circ} \cdot \tan 70^{\circ}}\)
⇒ tan 70° – tan 20°
= tan 50° [1 + tan 20° . tan (90° – 20°)]
= tan 70° – tan 20° = tan 50°[1 + tan 20° cot 20°]
⇒ tan 70° – tan 20° = 2 tan 50°.

Question 17.
If A + B = π/4, then prove that
i)(1 + tan A)(1 + tan B) = 2. (May ’11; Mar. ’07)
ii) (cot A – 1)(cot B – 1) = 2
Solution:
i) Given that A + B = π/4 (T.S) (Mar. ’16)
⇒ tan (A + B) = tan (π/4)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
Add 1 on both sides
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
⇒ (1 + tan A)(1 + tan B) = 2

ii) Given A + B = π/4
⇒ cot (A + B) = cot π/4
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = 1
⇒ cot A cot B – 1 = çot B + cot A
⇒ cot A cot B – cot A – cot B = 1
Again add 1 on both sides
cot A cot B – cot A – cot B + 1 = 1 + 1
∴ (cot A – 1) (cot B – 1) = 2.

Question 18.
If sin α = \(\frac{1}{\sqrt{10}}\), sin β = \(\frac{1}{\sqrt{5}}\) and α, β are acute, show that α + β = π/4.
Solution:
Given α is acute and sin α = \(\frac{1}{\sqrt{10}}\)
⇒ tan α = \(\frac{1}{3}\)
β is acute and sin β = \(\frac{1}{\sqrt{5}}\) ⇒ tan β = \(\frac{1}{2} .\)

Question 19.
If sin A = \(\frac{12}{13}\), cos B = \(\frac{3}{5}\) and neither A nor B is in the first quadrant, then find the quadrant in which A + B lies.
Solution:
sin A = \(\frac{12}{13}\) and A is not in first quadrant
⇒ A lies in second quadrant, ∵ sin A is +ve
cos B = \(\frac{3}{5}\) and B is not in first quadrant
⇒ B lies in fourth quadrant, cos B is +ve
∵ sin A = \(\frac{12}{13}\) ⇒ cos A = –\(\frac{5}{13}\)
cos B = \(\frac{3}{5}\) ⇒ sin B = –\(\frac{4}{5}\)
sin (A + B) = sin A cos B + cos A sin B

cos (A + B) = cos A cos B – sin A sin B

∵ sin (A + B) and cos (A + B) are positive
⇒ (A + B) lies in first quadrant.

Question 20.
Find (i) tan \(\left(\frac{\pi}{4}+\mathbf{A}\right)\) interms of tan A
and (ii) cot \(\left(\frac{\pi}{4}+\mathbf{A}\right)\) interms of cot A.
Solution:

Question 21.
Prove that \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = cot 36°. (A.P) (Mar ’15, Mar. ’11)
Solution:

Question 22.
Show that
cos 42° + cos 78° + cos 162° = 0 (May. ’11)
Solution:
L.H.S. = cos 42° + cos 78° + cos 162°
= 2 cos \(\left(\frac{42^{\circ}+78^{\circ}}{2}\right)\). cos \(\left(\frac{42^{\circ}-78^{\circ}}{2}\right)\) + cos (180° – 18°)
= 2 cos 60° . cos (-18°) + cos (180° – 18°)
= 2\(\left(\frac{1}{2}\right)\) cos 18° – cos 18° = 0
∴ cos 42° + cos 78° + cos 162° = 0

Question 23.
Express \(\sqrt{3}\) sin θ + cos θ as a sine of an angle.
Solution:
\(\sqrt{3}\) sin θ + cos θ = 2(\(\frac{\sqrt{3}}{2}\) sin θ + cos θ)
= 2(cos \(\frac{\pi}{6}\) sin θ + sin \(\frac{\pi}{6}\) cos θ)
= 2. sin[θ + \(\frac{\pi}{6}\)]

Question 24.
Prove that sin² θ + sin² [θ – \(\frac{\pi}{3}\)] + sin² [θ – \(\frac{\pi}{3}\)] = \(\frac{3}{2} .\)
Solution:

Question 25.
If A, B, C are angles of a triangle and if none of them is equal to \(\frac{\pi}{2}\), then prove that
i) tan A + tan B + tan C = tan A tan B tan C
ii) cot A cot B + cot B cot C + cot C cot A = 1
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180°
⇒ A + B = 180° – C
⇒ tan (A + B) = tan (180° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
⇒ tan A + tan B = -tan C + tan A tan Btan C
⇒ tan A + tan B + tan C = tan A tan B tan C

ii) ∵ A + B + C = 180°
⇒ A + B = 180° – C
⇒ cot (A + B) = cot (180° – C)
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = -cot C
⇒ cot A cot B – 1 = -cot B cot C – cot C. cot A
⇒ cot A cot B + cot B cot C + cot C cot A = 1

Question 26.
Let ABC be a triangle such that cot A + cot B + cot C = \(\sqrt{3}\). Then prove that ABC is an equilateral triangle.
Solution:
Given that A + B + C = 180°
Then we know that
cot A cot B + cot B cot C + cot C cot A = 1
i.e., Σ(cot A cot B) = 1 —– (1)
Now Σ (cot A – cot B)²
= Σ (cot² A + cot² B – 2 cot A cot B)
= 2 cot² A + 2 cot² B + 2 cot² C – 2 cot A cot B – 2 cot B cot C – 2 cot C cot A
= 2 [cot A + cot B + cot C]² – 6 (cot A cot B + cot B cot C + cot C cot A)
= 2
= 6 – 6 = 0
∴ Σ (cot A – cot B)² = 0 .
⇒ cot A = cot B = cot C
⇒ cot A = cot B = cot C = \(\frac{\sqrt{3}}{3}\) = \(\frac{1}{\sqrt{3}}\)
(∵ cot A + cot B + cot C = \(\sqrt{3})\))
⇒ A = B = C = 60°
∴ ΔABC is an equilateral triangle.

Question 27.
Suppose x = tan A, y = tan B, z = tan C. Suppose none of A, B, C, A – B, B – C, C – A is an odd multiple of \(\frac{\pi}{2}\).Then prove that \(\sum\left(\frac{x-y}{1+x y}\right)\) = \(\pi\left(\frac{x-y}{1+x y}\right)\)
Solution:
∵ x = tan A, y = tan B, z = tan C

Write P = A – B, Q = B – C, R = C – A.
Then P + Q + R = O
⇒ tan (P + Q) = tan (-R)
⇒ \(\frac{\tan P+\tan Q}{1-\tan P \tan Q}\) = -tan R
⇒ tan P + tan Q = -tan R + tan P tan Q tan R
⇒ tan P + tan Q + tan R = tan P tan Q tan R
⇒ Σ(tan P) = π (tan P)
⇒ Σtan (A – B) = π tan (A – B)
∴ Σ\(\left(\frac{x-y}{1+x y}\right)\) = π\(\left(\frac{x-y}{1+x y}\right)\)

Question 28.
Find the values of
i) sin 22\(\frac{1}{2}\)°
ii) cos 22\(\frac{1}{2}\)°
iii) tan 22\(\frac{1}{2}\)°
iv) cot 22\(\frac{1}{2}\)°
Solution:

Question 29.
Find the values of
i) sin 67\(\frac{1}{2}\)°
ii) cos 67\(\frac{1}{2}\)°
iii) tan 67\(\frac{1}{2}\)°
iv) cot 67\(\frac{1}{2}\)°
Solution:
Let A = 67\(\frac{1}{2}\)° ⇒ = 2A = 135°

Question 30.
Simplify: \(\frac{1-\cos 2 \theta}{\sin 2 \theta}\)
Solution:
\(\frac{1-\cos 2 \theta}{\sin 2 \theta}\) = \(\frac{2 \sin ^{2} \theta}{2 \sin \theta \cos \theta}\) = \(\frac{\sin \theta}{\cos \theta}\) = tan θ

Question 31.
If cos A = \(\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\), find the value of cos 2A.
Solution:
cos 2A = 2 cos²A – 1 = 2 \(\left(\frac{\sqrt{2}+1}{2 \sqrt{2}}\right)\) – 1
= \(\frac{\sqrt{2}+1}{\sqrt{2}}\) – 1 = \(\frac{1}{\sqrt{2}}\)

Question 32.
If cos θ = \(\frac{-5}{13}\) and \(\frac{\pi}{2}\) < θ < π, find the value of sin 2θ.
Solution:
\(\frac{\pi}{2}\) < θ < π ⇒ sin θ > 0 and cos θ = –\(\frac{5}{13}\)
⇒ Sin θ = \(\frac{12}{13}\)
∴ sin 2θ = 2 sin θ cos θ
= 2 . \(\frac{12}{13}\)(-\(\frac{5}{13}\)) = –\(\frac{120}{169}\)

Question 33.
For what values of x in the first quadrant \(\frac{2 \tan x}{1-\tan ^{2} x}\) is positive?
Solution:
\(\frac{2 \tan x}{1-\tan ^{2} x}\) > 0 ⇒ tan 2x > 0
⇒ 0 < 2x < \(\frac{\pi}{2}\) (since x is in the first quadrant)
⇒ 0 < x < \(\frac{\pi}{4}\)

Question 34.
If cos θ = \(\frac{-3}{5}\) and π < θ < \(\frac{3 \pi}{2}\), find the value of tan θ/2.
Solution:

Question 35.
If A is not an integral multiple of \(\frac{\pi}{2}\), prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:
i) tan A + cot A = \(\frac{\sin A}{\cos A}\) + \(\frac{\cos A}{\sin A}\)
= \(\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}\)

Question 36.
If A is not an integral multiple of prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:

Question 37.
If θ is not an integral multiple of \(\frac{\pi}{2}\), prove that tan θ + 2 tan 2θ + 4 tan 4θ +
8 cot 8θ = cot θ.
Solution:
From cot A – tan A = 2 cot 2A above tan A = cot A – 2 cot 2A ….(1)
∴ tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = (cot θ – 2 cot 2θ) + 2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ (by (1) above)
= cot θ

Question 38.
For A ∈ R, prove that
i) sin A.sin(π/3 +A).sin(π/3 – A) = \(\frac{1}{4}\) . sin3A
ii) cosA . cos (π/3 + A). cos(π/3 – A)
= \(\frac{1}{4}\) cos 3A and hence deduce that
iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
iv) cos \(\frac{\pi}{9}\), cos \(\frac{2 \pi}{9}\), cos \(\frac{3 \pi}{9}\). cos \(\frac{4 \pi}{9}\) = \(\frac{1}{16}\).
Solution:
i) sin A. sin (π/3 + A). sin (π/3 – A)
= sin A [sin² π/3 – sin² A]


iii) ∵ sin A. sin (60° + A). sin (60° – A)
= \(\frac{1}{4}\) sin 3A
Put A = 20°
⇒ sin 20°. sin (60° + 20°). sin (60° – 20°)
= \(\frac{1}{4}\). sin(3 × 20°) .
⇒ sin 20°. sin 40°. sin 80° = \(\frac{1}{4}\) sin 60°
Multiplying on both sides with sin 60°
We get, sin 20° sin 40° sin 60° sin 80°
= \(\frac{1}{4}\) sin² 60°
= \(\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)^{2}[latex] = [latex]\frac{3}{16}\)

iv) ∵ cos A. cos (60° + A). cos (60° – A) = \(\frac{1}{4}\)cos 3A
Put A = 20°
⇒ cos 20°. cos (60° + 20°) cos (60° – 20°)
= \(\frac{1}{4}\). cos(3 × 20°)
⇒ cos 20°. cos 40°. cos 80° = cos 60°
On multiplying both sides by cos 60°, we get cos 20°. cos 40°. cos 60°. cos 80°
= \(\frac{1}{4}\). cos² 60°

Question 39.
If 3A is not an odd multiple of \(\frac{\pi}{2}\), prove that tan A. tan (60° + A). tan (60° – A)
= tan 3A and hence find the value of tan 6°. tan 42°. tan 66°. tan 78°.
Solution:

∴ tan A . tan \(\left(\frac{\pi}{3}+\mathrm{A}\right)\) tan \(\left(\frac{\pi}{3}-A\right)\) = tan 3A
i.e., tan A tan (60° – A) tan (60° + A) = tan 3A —— (1)
Put A = 6°
⇒ tan 6° tan 54° tan 66° = tan 18° ——- (2)
Put A = 18° in (1)
tan 18° tan 42° tan 78° = tan 54° ——- (3)
put (2) in (3),
(tan 6° tan 54° tan 66°) tan 42° tan 78°
= tan 54
⇒ tan 6° tan 42° tan 66° tan 78° = 1

Question 40.
For α, β ∈ R, prove that (cos α + cos β)² + (sin α + sin β)² = 4 cos² \(\left(\frac{\alpha-\beta}{2}\right)\).
Solution:
LH.S. = (cos α + cos β)² + (sin α + sin β)²
= (cos² α + cos² β + 2 cos α cos β + (sin² α + sin² β + 2 sin α sin β)
= 1 + 1 + 2 (cos α cos β + sin α sin β)
= 2 + 2 cos (α – β) = 2 (1 + cos (α – β)]
= 2[2cos²\(\left(\frac{\alpha-\beta}{2}\right)\)]
= 4.cos²\(\left(\frac{\alpha-\beta}{2}\right)\); [∵ 1 + cos θ = 2 cos² \(\frac{\theta}{2}\)]

Question 43.
If a, b, c are non zero real numbers and α, β are solutions of the equation a cos θ + b sin θ = c then show that
(i) sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\) and
(ii) sin α. sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)
Solution:
∵ a cos θ + b sin θ = c
= a cos θ = c – b sin θ
⇒ (a cos θ)² = (c – b sin θ)²
⇒ a² cos² θ = c² + b² sin² θ – 2bc sinθ
⇒ a² (1 – sin² θ) = c² + b² sin² θ – 2bc sin θ
⇒ (a² + b²) sin² θ – 2bc sin θ + (c² – a²) = 0
This is a quadratic equatioñ in sin θ, whose roots are sin α and sin β (given that α, β are two solutions of θ)
∴ Sum of the roots sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
Product of the roots sin α sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)

Question 42.
If θ is not an odd multiple of \(\frac{\pi}{2}\) and cos θ ≠ \(\frac{-1}{2}\), prove that
\(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}\) = tan θ.
Solution:

Question 43.
Prove that sin⁴ \(\frac{\pi}{8}\) + sin⁴ \(\frac{3 \pi}{8}\) + sin⁴ \(\frac{5 \pi}{8}\) + sin⁴ \(\frac{7 \pi}{8}\) = \(\frac{3}{2}\)
Solution:

Question 44.
If none of 2A and 3A is an odd multiple of \(\frac{\pi}{2}\), then prove that tan 3A. tan 2A. tanA = tan 3A – tan 2A – tan A.
Solution:
∵ tan 3A = tan (2A + A)
= \(\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
⇒ tan 3A(1 – tan 2A tan A) = tan 2A + tan A
⇒ tan 3A – tan A tan 2A tan 3A
= tan 2A + tan A
∴ tan 3A – tan 2A – tan A
= tan A tan 2A tan 3A

Question 45.
Prove that sin 78° + cos 132° = \(\frac{\sqrt{5}-1}{4}\).
Solution:
L.H.S. = sin 78° + cos 132°
= sin 78° + cos (90° + 42°)
= sin 78° – sin 42°

Question 46.
Prove that sin 21° cos 9° – cos 84° cos 6° = \(\frac{1}{4}\).
Solution:
LH.S. = sin 21° cos 9° – cos 84° cos 6°
= \(\frac{1}{2}\)[2 sin 21° cos 9° – 2 cos 84° cos 6°]
= \(\frac{1}{2}\)[sin (21° + 9°) + sin (21° – 9°) – 2 cos (90° – 6°) cos 6°]
= \(\frac{1}{2}\)[sin 30° + sin 12° – 2 sin 6° cos 6°]
= \(\frac{1}{2}\)[\(\frac{1}{2}\) + sin 12° – sin (2 × 16°)
= \(\frac{1}{4}\) = R.H.S.

Question 47.
Find the value of sin 34° + cos 64° – cos 4°.
Solution:
sin 34° + (cos 64° – cos 4°)

Question 48.
Prove that cos² 76° + cos² 16° – cos 76° cos 16° = \(\frac{3}{4} .\)
Solution:
L.H.S. = cos² 76° + cos² 16° – cos 76° cos 16°
= cos² 76° + (1 – sin² 16°) – \(\frac{1}{2}\)(2 cos 76° cos 16°)
= 1 + (cos² 76° – sin² 16°) – \(\frac{1}{2}\)[cos (76° + 16°) + cos (76° – 16°)]
= 1 + cos (76° + 16°). cos (76° – 16°) – \(\frac{1}{2}\)[cos 92° + cos 60°]
= 1 + cos (92°). cos 60° – \(\frac{1}{2}\)cos 92° – \(\frac{1}{2}\)cos 60°
= 1 + \(\frac{1}{2}\). cos 92° – \(\frac{1}{2}\). cos 92° – \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\) = R.H.S.

Question 49.
If a, b, ≠ 0 and sin x + sin y = a and cos x + cos y = b, find two values of
i) tan \(\left(\frac{x+y}{2}\right)\)
ii) sin \(\left(\frac{x-y}{.2}\right)\) interms of a and b.
Solution:

First method:

ii) a² + b² = (sin x + sin y)² + (cos x + cos y)²
= sin² x + sin² y + 2 sin x sin y + cos² x + cos² y + 2 cos x cos y
= 2 + 2(cos x cos y + sin x sin y)
= 2 + 2 cos (x – y)
= 2[1 + cos (x – y)]
a² + b² – 2 = 2 cos (x – y)

Second method:

Question 50.
Prove that cos 12° + cos 84° + cos 132° + cos 156° = –\(\frac{1}{2}\).
Solution:
L.H.S. = cos 12° + cos 84° + cos 132° + cos 156°

Question 51.
Show that for any θ ∈ R
4 sin \(\frac{5 \theta}{2}\) cos \(\frac{3 \theta}{2}\) cos 3θ = sin θ – sin 2θ + sin 4θ + sin 7θ
Solution:
R.H.S. = 4 sin \(\frac{5 \theta}{2}\) cos \(\frac{3 \theta}{2}\) cos 3θ

= 2 [(sin 4θ + sin θ) cos 3θ]
= 2 sin 4θ cos 3θ + 2 sinθ cos 3θ
= sin (4θ + 3θ) + sin (4θ – 3θ) + sin (θ + 3θ) + sin(θ – 3θ)
= sin 7θ + sin θ + sin 4θ – sin 2θ = R.H.S.

Question 52.
If none of A, B, A + B is an integral multiple of π, then prove that

Solution:

Question 53.
For any α ∈ R, provethat cos² (α – π/4) + cos² (α + π/12) – cos² (α – π/12) = \(\frac{1}{2}\).
Solution:

Question 54.
Suppose (α – β) is not an odd multiple of \(\frac{\pi}{2}\), m is a non – zero real number such that m ≠ -1 and \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}\) = \(\frac{1-m}{1+m}\), Then prove that tan \(\left(\frac{\pi}{4}-\alpha\right)\) = m. tan \(\left(\frac{\pi}{4}+\beta\right)\)
Solution:

Question 55.
If A, B, C are the angles of a triangle, prove that
i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
ii) sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
L.H.S. = sin 2A + sin 2B + sin 2C

= 2 sin (A + B) cos (A – B) + sin 2C
= 2 sin (180° – C) cos (A – B) + sin 2C
By(1)
= 2 sin C. cos (A — B) + 2 sin C. cos C
= 2 sin C [cos (A — B) + cos C]
= 2 sin C [cos (A — B) + cos (180° — \(\overline{A+B}\))]
= 2 sin C [cos (A – B) — cos (A + B)]
= 2 sin C (2 sin A sin B)
= 4 sin A sin B sin C = R.H.S.

ii) LH.S. = sin 2A + sin 28 — sin 2C

= 2 sin(A+ B)cos(A—B)—sin2C
= 2 sin (180° — C) cos (A — B) — sin 2C
= 2 sin C. cos(A – B) – 2 sin C cos C
= 2 sin C [cos (A – B) – cos C]
= 2 sin C [cos (A — B)—cos (180°— \(\overline{A+B}\))]
By(1)
= 2 sin C [cos (A — B) + cos (A + B)]
= 2 sin C [2 cos A cas B]
= 4 cos A cos B sin C = R.H.S.

Question 56.
If A, B, C are angles of a triangle, prove that
i) cos 2A + cos 2B + cos 2C
= -4 cos A cos B cos C – 1

ii) cos 2A + cos 2B – cos 2C
= 1 – 4 sin A sin B cos C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ———— (1)
L.H.S. = (cos 2A + cos 2B) + cos 2C
= 2 cos(A + B) cos (A – B) + 2 cos² C – 1
= -2 cos C cos (A – B) – 2 cos C cos (A + B) – 1
[∵ cos (A + B) = -cos C]
= – 1 – 2 cos C[cos (A – B) + cos(A – B)]
= – 1 – 2 cos C[2 cos A cos B]
= -1 – 4 cos A cos B cos C = R.H.S.

ii) L.H.S. = cos 2A + cos 2B – cos 2C

= 2 cos (A + B). cos (A – B) – (2 cos² C – 1)
= 2 cos (180° – C) cos (A – B) – 2 cos² C + 1
= 1 – 2 cos C cos (A – B) – 2 cos² C
= 1 – 2 cos C [cos (A – B) + cos C]
= 1 – 2 cos C[cos(A – B) + cos(180° – \(\overline{\mathrm{A}+\mathrm{B}}\))]
= 1 – 2 cos C [cos (A – B) – cos (A + B)]
= 1 – 2 cos C [2 sin A sin B]
= 1 – 4 sin A sin B cos C = R.H.S.

Question 57.
If A B, C are angles in a triangle, then prove that
i) sin A + sin B + sin C = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)
ii) cos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)
Solution:

Question 58.
If A + B + C = π/2, then show that
i) sin² A + sin² B + sin² C = 1 – 2 sin A sin B sin C
ii) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Solution:
A + B + C = π/2 ——– (1)
L.H.S. = sin² A + sin² B + sin² C
= \(\frac{1}{2}\)(1 – cos2A + 1 – cos 2B + 1 – cos 2C]
= \(\frac{1}{2}\)[3 – (cos 2A + cos 2B + cos 2C)]
= \(\frac{1}{2}\)[3 – (1 + 4 sin A sin B sin C)
(By Problem No. 57(ii)]
(∵ 2A + 2B + 2C = 180°)
= \(\frac{1}{2}\)[2 – 4 sin A sin B sin C]
= 1 – 2 sin A sin B sin C

ii) A + B + C = 90°
⇒ 2A + 2B + 2C = 180°
2A + 2B + 2C 180°
sin 2A + sin 2B + sin 2C
= 4 cos A cos B cos C
By Problem No. 57(i).

Question 59.
If A + B + C = 3π/2, prove that cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C.
Solution:
A + B + C = 3π/2 —— (1)
L.H.S. = cos 2A + cos 2B + cos 2C
= 2 cos (A + B). cos(A – B) +1 – 2 sin² C
= 2 cos (270° – C). cos (A – B) + 1 – 2 sin² C
= 1 – 2 sin C cos (A – B) – 2 sin²C
= 1 – 2 sin C [cos(A – B) + sin C]
= 1 – 2 sin C[cos (A – B) + sin (270° – \(\overline{A+B}\))]
= 1 – 2 sin C[cos(A – B) – cos(A + B)]
= 1 – 2 sin C[2 sin A sin B]
= 1 – 4 sin A sin B sin C

Question 60.
If A B, C are angles of a triangle, then prove that
sin² \(\frac{A}{2}\) + sin² \(\frac{B}{2}\) – sin² \(\frac{C}{2}\) = 1 – 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\) (Mar. ’16, May ’12, ’11)
Solution:

Question 61
If A. B, C are angles of a triangle, then prove that sin \(\frac{\mathbf{A}}{\mathbf{2}}\) + sin \(\frac{\mathbf{B}}{\mathbf{2}}\) + sin \(\frac{\mathbf{C}}{\mathbf{2}}\) = 1 + 4 sin \(\frac{\pi-\mathbf{A}}{4}\) sin \(\frac{\pi-B}{4}\) sin \(\frac{\mathrm{C}}{2}\)
Solution:
A + B + C = 180° ——- (1)

Question 62.
If A + B + C = 0, then prove that cos² A + cos² B + cos² C = 1 + 2 cos A cos B cos C.
Solution:
A + B + C = 0 —— (1)
L.H.S. = cos² A + cos² B + cos² C
= cos² A + (1 – sin² B) + cos² C
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) cos (A – B) + cos² C
= 1 + cos(-C) cos(A – B) + cos² C
By (1)
= 1 + cos C cos (A – B) + cos² C
= 1 + cos C [cos (A – B) + cos c]
= 1 + cos C[cos(A – B) + cos(-B – A)]
By (1)
1 + cos C[cos (A – B) + cos(A + B)]
= 1 + cos C (2 cos A cos B)
= 1 + 2 cos A cos B cos C = R.H.S.

Question 63.
If A + B + C = 25, then prove that cos (S – A) + cos (S – B) + cos (S – C) + cos (S) = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\).
Solution:

Andhra Pradesh BIEAP AP Inter 1st Year Civics Study Material 1st Lesson Scope and Significance of Political Science Textbook Questions and Answers.

AP Inter 1st Year Civics Study Material 1st Lesson Scope and Significance of Political Science

Long Answer Questions

Question 1.
Define Political Science and explain its scope. [A.P. Mar, 17, 16, 15]
(or)
Explain the scope of Political Science.
Answer:
Introduction:
Political Science is a premier social science. It is mainly concerned with the study of the state in its relation to Society, Citizens, Associations, and the world at large. Aristotle is hailed as the Father of Political Science. He wrote famous book “THE POLITICS”.

Origin of the word Politics :
Aristotle, the Father of Political Science used the term “POLITICS” for the first time in his famous book “POUTICS”. The term “POLITICS” is derived from a greek word “POLIS” and latin word “POLITICUS” which means the city state.

Definitions of Political Science :
Political Scientists gave various definitions on Political Science. They are as follows :
1. J.W. Gamer:
“Political Science begins and ends with the state”.

2. Stephen Leacock:
“Political Science deals with the government”.

3. David Easton :
“Political Science is concerned with the authoritative allocation of values for a society”.

Scope of Political Science :
The scope of Political Science means the subject matter covered by it or the topics which are included in its study. It may be explained in the following ways :

i) Study of man in Relation to the Society and State :
Aristotle stated that “Man is a Social Animal”. Man can satisfy his basic needs like food, clothing, shelter and protection in the society. Political Science explains the relationship beween man and society. It examines how man should adjust himself with the societys. It is imperative that the modem man should develop proper attitude towards the society. This is possible only when he identifies himself with the society.

Political Science is concerned with the perennial and central issue of establishing proper relationship between the state and the individuals. It deals with many topics of state activity, such as limitations of Political Authority and sphere of Individual Freedoms.

ii) Study of State :
Aristotle also stated that man is a Political Animal. State is a human and political institution. It came into existence for the sake of man and continue in existence for providing happy and prosperous life for man. Individuals became members of the state since, its inception. We can’t imagine the life of individuals outside of the state. Political Science studies the intimate Relationship between the state and the citizens. It also studies the Nature, Functions and Various theories of state authority.

It also comprises a study of the various activities of the state from that of ancient police state to the modem welfare state. Thus, Political Science deals with the Present, Past and FUture aspects of the state.

iii) Study of the Government :
Government is an important essential element of Modem State. It is an instrument which fulfills aims and goals of the state. There can be no state with out a government. Government formulates, expresses and implements the will of the state. Government consists of three organs namely Legislature, Executive and Judiciary. Legislature makes the laws, Executive implements the laws and Judiciary interpretes the laws. Political Science studies the meaning, forms, structure, nature and functions of the government. It also discusses the relationship among the various organs of the government. Hence, Political Science is treated as a science of government.

iv) Study of Associations and Institutions :
Associations and Institutions help the Individuals for their moral, religious, cultural, scientific and technological progress. These carry on their activities at local, regional, national and international levels. Individuals join as members in these Associations out of their interests or purposes. There prevails a great linkage between these voluntary Associations and Institutions. Associations and Institutions in Modem times play a significant role in the Formulation and Implementation of policies of the state and government. Voluntary bodies such as trade unions, peasant groups, professional bodies etc., will have a great impact on the state and government. Political Science explains the nature, structure and functions of the various Associations and Insti-tutions.

v) Study of Rights and Responsibilities:
Scope of Political Science includes the study of Rights and Responsibilities of citizens. Citizens in democratic states enjoy certain rights such as right to life, right to liberty, right to property etc. Political Science enumerates the definition, classification and different theories of Rights. Similarly, citizens will have some Responsibilities towards the state. These include paying taxes, obeying the laws etc. It explains the significance of Rights and Responsibilities of the citizens. Hence, Political Science examines the Realtionship between Rights and Responsibilities.

vi) Study of National and International Issues:
The scope of Political Science covers various issues of Modem state in relation with other states in matters of safeguarding Ter-ritorial integrity and Sovereignty. It studies the topics like Cold war, Balance of power, Disarmament, Detente etc. Modem states are not isolated. They depend upon other states in many spheres like importing raw materials, exporting finished goods, transport, technology, services and communications. This requires close relations among the states in international sphere. Political Science discusses not only the domestic policies of the state but also the issues of international dimensions. It covers a wide range of topics such as diplomacy, international politics, international law, international organisations etc.

vii) Study of Power:
The behaviouralists of 20th century regarded Political Science as a study of sharing and shaping of power. They pointed out that Political Science discusses how power is grabbed, manipulated anti perpetuated to have a control over the society. Morgenthau defined the power as “Man’s control over the Minds and Actions of other Men”.

viii) Study of Public Policy:
Modem Political Scientists like David Easton and Gabriel Almond argued that Political Science is a “Policy Science”. They considered Political Science as the study Of formulation, execution and evaluation of Public Policy, with the advent of Public Policy the scope of Political Science has further widened to include the dimensions of vital topics such as Industrial Policy, Agricultural Policy, Land Reform Policy, Education Policy, Population Policy etc. Public Policy of a Nation in the context of International Relations plays a crucial role in the formulation of diplomatic, economic, military and scientific strategies.

Conclusion:
The above contents show the wide range of subjects that come under the purview of Political Science.

Question 2.
Discuss the significance of the study of Political Science. [A.P. Mar. 19]
Answer:
The study of Political Science is very useful and valuable. It’s knowledge is indispensible to the rulers as well as the ruled. Its significance or importance is analysed as follows.

i) Information about the State :
The primary aim of studying Political Science is to inculcate knowledge of the state. It’s origin, nature, Structure and functions. Knowledge of the state is of great significance to every one. Solutions to various political issues are found only when we have a proper understanding of the political institutions.

ii) Knowledge of Government and Administration :
The Administrators, political leaders and diplomats who manipulate the affairs of the state require a sound knowledge of political science in order to perform their functions with efficiency. Political Science creates awareness about the organisation, control and coordination of Administrative machinery. It also covers the study of local self-governments like Gram Panchayats, Mandal Parishads, Zilla Parishads, Municipalities, Corporations etc.

iii) Provides information about Democratic values :
Political Science provides accurate information about the various political terms such as State, Government, Nation, Con-stitution, Democracy, Sovereignty, Law etc., which are used commonly. Political Science also provides a good knowledge and awareness about the Democratic values like Liberty, Equality, Fraternity, Justice and Rights.

iv) Makes Democracy Successful:
At present Democracy is in vogue in several coun-tries of the world. It is the best form of government. People in democracy elect their repre-sentatives and are ruled by them. If honest, selfless and committed representatives are not elected, the expectations of the people will not be fulfilled. Political Science explains the significance of Franchise. It educates the common men on the conditions essential for the successful functioning of democratic government.

v) Awareness about rights and responsibilities:
The study of Political Science makes people conscious of their rights and responsibilities. It also enables the citizens to be familiar with their rights and responsibilities and the interrelationship between the two. Its study makes the citizens to realize the fact that a proper exercise of rights and responsibilities is a must for leading civilised life.

vi) Teaches the Qualities of Good Citizenship :
Political Science impacts the best civic knowledge by explaining the qualities of good citizenship like cooperation, sacrifice, patriotism, obedience to the state and to the laws, farsightedness, social service etc. It trains the people to become ideal citizens.

vii) Knowledge about World Affairs:
The study of Political Science enriches individual’s knowledge on World Affairs. Political Science is useful for observing and understanding the contemporary world affairs. It stimulates right thinking, broad vision and universal understanding of various phenomena of international politics. It’s study is useful for proper’ analysis and solution of various National and International Issues.

viii) Provides knowledge about International Organizations :
A citizen of the present is not only a member of the state but also a member of the world at large. The study of Political Science promotes the spirit of internationalism. It provides a good knowledge about International Organizations like United Nations Organisation, SAARC, ASEAN, NAM, OPEC etc. It teaches about the need for harmonious relations among the nations.

ix) Develops Political Awareness :
The study of Political Science is of great impor-tance in the sense that it imports best political knowledge. It explains about the structure and functioning of different political institutions like State Government, Political parties, Occupational Associations. Hence the study of Political Science is the best training ground for knowing, following and practising the art of leadership.

x) Explains the need for Co-Operation and Toleration :
National Integration has become a crucial factor in several states. Many obstacles like communalism, linguism, sub-national regional and sub-regional feelings etc., have been threatening the national inte-gration in these states. In this context, the study of Political Science teaches about the need for adjustment, cooperation and toleration.

xi) Knowledge of Political Science is Indispensable :
The study of Political Science helps everyone to understand the mechanism and constitutional system of modem govern-ments. It creates awareness about the contemporary issues in national and international spheres. If creates awareness about Rights and Responsibilities. It’s study is essentially indispensable for the people in developing nations like India. As the majority people in these states are poor, ignorant, illiterate and sentimental in their out look. The study of Political Science inculcates a good political knowledge among them.

Aristotle regarded Political Science as the supreme science and the master of all social sciences.

Question 3.
Define Political Science and explain its nature.
Answe:
Political Science is a premier Social Science. It is mainly concerned with the study of the state in it’s relation with society, citizens, associations and the world at large. Aristotle is hailed as the Father of Political Science. He wrote famous book ‘THE POLITICS”. The term “politics” is derived from a greek word” “POLIS” and latin word “POLiTICUS” which means the city state.

Definition of Political Science:
Political Scientists gave various definitions on Political Science. They are as follows :
1. J.W. Gamer:
“Political Science begins and ends with the state”.

2. Stephen Leacock:
“Political Science deals with the government”.

Nature of Political Science:
There prevailed a controversy among the political scien-tists in regard to the nature of Political Science. Some viewed Political Science as a science others treated it as an art. Let us examine the two aspects (Science and Art) of Political Science.

i) Political Science is a SCIENCE :
According to political philosophers like Aristotle, Bluntschli, Hobbes, Montesque and others Political Science is considered as a science on the following grounds.

1. Political Science is studied in a systematic manner.
2. Experimentation is possible in politics. Principles are applied in the actual organisation of Political Institutions.
3. Political Science, like other sciences, has absolute and universal laws.
4. Predictions are easily applicable in politics.
5. Certain generally agreed principles can be incorporated into the study of Political Science.
6. Political Science, like other sciences, gives scope for establishing relationship be-tween cause and effect.

ii) Political Science is an ART :
According to political writers like Barker, J.S. Mill, Laski and others Political Science is considered as an ART on the following grounds.

1. Political Science has no absolute and universal laws like physical sciences.
2. The phenomena studied in Political Science are interpreted in various ways de-pending upon the context and situation. So it lacks uniformity in the interpretation of various concepts.
3. Political Science is not evolutionary in nature as its concepts are not developed in a steady, regular and continuous manner.
4. Scientific methods of observation and experimentation are not applicable in political science.
5. Complete objectivity and detachment are not found in the case of various phenom-ena in Political Science.
6. Political Science gives no scope for accuracy.

Short Answer Questions

Question 1.
Write about the traditional definitions of Political Science.
Answer:
Traditional definitions of Political Science may be classified under three sub catego-ries. They are mentioned as follows.
i) Political Science – A Study of the State :
Political philosophers like J.W.Gamer, R.G.Gettle, Appadovai and others defined Political Science as a study of the state.

1. J.W.Gamer:
“Political Science begins and ends with the state”.

2. R.G.GettIe :
“Political Science is a historical investigation of what the state has been, an analytical study of what the state is and a politico-ettical discussion of what the state should be”.

3. Appadorai:
“Political Science is concerned with the conditions essential for the existence and development of the state”.

ii) Political Science – A study of the Government:
Some political philosophers like Stephen Leacock, J.R.Seely and others defined Political Science as a study of the govern-ment.

1. Stephen Leacock :
“Political Science deals with the Government”.

2. J.R.Seely :
“Political Science investigates the phenomenon of the Government”.

iii) Political Science – A study of the State and the Government:
Political philoso-phers like Paul Janet, Gilchrist, Catlin and others defined political science as a study of the state and the government.

1. Paul Janet:
“Political Science is that part of social science which treats the founda-tions of state and the principles of the government”.

2. R.N.Gilchrist :
“Political Science is a study of foundations of the state and the government”.

3. Catlin :
“Political Science is a study of the political activities of individuals and various organs of government”.

Question 2.
What are the various modem definitions of Political Science?
Answer:
Modem definitions of Political Science can be classified into two sub-categories. They are discussed as follows..
i) Political Science – A study of power:
1. Lasswell and Kaplan :
“Political Science as an empirical discipline, is the study of shaping and sharing of power”.

2. William A.Robson :
“Political Science is primarily concerned with the power in society”.

ii) Political Science – A study of Allocation of values :
1. David Easton :
“Political Science is concerned with the authoritative allocation of values for a society”.

2. Hill Man :
“Politics is the Science of who gets what, when, and why”.

Question 3.
Explain about the nature of Political Science.
Answer:
Political Science is a premier Social Science. It is mainly concerned with the study of the state in it’s relation with society, citizens, associations and the world at large. Aristotle is hailed as the Father of Political Science. He wrote famous book ‘THE POLITICS”. The term “politics” is derived from a greek word” “POLIS” and latin word “POLiTICUS” which means the city state.

Definition of Political Science:
Political Scientists gave various definitions on Political Science. They are as follows :
1. J.W. Gamer:
“Political Science begins and ends with the state”.

2. Stephen Leacock:
“Political Science deals with the government”.

Nature of Political Science:
There prevailed a controversy among the political scien-tists in regard to the nature of Political Science. Some viewed Political Science as a science others treated it as an art. Let us examine the two aspects (Science and Art) of Political Science.

i) Political Science is a SCIENCE :
According to political philosophers like Aristotle, Bluntschli, Hobbes, Montesque and others Political Science is considered as a science on the following grounds.

1. Political Science is studied in a systematic manner.
2. Experimentation is possible in politics. Principles are applied in the actual organisation of Political Institutions.
3. Political Science, like other sciences, has absolute and universal laws.
4. Predictions are easily applicable in politics.
5. Certain generally agreed principles can be incorporated into the study of Political Science.
6. Political Science, like other sciences, gives scope for establishing relationship be-tween cause and effect.

ii) Political Science is an ART :
According to political writers like Barker, J.S. Mill, Laski and others Political Science is considered as an ART on the following grounds.

1. Political Science has no absolute and universal laws like physical sciences.
2. The phenomena studied in Political Science are interpreted in various ways de-pending upon the context and situation. So it lacks uniformity in the interpretation of various concepts.
3. Political Science is not evolutionary in nature as its concepts are not developed in a steady, regular and continuous manner.
4. Scientific methods of observation and experimentation are not applicable in political science.
5. Complete objectivity and detachment are not found in the case of various phenom-ena in Political Science.
6. Political Science gives no scope for accuracy.

Question 4.
Mention any three topics coverd under the scope of Political Science.
Answer:
i) Study of man in Relation to the Society and State:
Aristotle stated that “Man is a Social Animal”. Man can satisfy his basic needs like food, clothing, shelter and protection in the society. Political Science explains the relationship beween man and society. It exam- ines how man should adjust himself with the societys. It is imperative that the modem man should develop proper attitude towards the society. This is possible only when he identifies himself with the society.

Political Science is concerned with the perennial and central issue of establishing proper relationship between the state and the individuals. It deals with many topics of state activity, such as limitations of Political Authority and sphere of Individual Freedoms.

ii) Study of State :
Aristotle also stated that man is a Political Animal. State is a hu-man and political institution. It came into existence for the sake of man and continue in existence for providing happy and prosperous life for man. Individuals became members of the state since its inception. We can’t imagine the life of individuals outside of the state. Political Science studies the intimate Relationship between the state and the citizens. It also studies the Nature, Functions and Various theories of state authority.

It also comprises a study of the various activities of the state from that of ancient police state to the modem welfare state. Thus, Political Science deals with the Present, Past and Future aspects of the state.

iii) Study of Associations and Institutions :
Associations and Institutions help the Individuals for their moral, religious, cultural, scientific and technological progress. These cany on their activities at local, regional, national and international levels. Individuals join as members in these Associations out of their interests or purposes. There prevails a great link-age between these voluntary Associations and Institutions. Associations and Institutions in Modem times play a significant role in the Formulation and Implementation of policies of the state and government. Voluntary bodies such as trade unions, peasant groups, professional bodies etc., will have a great impact on the state and government. Political Science explains the nature, structure and functions of the various Associations and Institutions.

Question 5.
Describe the scope of Political Science in the sphere of Government.
Ansnswer:
Scope of Political Science includes the study of government. Some political scientists like Stephen Leacock and John Richard Seeley confined the scope of the discipline to the government alone. Political Science mainly studies about the government Government is an agency of the state. There can be no State without a government. The state realises its aims through the instrument of government. Government formulates, expresses and implements the will of the state. There must be some men or body of men who are authorised to issue orders on behalf of the state. They are known as the government.

Political Science studies the meaning, forms, structure, nature and functions of the government. It discusses the relationship among the various organs of the government. It makes a differentiation between the State and Government. While dealing with the government, Political Science narrates the classification of various governments as given by Aristotle, Leacock and others. Political Science discusses the various merits and demerits, essential conditions and manifold activities of the above governments. Hence, political Science is treated as a science of government.

Question 6.
”Political Science is a study of the present, past and future of the state”. Analyse this statement
Answer:
Political scientists like Gamer and Paul Janet viewed Political Science as a study of the affairs of the state. They conceived the state as a Political Institution. The state is indispensable for every individual. Political Science studies the intimate relationship between the state and the citizens. Political Science studies the state in present, in the past and in future.

i) Study of state in the Present :
Political Science deals with the state as it exists today. It explains the meaning, nature, purpose, growth and functioning of the state. It also deals with public opinion, political parties and pressure groups which seek to capture the political power or influence public policies.

ii) Study of state in the Past:
Political Science explains about the origin and transfor-mation of the state. It also discusses about the diverse political institutions that existed within the state. It studies various factors that influenced the origin and evolution of the state. This sort of historical study is possible only in political science.

iii) Study of State in Future :
Political Science tries to determine the principles and concepts of a model state. It lays down the conditions under which a perfect state is real-ized. Political Scientists conceive the future with a view to improve the standards of political institutions and their activities in the light of changing conditions.

On the whole, the scope of Political Science includes the study of various activities; of the state from that of ancient police state to the modem welfare state. Thus, Political Science deals with the present, past and future aspects of the state.

Very Short Answer Questions

Question 1.
Write about Ancient city states.
Answer:
Ancient Greece consisted of a large number of city states. They were small both in size and population. For example, Athens,Sparta, Corinth. Each city state had its own gov-ernment. The greeks based their political philosophy on the concept of city-state. The popu-lation of the city-state was divided into three groups : 1) Citizens 2) Aliens and 3) Slaves.

Question 2.
Give any two traditional definitions of Political Science.
Answer:
Political scientists gave various definitions on Political Science. They are as follows.

1. J.W.Gamer:
“Political Science begins and ends with the State”.

2. Stephen Leacock:
“Political Science deals with the Government”.

3. David Easton :
“Political Science is concerned with the authoritative allocation of values for a society”.

Question 3.
Write about any two modem definitions of Political Science.
Answer:
Modem definitions of Political Science can be classified into two sub-categories. They are discussed as follows..
i) Political Science – A study of power:
1. Lasswell and Kaplan :
“Political Science as an empirical discipline, is the study of shaping and sharing of power”.

2. William A.Robson :
“Political Science is primarily concerned with the power in society”.

ii) Political Science – A study of Allocation of values :
1. David Easton :
“Political Science is concerned with the authoritative allocation of values for a society”.

2. Hill Man :
“Politics is the Science of who gets what, when and why”.

Question 4.
How does Political Science teach the qualities of good citizenship?
Answer:
Political Science imparts the best civic knowledge by explaining the qualities of good citizenship like cooperation, sacrifice, patriotism, obedience to the state and to the laws, forsightedness, social service etc. It trains the people to become ideal citizens.

Question 5.
Justify the statement that Political Science in an Art.
Answer:
Political Science is an ART:
According to political writers like Barker, J.S. Mill, Laski and others Political Science is considered as an ART on the following grounds.

1. Political,Science has no absolute and universal laws like physical sciences.
2. The phenomena studied in Political Science are interpreted in various ways de pending upon the context and situation. So it lacks uniformity in the interpretation of various concepts.
3. Political Science is not evolutionary in nature as its concepts are not developed in a steady, regular and continuous manner.
4. Scientific methods of observation and experimentation are not applicable in political science.
5. Complete objectivity and detachment are not found in the case of various phenom-ena in Political Science.
6. Political Science gives no scope for accuracy.

Question 6.
On what grounds is Political Science considered as a Science?
Answer:
Political Science is a SCIENCE :
According to political philosophers like Aristotle, Bluntschli, Hobbes, Montesque and others Political Science is considered as a science on the following grounds.

1. Political Science is studied in a systematic manner.
2. Experimentation is possible in politics. Principles are applied in the actual organisation of Political Institutions.
3. Political Science, like other sciences, has absolute and universal laws.

Question 7.
Name any four topics covered under the scope of Political Science.
Answer:
The scope of Political Science comprises the following points.

1. Study of man in relation to the society and state.
2. Study of the state.
3. Study of the government.
4. Study of Associations and Institutions.

Question 8.
In what way is Political Science considered as a study of Government?
Answer:
Study of the Government: Government is an important essential element of Modem State. It is an instrument which fulfills aims and goals of the state. There can be no state with out a government. Government formulates, expresses and implements the will of the state. Government consists of three organs namely Legislature, Executive and Judiciary. Legislature makes the laws, Executive implements the laws and Judiciary interpretes the laws. Political Science studies the meaning, forms, structure, nature and functions of the government. It also discusses the relationship among the various organs of the government. Hence, Political Science is treated as a science of government.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Equations Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Equations Solutions Exercise 7(a)

I.

Question 1.
Find the principal solutions of the angles in the equations
(i) 2 cos²θ = 1
Solution:
cos²θ = \(\frac{1}{2}\)
θ = 45°, 135°

(ii) √3 sec θ + 2 = 0
Solution:
sec θ = \(\frac{-2}{\sqrt{3}}\)
⇒ cos θ = \(\frac{-\sqrt{3}}{2}\)
⇒ θ = 150°

(iii) 3 tan²θ = 1
Solution:
tan²θ = \(\frac{1}{3}\)
θ = ±\(\frac{\pi}{6}\)

Question 2.
Solve the following equations.
(i) cos 2θ = \(\frac{\sqrt{5}+1}{4}\), θ ∈ [0, 2π]
Solution:
cos 2θ = \(\frac{\sqrt{5}+1}{4}\)
⇒ cos 2θ = cos 36° = cos (\(\frac{\pi}{5}\))
and \(\frac{\pi}{5}\) ∈ [0, 2π]
2θ = \(\frac{\pi}{5}\) ⇒ θ = \(\frac{\pi}{10}\) is the principal solution
and 2θ = 2nπ ± \(\frac{\pi}{5}\) where n ∈ Z is the general solution
⇒ θ = nπ ± \(\frac{\pi}{10}\)
The values of θ in [0, 2π] are \(\left\{\frac{\pi}{10}, \frac{9 \pi}{10}, \frac{11 \pi}{10}, \frac{19 \pi}{10}\right\}\)

(ii) tan²θ = 1, θ ∈ [-π, π]
Solution:
tan²θ = 1 ⇒ tan θ = ±1
tan θ = ±1 = tan (\(\pm \frac{\pi}{4}\))
The principal solutions are θ = \(\pm \frac{\pi}{4}\)
and the general solution is given by nπ ± \(\frac{\pi}{4}\), n ∈ Z
Put n = -1, 0, 1
\(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}\right\}\) is the solution set for the given equation in [-π, π]

(iii) sin 3θ = \(\frac{\sqrt{3}}{2}\), θ ∈ [-π, π]
Solution:
sin 3θ = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
and \(\frac{\pi}{3}\) ∈ [-π, π]
∴ 3θ = \(\frac{\pi}{3}\) is the principal solution
and 3θ = \(n \pi+(-1)^{n} \frac{\pi}{3}\), n ∈ Z
⇒ θ = \(\frac{n \pi}{3}+(-1)^{n} \cdot \frac{\pi}{9}\), n ∈ Z is the general solution.
The solution of θ in [-π, π] are \(\left\{\frac{-5 \pi}{9}, \frac{-4 \pi}{9}, \frac{\pi}{9}, \frac{2 \pi}{9}, \frac{7 \pi}{9}, \frac{8 \pi}{9}\right\}\)

(iv) cos²θ = \(\frac{3}{4}\), θ ∈ [0, π]
Solution:
cos²θ = \(\frac{3}{4}\)
⇒ cos θ = ±\(\frac{\sqrt{3}}{4}\)
The general solution is given by
θ = nπ ± \(\frac{\pi}{6}\), n ∈ Z
Put n = 0, 1
The solution set for the given equation in [0, π] is \(\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}\)

(v) 2 sin²θ = sin θ, θ ∈ (0, π)
Solution:
2 sin²θ – sin θ = 0
sin θ (2 sin θ – 1) = 0
sin θ = 0 or sin θ = \(\frac{1}{2}\)
since θ ∈ (0, π)
∴ The solution of θ = \(\left\{\frac{\pi}{6}+\frac{5 \pi}{6}\right\}\)

Question 3.
Find general solutions to the following equations.
(i) sin θ = \(\frac{\sqrt{3}}{2}\), cos θ = \(\frac{-1}{2}\)
Solution:
sin θ = \(\frac{\sqrt{3}}{2}\), cos θ = \(\frac{-1}{2}\)
∵ sin θ is +ve and cos θ is -ve
⇒ θ lies in II quadrant
sin θ = \(\frac{\sqrt{3}}{2}=\sin \left(\frac{2 \pi}{3}\right)\)
cos θ = \(\frac{-1}{2}=\cos \left(\frac{2 \pi}{3}\right)\)
⇒ θ = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ Z is the general solution.

(ii) tan x = \(\frac{-1}{\sqrt{3}}\), sec x = \(\frac{2}{\sqrt{3}}\)
Solution:
∵ tan x = \(\frac{-1}{\sqrt{3}}\) and sec x = \(\frac{2}{\sqrt{3}}\)
⇒ x lies in IV quadrant
tan x = \(\frac{-1}{\sqrt{3}}=\tan \left(\frac{-\pi}{6}\right)\)
sec x = \(\frac{2}{\sqrt{3}}=\sec \left(\frac{-\pi}{6}\right)\)
∴ θ = 2nπ + \(\left(\frac{-\pi}{6}\right)\), n ∈ Z is the general solution.

(iii) cosec θ = -2, cot θ = -√3
Solution:
cosec θ = -2, cot θ = -√3
⇒ θ lies in IV quadrant
cosec θ = -2
⇒ sin θ = \(-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right)\)
cot θ = -√3
⇒ tan θ = \(-\frac{1}{\sqrt{3}}=\tan \left(-\frac{\pi}{6}\right)\)
∴ θ = 2nπ + \(\left(\frac{-\pi}{6}\right)\), n ∈ Z is the general solution.

Question 4.
(i) If sin (270° – x) = cos 292°, then find x in (0, 360°).
Solution:
sin (270° – x) = cos (292°)
⇒ -cos x = cos (180° + 112°)
⇒ -cos x = -cos 112°
⇒ cos x = cos 112°
⇒ x = 112° or x = 360° – 112° = 248°

(ii) If x < 90°and sin (x + 28°) = cos (3x – 78°), then find x.
Solution:
sin (x + 28°) = cos (3x – 78°)
= sin (90° – 3x + 78°)
= sin (168° – 3x)
x + 28° = 168° – 3x + 28° (180°) or
= 180° – (168° – 3x) + 2x (180°)
⇔ there exists n ∈ Z such that
4x = 140° + 2x (180°)
2x = 16° – 2x (180°)
⇔ there exists n ∈ z such that
x = 35° + x(90°) or x = 8° – x (180°)
Hence x = 8° and x = 35° are the only values of x that lie (0, 90°) and satisfy the given equation.

Question 5.
Find general solutions to the following equations.
(i) 2 sin²θ = 3 cos θ
Solution:
2 sin²θ = 3 cos θ
⇒ 2(1 – cos²θ) = 3 cos θ
⇒ 2 cos²θ + 3 cos θ – 2 = 0
⇒ 2 cos²θ + 4 cos θ – cos θ – 2 = 0
⇒ 2 cos θ (cos θ + 2) – 1 (cos θ + 2) = 0
⇒ (2 cos θ – 1) (cos θ + 2) = 0
⇒ cos θ = \(\frac{1}{2}\) (or) cos θ = -2
∴ The range of cos θ is [-1, 1]
cos θ = -2 is not admissible
∴ cos θ = \(\frac{1}{2}=\cos \frac{\pi}{3}\)
⇒ θ = \(\frac{\pi}{3}\) is the principal solution and
θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.

(ii) sin²θ – cos θ = \(\frac{1}{4}\)
Solution:
sin²θ – cos θ = \(\frac{1}{4}\)
⇒ 4(1 – cos²θ) – 4 cos θ = 1
⇒ 4 cos²θ + 4 cos θ – 3 = 0
⇒ 4 cos²θ + 6 cos θ – 2 cos θ – 3 = 0
⇒ 2 cos θ (2 cos θ + 3) – (2 cos θ + 3) = 0
⇒ (2 cos θ – 1) (2 cos θ + 3) = 0
∴ cos θ = \(\frac{1}{2}\) (or) cos θ = \(\frac{-3}{2}\)
∵ The range of cos θ is [-1, 1]
cos θ = \(\frac{-3}{2}\) is not admisable
∴ cos θ = \(\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)\)
θ = \(\frac{\pi}{3}\) is the principal solution and
θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.

(iii) 5 cos²θ + 7 sin²θ = 6
Solution:
5 cos²θ + 7 sin²θ = 6
Dividing by cos²θ
⇒ 5 + 7 tan²θ = 6 sec²θ
⇒ 5 + 7 tan²θ = 6(1 + tan²θ)
⇒ tan²θ = 1
⇒ tan θ = ±1
∴ θ = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution.

(iv) 3 sin⁴x + cos⁴x = 1
Solution:
3 sin⁴x + cos⁴x = 1
⇒ 3 sin⁴x + (cos²x)² = 1
⇒ 3 sin⁴x + (1 – sin²x)² = 1
⇒ 3 sin⁴x + 1 + sin⁴x – 2 sin²x = 1
⇒ 4 sin⁴x – 2 sin²x = 0
⇒ 2 sin²x (2 sin²x – 1) = 0
⇒ sin x = 0 (or) sin x = ±\(\frac{1}{\sqrt{2}}\)
If sin x = 0
⇔ x = nπ, n ∈ Z is the general solution.
If sin x = ±\(\frac{1}{\sqrt{2}}\)
⇒ x = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution.
∴ General solution is x = nπ (or) nπ ± \(\frac{\pi}{4}\), n ∈ Z.

II.

Question 1.
Solve the following equations and write a general solution.
(i) 2 sin²θ – 4 = 5 cos θ
Solution:
2(1 – cos²θ) – 4 = 5 cos θ
2 – 2cos²θ – 4 = 5 cos θ
2 cos²θ + 5 cos θ + 2 = 0
2 cos²θ + 4 cos θ + cos θ + 2 = 0
2 cos θ (cos θ + 2) + 1 (cos θ + 2) = 0
(cos θ + 2) (2 cos θ + 1) = 0
cos θ = -2 or cos θ = \(-\frac{1}{2}\)
cos θ = -2 is not possible
∴ cos θ = \(-\frac{1}{2}\) = cos \(\frac{2 \pi}{3}\)
∴ General solution is θ = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ Z

(ii) 2 + √3 sec x – 4 cos x = 2√3
Solution:
2 + √3 sec x – 4 cos x = 2√3
\(\frac{2 \cos x+\sqrt{3}-4 \cos ^{2} x}{\cos x}\) = 2√3
2 cos x + √3 – 4 cos²x = 2√3 cos x
4 cos²x + 2√3 cos x – 2 cos x – √3 = 0
2 cos x (2 cos x + √3) – 1 (2 cos x + √3) = 0
(2 cos x – 1) (2 cos x + √3) = 0
cos x = \(\frac{1}{2}\) (or) cos x = \(\frac{-\sqrt{3}}{2}\)
If cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\), n ∈ Z
General solution x = 2nπ ± \(\frac{\pi}{3}\)
If cos x = \(\frac{-\sqrt{3}}{2}\) = cos \(\frac{5 \pi}{3}\)
General solution x = 2nπ ± \(\frac{5 \pi}{3}\), n ∈ Z

(iii) 2 cos²θ + 11 sin θ = 7
Solution:
2 (1 – sin²θ) + 11 sin θ = 7
2 – 2 sin²θ + 11 sin θ = 7
2 sin²θ – 11 sin θ + 5 = 0
2 sin²θ – 10 sin θ – sin θ + 5 = 0
2 sin θ (sin θ – 5) – 1(sin θ – 5) = 0
(sin θ – 5) (2 sin θ – 1) = 0
sin θ = 5 or sin θ = \(\frac{1}{2}\)
If sin θ = 5 is not possible
∴ sin θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
General solultion θ = nπ + (-1)ⁿ \(\frac{\pi}{6}\), n ∈ Z

(iv) 6 tan²x – 2 cos²x = cos 2x
Solution:
6 tan²x – 2 cos²x = cos 2x
⇒ 6(sec²x – 1) – 2 cos²x = 2 cos²x – 1
⇒ 6 sec²x – 6 – 4 cos²x + 1 = 0
⇒ 6 sec²x – 4 cos²x – 5 = 0
⇒ \(\frac{6}{\cos ^{2} x}\) – 4 cos²x – 5 = 0
⇒ 6 – 4 cos⁴x – 5 cos²x = 0
⇒ 4 cos⁴x + 5 cos²x – 6 = 0
⇒ 4 cos⁴x + 8 cos²x – 3 cos²x – 6 = 0
⇒ 4 cos²x (cos²x + 2) – 3(cos²x + 2) = 0
⇒ (4 cos²x – 3) (cos²x + 2) = 0
⇒ 4 cos²x = 3, cos²x ≠ -2
⇒ cos x = ±\(\frac{\sqrt{3}}{2}\)
∴ x = nπ ± \(\frac{\pi}{6}\), n ∈ Z is the general solution.

(v) 4 cos²θ + √3 = 2(√3 + 1) cos θ
Solution:
4 cos²θ – 2(√3 + 1) cos θ + √3 = 0
⇒ 4 cos²θ – 2√3 cos θ – 2 cos θ + √3 = 0
⇒ 2 cos θ (2 cos θ – √3) – 1(2 cos θ – √3) = 0
⇒ (2 cos θ – 1) (2 cos θ – √3) = 0
⇒ cos θ = \(\frac{1}{2}\) (or) cos θ = \(\frac{\sqrt{3}}{2}\)
If cos θ = \(\frac{1}{2}\) = cos \(\left(\frac{\pi}{3}\right)\)
∴ θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.
If cos θ = \(\frac{\sqrt{3}}{2}\) = cos \(\left(\frac{\pi}{6}\right)\)
∴ θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z is the general solution.

(vi) 1 + sin 2x – (sin 3x – cos 3x)²
Solution:
1 + sin 2x = (sin 3x – cos 3x)²
⇒ 1 + sin 2x = sin²3x + cos²3x – 2 sin 3x cos 3x
⇒ 1 + sin 2x = 1 – sin (2 × 3x)
⇒ sin 6x + sin 2x = 0
⇒ \(2 \sin \left(\frac{6 x+2 x}{2}\right) \cdot \cos \left(\frac{6 x-2 x}{2}\right)=0\)
⇒ sin (4x) . cos (2x) = 0
⇒ cos 2x = 0 (or) sin 4x = 0
If cos 2x = 0 = cos \(\frac{\pi}{2}\)
⇒ 2x = \(\frac{\pi}{2}\) is the principal solution and
2x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z is the general solution,
so that x = \(\frac{n \pi}{2}+\frac{\pi}{4}\), n ∈ Z
If sin 4x = 0 = sin(nπ), n ∈ Z
4x = nπ, n ∈ Z is the general solution.
So that 4x = nπ
⇒ x = \(\frac{n \pi}{4}\), n ∈ Z
∴ x = \(\frac{n \pi}{4}\); \(\frac{n \pi}{2}+\frac{\pi}{4}\), n ∈ Z is the general solution.

(vii) 2 sin²x + sin²2x = 2
Solution:
2 sin²x + sin²(2x) = 2
⇒ 2 sin2x + (2 sin x cos x)2 – 2 = 0
⇒ sin²x + 2 sin²x cos²x – 1 = 0
⇒ 2 sin²x cos²x – (1 – sin²x) = 0
⇒ 2 sin²x cos²x – cos²x = 0
⇒ cos²x (2 sin²x – 1) = 0
⇒ cos x = 0 (or) sin x = ±\(\frac{1}{\sqrt{2}}\)
If cos x = 0 = cos \(\frac{\pi}{2}\)
⇒ x = \(\frac{\pi}{2}\) is the principal solution
and cos x = 0 ⇔ \(\sin \left(x-\frac{\pi}{2}\right)=0\)
⇔ x – \(\frac{\pi}{2}\) = nπ, n ∈ Z
⇔ x = nπ + \(\frac{\pi}{2}\)
⇔ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
∴ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z is the general solution of cos x = 0
If sin x = ±\(\frac{1}{\sqrt{2}}\) = sin(±\(\frac{\pi}{4}\))
x = ±\(\frac{\pi}{4}\) are principal solutions
and x = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution
∴ The general solutions are \(\left[\left\{(2 n+1) \frac{\pi}{2}\right\},\left\{n \pi \pm \frac{\pi}{4}\right\} n \in Z\right]\)

Question 2.
Solve the following equations.
(i) √3 sin θ – cos θ = √2
Solution:

(ii) cot x + cosec x = √3
Solution:

(iii) sin x + √3 cos x = √2
Solution:

Question 3.
Solve the following equations:
(i) tan θ + sec θ = √3, 0 ≤ θ ≤ 2π
Solution:

(ii) cos 3x + cos 2x = sin \(\frac{3 x}{2}\) + sin \(\frac{x}{2}\), 0 ≤ x ≤ 2π
Solution:

(iii) cot²x – (√3 + 1) cot x + √3 = 0, 0 < x < \(\frac{\pi}{2}\)
Solution:
cot²x – (√3 + 1) cot x + √3 = 0
⇔ cot²x – √3 cot x – cot x + √3 = 0
⇔ cot x (cot x – √3) – 1(cot x – √3) = 0
⇔ cot x = √3 (or) cot x = 1
case (i): cot x = 1 ⇒ tan x = 1
∴ x = \(\left\{\frac{\pi}{4}\right\}\)
case (ii): cot x = √3 ⇒ tan x = \(\frac{1}{\sqrt{3}}\)
∴ x = \(\left\{\frac{\pi}{6}\right\}\)
∴ Solutions are \(\left\{\frac{\pi}{6}, \frac{\pi}{4}\right\}\)

(iv) sec x . cos 5x + 1 = 0; 0 < x < 2π
Solution:

III.

Question 1.
(i) Solve sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x
Solution:
(sin 3x + sin x) + sin 2x = (cos 3x + cos x) + cos 2x
⇔ 2 . sin(\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + sin 2x = 2 . cos (\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + cos 2x
⇔ 2 sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x
⇔ sin 2x (2 cos x + 1) = cos 2x (2 cos x + 1)
⇔ (2 cos x + 1) (sin 2x – cos 2x ) = 0
⇔ cos x = \(-\frac{1}{2}\) (or) sin 2x = cos 2x (i.e.,) tan (2x) = 1
Case (i):
cos x = \(-\frac{1}{2}\) = cos (\(\frac{2 \pi}{3}\))
Principal solution is x = \(\frac{2 \pi}{3}\)
and General solution is x = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ z
Case (ii):
tan 2x = 1 = tan \(\frac{\pi}{4}\)
∴ Principal solution is 2x = \(\frac{\pi}{4}\) (i.e.,) x = \(\frac{\pi}{2}\)
General solution is 2x = nπ + \(\frac{\pi}{4}\), n ∈ z
⇒ x = \(\frac{\mathrm{n} \pi}{2}+\frac{\pi}{8}\), n ∈ Z
∴ General solution is \(\left.\left[\left\{2 n \pi \pm \frac{2 \pi}{3}\right\},\left\{\frac{n \pi}{2}+\frac{\pi}{8}\right\} / n \in Z\right\}\right]\)

(ii) If x + y = \(\frac{2 \pi}{3}\) and sin x + sin y = \(\frac{3}{2}\), find x and y.
Solution:

(iii) If sin 3x + sin x + 2 cos x = sin 2x + 2 cos²x, find the general solution.
Solution:
Given sin 3x + sin x + 2 cos x = sin 2x + 2 cos²x
⇒ 2 . sin (\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + 2 cos x = 2 sin x cos x + 2 cos²x
⇒ 2 . sin 2x . cos x + 2 cos x = 2 cos x (sin x + cos x)
⇒ 2 cos x (sin 2x + 1) = 2 cos x (sin x + cos x)
⇒ 2 cos x [sin 2x + 1 – sin x – cos x] = 0
⇒ cos x = 0 (or) sin 2x – sin x + 1 – cos x = 0

(iv) Solve cos 3x – cos 4x = cos 5x – cos 6x
Solution:
-2 sin 5x . sin x = -2 sin 4x . sin x
⇒ 2 sin x [sin 5x – sin 4x] = 0
⇒ 4 sin x . cos \(\frac{9 x}{2}\) . sin \(\frac{x}{2}\) = 0

Question 2.
Solve the following equations.
(i) cos 2θ + cos 8θ = cos 5θ
Solution:
cos 2θ + cos 8θ = cos 5θ
2 cos (\(\frac{2 \theta+8 \theta}{2}\)) cos (\(\frac{2 \theta-8 \theta}{2}\)) – cos 5θ = 0
2 cos 5θ . cos 3θ – cos 5θ = 0
cos 5θ (2 cos 3θ – 1) = 0
If cos 5θ = 0
Solution is 5θ = (2n + 1) \(\frac{\pi}{2}\)
θ = (2n + 1) \(\frac{\pi}{10}\), n ∈ z
If 2 cos 3θ – 1 = 0
cos 3θ = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
Solution is 3θ = 2nπ ± \(\frac{\pi}{3}\)
θ = \(\frac{2 n \pi}{3} \pm \frac{\pi}{9}\), n ∈ Z

(ii) cos θ – cos 7θ = sin 4θ
Solution:
-2 sin (\(\frac{\theta+7 \theta}{2}\)) sin (\(\frac{\theta-7 \theta}{2}\)) – sin 4θ = 0
2 sin 4θ sin 3θ – sin 4θ = 0
sin 4θ (2 sin 3θ – 1) = 0
If sin 4θ = 0
∴ Solution is 4θ = nπ
θ = \(\frac{n \pi}{4}\), n ∈ z
If 2 sin 3θ – 1 = 0
sin 3θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
Solution is 3θ = nπ + (-1)ⁿ \(\frac{\pi}{6}\)
θ = \(\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}\), n ∈ z

(iii) sin θ + sin 5θ = sin 3θ, 0 < θ < π
Solution:
sin θ + sin 5θ = sin 3θ
sin θ + sin 5θ – sin 3θ = 0
sin θ + 2 cos (\(\frac{5 \theta+3 \theta}{2}\)) sin (\(\frac{5 \theta-3 \theta}{2}\)) = 0
sin θ + 2 cos 4θ . sin θ = 0
sin θ (1 + 2 cos 4θ) = 0
sin θ = 0, cos 4θ = \(\frac{-1}{2}\)
If sin θ = 0, solution is θ = nπ, n ∈ Z
If cos 4θ = \(\frac{-1}{2}\) = cos(\(\frac{2 \pi}{3}\))
Solution is 4θ = 2nπ ± \(\frac{2 \pi}{3}\)
θ = \(\frac{2 n \pi}{4} \pm \frac{2 \pi}{12}\), n ∈ Z
θ = \(\frac{n \pi}{2} \pm \frac{\pi}{6}\), n ∈ Z
Since 0 < θ < π
Then π = \(\frac{\pi}{6}, \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{5 \pi}{6}\)

Question 3.
(i) If tan pθ = cot qθ and p ≠ -q show that the solutions are in A.P. with common difference \(\frac{\pi}{p+q}\)
Solution:
tan pθ = cot qθ = tan \(\frac{\pi}{2}\) – qθ
pθ = nπ + \(\frac{\pi}{2}\) – qθ
(p + q) θ = (2n + 1) \(\frac{\pi}{2}\)
θ = \(\frac{(2 n+1)}{p+q} \frac{\pi}{2}\), n is an integer
The Solutions \(\frac{\pi}{2(p+q)}, \frac{3 \pi}{2(p+q)}, \frac{5 \pi}{2(p+q)}\) + …………
∴ The solution form an Arithematical proportion With common difference \(\frac{2 \pi}{2(p+q)}=\frac{\pi}{p+q}\)

(ii) Show that the solutions of cos pθ = sin qθ form two series each of which is an A.P. Find also the common difference of each A.P. (p ≠ ±q).
Solution:
cos pθ = sin qθ
cos pθ – sin qθ = 0
cos pθ + \(\cos \left[\frac{\pi}{2}+9 \theta\right]\) = 0

(iii) Find the number of solutions of the equation tan x + sec x = 2 cos x; cos x ≠ 0, lying in the interival (0, π).
Solution:
tan x + sec x = 2 cos x
\(\frac{\sin x}{\cos x}+\frac{1}{\cos x}\) = 2 cos x
sin x + 1 = 2 cos²x
sin x + 1 = 2(1 – sin²x)
sin x + 1 = (2 – 2 sin²x)
2 sin²x + sin x – 1 = 0
2 sin²x + 2 sin x – sin x – 1 = 0
2 sin x (sin x + 1) – 1 (sin x + 1) = 0
(sin x + 1)(2 sin x – 1) = 0
sin x = -1 (or) sin x = \(\frac{1}{2}\)
If sin x = -1
x = \(\frac{-\pi}{2} \text { (or) } \frac{3 \pi}{2}\)
If sin x = \(\frac{1}{2}\)
x = \(\frac{\pi}{6} \text { (or) } \frac{5 \pi}{6}\)
In the interval (0, π)
Number of solutions = 2

(iv) Solve sin 3α = 4 sin α sin(x + α) sin(x – α) where α ≠ nπ, n ∈ Z
Solution:
3 sin α – 4 sin³α = 4 sin α (sin²x – sin²α)
Dividing with sin α
3 – 4 sin²α = 4 (sin²x – sin²α)
3 – 4 sin²α = 4 sin²x – 4 sin²α
4 sin²x = 3
2 sin²x = \(\frac{3}{2}\)
1 – cos 2x = \(\frac{3}{2}\)
cos 2x = \(-\frac{1}{2}=\cos \frac{2 \pi}{3}\)
2x = 2nπ ± \(\frac{2 \pi}{3}\), ∀ n ∈ Z
x = nπ ± \(\frac{\pi}{3}\), n ∈ Z

Question 4.
(i) If tan(π cos θ) = cot(π sin θ), then prove that \(\cos \left(\theta-\frac{\pi}{4}\right)=\pm \frac{1}{2 \sqrt{2}}\)
Solution:

(ii) Find the range of θ if cos θ + sin θ is positive.
Solution:

Question 5.
If α, β are the solutions of the equation a cos θ + b sin θ = c, where a, b, c ∈ R and if a² + b² > 0, cos α ≠ cos β and sin α ≠ sin β, then show that
(i) sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
(ii) cos α + cos β = \(\frac{2 a c}{a^{2}+b^{2}}\)
(iii) cos α . cos β = \(\frac{c^{2}-b^{2}}{a^{2}+b^{2}}\)
(iv) sin α . sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)
Solution:
a cos θ = b sin θ = c
First write this as a quadratic equation in sin θ
⇔ a cos θ = c – b sin θ
By squaring on both sides, we get
a² cos²θ = (c – b sin θ)²
⇔ a² (1 – sin²θ) = c² + b² sin²θ – 2 bc sin θ
⇔ (a² + b²) sin²θ – 2bc sin θ + (c² – a²) = 0
It is a quadratic equation in sin θ,
It has sin α and sin β as roots since α and β are solutions for the given equation
(i) Sum of the roots = sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
Again a cos θ + b sin θ = c
Write this as a quadratic equation in cos θ
⇔ b sin θ = c – a cos θ
By squaring on both sides
⇔ b² sin²θ = (c – a cos θ)²
⇔ b²(1 – cos²θ) = c² + a² cos²θ – 2 ca cos θ
⇔ (a² + b²) cos²θ – 2 ca cos θ + (c² – b²) = 0
It is a quadratic equation in cos θ. It has cos α, cos β be its roots.
(ii) Sum of the roots = cos α + cos β = \(\frac{2 c a}{a^{2}+b^{2}}\)
(iii) Product of the roots = cos α . cos β = \(\frac{c^{2}-b^{2}}{a^{2}+b^{2}}\)
(iv) Product of the roots = sin α . sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)

Question 6.
(i) Find the common roots of the equations cos 2x + sin 2x = cot x and 2 cos²x + cos²2x = 1.
Solution:
Let tan x = A
Given that cos 2x + sin 2x = cot x
⇔ \(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}+\frac{2 \tan x}{1+\tan ^{2} x}=\frac{1}{\tan x}\)
⇔ \(\frac{1-A^{2}}{1+A^{2}}+\frac{2 A}{1+A^{2}}=\frac{1}{A}\)
⇔ (1 – A² + 2A) A = (1 + A2)
⇔ A – A³ + 2A² = 1 + A²
⇔ A³ – A² – A + 1 = 0
⇔ A = 1
A = 1 satisfy this equation

∴ A³ – A² – A + 1 = 0
⇔ (A – 1) (A² – 1) = 0
⇔ (A – 1) (A – 1) (A + 1) = 0
⇔ A = 1, A = -1
∴ tan x = ±1
⇒ x = (2n + 1) \(\frac{\pi}{4}\), n ∈ z
Given 2 cos²x + cos² 2x = 1
⇔ (2 cos²x – 1) + cos²(2x) = 0
⇔ cos 2x + cos²(2x) = 0
⇔ cos 2x (1 + cos 2x) = 0
⇔ cos 2x = 0 (or) cos 2x = -1
Case (i): cos 2x = 0
⇔ 2x = (2n + 1) \(\frac{\pi}{2}\)
∴ x = (2n + 1) \(\frac{\pi}{4}\), n ∈ z
∴ (2n + 1) \(\frac{\pi}{4}\), n ∈ z is the common root for the given two trigonometric equations.

(ii) Solve the equation \(\sqrt{6-\cos x+7 \sin ^{2} x}+\cos x=0\)
Solution:
\(\sqrt{6-\cos x+7 \sin ^{2} x}+\cos x=0\)
6 – cos x + 7 sin²x ≥ 0
⇒ 7(1 – cos²x) – cos x + 6 ≥ 0
⇒ 7 – 7 cos²x – cos x + 6 ≥ 0
⇒ 7 cos²x + cos x – 13 ≤ 0
Consider 7 cos²x + cos x – 13 = 0

cos x values do not lie in [-1, 1]
Hence the given equation has no solution.

(iii) If |tan x| = tan x + \(\frac{1}{\cos x}\) and x ∈ [0, 2π], find the value of x.
Solution:
Case (i):
|tan x| = tan x, if x lies either in I (or) in III quadrant
Then |tan x| = tan x + \(\frac{1}{\cos x}\)
⇒ tan x = tan x + sin x
⇒ sec x = 0 which is impossible, since sec x ∉ (-1, 1)
Case (ii):
|tan x| = -tan x, if x lies in II & IV quadrants
Then |tan x| = tan x + \(\frac{1}{\cos x}\)
⇒ -tan x = tan x + sec x
⇒ -2 tan x = sec x
⇒ \(-2 \frac{\sin x}{\cos x}-\frac{1}{\cos x}\) = 0
⇒ -2 sin x – 1 = 0
⇒ sin x = \(\frac{-1}{2}=\sin \left(\frac{-\pi}{6}\right)=\sin \left(2 \pi-\frac{\pi}{6}\right)\)
∴ x = \(\frac{11 \pi}{6}\)

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 11th Lesson Cell Cycle and Cell Division Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 11th Lesson Cell Cycle and Cell Division

Very Short Answer Questions

Question 1.
Between a Prokaryote and a eukaryote, which cell has a shorter cell division time?
Answer:
Prokaryotic cell.

Question 2.
Among Prokaryotes and eukaryotes, which one has a shorter duration of a cell cycle?
Answer:
Prokaryotic cell.

Question 3.
Which of the phases of the cell cycle is of the longest duration?
Answer:
Interphase.

Question 4.
Which tissue of animals and plants exhibits Meiosis?
Answer:
Meiosis occurs in diploid cells.

Question 5.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells take to become 32 cells?
Answer:
80 minutes.

Question 6.
Which part of the human body should one use to demonstrate stages in Mitosis?
Answer:
Cells lining the Gut and skin cells.

Question 7.
What attributes does a chromatid require to be classified as a chromosome?
Answer:
Each chromosome at Metaphase split and the two daughter chromatids now referred to as chromosoms.

Question 8.
Which of the four chromatids of a bivalent at prophase-I of Meiosis can involve in cross-over?
Answer:
Crossing over occurs between non-sisters chromatids of the homologous chromosomes.

Question 9.
If a tissue has at a given time 1024 cells. How many cycles of Mitosis had the original parental single cell undergone?
Answer:
10 Mitotic divisions.

Question 10.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Answer:
300.

Question 11.
At what stage of cell cycle does DNA synthesis occur?
Answer:
‘S’ phase (synthesis phase).

Question 12.
It is said that one cycle of cell division in human cells, (Eukaryotic cells) take 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Answer:
Interphase.

Question 13.
It is observed that heart cells do not exhibit cell-divisjon. Such cells do not divide further and exit phase to enter an inactive stage called of cell cycle. Fill in the blanks.

Question 14.
Identify the sub stages of Prophase n I in Meiosis in which synapse and desynapse are formed?
Answer:
Synthesis occurs in ‘zygotene’ a sub phase of Prophase I of Meiosis -I
Desynapse occurs in ‘diakinesis’ a sub phase of prophase I of Meiosis -I
G₁ Phase. Quiescent stage (G₀).

Question 15.
Name the stage of Meiosis in which actual reduction in chromosome number occurs.
Answer:
Anaphase – II

Question 16.
Mitochondria and plastids have their own DNA (genetic material). What is their fate during nuclear division like Mitosis?
Answer:
Mitochondria and plastids get distributed between the two daughter cells.

Question 17.
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number during Metaphase? What would be the DNA content (c) during anaphase?
Answer:
32 chromosomes. The DNA content during anaphase gets doubled.

Question 18.
The following events occur during the various phases of the cell cycle. Fill the blanks with suitable answer against each.
a) Disintegration of Nuclear Membrane
b) Appearance of Nucleolus
c) Division of centromere
d) Replication of DNA
Answer:
a) Prophase
b) Telophase
c) Anaphase
d) S-Phase

Question 19.
While examining the Mitotic stage in a tissue, one finds some cells with 16 chromosomes and some with 32 chromosomes. What possible reasons could you give to this difference in chromosome number ? Do you think cells with 16 chromosomes could have arisen from cells with 32 chromosomes or vice versa?
Answer:
In cells with 16 chromosomes Mitosis is over, in cells with 32 chromosomes, Mitosis is not started. No. Cells with 16 chromosomes have not arisen from the cells with 32 chromosomes.

Question 20.
Two key events take place during S phase in animal cells. DNA replication and duplication of Centriole. In which parts of the cell do these events occur?
Answer:
Nucleus, Cytoplasam.

Question 21.
Name a cell that is found arrested in diplotene stage for months and years. Comment in two or three sentences how it complete cell cycle?
Answer:
Oocytes of some vertebrates.

Short Answer Type Questions

Question 1.
In which phase of meiosis are the following formed? Choose the answers from hint points given below,
a) synaptonermal complex
b) Recombination nodules
c) Appearance / activation of a Enzyme recombinase
d) Termination of chliasmata
e) Interkinesis
f) Formation of dyad of cells.
Hints :
1) Zygotene
2) Pachytene
3) Pachytene
4) Diakinesis
5) After Telophase I / before Meiosis II
6) /Telophase I / After Meiosis I
Answer:
a) Synaptonemal complex = Zygotene
b) Recombination nucleus = Pachytene
c) Appearance / activation of = Pachytene Enzyme recombinase
d) Termination of chaismata = Diakinesis
e) Interkinesis = The stage between the two meiotic divisions (After telophase I / before meiosis II)
f) Formation of dyad of cells = Telophase I / After meiosis I.

Question 2.
Mitosis results in producing two cells which are similar to each other. What would be the consequence if each of the following irregularities occurs during Mitosis?
a) Nuclear membrane fails to disintegrate
b) Duplication of DNA does not occurs.
c) Centromeres do not divide
d) Cytokinesis does not occur.
Answer:
a) Nuclear membrane fails to disintegrate : The nuclear divisions takes place.
b) Duplication of DNA does not occurs : Of two daughter cells, one cannot get DNA.
c) Centromers do not divide : Chromosomes are not distributed to daughter cells.
d) Cytokinesis does not occur : Multinucleate condition arises leading to the formation of syneytium (liquid endoplasm of coconut).

Question 2.
Describe the events of prophase – I
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene :
The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres.

b) Zygotene :
The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene :
Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the non-sister chromatids of the homologous chromosomes. Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene :
The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata.

e) Diakinesis :
It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

The chromatids undergo condensation, contraction and thickening.

Question 4.
Mention the significance of Meiosis.
Answer:

1. It maintains the same chromosome number in the sexually reproducing organisms.
2. It restricts the multiplication of chromosome number and maintains the stability of the species.
3. Maternal and paternal genes get exchanged during crossing over. It results in variations among the offspring.
4. All the four chromatids of homologous pair of chromosomes seggregate and go over separately to four different daughter cells this leads to variation in the daughter cells genetically.
5. Variations play an important role in the process of evolution.

Question 5.
Which division is necessary to maintain constant chromosome number in all body cell of multicellular organism and why?
Answer:
Mitosis is necessary to maintain constant number of chromosomes in all body cells in multicellular organisms because mitosis results in the production of diploid daughter cells with genetic employment usually identical to that of the parent cell. The growth of multicellular organisms is due to mitosis. Cell growth results in distributing the ratio between the nucleus and cytoplasam.

It therefore essential for the cell division to restore the nucleo-cytoplasamic ratio. A very significant contribution of mitosis is cell repair. Mitotic divisions in the meristametic tissues, the apical and the lateral Meristems, results in a continuous growth of plants throughout their life.

Question 6.
Though redundantly described as a resting phase. Interphase does not really involve rest. Comment.
Answer:
The interphase also called phase of non-apparent division through called the resting phase is the time during which the cell is preparing for division by involving both cell growth and DNA replication. The interphase is divided into three further phases. They are :

i) ‘G₁‘ phase :
It corresponds to the internal between mitosis and initiation of DNA replication. In this the cell is metobolically active and continuously grows.

ii) ‘S’ phase :
Synthesis phase marks the period during which DNA synthesis or replication takes place. The amount of DNA per cell doubles. However, there is no increase in the number of chromosomes.

iii) ‘G₂‘ phase :
Proteins are synthesized.

Question 7.
Comment on the statement – Meiosis enables the conservation of specific chromosome number of each species even though the process per se. results in reduction of chromosome number.
Answer:
Meiosis is the mechanism by which conservation of specific chromosome number in each species is achieved across generations in sexually reproducing organisms, even though the process per sec periodically results in the reproduction of chromosome number by half. It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
In animal cells, cytokineis is achieved by the appearence of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasam into two plant cells, are enclosed by a relatively inextensible cell wall. In plant cells, wall fromation starts in the centre of the cell and grows outward to meet the exiting lateral walls. The formation of the new cell wall begins with the formation of cell plate that represents the middle lamella between the walls of two adjacent cells.

Long Answer Type Questions

Question 1.
Discuss on the statement – Telophase is reverse of prophase.
Answer:
Telophase is the final stage of Mitosis in which the chromosomes that have reached their respective poles decondenses and loose their individuality. Chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements. Nuclear membrane assimilates around the chromosome clusters. Nucleolus, Golgi complex and endoplasmic reticulum reform where as in prophase, chromosomal material condenses and organises to form compact chromosomes. Nuclear membrane, Nucleolus, Golgi complexes. Endoplasmic reticulum disappears. Thats why Telophase is the reverse phase to prophase.

Question 2.
What are the various stages of meiotic prophase – I? Enumerate the chromosomal events during each stage?
Answer:
Meiosis I is longer phase and consists of 5 sub phases namely Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

a) Leptotene :
The nucleus increases in size by absorbing water from the cytoplasm. The chromatin material organises into a constant number of chromosomes. The chromosomes are long, slender and show bead like structures called chromomeres.

b) Zygotene :
The chromosomes become shorter and thicker. They approach each other and form pairs. This homologous pair is called bivalent and the process of pairing is called synapsis. It is accompanied by the formation of Synaptonemal complex. The synapsis occurs at proterminal point or procentric or random means.

c) Pachytene :
Bivalent chromosomes now clearly appear as tetrads. This stage is characterised by the presence of recombination modules, the sites of which crossing over occurs between the non-sister chromatids of the homologous chromosomes. Crossing – over is the exchange of genetic material between the two homologous chromosomes. It is also an enzyme mediated process by ‘recombinase’ crossing over leads to recombination of genetic material on the two chromosomes.

d) Diplotene :
The homologous chromosomes of a bivalent begin to separate from each other except at the sites of cross overs to dissolution of synaptonemal complex. The ‘X’ shaped structures are called chaismata. The chromatids undergo condensation, contraction and thickening.

e) Diakinesis :
It is marked by terminalisation of chaismata. In this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of this phase, the nuclear membrane breaks down the nucleolus disappears.

Question 3.
Differentiate between the events of mitosis and meiosis.
Answer:

Question 4.
Write brief note on the following :
a) Synaptonemal complex b) Metaphase plate
Answer:
a) Synaptonemal complex :
The homologous chromosomes approach each other and form pairs called Bivalents and the process is called synapsis. The synapsis occurs at the both ends and progresses towards their centromeres called proterminal or the synapsis starts from their centromeres and the pairing progresses towards the end of the chromosomes called procentric or the pairing occurs at various points of homologous chromosomes called Randon synapsis. The paired homologous chromosomes are joined by a thick protein containing frame work called synaptonemal complex (Sc).Sc stabilizes the pairing of the homologous chromosomes and facilitates crossing over and recombination.

b) Metaphase plate :
Metaphase chromosome is made up of two sister chromatids’which were held together by the centromere. Small disc shaped structures at the surface of of the centromeres are called Kinetochores. These serve as the sites of attachment of spindle fibres to the chromosomes that are moved into centre of the cell. Hence all chromosomes lie at the equator with one chromatid of each chromosome connected by its kinetochore to spindle fibres from one pole and its sister chromatid connected by its kinetochore to spindle fibres from the opposite pole. It is called ‘Metaphase Plate’.

Question 5.
Write briefly the significance of mitosis and meiosis in multicellular organism.
Answer:
Significance of Mitosis :

1. Growth in organisms is caused by Mitosis.
2. The daughter cells formed by Mitosis are identical with the mother cell in characters. Hence it it important in conserving the genetic diversity of the organisms.
3. In unicellular organisms, Mitosis help in reproduction.
4. The old dead and decaying cells of body are replaced with the help of Mitosis.
5. It is useful in regeneration of lost parts and for grafting in vegetative reproduction.

Significance of Meiosis :

1. Meiosis helps in the maintenance of a constant chromosome number from one generation to the next.
2. Due to crossing over, genetic recombinations are caused which help in the origin of new species and lead to evolution.
3. It helps in the formation of gametes and is thus useful in sexual reproduction.

Intext Questions

Question 1.
Name a stain commonly used to colour chromosome.
Answer:
Giemsa strain is used.

Question 2.
Name the patholqgical condition when uncontrolled cell division occurs.
Answer:
Cancer.

Question 3.
An organism has two parts of chromosomes (i.e., chromosome number = 4) Diagra-matically represent the chromosomal arrangement during different phases of Meiosis II.
Answer:
An organism has two pairs of chromosomes (i.e., chromosome number = 4) Diagrammatically represent the chromosomal arrangement during phases of meiosis – II.

Question 4.
Meiosis has events that lead to both gene recombinations as well as mendelian recombinations. Discuss.
Answer:
An organism often many phenotypic fruits in its body; are determined by at least pair of alleles (or) gener During the events of meiosis – I (crossing over) recombination events occur to pass different combination of chromosomes and consequently different combination of characters in both daughter cells.

The random assortment of the genes is due not only to crossing over but also to the random distribution of the chromosomes in first and second division. How ever, since this separation is a random process, the resulting cells will contain different chromosomal combinations even in the abscence of crossing over.

Question 5.
Both unicellular and multicellular organisms undergo mitosis. What are the differences if any observed between the two processes?
Answer:

1. In unicellulars it is referred as binary fission and in multicellulars it is referred as mitosis.
2. Mitosis allows unicellular organisms to reproduce and create more (identical) organisms.
3. In unicellular only one cell undergoes mitosis whereas in multicellular all cells undergo mitosis.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
“Within the elastic limit stress directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
k = \(\frac{\text { Stress }}{\text { Strain }}\)
Where k is the modulus of elasticity.

Question 2.
State the units and dimensions of stress.
Answer:

1. Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
   S.l units → N/m² (or) Pascal
2. Dimensional formula
   Stress = \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\) = [ML^-1T^-2].

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:
Modulus of elasticity (k) = \(\frac{\text { Stress }}{\text { Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 4.
State the units and dimensions of Young’s modulus.
Answer:
Young’s modules (y) = \(\frac{\text { LongitudinalStress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity (G) = \(\frac{F}{A \theta}=\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 6.
State the units and dimensions of Bulk modulus.
Answer:
Bulk modulus (B) = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}=\frac{-P V}{\Delta V}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:

• Nearly perfect elastic bodies are quartz fibre.
• Nearly perfect plastic bodies are dough and day.

Short Answer Questions

Question 1.
Define Hooke’s Law of elasticity, proportionality, permanent set and breaking stress.
Answer:
Hooke’s law : “With in the elastic limit stress is directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Proportionality limit: The maximum stress developed in a body till it obeys Hookes law is called proportionality limit.
Permanent Set : Permanent deformation produced when a body is stretched beyond its elastic limit.
Breaking stress : The maximum stress a body can bear before it breaks.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
Modulus of elasticity : It is the ratio stress applied on a body to the strain produced in the body.
k = \(\frac{\text { Stress }}{\text { Strain }}\)
S.I unit → N/m² (or) Pascal
Stress : When a body is subjected to an external force, the force per unit area is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
S.I unit → N/m² (or) Pascal
Strain : When deforming forces act on a body, the fractional deformation produced in the body. It has no units

Poisson’s ratio (σ) : The ratio between lateral strain to longitudinal strain of a body is called poisson’s ratio.
σ = \(\frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}=\frac{\frac{-\Delta \mathrm{r}}{\mathrm{r}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)

Question 3.
Define Young’s modulus. Bulk modulus and Shear modulus.
Answer:
Young’s modulus (y) : With in the elastic limit, the ratio of longitudinal stress to longitudinal strain is called young’s modulus.
y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
y = \(\frac{\mathrm{FL}}{\mathrm{A} \cdot \mathrm{e}}\)
S.I unit → N/m² (or) Pascal

Bulk modulus (B) : With in the elastic limit, it is defined as the ratio of Bulk stress to Bulk strain
B = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}\)
B = \(\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{-\Delta \mathrm{V}}{\mathrm{V}}}=\frac{-\mathrm{PV}}{\Delta \mathrm{V}}\) (∵ -ve sign indicates volume decreases)
S.I unit → N/m² (or) Pascal

Rigidity modulus (G) : With in the elastic limit, it is defined as the ratio of shearing stress to shearing strain.
G = \(\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
G = \(\frac{\frac{F}{A}}{\theta}=\frac{F}{A \theta}\)
S.I unit → N/m² (or) Pascal

Question 4.
Define stress and explain the types of stress. [T.S. Mar. 16]
Answer:
Stress : The restoring force per unit area is called stress
∴ Stress = \(\frac{\text { Restoring Force }}{\text { Area }}=\frac{F}{A}\)
Stress is classified into three types.

1. Longitudinal stress
2. Volume (or) Bulk stress
3. Tangential (or) shearing stress

1. Longitudinal stress (or) Linear stress : When a normal stress changes the length of a body, then it is called longitudinal stress.
Longitudinal stress = \(\frac{F}{A}\)

2. Volume (or) Bulk stress : When a normal stresschanges the volume of a body, then it is called volume stress.
Volume stress = \(\frac{\text { Force }}{\text { Area }}\) = pressure.

3. Tangential (or) shearing stress : When the stress is tangential to the surface due to the application of forces parallel to the surface, then the stress is called tangential stress.
Tangential stress = \(\frac{F}{A}\).

Question 5.
Define strain and explain the types of strain.
Answer:
Strain : It is the ratio of change in dimension to its original dimension.
Strain = \(\frac{\text { Changes in dimension }}{\text { Original dimension }}\)
Strain is of three types.
1. Longitudinal strain : It is the ratio of change in length to its original length.
Longitudinal strain = \(\frac{\text { Changes in length }}{\text { Original length }}=\frac{e}{L}\)

2. Shearing strain (or) Tangential strain : When simultaneous compression and extension in mutually perpendicular direction takes place in a body, the change of shape it under goes, is called shearing strain.

Shearing strain (θ) = \(\frac{1}{L}\)

3. Bulk (or) volume strain : It is the ratio of change in volume to its original volume is called bulk strain. It is called Bulk (or) volume strain.
Bulk strain = \(\frac{\text { Change in Volume }}{\text { Original Volume }}=\frac{\Delta V}{V}\)

Question 6.
Define strain energy and derive the equation for the same. [Mar. 14]
Answer:
The potential energy stored in a body when stretched is called strain energy.
Let us consider a wire of length L and cross – sectional area A. Let x be the change in length of the wire by the application of stretching force F.

Strain energy per unit volume = \(\frac{1}{2} \times \frac{F}{A} \cdot \frac{x}{L}\)
= \(\frac{1}{2}\) × stress × strain.

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
The elastic behavior of materials plays an important role in everyday life. Designing of buildings, the structural design of the columns, beams and supports require knowledge of strength of material used. The elasticity of the material is due to stress developed with in the body, when extenal force acts on it. A material is of more elastic nature if it develops more stress (or) restoring force. Steel develops more stress than copper, brass, aluminium for same strain. So steel is more elastic.
y = \(\frac{\text { Stress }}{\text { Strain }}\)

Question 8.
Describe the behaviour of a wire under gradually increasing load. [A.P. – Mar. ’18, ’16, ’15; TS – Mar. ’18, ’15, ’13]
Answer:
When the load is increased in steps, a graph is drawn between stress on y-axis and corresponding strain on x-axis.

1. Proportionality limit : In the linear position OA, stress is proportional to strain, i.e. Hookes law is obeyed by the wire upto point A. The graph is a straight lint. When ever the stretching force at A is removed, the wire regains its original length.
A is called proportionality limit.

2. Elastic limit : In the graph B is the elastic limit.
Through the wire doesnot obey Hooke’s law at B. The wire regains its original length after removing the stretching force at B. upto point B the wire is under elastic behaviour.

3. Permanent set (or) yield point: In the graph c is the yield point. If the stretching force at c is removed, the wire doesnot regain its original length and the length of the wire changes permanently. In this position the wire flows like a viscous liquid. After the point c, the wire is under plastic behavior, c is called permanent set (or) yield point.

4. Breaking point: When the stress increased, the wire becomes thinner and thinner. When the stress increases to a certain limit the wire breaks. The stress at which the wire breaks is called breaking stress and the point D is called breaking point.

5. Elastic fatigue : The state of temperary loss of elastic nature of a body due to continuous strain is called elastic fatigue. When a body is subjected to continuous strain with in the elastic limit, it appears to have lost Hastic property temporarily to some extent and becomes weak.

Question 9.
Two identical solid balls, one of ivory and the other of wet-day are dropped from the same height onto the floor. Which one will rise to greater height after striking the floor and why ?
Answer:
Ivory ball rise to greater height after striking the floor. The ivory ball regain its original shape after striking the floor. The elastic property of ivory ball is more. Where as wet-day ball does not regain its original shape after striking the floor.

So wet-day ball acts like plastic body.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why ?
Answer:
Use of pillars (or) columns is also very common in buildings and bridges. A pillar with rounded ends supports less load than that with a distributed shape at the ends. The precise design of a bridge (or) a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable materials.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km ?
Answer:
The maximum height of a mountain on earth is 10 km, can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and this provides some shearing stress to rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg. Where ρ is the density of the mountain. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free.
There is a shear component, approximately hρg itself.
Elastic limit for 3 typical rock is 30 × 10⁷ N/m²
hρg = 30 × 10⁷ (ρ = 3 × 10³ kg/m³)
h = \(\frac{30 \times 10^7}{3 \times 10^3 \times 10}\)
h = 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it.
Answer:
“When a wire is put under a tensile stress, work is done against the inter-atomic forces. The work is stored in the wire in the form of elastic potential energy”.

Expression for elastic potential energy : Consider a wire of length L and area of cross section A is subjected to a deforming force F along the length of the wire. Let the length of the wire is elongated by l.
Young’s modulus (y) = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
F = \(\frac{\mathrm{yAl}}{\mathrm{L}}\) ……………. (1)
Work done due to further elongation of small length dl
Work done (dw) = F × dl = (\(\frac{\mathrm{yAl}}{\mathrm{L}}\))dl ……………… (2)
Total work done in increasing the length of the wire from L to (L + l)
w = \(\int_0^1 \frac{\mathrm{yAl} }{\mathrm{L}} \mathrm{dl}=\frac{\mathrm{yA}}{2} \times \frac{l^2}{\mathrm{~L}}\)
w = \(\frac{1}{2} \times \mathrm{y} \times\left(\frac{l}{\mathrm{~L}}\right)^2 \times \mathrm{Al}\)
= \(\frac{1}{2}\) y × stress² × volume of the wire
w = \(\frac{1}{2}\) × stress × strain × volume of the wire.
This work is stored in the wire in elastic potential energy (u).

Long Answer Question

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s law : With in the elastic limit, stress is directly proportional to the strain.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Determination of young’s modulus of the material of a wire:

Young’s Modulus of the Material of a wire

1. It consists of two long straight wires of same length and same area of cross-section suspended side by side from a rigid .support.
2. The wire A (reference wire) carries a metre scale M and a pan to place a weight.
3. The wire B (experimental wire) carries a pan in which known weights can be placed.
4. A vernier scale v is attached to a pointer at the bottom of the experimental wire B and the main scale M is fixed to the wire A.
5. The weights placed in the pan, the elongation of the wire is measured by the vernier arrangement.
6. The reference wire is used to compensate for any change in length that may occur due to change in room temperature.
7. Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted.
8. Now the experimental wire is gradually loaded with more weights, the vernier reading is noted again.
9. The difference between two vernier readings gives the elongation produced in the wire.
10. Let r and L be the radius and initial length of the experimental wire. Let M be the mass that produced an elongation ∆L in the wire.
    Young’s modulus of the material of the experimental wire is given by
    y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)
    y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
    y = \(\frac{\mathrm{MgL}}{\pi r^2 \times \Delta \mathrm{L}}\)
    From above equation young’s modulus of the material of the wire is determined.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
D = 1 m.m = 10^-3m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m.
F = 10 N
Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{10}{3.14 \times\left(0.5 \times 10^{-3}\right)^2}\)
= 1.273 × 10⁷ N/m².

Question 2.
A tungsten wire of length 20 cm is stretched by 0.1 cm. Find the strain on the wire.
Solution:
L = 20 × 10^-2 m, ∆L = 0.1 × 10^-2 m
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{0.1 \times 10^{-2}}{20 \times 10^{-2}}\) = 0.005.

Question 3.
If an iron wire is stretched by 1 %, what is the strain on the Wire ?
Solution:
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) = 1 %
Strain = \(\frac{1}{100}\) = 0.01

Question 4.
A brass wire of diameter 1mm and length 2 m is streched by applying a force of 20N. If the increase in length is 0.51 mm. find
(i) the stress,
(ii) the strain and
(iii) the Young’s modulus of the wire.
Solution:
D = 1 m.m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m
L = 2 m, F = 20 N, ∆L = 0.51 m.m = 0.51 × 10^-3 m

= 9.984 × 10¹⁰ N/m²

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹ Nm^-2).
Solution:

= \(\frac{4}{5} \times \frac{3}{2} \times\left(\frac{0.7 \times 10^{11}}{1.1 \times 10^{11}}\right) \times\left(\frac{3}{2}\right)^2\)
\(\frac{\Delta \mathrm{L}_1}{\Delta \mathrm{L}_2}=\frac{189}{110}\)

Question 6.
A brass wire of cross-sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³ kg m^-3).
Solution:
A = πr² = 2 × 10^-6 m², V = 100 × 10^-6 = 10^-4 m³
∆L = 0.11 × 10^-3 m, y_Brass = 0.91 × 10¹¹ N/m², ρ = 10³ kg/m³
y = \(\frac{M g L}{A \times \Delta L}=\frac{v \rho g L}{A \times \Delta L}\)
L = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{v \rho g}}=\frac{0.91 \times 10^{11} \times 2 \times 10^{-6} \times 0.11 \times 10^{-3}}{10^{-4} \times 10^3 \times 9.8}\)
L = 2.04 m.

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1:2. If the extensions produced are equal, what is the ratio of loads ?
Solution:

Question 8.
Two wires of different material have same lengths and areas of cross¬section. What is the ratio of their increase in length when forces applied are the same ?
(Y₁ = 0.9 × 10¹¹ Nm^-2, Y₂ = 3.6 × 10¹¹ Nm^-2)
Solution:
y₁ = 0.9 × 10¹¹ Nm^-2
y₂ = 3.6 × 10¹¹ Nm^-2
y = \(\frac{F L}{A \times \Delta L}\)
∆L ∝ \(\frac{1}{y}\)
\(\frac{(\Delta L)_1}{(\Delta L)_2}=\frac{y_2}{y_1}=\frac{3.6 \times 10^{11}}{0.9 \times 10^{11}}=\frac{4}{1}\)

Question 9.
A metal wire of length 2.5 m and area of cross-section 1.5 × 10^-6 m² is stretched through 2 mm. if its Young’s modulus is 1.25 × 10¹¹ N.m², find the tension in the wire.
Solution:
L = 2.5 m, A = 1.5 × 10^-6 m²
∆L = 2 × 10^-9 m
y = 1.25 × 10¹¹ N.m²
y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
F = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{L}}\)
= \(\frac{1.25 \times 10^{11} \times 1.5 \times 10^{-6} \times 2 \times 10^{-3}}{2.5}\)
F = 150 N

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the
(i) stress in the two wires and
(ii) strain in the two wires.
(Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²).
Solution:
Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²
∆L₁ + ∆L₂ = 1.35 mm ……………… (1)

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed, Calculate the rigidity modulus of the substance.
Solution:
L = 2 × 10^-2 m, A = L² = 4 × 10^-4 m²
∆x = 0.15 × 10^-2 m
F = 0.3 N
G = \(\frac{\frac{F}{A}}{\frac{\Delta x}{L}}=\frac{F L}{A \Delta x}\) (∵ θ = \(\frac{\Delta x}{L}\))
G = \(\frac{0.3 \times 2 \times 10^{-2}}{4 \times 10^{-4} \times 0.15^6 \times 10^{-2}}\)
G = 10⁴ N/m²

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 Atmosphere. The change in volume is 10^-2 cm³. If the ball is made of iron, find its bulk modulus.
(1 atmosphere = 1 × 10⁵ Nm^-2).
Solution:
v = 1000 cm³ = 1000 × 10^-6 = 10^-3 m³
p = 1 atm = 1 × 10⁵ = 10⁵ N/m²
-∆v = 10^-2 cm³ = 10^-2 × 10^-6 = 10^-8 m³
Bulk modulus (B) = \(\frac{-p v}{\Delta v}\)
= \(\frac{10^5 \times 10^{-3}}{10^{-8}}\)
B = 10¹⁰ N/m².

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2. (1 atm = 1 × 10⁵ Nm^-2).
Solution:
l = 1 cm = 10^-2 m
V = Volume of the cube = l³ = 1cm³
= 10^-6 m³
P = 100 atm = 100 × 10⁵ = 10⁷ N/m²
B = 1.4 × 10¹¹ N/m²
B = \(\frac{-P V}{\Delta V}\)
-∆V = \(\frac{P V}{B}=\frac{10^7 \times 10^{-6}}{1.4 \times 10^{11}}\)
-∆V = 0.7143 × 10^-10 m³.

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2.
Solution:
\(\frac{-\Delta V}{V}\) = 2 % = \(\frac{2}{100}\)
B = 2.2 × 10⁹ Nm²
B = \(\frac{-P V}{\Delta V}\)
P = -B × \(\frac{\Delta V}{V}\)
= 2.2 × 10⁹ × \(\frac{2}{100}\)
P = 4.4 × 10⁷ N/m².

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
Solution:
L = 20 cm = 20 × 10^-2 m
∆L = 0.2 × 10^-2 m
σ = 0.19
σ = \(\frac{\text { Lateral strain }}{\text { Longitudinal strain }\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)}\)
Lateral strain = σ × \(\frac{\Delta L}{L}\)
= \(\frac{0.19 \times 0.2 \times 10^{-2}}{20 \times 10^{-2}}\)
= 0.0019

Additional Problems

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^-5 m²,
Solution:
Given, for steel wire, a₁ = 3.0 × 10^-5 m², l₁ = 4.7 m, ∆l₁ = ∆l, F₁ = F
For copper wire, a₂ = 4.0 × 10^-5 m², l₂ = 3.5 m, ∆l₂ = ∆l, F₂ = F .
Let y₁, y₂ be the young modulus of steel wire and copper wire respectively.
∴ y₁ = \(\frac{F_1}{a_1} \times \frac{l_1}{\Delta l_2}=\frac{F}{3.0 \times 10^{-5}} \times \frac{4.7}{\Delta l}\) ………….. (i)
and y₂ = \(\frac{F_2 \times l_2}{a_2 \times \Delta l_2}=\frac{F \times 3.2}{4 \times 10^{-5} \times \Delta l}\)
\(\frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}}\) = 1.8
Here y₁ : y₁ = 1.8 : 1.

Question 2.
Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material ?

Solution:
a) From graph, for stress = 150 × 10⁶ Nm^-2 the corresponding strain = 0.002
young’s modulus y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^6}{0.002}\)
= 7.5 × 10¹⁰ Nm^-2

b) Approximate yeild strength will be equal to the maximum stress it can substain with out crossing the elastic limit. Therefore, the approximate yeild strength
= 300 × 10⁶ Nm^-2
= 3 × 10⁸ Nm^-2

Question 3.
The stress-strain graphs for materials A and B are shown in Fig.

The graphs are drawn to the same scale.
a) Which of the materials has the greater Young’s modulus ?
b) Which of the two is the stronger material ?
Solution:
a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence young’s modulus (= stress/ strain) is greater for A than that of B.

b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
a) The Young’s modulus of rubber is greater than that of steel;
b) The stretching of a coil is determined by its shear modulus.
Solution:
a) False, because for a given stress there is more strain in rubber than steel and modulus of elasticity is inversly proportional to strain.

b) True because the strecting of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass Wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution:
For steel wire, total force on steel wire,
F₁ = 4 + 6 = 10 kg, f = 10 × 9.8 N
l₁ = 1.5 m, ∆l₁ = ?, 2r₁ = 0.25cm or r₁ =(0.25/2)cm = 0.125 × 10^-2 m
y₁ = 2.0 × 10¹¹ pa
For brass wire, F₂ = 6.0 kg, f = 6 × 9.8 N
2r₂ = 0.25 cm or r₂ = (0.25/2) cm = 0.125 × 10^-2 m,
y₂ = 0.91 × 10¹¹ pa, l₂ = 1.0 m, ∆l₂ = ?

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a verticle wall. A mass of 100 kdis then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
A = 0.10 × 0.10 = 10^-2 m², F = mg = 100 × 10 N
Shearing strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}{\eta}\)
or ∆L = \(\frac{F L}{A \eta}\)
= \(\frac{(100 \times 10) \times(0.10)}{10^{-2} \times\left(25 \times 10^9\right)}\) = 4 × 10^-7 m.

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the corn pressional strain of each column.
Solution:
Load on each column, F = \(\frac{50,000}{4}\) kgwt
= \(\frac{50,000 \times 9.8}{4}\) N
A = π(r₂² – r₁²) = \(\frac{22}{7}\)(0.60)² – (0.30)²]
= \(\frac{22}{7}\) 0.27 m²
Compression strain = \(\frac{\frac{F}{A}}{y}=\frac{F}{A y}\)
= \(\frac{50,000 \times 9.8}{4 \times\left(\frac{22}{7} \times 0.27\right) \times 2.0 \times 10^{11}}\)
= 7.21 × 10^-7.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Here, A = 15.2 × 19.2 × 10^-6 m², F = 44,
500 N, η = 42 × 10⁹ Nm^-2
Strain = \(\frac{\text { Stress }}{\text { Modulus of elasticity }}\)
= \(\frac{\frac{F}{A}}{\eta}=\frac{F}{A \eta}=\frac{44500}{\left(15.2 \times 19.2 \times 10^{-6}\right) \times 42 \times 10^9}\)
= 3.65 × 10^-3.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress Is not to exceed 10⁸ N m^-2, what is the maximum load the cable can support ?
Solution:
Maximum load macimum stress × area of cross-section
= 10⁸πr²
= 10⁸ × \(\frac{22}{7}\) × (1.5 × 10^-2)²
= 7.07 × 104 N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Solution:
As each wire has same tension F, so each wire has same extansion due to mass of rigid bar. As each wire is of same length, hence each wire has same strain, if D is the diameter of wire, then
y = \(\frac{4 \mathrm{~F} / \pi \mathrm{D}^2}{\text { Strain }}\) or D² ∝ 1/y
\(\frac{D_{\mathrm{cu}}}{\mathrm{D}_{\mathrm{iron}}}=\sqrt{\frac{\mathrm{y}_{\mathrm{iron}}}{\mathrm{y}_{\mathrm{cu}}}}\)
= \(\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}\) = 1.31.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Here, m = 14.5 kg, l = r = 1m, v = 2 rps, A = 0.065 × 10^-4 m²
Total pulling force on mass, when it is at the lowest position of the vertical circle is
F = mg + mrω²
= mg + mr4πv²
= 14.5 × 9.8 + 14.5 × 1 × 4 × (\(\frac{22}{7}\))² × 2²
= 142.1 +2291.6
= 2433.7 N
y = \(\frac{F}{A} \times \frac{l}{\Delta l}\) or ∆l = \(\frac{F l}{A y}\)
= \(\frac{2433.7 \times 1}{\left(0.065 \times 10^{-4}\right) \times\left(2 \times 10^{11}\right)}\)
= 1.87 × 10^-3 m
= 1.87 mm.

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Here, V = 100 lit = 100 × 10^-3 m³, P = 100 atm = 100 × 1.013 × 10⁵ Pa
V + ∆V = 100.5 litre or ∆V= (V + ∆V) – V
= 100.5 – 100
= 0.5 litre = 0.5 × 10^-3 m³
We known that bulk modulus, B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}\)
= \(\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\)
= 2.026 × 10⁹ Pa
Bulk modulus of air = 1.0 × 10⁵ Pa
\(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^9}{1.0 \times 10^5}\)
= 2.026 × 10¹⁴.
It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3 ?
Solution:
Here, P = 80.0 atm = 80.0 × 1.013 × 10⁵ pa,
compressibility = \(\left(\frac{1}{B}\right)\) = 45.8 × 10^-11 pa^-1
Density of water at surface,
ρ = 1.03 × 10³ kg m^-3
Let p be the density of water at the given depth, if v and v’ are volumes of certain mass M of ocean water at surface and at a given

Putting this value in (i) we get
1 – \(\frac{1.03 \times 10^3}{\rho^{\prime}}\) = 3.712 × 10^-3 or
ρ’ = \(\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}\) = 1.034 × 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Here, P = 10 atm = 10 × 1.013 × 10⁵ pa,
B = 37 × 10⁹ Nm^-2
Volumetric strain = \(\frac{\Delta V}{V}=\frac{P}{B}\)
= \(\frac{10 \times 1.013 \times 10^5}{37 \times 10^9}\) = 2.74 × 10^-5
∴ Fractional change in volume = \(\frac{\Delta V}{V}\)
= 2.74 × 10^-5

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
Solution:
Here, L = 10 cm = 0.10m; P = 7 × 10⁶ pa B = 140 Gpa = 140 × 10⁹ pa
As B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}=\frac{\mathrm{Pl}^3}{\Delta \mathrm{V}}\) or ∆V = \(\frac{\mathrm{Pl}^3}{\mathrm{~B}}\)
= \(\frac{\left(7 \times 10^6\right) \times(0.10)^3}{140 \times 10^9}\) = 5 × 10^-8 m³
= 5 × 10^-2 mm³

Question 16.
How much should be pressure on a litre of water be changed to compress it by 0.10% ?
Solution:
Here, V = 1 litre = 10^-3m³;
∆V/V = 0,10/100 = 10^-3
B = \(\frac{P V}{\Delta V}\) or P = B \(\frac{\Delta V}{V}\)
= (2.2 × 10⁹) × 10^-3 = 2.2 × 10⁶pa

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil ?

Solution:
Here, D = 0.5 mm = 0.5 × 10^-3m = 5 × 10^-4m
F = 50,000 N = 5 × 10⁴N
Pressure at the tip of anvil.
P = \(\frac{F}{\pi D^2 / 4}=\frac{4 F}{\pi D^2}\)
P = \(\frac{4 \times\left(5 \times 10^4\right)}{(22 / 7) \times\left(5 \times 10^{-4}\right)^2}\) = 2.5 × 10¹¹pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its end by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
For Steel wire A, l₁ = l, A₁ = 1 mm²
Y₁ = 2 × 10¹¹Nm^-2
For aluminium wire B, l₂ = l;
A₂ = 2mm²; y₂ = 7 × 10¹⁰ Nm^-2
a) Let mass m be suspended from the rod at the distance × from the end where wires A is connected. Let F₁ and F₂ be the tension in two wires and there is equal stress in two wires, then
\(\frac{F_1}{A_1}=\frac{F_2}{A_2} \text { or } \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1}{2}\) …………………. (i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂ (1.05 – x) or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2}\)
or 2.10 – 2x = x or x = 0.70m = 70 cm

b) Let mass m be supended from the rod at distance × from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.
\(\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2}\) or \(\frac{F_1}{F_2}=\frac{A_1}{A_2} \frac{Y_1}{Y_2}\)
= \(\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}\)
As the rod is stationary, so F₁x = F₂(1.05 – x)
or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{10}{7}\)
or 10 x = 7.35 – 7x
or x = 0.4324 m
x = 43.2cm.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the- mid-point.
Solution:
Refer the figure, let x be the depression at the mid point i.e CD = x
In fig. AC = CB = Z = 0.5m
m = 100g = 0.100 kg
AD = BD = (l² + x²)^1/2
Increase in length, ∆l = AD + DB – AB
= 2 AD – AB
= 2 (l² + x²)^1/2 – 2l
= 2l(1 + \(\frac{x^2}{l^2}\))^1/2 – 2l
= 2l (1 + \(\frac{x^2}{2 l^2}\)) – 2l = \(\frac{x^2}{l}\)
Strain = \(\frac{\Delta l}{2 l}=\frac{x^2}{2 l^2}\)
If T is the tension in the wire, then 2T cos θ
= mg or T = \(\frac{\mathrm{mg}}{2 \cos \theta}\)

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa ? Assume that each rivet is to carry one quarter of the load.
Solution:
Here, r = 6/2 = 3mm = 3 × 10^-3 m, max.
stress = 6.9 × 10⁷ Pa
Max . load on a rivet = Max stress × area of cross section
= 6.9 × 10⁷ × (22/7) × (3 × 10^-3)²
∴ Maximum tension
= 4 (6.9 × 10⁷ × \(\frac{22}{7}\) × 9 × 10^-6)
= 7.8 × 10³N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?
Solution:
Here, P = 1.1 × 10⁸ Pa, V = 0.32 m³,
B = 16 × 10¹¹Pa
∆V = \(\frac{\mathrm{PV}}{\mathrm{B}}\)
= \(\frac{\left(1.1 \times 10^8\right) \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 × 10^-4m³

Textual Examples

Question 1.
A structural steel rod has a radius of 10 mm and a length of 10 m. A 100 KN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel ¡s 2.0 × 10¹¹ Nm².
Answer:
a) Given Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{100 \times 10^3 \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^2}\) = 3.18 × 10⁸ Nm^-2

b) The elongation
∆L = \(\frac{(\mathrm{F} / \mathrm{A}) \mathrm{L}}{\mathrm{Y}}\)
= \(\frac{\left(3.18 \times 10^8 \mathrm{Nm}^2\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{Nm}^{-2}}\)
= 1.59 × 10^-3 m
= 1.59 mm

c) The strain is given by
Strain = ∆L/L = (1.59 × 10^-3) km
= 1.59 × 10^-3 = 0.16%

Question 2.
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer:
From y = \(\frac{\sigma}{\varepsilon}\)
we have stress = strain × Young’s modulus.
W/A = Y_c × (∆L_c/L_c) = Y_s × (∆L_s/L_s)
where the subscripts c and s refer to copper and stainless steel respectively,
∆L_c/∆L_s = (Y_s/Y_s) () (L_c/L_s)
Given L_c = 2.2 m, L_s = 1.6 m,
Y_c = 1.1 × 10¹¹ N.m^-2 and Y_s = 2.0 × 10¹¹ N.m^-2.
∆L_c/∆L_s = \(\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}=\frac{2.2}{1.6}\) = 2.5
∆L_c + ∆L_s = 7.0 × 10^-4 m
Solving the above equations,
∆L_c = 5.0 × 10^-4 m and ∆L_s = 2.0 × 10^-4 m.
∴ W = (A × Y_c × ∆L_c)L_c
= π(1.5 × 10^-3)² × \(\left[\frac{\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right)}{2.2}\right]\) = 1.8 × 10² N.

Question 3.
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig.) The combined mass of all the persons performing the act and the tables, planks etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Answer:
Total mass of all the performers, tables, plaques = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60
= 220 kg
Weight of this supported mass = 220 kg wt.
= 220 × 9.8 N = 2156 N
Weight supported by each thighbone of the performer = \(\frac{1}{2}\) (2156) N = 1078 N.
The Young’s modulus for bone is Y = 9.4 × 10⁹ Nm^-2 (compressive)
Length of each thighbone L = 0.5 m the radius of thigbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10^-2)² m²
= 1.26 × 10^-3 m²
Using \(\frac{(F \times L)}{(A \times \Delta L)}\) the compression in each thigbone (∆L) can be computed as
∆L = \(\frac{F \times L}{Y \times A}\)
= [(1078 × 0.5)/(9.4 9 × 10⁹ × 1.26 × 10^-3)]
= 4.55 × 10^-5 m or 4.55 × 10^-3 cm.
This is a very small change ! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.

Question 4.
A square slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
Answer:
The area (A) = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m²
Therefore, the stress appIid is
= (9.4 × 10⁴N/0.05 m²)
= 1.80 × 10⁶ N.m²
We know that shearing strain = (∆x/L)
= Stress/G.
Therefore the displacement
∆x = (Stress × L)/G = \(\frac{\left(1.8 \times 10^6 \mathrm{Nm}^{-2} \times 0.5 \mathrm{~m}\right)}{5.6 \times 10^9 \mathrm{Nm}^{-2}}\)
= 1.6 × 10^-4 m = 0.16 mm.

Question 5.
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression. ∆V/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 Nm^-2. (Take g = 10 ms^-2)
Answer:
The pressure exerted by a 3000 m column of water on the bottom layer
ρ = hρg
= 3000 m × 1000 kg m^-3× 10 ms^-2
= 3 × 10⁷ Nm^-2
Fractional compression ∆V/V, is
∆V/V = stress \(\frac{\left(3 \times 10^7 \mathrm{Nm}^{-2}\right)}{2.2 \times 109 \mathrm{Nm}^{-2}}\)
= 1.36 × 10^-2 or 1.36%

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 8 Limits and Continuity to solve questions creatively.

Intermediate 1st Year Maths 1B Limits and Continuity Formulas

Right Limit :
Suppose f is defined on (a, b) and Z ∈ R. Given ε < 0 , there exists δ > 0 such that a < x < a + δ => |f(x) – l| < ε, then l is said to be the right limit of’ f’ at ‘a’.
It is denoted by \({Lt}_{x \rightarrow a+} f(x)\)f(x) = l

Left Limit :
Suppose ‘ f’ is defined on (a, b) and Z e R. Given e > 0, there exists δ > 0 such that a – δ < x < a ⇒ |f(x) – l| < ε, then l is said to be the left limit of’ f’ at ’a’ and is denoted by \({Lt}_{x \rightarrow a-} \dot{f(x)}\) = l

Suppose f is defined in a deleted neighbourhood of a and l ∈ R
\({Lt}_{x \rightarrow a} f(x)=l \Leftrightarrow \underset{x \rightarrow a+}{L t} f(x)={Lt}_{x \rightarrow a-} \quad f(x)=l\)

Standard limits:

• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = na^n-1
• \({Li}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1 (x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan x}{x}\) = 1
• \({lt}_{x \rightarrow 0}(1+x)^{1 / x}\) = e
• \({Lit}_{x \rightarrow 0}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)\) = log_ea
• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x^{m}-a^{m}}=\frac{n}{m}\)a^{n – m}
• \({Lt}_{x \rightarrow 0} \frac{\sin a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{Q x}\) = e^PQ
• \({lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0}(1+p x)^{\frac{Q}{x}}\) = e^PQ

Intervals
Definition:
Let a, b ∈ R and a < b. Then the set {x ∈ R: a ≤ x ≤ b} is called a closed interval. It is denoted by [a, b]. Thus
Closed interval [a, b] = {x ∈ R: a ≤ x ≤ b}. It is geometrically represented by

Open interval (a,b) = {x ∈ R: a < x < b} It is geometrically represented by

Left open interval
(a, b] = {x ∈ R: a < x ≤ b}. It is geometrically represented by

Right open interval
[a, b) = {x ∈ R: a ≤ x < b}. It is geometrically represented by

[a, ∞) = {x ∈ R : x ≥ a} = {x ∈ R : a ≤ x < ∞} It is geometrically represented by

(a, ∞) = {x ∈ R : x > a} = {x ∈ R : a < x < ∞}

(-∞, a] = {x ∈ R : x ≤ a} = {xe R : -∞ < x < a}

Neighbourhood of A Point: Definition: Let ae R. If δ > 0 then the open interval (a – δ, a + δ) is called the neighbourhood (δ – nbd) of the point a. It is denoted by N_δ (a) . a is called the centre and δ is called the radius of the neighbourhood .
∴ N_δ(a) = (a – δ, a + δ) = {x ∈ R: a – δ< x < a + δ} = {x ∈ R: |x – a| < δ}

The set N_δ(a) – {a} is called a deleted δ – neighbourhood of the point a.
∴ N_δ(a) – {a} = (a – δ, a) ∪ (a, a + δ) = {x ∈ R :0 < | x – a | < δ}
Note: (a – δ, a) is called left δ -neighbourhood, (a, a + δ) is called right δ – neighbourhood of a

Graph of A Function:

Mod function:
The function f: R-R defined by f(x) = |x| is called the mod function or modulus function or absolute value function.
Dom f R. Range f [0, )

Reciprocal function :
The function f: R – {0} – R defined by
f(x) = \(\frac{1}{x}\) is called the reciprocal function,
Dom f = R – {0}= Range f = R

Identity function:
The function f R-R defined by f(x) = x is called the identity
It is denoted by I(x)

Limit of A Function
Concept of limit:
Before giving the formal definition of limit consider the following example.
Let f be a function defined by f (x) = \(\frac{x^{2}-4}{x-2}\). clearly, f is not defined at x = 2.
When x ≠ 2, x – 2 ≠ 0 and f(x) = \(\frac{(x-2)(x+2)}{x-2}\) = x + 2

Now consider the values of f(x) when x ≠ 2, but very very close to 2 and <2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values less than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that left hand limit of f(x) as x → -2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{-}} f(x)\) = 4

Again we consider the values of f(x) when x ≠ 2, but is very-very close to 2 and x > 2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values greater than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that right hand
limit of f(x) as x → 2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{+}} f(x)\) = 4

Thus we see that f(x) is not defined at x = 2 but its left hand and right hand limits as x → 2 exist and are equal.
When \({lt}_{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x}), \mathrm{lt}_{\mathrm{x} \rightarrow \mathrm{a}^{-}}^{\mathrm{f}(\mathrm{x})}\) are equal to the same number l, we say that \(\begin{array}{ll}
l_{x \rightarrow a} & f(x)
\end{array}\) exist and equal to 1.

Thus, in above example,

One Sided Limits Definition Of Left Hand Limit:
Let f be a function defined on (a – h, a), h > 0. A number l1 is said to be the left hand limit (LHL) or left limit (LL) of f at a if to each
ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₁| < ε.
In this case we write \(\underset{x \rightarrow a-}{L t} f(x)\) = l₁ (or) \({Lt}_{x \rightarrow a-0} f(x)\) = l₁

Definition of Right Limit:
Let f be a function defined on (a, a + h), h > 0. A number l ₂is said to the right hand limit (RHL) or right limit (RL) of f at a if to each ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₂| < ε.
In this case we write \(\underset{x \rightarrow a+}{L t} f(x)\) = l₂ (or) \({Lt}_{x \rightarrow a+0} f(x)\) = l₂

Definition of Limit:
Let A ∈ R, a be a limit point of A and
f : A → R. A real number l is said to be the limit of f at a if to each ε > 0, ∃ a δ > 0 such that x ∈ A, 0 < |x – a| < δ ⇒ | f(x) – l| < ε. In this case we write f (x) → 2 (or) \({Lt}_{x \rightarrow a} f(x)\) = l. Note: 1.If a function f is defined on (a – h, a) for some h > 0 and is not defined on (a, a + h) and if \(\underset{x \rightarrow a-}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a-} f(x)\).
2. If a function f is defined on (a, a + h) for some h > 0 and is not defined on (a – h, a) and if \(\underset{x \rightarrow a+}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a+} f(x)\).

Theorem:
If \({Lt}_{x \rightarrow a} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow 0} f(x+a)={Lt}_{x \rightarrow 0} f(a-x)\)

Theorems on Limits WithOut Proofs
1. If f : R → R defined by f(x) = c, a constant then \({Lt}_{x \rightarrow a} f(x)\) = c for any a ∈ R.

2. If f: R → R defined by f(x) = x, then \({Lt}_{x \rightarrow a} f(x)\) = a i.e., \({Lt}_{x \rightarrow a} f(x)\) = a (a ∈ R)

3. Algebra of limits

vii) If f(x) ≤ g(x) in some deleted neighbourhood of a, then \({Lt}_{x \rightarrow a} f(x) \leq {Lt}_{x \rightarrow a} g(x)\)
viii) If f(x) ≤ h(x) < g(x) in a deleted nbd of a and \({Lt}_{x \rightarrow a} f(x)\) = l = \({Lt}_{x \rightarrow a} g(x)\) then \({Lt}_{x \rightarrow a} h(x)\) = l
ix) If \({Lt}_{x \rightarrow a} f(x)\) = 0 and g(x) is a bounded function in a deleted nbd of a then \({Lt}_{x \rightarrow a}\) f(x) g(x) = 0.

Theorem
If n is a positive integer then \({Lt}_{x \rightarrow a} x^{n}\) = aⁿ, a ∈ R

Theorem
If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a)

Evaluation of Limits:
Evaluation of limits involving algebraic functions.
To evaluate the limits involving algebraic functions we use the following methods:

• Direct substitution method
• Factorisation method
• Rationalisation method
• Application of the standard limits.

1. Direct substitution method:
This method can be used in the following cases:

• If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a).
• If f (x) = \(\frac{P(a)}{Q(a)}\) where P(x) and Q(x) are polynomial functions then \({Lt}_{x \rightarrow a}\)f(x) = \(\frac{P(a)}{Q(a)}\) provided Q(a) ≠ 0.

2. Factiorisation Method:
This method is used when \({Lt}_{x \rightarrow a}\)f (x) is taking the indeterminate form of the type 0 by the substitution of x = a.
In such a case the numerator (Nr.) and the denominator (Dr.) are factorized and the common factor (x – a) is cancelled. After eliminating the common factor the substitution x = a gives the limit, if it exists.

3. Rationalisation Method : This method is used when \({Lt}_{x \rightarrow a} \frac{f(x)}{g(x)}\) is a \(\frac{0}{0}\) form and either the Nr. or Dr. consists of expressions involving radical signs.

4. Application of the standard limits.
In order to evaluate the given limits , we reduce the given limits into standard limits form and then we apply the standard limits.

Theorem 1.
If n is a rational number and a > 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1

Note:

• If n is a positive integer, then for any a ∈ R. \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If n is a real number and a> 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If m and n are any real numbers and a > 0, then \({Lt}_{x \rightarrow a} \frac{x^{m}-a^{m}}{x^{n}-a^{n}}=\frac{m}{n}\)a^m-n

Theorem 2.
If 0 < x < \(\frac{\pi}{2}\) then sin x < x < tan x.

Corollary 1:
If – \(\frac{\pi}{2}\) < x < 0 then tan x < x < sin x

Corollary 2:
If 0 < |x| < \(\frac{\pi}{2}\) then |sin x| < |x| < |tan x|

Standard Limits:

• \({Lt}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1
• \(\underset{x \rightarrow 0}{L t} \frac{\tan x}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{a^{x}-1}{x}\) = log_ea
• \({Lt}_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e

Limits At Infinity
Definition:
Let f(x) be a function defined on A = (K,).
(i) A real number l is said to be the limit of f(x) at ∞ if to each δ > 0, ∃ , an M > 0 (however large M may be) such that x ∈ A and x > M ⇒ |f (x) – l|< δ.
In this case we write f (x) → l as x → +∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

(ii) A real number l is said to be the limit of f(x) at -∞ if to each δ > 0, ∃ > 0, an M > 0 (however large it may be) such that x ∈ A and x < – M ⇒ |f (x) – l| < δ. In this case, we write f (x) → l as x → -∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

Infinite Limits Definition:
(i) Let f be a function defined is a deleted neighbourhood of D of a. (i) The limit of f at a is said to be ∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < |x – a| < δ ⇒ f(x) > M. In this case we write f(x) as x → a or \({Lt}_{x \rightarrow a} f(x)\) = +∞

(ii) The limit of f(x) at a is said be -∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < | x – a | < δ ⇒ f(x) < – M. In thise case we write f(x) → -∞ as x → a or \({Lt}_{x \rightarrow a} f(x)\) = -∞

Indeterminate Forms:
While evaluation limits of functions, we often get forms of the type \(\frac{0}{0}, \frac{\infty}{\infty}\), 0 × -∞, 0⁰, 1^∞, ∞⁰ which are termed as indeterminate forms.

Continuity At A Point:
Let f be a function defined in a neighbourhood of a point a. Then f is said to be continuous at the point a if and only if \({Lt}_{x \rightarrow a} f(x)\) = f (a).
In other words, f is continuous at a iff the limit of f at a is equal to the value of f at a.

Note:

• If f is not continuous at a it is said to be discontinuous at a, and a is called a point of discontinuity of f.
• Let f be a function defined in a nbd of a point a. Then f is said to be
  • Left continuous at a iff \({Lt}_{x \rightarrow a-} f(x)\) = f (a).
  • Right continuous at a iff \({Lt}_{x \rightarrow a+} f(x)\) = f (a).
• f is continuous at a iff f is both left continuous and right continuous at a
  i.e, \({Lt}_{x \rightarrow a} f(x)=f(a) \Leftrightarrow {Lt}_{x \rightarrow a^{-}} f(x)=f(a)={Lt}_{x \rightarrow a+} f(x)\)

Continuity of A Function Over An Interval:
A function f defined on (a, b) is said to be continuous (a,b) if it is continuous at everypoint of (a, b) i.e., if \({Lt}_{x \rightarrow c} f(x)\) =f (c) ∀c ∈ (a, b)
II) A function f defined on [a, b] is said to be continuous on [a, b] if

• f is continuous on (a, b) i.e., \({Lt}_{x \rightarrow c} f(x)\) = f (c) ∀c ∈(a, b)
• f is right continuous at a i.e., \({Lt}_{x \rightarrow a+} f(x)\) = f (a)
• f is left continuous at b i.e., \({Lt}_{x \rightarrow b-} f(x)\) = f (b).

Note :

• Let the functions f and g be continuous at a and k€R. Then f + g, f – g, kf , kf + lg, f.g are continuous at a and \(\frac{f}{g}\) is continuous at a provided g(a) ≠ 0.
• All trigonometric functions, Inverse trigonometric functions, hyperbolic functions and inverse hyperbolic functions are continuous in their domains of definition.
• A constant function is continuous on R
• The identity function is continuous on R.
• Every polynomial function is continuous on R.

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 4 Pair of Straight Lines to solve questions creatively.

Intermediate 1st Year Maths 1B Pair of Straight Lines Formulas

→ If ax² + 2hxy + by² = 0 represents a pair of lines, then the sum of the slopes of lines is \(-\frac{2 h}{b}\) and the product of the slopes of lines is \(\frac{a}{b}\).

→ If ‘θ’ is an angle between the lines represented by ax² + 2hxy + by² = 0
then cos θ = \(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{a+b}\)
If ‘θ’ is accute, cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\); tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{|a+b|}\)

→ If h² = ab, then ax² + 2hxy + by² = 0 represents coincident or parallel lines.

→ ax² + 2hxy + by² = 0 represents a pair of ⊥^lr lines ⇒ a+ b = 0 i.e., coeft. of x² + coeff. of y² = 0.

→ The equation of pair of lines passing through origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and perpendicular to ax² + 2hxy + by² = 0 isb (x – x₁)² – 2h(x – x₁) (y – y₁) + a(y – y₁)² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is a(x-x,)² + 2h(x-x1)(y-y1) + b(y – y₁)² = 0.

→ The equations of bisectors of angles between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, is \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{\left(a_{2} x+b_{2} y+c_{2}\right)}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

→ The equation to the pair of bisectors of angles between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy.

→ The product of the perpendicular is from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

→ The area of the triangle formed by ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h / m+b\right|^{2} \mid}\)

→ The line ax + by + c = 0 and pair of lines (ax + by)² – 3(bx – ay)² =0 form an equilateral triangle and the area is \(\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)}\) sq.units

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines, then

• abc + 2fgh – af² – bg² – ch² = 0
• h² ≥ ab
• g² ≥ ac
• f² ≥ be

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines and h² > ab, then the point of intersection of the lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel lines then h² = ab and af² = bg²
The distance between the parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}=2 \sqrt{\frac{f^{2}-b c}{b(a+b)}}\)

Pair of Straight Lines:
Let L₁ = 0, L₂ = 0 be the equations of two straight lines. If P(x₁, y₁) is a point on L₁ then it satisfies the equation L₁ = 0. Similarly, if P(x₁, y₁) is a point on L₂ = 0 then it satisfies the equation.

If P(x₁, y₁) lies on L₁ or L₂, then P(x₁,y₁) satisfies the equation L₁L₂= 0.
L₁L₂ = 0 represents the pair of straight lines L₁ = 0 and L₂ = 0 and the joint equation of L₁ = 0 and L₂ = 0 is given by L₁ L₂= 0. ……………(1)
On expanding equation (1) we get and equation of the form ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 which is a second degree (non – homogeneous) equation in x and y.
Definition: If a, b, h are not all zero,then ax² + 2hxy + by² = 0 is the general form of a second degree homogeneous equation in x and y.
Definition: If a, b, h are not all zer, then ax² + 2hxy + by² + 2gx + 2 fy + c = 0 is the general form of a second degree non – homogeneous equation in x and y.

Theorem:
If a, b, h are not all zero and h² ≥ ab then ax² + 2hxy + by² = 0 represents a pair of straight lines passing through the origin.
Proof:
Case (i) : Suppose a = 0.
Given equation ax² + 2hxy + by² = 0 reduces to 2hxy + by² = 0 ^ y(2hx + by) = 0 .
Given equation represents two straight lines y = 0 ………..(1) and 2hx + by = 0 ………(2) which pass through the origin.
Case (ii): Suppose a ≠ 0.
Given equation ax² + 2hxy + by² = 0
⇒ a²x² + 2ahxy + aby² = 0
⇒ (ax)² + 2(ax)(hy) + (hy)² – (h² – ab)y² = 0
⇒ (ax + hy)² – (y\(\sqrt{h^{2}-a b}\))² = 0
[ax + y (h + \(\sqrt{h^{2}-a b}\))][ax + y (h – \(\sqrt{h^{2}-a b}\)] = 0

∴ Given equation represents the two lines
ax + hy + y\(\sqrt{h^{2}-a b}\) = 0, ax + hy – y\(\sqrt{h^{2}-a b}\) = 0 which pass through the origin.

Note 1:

• If h² > ab , the two lines are distinct.
• If h² = ab , the two lines are coincident.
• If h² < ab , the two lines are not real but intersect at a real point (the origin).
• If the two lines represented by ax² + 2hxy + by² = 0 are taken as l₁x + m₁y = 0 and l₂x + m₂y = 0 then
  ax² + 2kxy + by² = (4 x + m₁y) (x + m₂y) = 2 x² + (l₁m₂ + l₂m₁) xy + m₁m₂ y²
• Equating the coefficients of x², xy and y² on both sides, we get l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Theorem:
If ax² + 2hxy + by² = 0 represent a pair of straight lines, then the sum of slopes of lines is \(\frac{-2 h}{b}\) product of the slopes is \(\frac{a}{b}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………….(1) and l₂x + m₂y = 0 ………….(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
Slopes of the lines (1) and (2) are –\(\frac{l_{1}}{m_{1}}\) and \(-\frac{l_{2}}{m_{2}}\).
sum of the slopes = \(\frac{-l_{1}}{m_{1}}+\frac{-l_{2}}{m_{2}}=-\frac{l_{1} m_{2}+l_{2} m_{1}}{m_{1} m_{2}}=-\frac{2 h}{b}\)
Product of the slopes = \(\left(\frac{-l_{1}}{m_{1}}\right)\left(-\frac{l_{2}}{m_{2}}\right)=\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{a}{b}\)

Angle Between A Pair of Lines:
Theorem :
If θ is the angle between the lines represented by ax² + 2hxy + by² = 0, then cos θ = ±\(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁ x + m₁ y = 0 ………..(1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Let θ be the angle between the lines (1) and (2). Then cos θ = ±\(\frac{l_{1} l_{2}+m_{1} m_{2}}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)

Note 1:
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then tan θ = ±\(\frac{2 \sqrt{h^{2}-a b}}{a+b}\) and sin θ = \(\frac{2 \sqrt{h^{2}-a b}}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Conditions For Perpendicular And Coincident Lines:

• If the lines ax² + 2hxy + by² = 0 are perpendicular to each other then θ = π/ 2 and cos θ = 0 ⇒ a + b = 0 i.e., co-efficient of x² + coefficient of y² = 0.
• If the two lines are parallel to each other then 0 = 0.
  ⇒ The two lines are coincident ⇒ h² = ab

Bisectors of Angles:
Theorem:
The equations of bisectors of angles between the lines a₁ x + b₁ y + c₁ = 0, a₂ x + b₂ y + c₂ = 0 are \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\) = ±\(\frac{a_{2} x+b_{2} y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

Pair of Bisectors of Angles:
The equation to the pair bisectors of the angle between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy (or) \(\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……….(1)
and l₂x + m₂y = 0 ………(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The equations of bisectors of angles between (1) and (2) are \(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}-\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0 and
\(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}+\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0

The combined equation of the bisectors is

Theorem
The equation to the pair of lines passing through (x₀, y₀) and parallel ax² + 2hxy + by² = 0
is a( x – x₀)² + 2h( x – x₀)(y – y₀) + b( y – y₀)² = 0
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……(1) and l₂x + m₂ y = 0 …….. (2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of line parallel to (1) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(3)
The equation of line parallel to (2) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(4)

The combined equation of (3), (4) is
[l₁( x – x₀) + m₁( y – y₀)][l₂( x – x₀) + m₂(y – y₀)] = 0
⇒ l₁l₂(x – x₀)² + (l₁m₂ + l₂m₁)(x – x₀)(y – y₀) + m₁m₂(y – y₀)² = 0
⇒ a( x – x₀)² + 2h( x – x₀)( y – y₀) + b( y – y₀)2² = 0

Theorem:
The equation to the pair of lines passing through the origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0 .
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of the line perpendicular to (1) and passing through the origin is m₁x – l₁y = 0 ……….(3)
The equation of the line perpendicular to (2) and passing through the origin is m₂ x – l₂ y = 0 — (4)
The combined equation of (3) and (4) is
⇒ (m₁x – l₁ y)(m₂ x – l₂ y) = 0
⇒ m₁m₂x – (l₁m₂ + l₂m₁ )ny + l₁l₂ y = 0
bx² – 2hxy + ay² = 0

Theorem:
The equation to the lines passing through (x₀, y₀) and Perpendicular to ax² + 2hxy + by² = 0 is b(x – x₀)² – 2h(x – x₀)(y – y₀) + a(y – y₀)² = 0

Area of the triangle:
Theorem:
The area of triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h \ell m+b \ell^{2}\right|}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The given straight line is lx + my + n = 0 ………(3)
Clearly (1) and (2) intersect at the origin.
Let A be the point of intersection of (1) and (3). Then

Theorem:
The product of the perpendiculars from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 — (1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The lengths of perpendiculars from (α, β) to
the line (1) is p = \(\frac{\left|l_{1} \alpha+m_{1} \beta\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}\)
and to the line (2) is q = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)

∴ The product of perpendiculars is

Pair of Lines-Second Degree General Equation:
Theorem:
If the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines then
(i) Δ ≡ abc + 2fgh – af² – bg² – ch² =0 and
(ii) h² ≥ ab, g² ≥ ac, f² ≥ bc
Proof:
Let the equation S = 0 represent the two lines l₁x + m₁ y + n₁ = 0 and l₂ x + m₂ y + n₂ = 0. Then
ax2 + 2hxy + by2 + 2 gx + 2 fy + c
≡ (l₁x + m₁y + n₁)(l₂ x + m₂ y + n₂) = 0

Equating the co-efficients of like terms, we get
l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b, and l₁n₂ + l₂n₁ = 2g , m₁n₂ + m₂n₁ = 2 f , n₁n₂ = c

(i) Consider the product(2h)(2g)(2f)
= (l₁m₂ + l₂m₁)(l₁n₂ + l₂n₁)(m₁n₂ + m₂n₁)
= l₁l₂ (m₁²n₂ + m₂²n₁²) + m₁m₂ (l₁²n₂² + l₂²n₁²) + n₁n₂ (l₁²m₂² + l₂²m₁²) + 2l₁l₂m₁m₂n₁n₂
= l₁l₂[(m₁n₂ + m₂n₁) – 2m₁m₂n₁n₂] + m₁m₂[(l₁n₂ + l₂n₁) – 2l₁l₂n₁n₂] + n₁n₂[(l₁m₂ + l₂m₁) – 2l₁l₂m₁m₂] + 2l₁l₂m₁m₂n₁n₂
= a(4 f² – 2bc) + b(4g² – 2ac) + c(4h² – 2ab)
8 fgh = 4[af² + bg² + ch² – abc]
abc + 2 fgh – af² – bg² – ch² = 0

(ii) h² – ab = \(\left(\frac{l_{1} m_{2}+l_{2} m_{1}}{2}\right)^{2}\) – l1l2m1m2 = \(\frac{\left(l_{1} m_{2}+l_{2} m_{1}\right)^{2}-4-l_{1} l_{2} m_{1} m_{2}}{4}\)
= \(\frac{\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}{4}\) ≥ 0
Similarly we can prove g² > ac and f² ≥ bc

Note :
If A = abc + 2 fgh – af² – bg² – ch² = 0 , h² ≥ ab, g² ≥ ac and f² ≥ bc, then the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines

Conditions For Parallel Lines-Distance Between Them:
Theorem:
If S = ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of parallel lines then h² = ab and bg² = af². Also the distance between the two parallel lines is 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) (or) 2\(\sqrt{\frac{f^{2}-b c}{b(a+b)}}\)
Proof:
Let the parallel lines represented by S = 0 be
lx + my + n₁ = 0 ……….(1) lx + my + n₂ = 0 ………..(2)
ax² + 2hxy + 2gx + 2 fy + c
= (lx + my + n₁)(lx + my + n₂)

Equating the like terms
l² = a ………(3)
2lm = 2h …………(4)
m² = b ………..(5)
l(n₁ + n₂) = 2g …….(6)
m(n₁ + n₂) = 2 f ….(7)
n₁n₂ = c …….(8)

From (3) and (5), l²m² = ab and from (4) h² = ab .

Point of Intersection of Pair of Lines:
Theorem:
The point of intersection of the pair of lines represented by
a² + 2hxy + by² + 2gx + 2fy + c = 0 when h² > ab is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
Proof:
Let the point of intersection of the given pair of lines be (x₁, y₁).
Transfer the origin to (x₁, y₁) without changing the direction of the axes.
Let (X, Y) represent the new coordinates of (x, y). Then x = X + x₁ and y = Y + y₁.
Now the given equation referred to new axes will be
a( X + x₁)2 + 2h( X + x₁)(Y + y₁) +b(Y + y₁)2 + 2 g (X + x₁) + 2 f (Y + y₁) + c = 0
⇒ aX² + 2hXY + bY² + 2 X (ax₁ + hy₁ + g) + 2Y(hx₁ + by₁ + f) +(ax₁² + 2hx₁y₁ + by² + 2 gx₁ + 2 fy₁ + c) = 0

Since this equation represents a pair of lines passing through the origin it should be a homogeneous second degree equation in X and Y. Hence the first degree terms and the constant term must be zero. Therefore,
ax₁ + hy₁ + g = 0
hx₁ + by₁ + f = 0
ax₁² + 2hx₁ y₁ + by₁² + 2 gx₁ + 2 fyx + c = 0
But (3) can be rearranged as
x₁(ax₁ + hy + g) + y (hx₁ + byx + f) + (gx₁ + fq + c) = 0
⇒ gx₁ + fy₁ + c = 0 ………..(4)

Solving (1) and (2) for x₁ and y₁

Hence the point of intersection of the given pair of lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Theorem:
If the pair of lines ax² + 2hxy + by² = 0 and the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus then (a – b) fg+h(f² – g²) = 0.
Proof:
The pair of lines ax² + 2hxy + by² = 0 …………(1) is parallel to the lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 ……….. (2)

Now the equation
ax² + 2hxy + by² + 2gx + 2 fy + c + λ(ax² + 2hxy + by²) = 0

Represents a curve passing through the points of intersection of (1) and (2).
Substituting λ = -1, in (3) we obtain 2gx + 2fy + c = 0 …(4)
Equation (4) is a straight line passing through A and B and it is the diagonal \(\overline{A B}\)

The point of intersection of (2) is C = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
⇒ Slope of \(\overline{O C}=\frac{g h-a f}{h f-b g}\)
In a rhombus the diagonals are perpendicular ⇒ (Slope of \(\overline{O C}\)) (Slope of \(\overline{A B}\)) = -1
⇒ \(\left(\frac{g h-a f}{h f-b g}\right)\left(-\frac{g}{f}\right)\) = -1
⇒ g²h – afg = hf² – bfg
⇒ (a – b)fg + h(f² – g²) = 0
\(\frac{g^{2}-f^{2}}{a-b}=\frac{f g}{h}\)

Theorem:
If ax² + 2hxy + by² = 0 be two sides of a parallelogram and px + qy = 1 is one diagonal, then the other diagonal is y(bp – hq) = x(aq – hp)
proof:
Let P(x₁, y₁) and Q(x₂, y₂) be the points where the digonal

px + qy = 1 meets the pair of lines.
\(\overline{O R}\) and \(\overline{P Q}\) biset each other at M(α, β)
∴ α = \(\frac{x_{1}+x_{2}}{2}\) and β = \(\frac{y_{1}+y_{2}}{2}\)

Eliminating y from ax² + 2hxy+by² = 0
and px + qy = 1 ………..(2)
ax² + 2hx\(\left(\frac{1-p x}{q}\right)\) + b\(\left(\frac{1-p x}{q}\right)^{2}\) = 0
⇒ x² (aq² – 2hpq + bp²) + 2 x(hp – bp) + b = 0

The roots of this quadratic equation are x₁ and x₂ where
x₁ + x₂ = \(-\frac{2(h q-b p)}{a q^{2}-2 h p q-b p^{2}}\)
⇒ α = \(\frac{(b p-h q)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)

Similarly, by eliminating x from (1) and (2) a quadratic equation in y is obtained and y₁,
y₂ are its roots where
y₁ + y₂ = \(-\frac{2(h p-a q)}{a q^{2}-2 h p q-n p^{2}}\) ⇒ β = \(\frac{(a q-h p)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)
Now the equation to the join of O(0, 0) and M(α, β) is (y – 0)(0 – α) = (x – 0)(0 – β)
⇒ αy = βx
Substituting the values of α and β, the equation of the diagonal OR
is y(bp – hq) = x(aq – hp).

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a)

I.

Question 1.
ABCD is a Parallelogram. If L and M are the middle points of BC and CD, respectively, then find (i) AL and AM in terms of AB and AD (ii) λ, if AM = λ AD – LM.
Solution:

Question 2.
In ∆ABC, P, Q, and R are the midpoints of the sides AB, BC, and CA respectively. If D is any point.
(i) then express \(\overline{\mathrm{DA}}+\overline{\mathrm{DB}}+\overline{\mathrm{DC}}\) interms of \(\overline{D P}\), \(\overline{D Q}\) and \(\overline{D R}\).
(ii) If \(\overline{\mathbf{P A}}+\overline{\mathbf{Q B}}+\overline{\mathbf{R C}}=\bar{\alpha}\) then find \(\bar{\alpha}\)
Solution:

Question 3.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\mathbf{3} \overline{\mathbf{i}}+\overline{\mathbf{j}}\). Find the unit vector in the direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}\).
Solution:

Question 4.
If the vectors \(-3 \overline{\mathbf{i}}+4 \bar{j}+\lambda \overline{\mathbf{k}}\) and \(\mu \bar{i}+8 \bar{i}+6 \bar{k}\) are coilinear vectors , then find λ and µ.
Solution:

Question 5.
ABCDE is a pentagon. If the sum of the vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{BC}}, \overline{\mathrm{DC}}, \overline{\mathrm{ED}}\) and \(\overline{\mathbf{A C}}\) is λ \(\overline{\mathbf{A C}}\), then find the value of λ.
Solution:

Question 6.
If the position vectors of the points A, B and C are \(-2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}},-4 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and \(6 \bar{i}-3 \bar{j}-13 \bar{k}\) respectively and \(\overline{\mathbf{A B}}=\lambda \overline{\mathrm{AC}}\), then find the value of λ.
Solution:

Question 7.
If \(\overline{\mathrm{OA}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathrm{AB}}=3 \bar{i}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\overline{B C}=\bar{i}+2 \bar{j}-2 \bar{k}\) and \(\overline{C D}=2 \bar{i}+\bar{j}+3 \bar{k}\), then find the vector \(\overline{O D}\).
Solution:

Question 8.
\(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\bar{b}=4 \bar{i}+m \bar{j}+n \bar{k}\) are collinear vectors, then find m and n.
Solution:

Question 9.
Let \(\bar{a}=2 \bar{i}+4 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}, \bar{b}=\hat{i}+\bar{j}+\bar{k}\) and \(\bar{c}=\bar{j}+2 \bar{k}\). Find the unit vector in the opposite direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\).
Solution:

Question 10.
Is the triangle formed by the vectors \(3 \bar{i}+5 \bar{j}+2 \bar{k}, 2 \bar{i}-3 \bar{j}-5 \bar{k}\) and \(-5 \bar{i}-2 \bar{j}+3 \bar{k}\) equilateral?
Solution:

Question 11.
If α, β and γ be the angles made by the vector \(3 \bar{i}-6 \bar{i}+2 \bar{k}\) with the positive directions of the co-ordinate axes, then find cos α, cos β, cos γ.
Solution:
Unit vectors along the co-ordinate axes are respectively \(\bar{i}, \bar{j}, \bar{k}\).

Question 12.
Find the angles made by the straight line passing through the points (1, -3, 2) and (3, -5, 1) with the co-ordinate axes.
Solution:
Unit vectors along the co-ordinate axes are respectively \(\bar{i}, \bar{j}, \bar{k}\).
Let A(1, -3, 2) and B(3, -5, 1) be two given points.
Let ‘O’ be the origin. Then

II.

Question 1.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}=\alpha \overline{\mathbf{d}}, \overline{\mathbf{b}}+\overline{\mathbf{c}}+\overline{\mathbf{d}}=\beta \overline{\mathbf{a}}\) and \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathrm{c}}\) are non-coplanar vectors, then show that \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}+\overline{\mathbf{d}}=\mathbf{0}\).
Solution:

Question 2.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-coplanar vectors. Prove that the following four points are coplanar.
(i) \(-\overline{\mathbf{a}}+4 \overline{\mathbf{b}}-3 \bar{c}, \quad 3 \bar{a}+2 \bar{b}-5 \bar{c}\), \(-3 \overline{\mathbf{a}}+8 \overline{\mathbf{b}}-5 \overline{\mathbf{c}},-3 \overline{\mathbf{a}}+2 \overline{\mathbf{b}}+\overline{\mathbf{c}}\)
Solution:
Let ‘O’ be the origin and A, B, C, D be the four points.


4 + 2x + 2y = 0 ……..(1)
-2 – 4x + 2y = 0 ……..(2)
-2 + 2x – 4y = 0 …….(3)
Solve (1) and (3)
6y + 6 = 0 ⇒ y = -1
Substitute in (1)
2x + 2 (-1) + 4 = 0
⇒ 2x + 2 = 0
⇒ x = -1
Substitute x = -1, y = -1 in (2)
-2 – 4(-1) + 2(-1) = -4 + 4 = 0
∴ The vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar
⇒ A, B, C, D are coplanar
Hence the given points are coplanar.

(ii) \(6 \bar{a}+2 \bar{b}-\bar{c}, 2 \bar{a}-\bar{b}+3 \bar{c},-\bar{a}+2 \bar{b}-4 \bar{c},\)\(-12 \bar{a}-\bar{b}-3 \bar{c}\)
Solution:
Let O be the origin. Let A, B, C, D be the given points.

Let us suppose that one vector can be expressed as a linear combination of the other two.

∵ \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors.
7x + 18y – 4 = 0 ………(1)
-3 + 3y = 0 ⇒ y = 1 …….(2)
3x + 2y + 4 = 0 ………(3)
Substitute y =1 in (3)
3x + 2 + 4 = 0 ⇒ x = -2
Substitute x = -2 and y = 1 in (1)
7(-2) + 18(1) – 4 = 0 ⇒ 0 = 0
Hence \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) and \(\overline{\mathrm{AD}}\) are coplanar.
⇒ The points A, B, C, D are coplanar.

Question 3.
If \(\overline{\mathbf{i}}, \overline{\mathbf{j}}, \overline{\mathbf{k}}\) are unit vectors along the positive directions of the coordinate axes, then show that the four points \(4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\overline{\mathbf{k}},-\overline{\mathbf{j}}-\overline{\mathbf{k}}, 3 \overline{\mathbf{i}}+9 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}\) and \(-4 \bar{i}+4 \bar{j}+4 \bar{k}\) are coplanar.
Solution:
Let ‘O’ be the origin and let A, B, C, D be the given points.


⇒ The given points A, B, C, D are coplanar.
Second Method:
\(\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]\) = \(\left|\begin{array}{ccc}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)
= -4(12 + 3) + 6(-3 + 24) – 2(1 + 32)
= -60 + 126 – 66
= 0
Hence the vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) and \(\overline{\mathrm{AD}}\) are coplanar.
⇒ The given points A, B, C, D are coplanar.

Question 4.
If a, b, c are non-coplanar vectors, then test for the collinearity of the following points whose position vectors are given by
(i) \(\bar{a}-2 \bar{b}+3 \bar{c}, 2 \bar{a}+3 \bar{b}-4 \bar{c},-7 \bar{b}+10 \bar{c}\)
Solution:
Let ‘O’ be the origin. A, B, C be the given points.
Then \(\overline{\mathrm{OA}}=\overline{\mathrm{a}}-2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}\)

(ii) \(3 \bar{a}-4 \bar{b}+3 \bar{c}\), \(-4 \bar{a}+5 \bar{b}-6 \bar{c}\), \(4 \overline{\mathbf{a}}-7 \overline{\mathbf{b}}+6 \overline{\mathbf{c}}\)
Solution:

(iii) \(\begin{aligned}
&2 \bar{a}+5 \bar{b}-4 \bar{c}, \bar{a}+4 \bar{b}-3 \bar{c}, \\
&4 \bar{a}+7 \bar{b}-6 \bar{c}
\end{aligned}\)
Solution:
Let ‘O’ be the origin and A, B, C be the given points.
Then \(\overline{\mathrm{OA}}=2 \overline{\mathrm{a}}+5 \overline{\mathrm{b}}-4 \overline{\mathrm{c}}\), \(\overline{\mathrm{OB}}=\overline{\mathrm{a}}+4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}}\)

∴ The points A, B, C are collinear.

III.

Question 1.
In the Cartesian plane, O is the origin of the coordinate axes. A person starts at O and walks a distance of 3 units in the NORTH-EAST direction and reaches point P. From P he walks 4 units of distance parallel to NORTH-WEST direction and reaches the point Q. Express the vector \(\overline{\mathbf{O Q}}\) in terms of \(\overline{\mathbf{i}}\) and \(\overline{\mathbf{j}}\) (observe that ∠XOP = 45°)
Solution:
‘O’ the origin of co-ordinate axes.

Question 2.
The points O, A, B, X and Y are such that \(\overline{\mathbf{O A}}=\overline{\mathbf{a}}, \overline{\mathbf{O B}}=\overline{\mathbf{b}}, \overline{\mathbf{O X}}=\mathbf{3} \overline{\mathbf{a}}\) and \(\overline{\mathbf{O Y}}=\mathbf{3} \overline{\mathbf{b}}\). Find \(\overline{\mathbf{B X}}\) and \(\overline{\mathbf{A Y}}\) interms of \(\bar{a}\) and \(\bar{b}\). Futher, if the point P divides AY in the ratio 1 : 3, then express \(\overline{\mathrm{BP}}\) interms of \(\bar{a}\) and \(\bar{a}\).
Solution:

Question 3.
If ∆OAB, E is the midpoint of AB and F is a point on OA such that OF = 2(FA). If C is the point of intersection of \(\overline{\mathrm{OE}}\) and \(\overline{\mathrm{BF}}\), then find the ratios OC : CE and BC : CF.
Solution:

Question 4.
Point E divides the segment PQ internally in the ratio 1 : 2 and R is any point not on the line PQ. If F is a point on QR such that QF : FR = 2 : 1, then show that EF is parallel to PR.
Solution: