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Students get through AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules

Very Short Answer Questions

Question 1.
Define Carbohydrates.
Answer:
The compounds which are primarily produced by plants and form a very large group of naturally occuring organic compounds are called Carbohydrates.
Eg: Glucose, Fructose, Starch. Carbohydrates are the polyhydroxy aldehydes (or) ketones.

Question 2.
Name the different types of carbohydrates on the basis of their hydrolysis. Give one example for each.
Answer:
On the basis of the hydrolysis, carbohydrates are classified as

1. Monosaccharides, Eg : Glucose, fructose No saccharides
2. Oligosaccharides, Eg : Sucrose, maltose two monosaccharides
3. Polysaccharides, Eg : Starch, cellulose large number of monosaccharides.

Question 3.
How is Glucose prepared ? [IPE – 2014]
Answer:
Glucose is prepared by the hydrolysis of starch in presence of a little acid.

Question 4.
Why are sugars classified as reducing and non-reducing sugars ?
Answer:

• Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
  Eg: Glucose.
• Carbohydrates that doesnot reduce Fehling’s reagent, Tollen’s reagent are called non reducing sugars.
  Eg : Sucrose.

Question 5.
What do you understand by invert sugars ?
Answer:
During the hydrolysis of sucrose there is a change in the sign of rotation, from dextro (+) to laevo (-) and the product is named as invert sugar.

Question 6.
What do you mean by essential amino acids ? Give two examples for non essential amino acids ? [IPE 2014] [T.S. Mar. 16]
Answer:
Essential amino acids : The amino acids which cannot be synthesized in the body and must be obtained through diet are known as essential aminoacids.
Eg : valine, leucine etc.
Examples of non essential amino acids are Alanine, Glycine, Aspartic acid.

Question 7.
What is zwitter ion ? Give an example. [IPE 2015 (AP)]
Answer:
Zwitter ion: In aqueous solution of amino acids, the carboxyl group can lose a proton and amino acid can accept that proton to form a dipolar ion. This ion is called as zwitter ion.

Question 8.
What are reducing sugars ?
Answer:
Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
Eg: glucose.

Question 9.
What are proteins ? Give an example.
Answer:
Proteins : A poly peptide with more than hundred amino acid residues, having molecular mass higher than 10,000 units is called a protein.
Eg : keratin, myosin, insulin.

Question 10.
What are the components of a nucleic acid ?
Answer:

• Nucleic acids are long chain polymers of nucleotides i.e., poly nucleotides.
• Nucleic acids are constituted by pentose sugar, phosphoric acid, and nitrogenous hetero cyclic base (purine (or) pyrimidine).

Question 11.
What are essential and non – essential amino acids ? Give one example for each. [IPE – 2016 (TS)]
Answer:

• Essential amino acids : The amino acids that cannot be synthesised by the body and must be supplied in the diet are called essential amino acids.
  Eg : valine, leucine, phenyl alanine etc.
• Non-essential amino acids : Other amino acids synthesised by the tissues of the body are called non-essential amino acids..
  Eg: Glycine, alanine etc.

Question 12.
Differentiate between globular and fibrous proteins.
Answer:
Globular proteins

1. In this proteins the chains of poly peptides coil around to give a spherical shape.
2. hese are soluble in water.
3. Eg : Insulin

Fibrous proteins

1. In this proteins the poly peptides run parallel and are held together by hydrogens and disulphite – bonding.
2. These are insoluble in water.
3. Eg: keratin

Question 13.
Why are vitamin A and vitamin C essential to us ? Give their important sources.
Answer:
Vitamin A and Vitamin C are essential to us.
Explanation :

• Deficiency of vitamin A causes night blindness, redness in eyes, xerophthalnia.
• Deficiency of vitamin C causes pernicious anaemia (RBC deficient in haemoglobin).

Sources:

• Vitamin – A : Fish liver oil, carrots, butter and milk.
• Vitamin – C : Citrous fruits, amla, green leafy vegetables

Question 14.
What do you understand from the names (a) aldo pentose and (b) ketoheptose ?
Answer:
a) Aldo pentose : If a monosaccharide contains 5 carbon atoms with aldehyde group then it is known as aldo pentose.
b) Ketoheptose : If a monosaccharide contain seven carbons with a ketone group then it is called ketoheptose.

Question 15.
What are anomers ?
Answer:
Anomers : The two isomeric structures of a compound which differ in configuratiofi at C-l only are called Anomers.

Question 16.
What are amino acids ? Give two examples.
Answer:
The organic compounds which contain amino (-NH₂) functional group and carboxyl (-COOH) functional group are called amino acids.
Eg : Glycine, Alanine etc.

Question 17.
What are fibrous proteins ? Give examples.
Answer:
When the poly peptide chains run parallel and are held together by hydrogen and disulphide bonds then fibre – like structure is formed. These are called fibrous proteins. These are insoluble in water.
Eg : keratin, myosin.

Question 18.
What are globular proteins ? Give examples.
Answer:
When the chains of polypeptides coil around to give a spherical shape then globular proteins are formed. These are usually soluble in water.
Eg : insulin and albumins.

Short Answer Questions

Question 1.
Explain the denaturation of proteins.
Answer:
The phenomenon of disorganization of native protein structure is known as denaturation. Denaturation results in the loss of secondary, tertiary and quaternary structure of proteins. This involves a change in physical, chemical and biological properties of protein molecules.
Agents of denaturation:
Physical agents – Heat, violent shaking, X – rays, UV – radiation.
Chemical gents – Acids, alkalies, organic solvents, urea, salts of heavy metals.

Question 2.
What are enzymes ? Give examples ?
Answer:
The group of complex proteinoid organic compounds, elaborated by living organism which catalyse specific organic reactions are called enzymes.
Eg.: Lipases, Rennin, Maltase, Invertase etc. Practically all biological processes such as digestion, respiration etc., are carried on through the agency of enzymes.
Enzymes may be defined as biocatalysts synthesised by living cells.
The functional unit of enzyme is known as holo enzyme made up at apo enzyme (protein part) and co enzyme (non-protein part).

Question 3.
Write notes on vitamins.
Answer:
Vitamin is defined as an “accessory food factor which is essential for growth and healthy maintenance of the body.”
Classification : Vitamins are broadly classified into two major groups.
a) the fat soluble Eg : vitamin A, D, E and K.
b) water soluble Eg : vitamin B – complex and C.

Question 4.
What are harmones ? Give one example for each. [IPE – 2016 (TS)] [Mar. 14]

1. steroid hormones
2. Poly peptide hormones and
3. amino acid derivatives.

Answer:
Hormones: Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”.
Eg : testosterone, oestrogen.

1. Example for steroid hormones : Testosterone, oestrogen
2. Example for poly peptide hormones: Insulin
3. Example for Amino acid derivative : Thyroidal hormones thyroxine.

Question 5.
Write the importance of carbohydrates.
Answer:
Importance of carbohydrates:

• Carbohydrates are essential for life of plants.
• Carbohydrates are used as storage molecules as starch in plants.
• Cell wall of plants is made up of cellulose.
• Carbohydrates are also essential for life of animals. Carbohydrates are used as storage molecules as glycogen in animals.
• Carbohydrate source honey is used for a long time as an instant source of energy.

Question 6.
Give the sources of the following vitamins and name the diseases caused by their- deficiency [T.S. Mar. 17] [IPE AP & TS (Mar. 15) BMP, 2016 (AP]
(a) A
(b) D
(c) E and
(d) K
Answer:
Vitamin : A
Sources : Fish oils, liver, kidney
Deficiency diseases : Night blindness, Redness in eyes.

Vitamin : D
Sources : Fish oils, butter, milk, egg
Deficiency diseases : Rickets in children, osteomalacia in adults

Vitamin : E
Sources : Wheat germ oil, egg yolk, green vegetables
Deficiency diseases : Sterility

Vitamin : K
Sources : Green vegetables, Intestinal flora.
Deficiency diseases : Blood coagulation is prevented

Question 7.
Write notes on proteins. [IPE – 2016 (TS)]
Answer:
Proteins are polypeptide chains of amino acids.
Classification of Proteins : Proteins can be classified into two types on the basis of their molecular shape.
a) Fibrous Proteins : These are fibre like proteins, the polypeptide chains are parallel which are held together by hydrogen and disulphide bonds. These are insoluble in water.
Ex : Keratin present in hair, wool, silk etc., and myosin present in muscles.

b) Globular proteins: In these proteins, the polypeptide chains coil around to give a spherical shape. These are soluble in water.
Ex : insulin and albumin.

The structure of proteins is explained in four different levels
a) Primary structure
b) Secondary structure
c) Tertiary structure
d) Quaternary structure

Denaturation of proteins : A protein in a biological system with a specific structure and biological activity is called a native protein. The process in which a protein loses its activity when subjected to heating, change in pH, addition of reagents is called denaturation of protein. Denaturation may be reversible or irreversible.
Ex : Coagulation of egg white on boiling is an irresersible denaturation.
Renaturation is the reverse process of denaturation.

Question 8.
Write about polysaccharides with starch and cellulose as examples.
Answer:
Polysaccharides : The saccharides which on hydrolysis to form large number of monosaccharides are called polysaccharides.
Eg : Starch and cellulose

Starch :

• Starch is the most important dietary source for human beings.
• Vegetables, roots, cereals are important sources of starch.
• It is a polymer ofα – glucose.
• It is constituted by two components Amylose and amylopectin.

Amylose :

• It constitutes 15 – 20% of starch.
• Amylose is water soluble component.
• Amylose is a branched chain with 200 – 1000 α – D – glucose units held by C – 1 to C – 4 glycosidic linkage.

Amylopectin :

• Amylopectin constitutes 80 – 85% of starch.
• It is a branched chain polymer of a – glucose units in which chain is formed by C. – 1 to C – 4 glycosidic linkage whereas branching occurs by C – 1 to C – 6 glycosidic linkage.

Cellulose :

• Cellulose occurs in plants and it is the most abundant organic substance.
• It is a major constituent of cell wall of plant cells.
• Cellulose is a straight chain polysaccharide composed only of β – D – glucose units . which are joined by glycosidic linkage between C – 1 of one glucose and C – 4 of the next glucose.

Question 9.
Write notes on the functions of different hormones in the body. [IPE 2014]
Answer:
Functions of Hormones:

• Hormones help to maintain the balance of biological activities in the body.
• Insulin maintains the blood glucose level within the limit.
• Growth hormones and sex hormones play role in growth and development.
• Low level of thyroxine (produced from thyroid gland) causes hypothyroidism. High level of thyroxine causes hyper thyroidism.
• Gluco corticoids control the carbohydrate metabolism, modulates the inflammatory reactions.
• The mineralo corticoids control the level of excretion of water and salt by the kidney.
• Adrenal cortex does not function properly then results in Addison’s disease.
• Hormones released by gonads are responsible for development of secondary sex characters.
• Testosterone is responsible for development of secondary sex hormone produced in male.
• Estradiol is the main female sex hormone responsible for development of secondary female characterstics like control of menstrual cycle.
• Progesterone is responsible for preparing the uterus for implantation of fertiised egg.

Question 10.
Write the sources of vitamin and diseases due to vitamin deficIency. [AP Mar. 20]
Answer:

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(d) Group-18 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(d) Group-18 Elements

Very Short Answer Questions

Question 1.
List out the uses of Neon.
Answer:
Uses of Ne:

1. Ne is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
2. ‘Ne’ – bulbs are used in botanical gardens and in greenhouses.

Question 2.
Write any two uses of argon.
Answer:
Uses of Ar:

1. ‘Ar’ is used to create an inert atmosphere in a high-temperature metallurgical process.
2. ‘Ar’ is used in filling electric bulbs.

Question 3.
In modern diving apparatus, a mixture of He ánd O₂ is used – Why? (IPE 2016 (AP))
Answer:
In modem diving apparatus, a mixture of He and O₂ is used because He is very low soluble in blood.

Question 4.
Helium is heavier than hydrogen. Yet helium is used (instead of H₂) in filling balloons for meteorological observations – Why?
Answer:
‘He’ is a non—inflammable and light gas. Hence it is used in filling balloons for meterological observations.

Question 5.
How is XeO₃ prepared? (IPE May – 2015(AP), 2014)
Answer:
XeF₆ on hydrolysis produce XeO₃.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 6.
Give the preparation of
a) XeOF₄ and
b) XeO₂F₂. (TS Mar. 17; IPE 2014)
Answer:
Partial hydrolysis of XeF₆ gives oxy fluorides XeOF₄ and XeO₂F₂
XeF₆ + H₂O → XeOF₄ + 2HF
XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 7.
Explain the structure of XeO₃. (TS Mar. 17; IPE 16, 15’ 14 (TS))
Answer:
Structure of XeO₃:

1. Central atom is Xe’.
2. ‘Xe undergoes sp³ hybridisation in 3^rd excited state.
   
3. ‘Xe’ forms 3σ-bonds and 3π-bonds with three oxygens.
4. Shape of molecule is pyramidal with bond angle 103°.
   

Question 8.
Explain the shape of XeF₄ on the basis of VSEPR theory.
Answer:
Shape of XeF₄:

1. Central atom in XeF₄ is
   
2. Xe undergoes sp³d² hybridisation in its 2^nd excited state.
3. Shape of the molecule is squãre planar with bond angle 90° and bond length 1.95A
   
4. Xe – forms four σ-bonds by the overlap of sp³d² – 2p_z(F) orbitals.

Question 9.
Which noble gas is radio active ? How is it formed?
Answer:
Radon (Rn) is radio active noble gas. Radon is obtained as decay product of ₈₆R²²⁶.
₈₆Ra²²⁶ → ₈₆Rn²²² + ₂He⁴

Question 10.
How are XeF₂, XeF₄, XeF₆ prepared? Give equation.
Answer:

Question 11.
Noble gases are inert – explain.
Answer:
Noble gases are chemically inert:

1. Noble gases have stable electronic configuration octet configuration except He.
2. Noble gases have high Ionisation energy values and have large positive values of electron gain enthalpy.

Question 12.
Write the name and formula of the first noble gas compound prepared by Bertlett.
Answer:
The first noble gas compound prepared by Bertlett is XePtF₆. Name of the compound is xenon hexa fluoro platinate.

Question 13.
Why do noble gases form compounds with fluorine and oxygen only?
Answer:
Noble gases form compounds with flourine and oxygen only.
Reason: Oxygen and Fluorine are most electronegative elements.

Short Answer Questions

Question 1.
Explain the structures of
a) XeF₂ and
b) XeF₄. (AP Mar. ‘17, IPE 16, 15, ‘14 (TS)) (TS Mar. ’14)
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Structure of XeF₂:

1. In XeF₂ central atom is ‘Xe’.
2. ‘Xe’ undergoes sp³d hybridisation in its 1^st excited state.
   
3. Shape of molecule is linear.
4. Xe form two σ-bonds with two fluorines.
   

b) Structure of XeF₄:

1. Central atom in XeF₄ is ‘Xe’.
2. Xe undergoes sp³d² hybridisation in it’s 2^nd excited state.
   
3. Shape of the molecule is square planar with bond angle 90° and bond length 1.95A
   
4. Xe forms four σ-bonds by the overlap of sp³d² – 2p_z (F) orbitals.

Question 2.
Explain the structures of
a) XeF₆ and
b) XeOF₄ (IPE Mar & May – 2015, 14)
Answer:
Structure òf XeF₆ is ‘Xe’.

1. Central atom in XeF₆ is ‘Xe’.
2. Xe undergoes sp³d³ hybridisation in it’s 3^rd excited state.
   
3. Shape of molecule is distorted octahedral.
   

b) Structure of XeOF₄:

1. In XeOF₄ molecule ‘Xe’ undergoes sp³d² hybridisation.
2. Shape of the molecule is square pyramid.
3. There is one Xe-O double bond containing.
   
   pπ = dπ overlapping.
   Partial hydrolysis of XeF₆ gives XeOF₄
   XeF₆ + H₂O → XeOF₄ + 2HF
   XeOF₄ is a colourless volatile liquid. It has a square pyramidal shape.

Question 3.
Explain the structure of
a) XeO₃ and
b) XeO₄ (T.S. Mar. ‘16)
Answer:
a) Structure of XeO₃:

1. Central atom is ‘Xe’
2. ‘Xe’ undergoes sp³ hybridisation in 3^rd excited state.
   
3. ‘Xe forms 3σ-bonds and 3π-bonds with three oxygens.
4. Shape of molecule is pyramidal with bond angle 103°.
   

b) Structure of XeO₄:
Xe is in sp₃ hybridisation four singma bonds and four dπ – pπ bonds. Shape of XeO₄ is tetrahedral.

Question 4.
Write the preparations of Xenon flourides.
Answer:
Xenon flourides : Xenon forms three binary fluorides, Xe F₂, XeF₄, and XeF₆ by the direct reaction of Xenon with fluorine under suitable conditions.

XeF₆ can also be prepared by the interaction of XeF₄ and O₂F₂ at 143 K.
XeF₄ + O₂F₂ → XeF₆ + O₂

Question 5.
Write the preparations of Xenon Oxides.
Answer:
Xenon Oxides: Xenon forms two oxides XeO₃ and XeO₄ with oxygen.
These oxides are formed by the hydrolysis of xenon fluorides.
6XeF₄ + 12 H₂O → 4 Xe + 2 XeO₃ + 24 HF + 3O₂
XeF₆ + 3 H₂O → XeO₃ + 6 HF

Question 6.
Give the formulae and describe the structures of a noble gas species, isostructural with
a) \(\mathrm{ICl}_4^{-}\)
b) \(\mathbf{I B r}_2^{-}\)
c) \(\mathrm{BrO}_3^{-}\)
Answer:
a) \(\mathrm{ICl}_4^{-}\) is also structural with XeF₄ and it has square planar shape.
b) \(\mathbf{I B r}_2^{-}\) is also structural with XeF₂ and it has linear shape.
c) \(\mathrm{BrO}_3^{-}\) is iso-structural with XeO₄ and it has tetrahedral šhape.

Question 7.
Write any two uses of Helium.
Answer:
Uses of Helium:

1. It is non inflammable and very light gas. Hence it is used in the balloons of Meteorological observations.
2. It is used in gas cooled nuclear reactors and used as cryogenic agent.
3. It is used as diluent for oxygen in modern diving apparatus.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d)

Question 1.
Find the coefficient of x³ in the power series expansion of \(\frac{5 x+6}{(x+2)(1-x)}\) specifying the region in which the expansion is valid.
Solution:

Question 2.
Find is the coefficient of x⁴ in the power series expansion of \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}\) specifying the interval in which the expansion is valid.
Solution:
Let \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}=\frac{A}{x-3}+\frac{B x+C}{x^2+2}\)
Multiplying with (x² + 2) (x – 3)
3x² + 2x = A(x² + 2) + (Bx + C) (x – 3)
x = 3
⇒ 27 + 6 = A(9 + 2)
⇒ 33 = 11A
⇒ A = 3
Equating the coefficients of x²
3 = A + B
⇒ B = 3 – A = 3 – 3 = 0
Equating the constants,
2A – 3C = 0
⇒ 3C = 2A = 6
⇒ C = 2

Question 3.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x-4}{x^2-5 x+6}\) specifying the region in which the expansion is valid.
Solution:
Let \(\frac{x-4}{x^2-5 x+6}=\frac{A}{x-2}+\frac{B}{x-3}\)
Multiplying with (x – 2) (x – 3)
x – 4 = A(x – 3) + B(x – 2)
x = 2
⇒ -2 = A(2 – 3) = -A
⇒ A = 2
x = 3
⇒ -1 = B(3 – 2) = B
⇒ B = -1

Question 4.
Find the coefficient of xⁿ in the power series expansion of \(\frac{3 x}{(x-1)(x-2)^2}\)
Solution:

∴ 3x = A(x – 2)² + B(x – 1) (x – 2) + C(x – 1) ……..(1)
putting x = 1,
3 = A(1 – 2)²
⇒ A = 3
putting x = 2,
6 = C(2 – 1)
⇒ C = 6
Now equating the co-efficient of x² terms in (1)
0 = A + B
⇒ B = -A
⇒ B = -3

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c)

Resolve the following into partial fractions.

Question 1.
\(\frac{x^2}{(x-1)(x-2)}\)
Solution:
Let \(\frac{x^2}{(x-1)(x-2)}=1+\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with (x – 1) (x – 2)
x² = (x – 1) (x – 2) + A(x – 2) + B(x – 1)
Put x = 1, 1 = A(-1) ⇒ A = -1
Put x = 2, 4 = B(1) ⇒ B = 4
∴ \(\frac{x^2}{(x-1)(x-2)}=1-\frac{1}{x-1}+\frac{4}{x-2}\)

Question 2.
\(\frac{x^3}{(x-1)(x+2)}\)
Solution:

Question 3.
\(\frac{x^3}{(2 x-1)(x-1)^2}\)
Solution:
Let \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{A}{2 x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with 2(2x – 1) (x – 1)²
2x³ = (2x – 1) (x – 1)² + 2A(x – 1)² + 2B(2x – 1) (x – 1) + 2C(2x – 1)
Put x = \(\frac{1}{2}\),
⇒ 2(\(\frac{1}{8}\)) = 2A(\(\frac{1}{4}\))
⇒ A = \(\frac{1}{2}\)
Put x = 1,
⇒ 2(1) = 2C(1)
⇒ C = 1
Put x = 0,
0 = (-1) (1) + 2A(1) + 2B(-1) (-1) + 2C(-1)
⇒ 2A + 2B – 2C = 1
⇒ 2B = 1 + 2C – 2A
⇒ 2B = 1 + 2 – 1 = 2
⇒ B = 1
∴ \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{1}{2(2 x-1)}+\frac{1}{(x-1)}+\frac{1}{(x-1)^2}\)

Question 4.
\(\frac{x^3}{(x-a)(x-b)(x-c)}\)
Solution:
Let \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\)
Multiplying with (x – a)(x – b) (x – c),
x³ = (x – a)(x – b) (x – c) + A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x – b)
Put x = a,
a³ = A(a – b) (a – c)
⇒ A = \(\frac{a^3}{(a-b)(a-c)}\)
Put x = b,
b³ = B(b – a) (b – c)
⇒ B = \(\frac{b^3}{(b-a)(b-c)}\)
Put x = c, c³ = C(c – a) (c – b)
⇒ C = \(\frac{c^3}{(c-a)(c-b)}\)
∴ \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-a)(b-c)(x-b)}\) + \(\frac{c^3}{(c-a)(c-b)(x-c)}\)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b)

Resolve the following into partial fractions.

Question 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)
Multiplying with (x – 1) (x² + 2)
2x² + 3x + 4 = A(x² + 2) + (Bx + C) (x – 1)
x = 1
⇒ 2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
Equating the coefficients of x²
2 = A + B
⇒ B = 2 – A = 2 – 3 = -1
Equating constants
4 = 2A – C
⇒ C = 2A – 4 = 6 – 4 = 2
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)

Question 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{A}{2+x}+\frac{B x+C}{1-x+x^2}\)
Multiplying with (2 + x) (1 – x + x²)
3x – 1 = A(1 – x + x²) (Bx + C) (2 + x)
x = -2
⇒ -7 = A(1 + 2 + 4) = 7A
⇒ A = -1
Equating the coefficients of x²
0 = A + B ⇒ B = -A = 1
Equating the constants
-1 = A + 2C
⇒ 2C = -1 – A = -1 + 1 = 0
⇒ C = 0
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(2+x)}=-\frac{1}{2+x}+\frac{x}{1-x+x^2}\)

Question 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
Solution:
Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
Multiplying with (x + 2) (x² + 1)
x² – 3 = A(x² + 1) + (Bx + C) (x + 2)
x = -2
⇒ 4 – 3 = A(4 + 1)
⇒ 1 = 5A
⇒ A = \(\frac{1}{5}\)
Equating the coefficients of x²
1 = A + B
⇒ B = 1 – A = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Equating the constants
-3 = A + 2C
⇒ 2C = -3 – A
⇒ 2C = -3 – \(\frac{1}{5}\)
⇒ 2C = \(-\frac{16}{5}\)
⇒ C = \(-\frac{8}{5}\)
∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

Question 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
Solution:
Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)
Multiplying with (x² + x + 1)²
x² + 1 = (Ax + B) (x² + x + 1) + (Cx + D)
Equating the coefficients of x³,
A = 0
Equating the coefficients of x²,
A + B = 1 ⇒ B = 1
Equating the coefficients of x,
A + B + C = 0
⇒ 1 + C = 0
⇒ C = -1
Equating the constants,
B + D = 1
⇒ D = 1 – B = 1 – 1 = 0
∴ Ax + B = 1, Cx + D = -x
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\)

Question 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
Solution:

∴ x³ + x² + 1 = A(x – 1) (x² + x + 1) + B(x² + x + 1) + (Cx + D) (x – 1)² …….(2)
Put x = 1 in (2)
1 + 1 + 1 = A(0) + B(1 + 1 + 1) + (C(1) + D) (0)
⇒ 3B = 3
⇒ B = 1
Equating the coefficients of x³ in (2)
1 = A + C ………(3)
Equating the coefficients of x² in (2)
1 = A(1 – 1) + B(1) + C(-2) + D(1)
⇒ 1 = B – 2C + D
∵ B = 1,
⇒ 1 = 1 – 2C + D
⇒ 2C = D ………(4)
Put x = 0 in (2)
1 = A(-1)(1) + B(1) + D(-1)²
⇒ -A + B + D = 1
⇒ -A + 1 + D = 1
⇒ A = D ………(5)
From (3), (4) and (5)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a)

Resolve the following into partial fractions.

I.

Question 1.
\(\frac{2 x+3}{(x+1)(x-3)}\)
Solution:
Let \(\frac{2 x+3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}\)
Multiplying with (x + 1) (x – 3)
2x + 3 = A(x – 3) + B(x + 1)
x = -1 ⇒ 1 = A(-4) ⇒ A = \(-\frac{1}{4}\)
x = 3 ⇒ 9 = B(4) ⇒ B = \(\frac{9}{4}\)
\(\frac{2 x+3}{(x+1)(x-3)}=\frac{-1}{4(x+1)}+\frac{9}{4(x-3)}\)

Question 2.
\(\frac{5 x+6}{(2+x)(1-x)}\)
Solution:
Let \(\frac{5 x+6}{(2+x)(1-x)}=\frac{A}{2+x}+\frac{B}{1-x}\)
Multiplying with (2 + x) (1 – x)
5x + 6 = A(1 – x) + B(2 + x)
Put x = -2,
-10 + 6 = A(1 + 2)
⇒ A = \(-\frac{4}{3}\)
Put x = 1,
5 + 6 = B(2 + 1)
⇒ B = \(\frac{11}{3}\)
∴ \(\frac{5 x+6}{(2+x)(1-x)}=-\frac{4}{3(2+x)}+\frac{11}{3(1-x)}\)

II.

Question 1.
\(\frac{3 x+7}{x^2-3 x+2}\)
Solution:
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with x² – 3x + 2
3x + 7 = A(x – 2) + B(x – 1)
x = 1 ⇒ 10 = -A ⇒ A = -10
x = 2 ⇒ 13 = B ⇒ B = 13
∴ \(\frac{3 x+7}{x^2-3 x+2}=\frac{-10}{x-1}+\frac{13}{x-2}\)

Question 2.
\(\frac{x+4}{\left(x^2-4\right)(x+1)}\)
Solution:
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-2}\)
Multiplying with (x² – 4) (x + 1)
x + 4 = A(x² – 4) + B(x + 1) (x – 2) + C(x + 1) (x + 2)
x = -1
⇒ 3 = A(1 – 4)
⇒ 3 = -3A
⇒ A = -1
x = -2
⇒ 2 = B(-2 + 1) (-2 – 2)
⇒ 2 = 4B
⇒ B = \(\frac{1}{2}\)
x = 2
⇒ 6 = C(2 + 1)(2 + 2)
⇒ 6 = 12C
⇒ C = \(\frac{1}{2}\)
∴ \(\frac{x+4}{\left(x^2-4\right)(x+1)}=-\frac{1}{x+1}+\frac{1}{2(x+2)}\) + \(\frac{1}{2(x-2)}\)

Question 3.
\(\frac{2 x^2+2 x+1}{x^3+x^2}\)
Solution:
Let \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
Multiplying with x²(x + 1)
2x² + 2x + 1 = Ax(x + 1) + B(x + 1) + Cx²
Put x = 0, 1 = B
Put x = -1, 2 – 2 + 1 = C(1) ⇒ C = 1
Equating the coefficients of x²,
2 = A + C
⇒ A = 2 – C = 2 – 1 = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\)

Question 4.
\(\frac{2 x+3}{(x-1)^3}\)
Solution:
\(\frac{2 x+3}{(x-1)^3}\)
Put x – 1 = y ⇒ x = y + 1
⇒ \(\frac{2 x+3}{(x-1)^3}=\frac{2(y+1)+3}{y^3}=\frac{2 y+5}{y^3}\)
⇒ \(\frac{2 x+3}{(x-1)^3}\) = \(\frac{2}{y^2}+\frac{5}{y^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)
∴ \(\frac{2 x+3}{(x-1)^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)

Question 5.
\(\frac{x^2-2 x+6}{(x-2)^3}\)
Solution:
Let x – 2 = y then x = y + 2

III.

Question 1.
\(\frac{x^2-x+1}{(x+1)(x-1)^2}\)
Solution:
Let \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with (x + 1) (x – 1)²
x² – x + 1 = A(x – 1)² + B(x + 1) (x – 1) + C(x + 1)
Put x = -1,
1 + 1 + 1 = A(4)
⇒ A = \(\frac{3}{4}\)
Put x = 1,
1 – 1 + 1 = C(2)
⇒ C = +\(\frac{1}{2}\)
Equating the coefficients of x²,
A + B = 1
⇒ B = 1 – A
⇒ B = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}\) + \(\frac{1}{2(x-1)^2}\)

Question 2.
\(\frac{9}{(x-1)(x+2)^2}\)
Solution:
Let \(\frac{9}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
Multiplying with (x – 1) (x + 2)²
9 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1)
x = 1
⇒ 9 = 9A
⇒ A = 1
x = -2
⇒ 9 = -3C
⇒ C = -3
Equating the coefficients of x²
A + B = 0 ⇒ B = -A = -1
∴ \(\frac{9}{(x-1)(x+2)^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{(x+2)^2}\)

Question 3.
\(\frac{1}{(1-2 x)^2(1-3 x)}\)
Solution:
Let \(\frac{1}{(1-2 x)^2(1-3 x)}=\frac{A}{1-3 x}+\frac{B}{1-2 x}+\frac{C}{(1-2 x)^2}\)
Multiplying with (1 – 2x)² (1 – 3x)
1 = A(1 – 2x)² + B(1 – 3x) (1 – 2x) + C(1 – 3x)

Question 4.
\(\frac{1}{x^3(x+a)}\)
Solution:
Let \(\frac{1}{x^3(x+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+a}\) = \(\frac{A \cdot x^2(x+a)+B(x)(x+a)+C(x+a)+D x^3}{x^3(x+a)}\)
∴ 1 = A (x²) (x + a) + Bx (x + a) + C(x + a) + Dx³ ……..(1)
Put x = 0 in (1)
1 = A(0) + B(0) + C(0 + a) + D(0)
⇒ 1 = C(a)
⇒ C = \(\frac{1}{a}\)

Question 5.
\(\frac{x^2+5 x+7}{(x-3)^3}\)
Solution:
Let x – 3 = y ⇒ x = y + 3
\(\frac{x^2+5 x+7}{(x-3)^3}=\frac{(y+3)^2+5(y+3)+7}{y^3}\)

Question 6.
\(\frac{3 x^3-8 x^2+10}{(x-1)^4}\)
Solution:
Put x – 1 = y ⇒ x = y + 1

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(c)

Question 1.
Find an approximate value of the following corrected to 4 decimal places.
(i) \(\sqrt[5]{242}\)
Solution:

(ii) \(\sqrt[7]{127}\)
Solution:

(iii) \(\sqrt[5]{32.16}\)
Solution:

(iv) \(\sqrt{199}\)
Solution:

(v) \(\sqrt[3]{1002}-\sqrt[3]{998}\)
Solution:

(vi) \((1.02)^{3 / 2}-(0.98)^{3 / 2}\)
Solution:

Question 2.
If |x| is so small that x² and higher powers of x may be neglected then find the approximate values of the following.
(i) \(\frac{(4+3 x)^{1 / 2}}{(3-2 x)^2}\)
Solution:

(ii) \(\frac{\left(1-\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{1 / 5}}{(3-x)^3}\)
Solution:

(iii) \(\sqrt{4-x}\left(3-\frac{x}{2}\right)^{-1}\)
Solution:

(iv) \(\frac{\sqrt{4+x}+\sqrt[3]{8+x}}{(1+2 x)+(1-2 x)^{-1 / 3}}\)
Solution:

(v) \(\frac{(8+3 x)^{2 / 3}}{(2+3 x) \sqrt{4-5 x}}\)
Solution:

Question 3.
Suppose s and t are positive and t is very small when compared to s. Then find an approximate value of \(\left(\frac{s}{s+t}\right)^{1 / 3}-\left(\frac{s}{s-t}\right)^{1 / 3}\)
Solution:
Since t is very small when compared with s, \(\frac{t}{s}\) is very very small.

Question 4.
Suppose p, q are positive and p is very small when compared to q. Then find an approximate value of \(\left(\frac{q}{q+p}\right)^{1 / 2}+\left(\frac{q}{q-p}\right)^{1 / 2}\)
Solution:

Question 5.
By neglecting x⁴ and higher powers of x, find an approximate value of \(\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}\)
Solution:

Question 6.
Expand 3√3 in increasing powers of \(\frac{2}{3}\).
Solution:

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(b)

I.

Question 1.
Find the set of values of x for which the binomial expansions of the following are valid.
(i) (2 + 3x)^-2/3
(ii) (5 + x)^3/2
(iii) (7 + 3x)^-5
(iv) \(\left(4-\frac{x}{3}\right)^{-1 / 2}\)
Solution:
(i) (2 + 3x)^-2/3 = \(\left[2\left(1+\frac{3}{2} x\right)\right]^{-2 / 3}\)

Question 2.
Find the
(i) 6th term of \(\left(1+\frac{x}{2}\right)^{-5}\)
Solution:

(ii) 7th term of \(\left(1-\frac{x^2}{3}\right)^{-4}\)
Solution:

(iii) 10th term of (3 – 4x)^-2/3
Solution:

(iv) 5th term of \(\left(7+\frac{8 y}{3}\right)^{7 / 4}\)
Solution:

Question 3.
Write down the first 3 terms in the expansion of
(i) (3 + 5x)^-7/3
Solution:

(ii) (1 + 4x)^-4
Solution:

(iii) (8 – 5x)^2/3
Solution:

(iv) (2 – 7x)^-3/4
Solution:

Question 4.
Find the general term (r + 1)^th term in the expansion of
(i) (4 + 5x)^-3/2
Solution:

(ii) \(\left(1-\frac{5 x}{3}\right)^{-3}\)
Solution:

(iii) \(\left(1+\frac{4 x}{5}\right)^{5 / 2}\)
Solution:

(iv) \(\left(3-\frac{5 x}{4}\right)^{-1 / 2}\)
Solution:

II.

Question 1.
Find the coefficient of x¹⁰ in the expansion of \(\frac{1+2 x}{(1-2 x)^2}\)
Solution:

Question 2.
Find the coefficient of x⁴ in the expansion of (1 – 4x)^-3/5
Solution:

Question 3.
(i) Find the coefficient of x⁵ in \(\frac{(1-3 x)^2}{(3-x)^{3 / 2}}\)
Solution:

(ii) Find the coefficient of x⁸ in \(\frac{(1+x)^2}{\left(1-\frac{2}{3} x\right)^3}\)
Solution:

(iii) Find the coefficient of x⁷ in \(\frac{(2+3 x)^3}{(1-3 x)^4}\)
Solution:

Question 4.
Find the coefficient of x³ in the expansion of \(\frac{\left(1+3 x^2\right)^{3 / 2}}{(3+4 x)^{1 / 3}}\)
Solution:

III.

Question 1.
Find the sum of the infinite series
(i) \(1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\ldots\)
Solution:

(ii) \(1-\frac{4}{5}+\frac{4.7}{5.10}-\frac{4.7 .10}{5.10 .15}+\ldots \ldots\)
Solution:

(iii) \(\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5 .7}{4.8 .12}+\ldots\)
Solution:

(iv) \(\frac{3}{4.8}-\frac{3.5}{4.8 .12}+\frac{3.5 .7}{4.8 .12 .16}-\ldots \ldots\)
Solution:

Question 2.
If t = \(\frac{4}{5}+\frac{4.6}{5.10}+\frac{4.6 .8}{5.10 .15}+\ldots \ldots \ldots \infty\), then prove that 9t = 16.
Solution:

Question 3.
If x = \(\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\frac{1.3 .5 .7}{3.6 .9 .12}+\ldots \ldots\) then prove that 9x² + 24x = 11.
Solution:

⇒ 3x + 4 = 3√3
Squaring on both sides
(3x + 4)² = (3√3)²
⇒ 9x² + 24x + 16 = 27
⇒ 9x² + 24x = 11

Question 4.
If x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots\) then find the value of x² + 4x.
Solution:

Question 5.
Find the sum of the infinite series \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1.3}{1.2} \cdot \frac{1}{10^4}+\frac{1.3 .5}{1.2 .3} \cdot \frac{1}{10^6}+\ldots .\right)\)
Solution:

Question 6.
Show that \(1+\frac{x}{2}+\frac{x(x-1)}{2.4}+\frac{x(x-1)(x-2)}{2.4 .6}+\ldots .\) = \(1+\frac{x}{3}+\frac{x(x+1)}{3.6}+\frac{x(x+1)(x+2)}{3.6 .9}+\ldots\)
Solution:

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Binomial Theorem Solutions Exercise 6(a)

I.

Question 1.
Expand the following using the binomial theorem.
(i) (4x + 5y)⁷
Solution:

(ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
Solution:

(iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:

\(\sum_{r=0}^6(-1)^{r \cdot 6} C_r\left(\frac{2 p}{5}\right)^{6-r}\left(\frac{3 q}{7}\right)^r\)

(iv) (3 + x – x²)⁴
Solution:

Question 2.
Write down and simplify
(i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is

(ii) 7th term in (3x – 4y)¹⁰
Solution:
7th term in (3x – 4y)¹⁰
The general term in (3x – 4y)¹⁰ is

(iii) 10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
10th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
General term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is

(iv) r^th term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 9)
Solution:
r^th term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is

Question 3.
Find the number of terms in the expansion of
(i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

(ii) (3p + 4q)¹⁴
Solution:
Number of terms in (3p + 4q)¹⁴ are 14 + 1 = 15

(iii) (2x + 3y + z)⁷
Solution:
Number of terms in (a + b + c)ⁿ are \(\frac{(n+1)(n+2)}{2}\), where n is a positive integer.
Hence number of terms in (2x + 3y + z)⁷ are = \(\frac{(7+1)(7+2)}{2}=\frac{8 \times 9}{2}\) = 36

Question 4.
Find the number of terms with non-zero coefficients in (4x – 7y)⁴⁹ + (4x + 7y)⁴⁹.
Solution:

∴ The number of terms with non-zero coefficient in (4x – 7y)⁴⁹ + (4x + 7y)⁴⁹ is 25.

Question 5.
Find the sum of the last 20 coefficients in the expansions of (1 + x)³⁹.
Solution:

∴ The sum of the last 20 coefficients in the expansion of (1 + x)³⁹ is 238.

Question 6.
If A and B are coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^2n-1 respectively, then find the value of \(\frac{A}{B}\)
Solution:
Given A and B are the coefficient of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^2n-1 respectively.

II.

Question 1.
Find the coefficient of
(i) x^-6 in \(\left(3 x-\frac{4}{x}\right)^{10}\)
Solution:

(ii) x¹¹ in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Solution:

(iii) x² in \(\left(7 x^3-\frac{2}{x^2}\right)^9\)
Solution:

(iv) x^-7 in \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\)
Solution:

Question 2.
Find the term independent of x in the expansion of
(i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)
Solution:

(ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)
Solution:

(iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)
Solution:
The general term in \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is

For term independent of x,
put 42 – 5r = 0
⇒ r = \(\frac{42}{5}\) which is not an integer.
Hence term independent of x in the given expansion is zero.

(iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)
Solution:

Question 3.
Find the middle term(s) in the expansion of
(i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)
Solution:
The middle term in (x + a)ⁿ when n is even and is \(\frac{T_{n+1}}{2}\), when n is odd, we have two middle terms, i.e., \(\frac{T_{n+1}}{2}\) and \(\frac{T_{n+3}}{2}\)
∵ n = 10 is even,
we have only one middle term (i.e.,) \(\frac{10}{2}\) + 1 = 6th term.

(ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)
Solution:
Here n = 11 is an odd integer,
we have two middle terms, i.e., \(\frac{n+1}{2}\) and \(\frac{n+3}{2}\) terms
= 6th and 7th terms are middle terms.
T₆ in \(\left(4 a+\frac{3}{2} b\right)^{11}\) is \({ }^{11} C_5(4 a)^6\left(\frac{3}{2} b\right)^5\)

(iii) (4x² + 5x³)¹⁷
Solution:
(4x² + 5x³)¹⁷ = [x²(4 + 5x)]¹⁷ = x³⁴(4 + 5x)¹⁷ ……..(1)
Consider (4 + 5x)¹⁷
∵ n = 17 is an odd positive integer, we have two middle terms.
They are \(\left(\frac{17+1}{2}\right)^{\text {th }}\) and \(\left(\frac{17+3}{2}\right)^{\text {th }}\) (i.e.,) 9th and 10th terms are middle terms.

(iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Solution:
Here n = 20 is an even positive integer, we have only one middle term

Question 4.
Find the numerically greatest term(s) in the expansion of
(i) (4 + 3x)¹⁵ when x = \(\frac{7}{2}\)
Solution:

(ii) (3x + 5y)¹² when x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)
Solution:

(iii) (4a – 6b)¹³ when a = 3, b = 5
Solution:

(iv) (3 + 7x)ⁿ when x = \(\frac{4}{5}\), n = 15
Solution:

Question 5.
Prove the following.
(i) 2 . C₀ + 5 . C₁ + 8 . C₂ + ……… + (3n+2) . C_n = (3n + 4) . 2^n-1
Solution:

(ii) C₀ – 4 . C₁ + 7 . C₂ – 10 . C₃ + ……… = 0, if n is an even positive integer.
Solution:

(iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)
Solution:

(iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3\) + ……… + \(\frac{3^n}{n+1} \cdot C_n=\frac{4^{n+1}-1}{3(n+1)}\)
Solution:

(v) C₀ + 2 . C₁ + 4 . C₂ + 8 . C₃ + ….. + 2ⁿ . C_n = 3ⁿ
Solution:

Question 6.
Find the sum of the following.
(i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \frac{{ }^{15} C_3}{{ }^{15} C_2}\) + …….. + \(15 \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)
Solution:

(ii) C₀ . C₃ + C₁ . C₄ + C₂ . C₅ + …….. + C_n-3 . C_n
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁ x + C₂ x² + ……. + C_n . xⁿ ……….(1)
On replacing x by \(\frac{1}{x}\), we get

(iii) 2² . C₀ + 3² . C₁ + 4² . C₂ + ……… + (n + 2)² C_n
Solution:

(iv) 3C₀ + 6C₁ + 12C₂ + ……… + 3 . 2ⁿ . C_n
Solution:

Question 7.
Using the binomial theorem, prove that 50ⁿ – 49n – 1 is divisible by 492 for all positive integers n.
Solution:

= 49² [a positive integer]
Hence 50ⁿ – 49n – 1 is divisible by 49² for all positive integers of n.

Question 8.
Using the binomial theorem, prove that 5⁴ⁿ + 52n – 1 is divisible by 676 for all positive integers n.
Solution:

∴ 5⁴ⁿ + 52n – 1 is divisible by 676, for all positive integers n.

Question 9.
If (1 + x + x²)ⁿ = a₀ + a₁ x + a₂ x² + ……… + a_2n x²ⁿ, then prove that
(i) a₀ + a₁ + a₂ + ……… + a_2n = 3ⁿ
(ii) a₀ + a₂ + a₄ + …… + a_2n = \(\frac{3^n+1}{2}\)
(iii) a₁ + a₃ + a₅ + ……… + a_2n-1 = \(\frac{3^n-1}{2}\)
(iv) a₀ + a₃ + a₆ + a₉ + ……….. = 3^n-1
Solution:

Question 10.
If (1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………. b₂₁ x²¹, then find the value of
(i) b₀ + b₂ + b₄ + …….. + b₂₀
(ii) b₁ + b₃ + b₅ + ………. + b₂₁
Solution:

Question 11.
If the coefficient of x¹¹ and x¹² in the binomial expansion of \(\left(2+\frac{8 x}{3}\right)^n\) are equal, find n.
Solution:
The general term of \(\left(2+\frac{8 x}{3}\right)^n\) is \(T_{r+1}={ }^n C_r(2)^{n-r}\left(\frac{8 x}{3}\right)^r\)

Question 12.
Find the remainder when 22013 is divided by 17.
Solution:
We know 2⁴ = 16
The remainder when 24 is divided by 17 is 1
22013 = (24)⁵⁰³ . 2¹
∴ The remainder when 22013 is divided by 17 is (-1)⁵⁰³ . 2 = (-1) . 2 = -2

Question 13.
If the coefficients of (2r + 4)^th term and (3r + 4)^th term in the expansion of (1 + x)²¹ are equal, find r.
Solution:

III.

Question 1.
If the coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
The coefficients of x⁹, x¹⁰, x¹¹ in (1 + x)ⁿ are

⇒ (n – 9) (n – 21) = 11(n – 19)
⇒ n² – 9n – 21n + 189 = 11n – 209
⇒ n² – 41n + 398 = 0

Question 2.
If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + x)ⁿ, find n.
Solution:
Let ⁿC_r-1, ⁿC_r, ⁿC_r+1 are three successive binomial coefficients in (1 + x)ⁿ.
Then ⁿC_r-1 = 36; ⁿC_r = 84 and ⁿC_r+1 = 126

Question 3.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)ⁿ are respectively 240, 720, 1080, find a, x, n.
Solution:

Question 4.
If the coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)ⁿ are in A.P. then show that n² – (4r + 1)n + 4r² – 2 = 0.
Solution:
Coefficient of Tr = ⁿC_r-1

Question 5.
Find the sum of the coefficients of x³² and x^-18 in the expansion of \(\left(2 x^3-\frac{3}{x^2}\right)^{14}\)
Solution:

Question 6.
If P and Q are the sums of odd terms and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
(i) P² – Q² = (x² – a²)ⁿ
(ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ
Solution:

Question 7.
If the coefficients of 4 consecutive terms in the expansion of (1 + x)ⁿ are a₁, a₂, a₃, a₄ respectively, then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)
Solution:
Given a₁, a₂, a₃, a₄ are the coefficients of 4 consecutive terms in (1 + x)ⁿ respectively.
Let a₁ = ⁿC_r-1, a₂ = ⁿC_r, a₃ = ⁿC_r+1, a₄ = ⁿC_r+2

Question 8.
Prove that (²ⁿC₀)² – (²ⁿC₁)² + (²ⁿC₂)² – (²ⁿC₃)² + ……… + (²ⁿC_2n)² = (-1)^{n 2n}C_n
Solution:

Question 9.
Prove that (C₀ + C₁)(C₁ + C₂)(C₂ + C₃) ………… (C_n-1 + C_n) = \(\frac{(n+1)^n}{n !}\) . C₀ . C₁ . C₂ ……… C_n
Solution:

Question 10.
Find the term independent of x in \((1+3 x)^n\left(1+\frac{1}{3 x}\right)^n\)
Solution:

Question 11.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1.3 .5 \ldots(2 n-1)}{n !}(2 x)^n\)
Solution:
The expansion of (1 + x)²ⁿ contains (2n + 1) terms.
middle term = ²ⁿC_n xⁿ

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + …….. + a₂₀ x²⁰ then prove that
(i) a₀ + a₁ + a₂ + ……… + a₂₀ = 2¹⁰
(ii) a₀ – a₁ + a₂ – a₃ + ……….. + a₂₀ = 4¹⁰
Solution:
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……… + a₂₀ x²⁰

Question 13.
If (3√3 + 5)²ⁿ⁺¹ = x and f = x – [x] where ([x] is the integral part of x), find the value of x.f.
Solution:

Question 14.
If R, n are positive integers, n is odd, 0 < F < 1 and if (5√5 + 11)ⁿ = R + F, then prove that
(i) R is an even integer and
(ii) (R + F) . F = 4ⁿ
Solution:
(i) Since R, n are positive integers, 0 < F < 1 and (5√5 + 11)ⁿ = R + F
Let (5√5 – 11)ⁿ = f
Now, 11 < 5√5 < 12
⇒ 0 < 5√5 – 11 < 1
⇒ 0 < (5√5 – 11)ⁿ < 1
⇒ 0 < f < 1

Question 15.
If I, n are positive integers, 0 < f < 1 and if (7 + 4√3 )ⁿ = I + f, then show that
(i) I is an odd integer and
(ii) (I + f) (1 – f) = 1
Solution:
Given I, n are positive integers and
(7 + 4√3)ⁿ = I + f, 0 < f < 1
Let 7 – 4√3 = F
Now 6 < 4√3 < 7
⇒ -6 > -4√3 > -7
⇒ 1 > 7 – 4√3 > 0
⇒ 0 < (7 – 4√3)ⁿ < 1
∴ 0 < F < 1
I + f + F = (7 + 4√3)ⁿ + (7 – 4√3)ⁿ

= 2k where k is an integer.
∴ I + f + F is an even integer.
⇒ f + F is an integer since I is an integer.
But 0 < f < 1 and 0 < F < 1
⇒ 0 < f + F < 2
∴ f + F = 1 ………..(1)
⇒ I + 1 is an even integer.
∴ I is an odd integer.
(I + f) (I – f) = (I + f) F …..[By (1)]
= (7 + 4√3)ⁿ (7 – 4√3)ⁿ
= [(7 + 4√3) (7 – 4√3)]ⁿ
= (49 – 48)ⁿ
= 1

Question 16.
If n is a positive integer, prove that \(\sum_{r=1}^n r^3\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{(n)(n+1)^2(n+2)}{12}\)
Solution:

Question 17.
Find the number of irrational terms in the expansion of (5^1/6 + 2^1/8)¹⁰⁰.
Solution:
General term
T_r+1 = \({ }^{100} C_r\left(5^{1 / 6}\right)^{100-r}\left(2^{1 / 8}\right)^r\) = \({ }^{100} C_r 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)
\(\frac{100-r}{6}\) is an integer in the span
or 0 ≤ r ≤ 100 if r = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100
\(\frac{r}{8}\) is an integer in the span of 0 ≤ r ≤ 100
if r = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, \(\frac{100-r}{6}\), \(\frac{r}{8}\) both an integers
If r = 16, 40, 64, 88
∴ The number of rational terms in the expansion of (5^1/6 + 2^1/8)^r is 4.
∴ The number of irrational terms in the expansion of (5^1/6 + 2^1/8)^r is 101 – 4 = 97 terms.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(c)

I.

Question 1.
Find the number of ways of arranging 7 persons around a circle.
Solution:
Number of persons, n = 7
∴ The number of ways of arranging 7 persons around a circle = (n – 1)!
= 6!
= 720
Hint: The no. of circular permutations of n dissimilar things taken all at a time is (n – 1)!

Question 2.
Find the number of ways of arranging the chief minister and 10 cabinet ministers at a circular table so that the chief minister always sits in a particular seat.
Solution:
Total number of persons = 11
The chief minister can be occupied a separate seat in one way and the remaining 10 seats can be occupied by the 10 cabinet ministers in (10)! ways.
∴ The number of required arrangements = (10)! × 1
= (10)!
= 36,28,800

Question 3.
Find the number of ways to prepare a chain with 6 different coloured beads.
Solution:
Hint: The number of circular permutations like the garlands of flowers, chains of beads, etc., of n things = \(\frac{1}{2}\)(n – 1)!
The number of ways of preparing a chain with 6 different coloured beads = \(\frac{1}{2}\)(6 – 1)!
= \(\frac{1}{2}\) × 5!
= \(\frac{1}{2}\) × 120
= 60

II.

Question 1.
Find the number of ways of arranging 4 boys and 3 girls around a circle so that all the girls sit together.
Solution:
Treat all the 3 girls as one unit. Then we have 4 boys and 1 unit of girls. They can be arranged around a circle in 4! ways. Now, the 3 girls can be arranged among themselves in 3! ways.
∴ The number of required arrangements = 4! × 3!
= 24 × 6
= 144

Question 2.
Find the number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladies wish to sit together.
Solution:
First, arrange the 7 gents around a circular table in 6! ways.
Then we can find 7 gaps between them. The 4 ladies can be arranged in these 7 gaps in ⁷P₄ ways.

∴ The number of required arrangements = 6! × ⁷P₄
= 720 × 7 × 6 × 5 × 4
= 6,04,800

Question 3.
Find the number of ways of arranging 7 guests and a host around a circle if 2 particular guests wish to sit on either side of the host.
Solution:
Number of guests = 7
Treat the two particular guests along with the host as one unit. Then we have 5 guests and one unit of 2 particular guests along with the host.
They can be arranged around a circle in 5! ways.
The two particular guests can be arranged on either side of the host in 2! ways.
∴ The number of required arrangements = 5! × 2!
= 120 × 2
= 240

Question 4.
Find the number of ways of preparing a garland with 3 yellow, 4 white, and 2 red roses of different sizes such that the two red roses come together.
Solution:
Treat that 2nd rose of different sizes as one unit. Then we have 3 yellow, 4 white, and one unit of red roses.
Then they can be arranged in garland form in \(\frac{1}{2}\) (8 – 1)! = \(\frac{1}{2}\) (7!) ways.
Now 2 red roses in one unit can be arranged among themselves in 2! ways.
∴ The number of ways of preparing a garland = \(\frac{1}{2}\) (7!) × (2!)
= \(\frac{1}{2}\) × 5040 × 2
= 5040

III.

Question 1.
Find the number of ways of arranging 6 boys and 6 girls around a circular table so that
(i) all the girls sit together
(ii) no two girls sit together
(iii) boys and girls sit alternately
Solution:
(i) Treat all 5 girls as one unit. Then we have 6 boys and 1 unit of girls. They can be arranged around a circular table in 6! ways.
Now, the 6 girls can be arranged among themselves in 6! ways.
∴ The number of required arrangements = 6! × 6!
= 720 × 720
= 5,18,400

(ii) First arrange the 6 boys around a circular table in 5! ways. Then we can find 6 gaps between them.
The 6 girls can be arranged in these 6 gaps in 6! ways.

∴ The number of required arrangements = 5! × 6!
= 120 × 720
= 86,400

(iii) Here the number of girls and number of boys are the same.

Hence the arrangements of boys and girls sit alternatively in the same as the arrangements of no two girls sitting together or arrangements of no two boys sitting together.
First, arrange the 6 girls around a circular table in 5! ways. Then we can find 6 gaps between them.
The 6 boys can be arranged in these 6 gaps in 6! ways.
∴ The number of required arrangements = 5! × 6!
= 120 × 720
= 86,400

Question 2.
Find the number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland. In how many of them
(i) all the yellow roses are together
(ii) no two yellow roses are together
Solution:
Hint: The number of circular permutations like the garlands of flowers, chains of beads, etc., of n things = \(\frac{1}{2}\)(n – 1)!
Total number of roses = 6 + 3 = 9
∴ The number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland = \(\frac{1}{2}\)(9 – 1)!
= \(\frac{1}{2}\) × 8!
= \(\frac{1}{2}\) × 40,320
= 20,160
(i) Treat all the 3 yellow roses as one unit. Then we have 6 red roses and one unit of yellow roses. They can be arranged in garland form in (7 – 1)! = 6! ways.
Now, the 3 yellow roses can be arranged among themselves in 3! ways.
But in the case of garlands, clockwise arrangements look alike.
∴ The number of required arrangements = \(\frac{1}{2}\) × 6! × 3!
= \(\frac{1}{2}\) × 720 × 6
= 2160

(ii) First arrange the 6 red roses in garland form in 5! ways. Then we can find 6 gaps between them.
The 3 yellow roses can be arranged in these 6 gaps in ⁶P₃ ways.
But in the case of garlands, clockwise and anti-clockwise arrangements look alike.
∴ The number of required arrangements = \(\frac{1}{2}\) × 5! × ⁶P₃
= \(\frac{1}{2}\) × 120 × 6 × 5 × 4
= 7200

Question 3.
A round table conference is attended by 3 Indians, 3 Chinese, 3 Canadians, and 2 Americans. Find a number of ways of arranging them at the round table so that the delegates belonging to the same country sit together.
Solution:
Since the delegates belonging to the same country sit together, first arrange the 4 countries in a round table in 3! ways.
Now, 3 Indians can be arranged among themselves in 3! ways,
3 Chinese can be arranged among themselves in 3! ways,
3 Canadians can be arranged among themselves in 3! ways,
and 2 Americans can be arranged among themselves in 2! ways.
∴ The number of required arrangements = 3! × 3! × 3! × 3! × 2!
= 6 × 6 × 6 × 6 × 2
= 2592

Question 4.
A chain of beads is to be prepared using 6 different red coloured beads and 3 different blue coloured beads. In how many ways can this be done so that no two blue-coloured beads come together?
Solution:
First, arrange the 6 red-coloured beads in the form of a chain of beads in (6 – 1)! = 5! ways.
Then there are 6 gaps between them. The 3 blue coloured beads can be arranged in these 6 gaps in ⁶P₃ ways.
Then the total number of circular permutations = 5! × ⁶P₃
But in the case of a chain of beads, clockwise and anti-clockwise arrangements look alike.
∴ The number of required arrangements = \(\frac{1}{2}\) × 5! × ⁶P₃
= \(\frac{1}{2}\) × 120 × 6 × 5 × 4
= 7200

Question 5.
A family consists of a father, a mother, 2 daughters, and 2 sons. In how many different ways can they sit at a round table if the 2 daughters wish to sit on either side of the father?
Solution:
Total number of persons in a family = 6
Treat the 2 daughters along with a father as one unit. Then we have a mother, 2 sons, and one unit of daughters along with the father in a family.
They can be seated around a table in (4 – 1)! = 3! ways.
The 2 daughters can be arranged on either side of the father in 2! ways.
∴ The number of required arrangements = 3! × 2!
= 6 × 2
= 12

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(b)

I.

Question 1.
Find the number of 4-digited numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed.
Solution:
The number of 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed = 6⁴ = 1296

Question 2.
Find the number of 5-letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times.
Solution:
The number of 5 letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times = 5⁵ = 3125

Question 3.
Find the number of functions from a set A containing 5 elements into a set B containing 4 elements.
Solution:
Set A contains 5 elements.
Set 8 contains 4 elements.
Hint: The total number of functions from set A containing m elements to set B containing n elements is n^m.
For the image of each of the 5 elements of the set, A has 4 choices.
∴ The number of functions from set A containing 5 elements into a set B containing 4 elements = 4 × 4 × 4 × 4 × 4 (5 times)
= 4⁵
= 1024

II.

Question 1.
Find the number of palindromes with 6 digits that can be formed using the digits
(i) 0, 2, 4, 6, 8
(iii) 1, 3, 5, 7, 9
Solution:
Palindromes mean first digit, sixth digit and second digit, fifth digit and third digit, and the fourth digit are the same numbers.
That is we have filled the first three digits only and then the remaining digits are the same.

(i) Given numbers are 0, 2, 4, 6, 8
For the required 6-digit palindromes first digit can be filled 4 ways except ‘0’.
The second digit can be filled in 5 ways and the third can be filled in 5 ways.
∴ The number of 6-digit palindromes using the digits 0, 2, 4, 6, 8 are 4 × 5² = 100

(ii) Given numbers are 1, 3, 5, 7, 9
For the required 6-digit palindromes first digit can be filled in 5 ways, the second digit can be filled in 5 ways and the third digit can be filled in 5 ways.
∴ The number of 6-digit palindromes using the digits 1, 3, 5, 7, 9 are 5³ = 125

Question 2.
Find the number of 4-digit telephone numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 with atleast one digit repeated.
Solution:
The number of 4 digited numbers formed using the digits 1, 2, 3, 4, 5, 6 when repetition is allowed = 6⁴
The number of 4 digited numbers formed using the digits 1, 2, 3, 4, 5, 6 when repetition is not allowed = ⁶P₄
The number of 4 digited telephone numbers in which atleast one digit is repeated = 6⁴ – ⁶P₄
= 6⁴ – 6 × 5 × 4 × 3
= 1296 – 360
= 936

Question 3.
Find the number of bijections from a set A containing 7 elements onto itself.
Solution:
Hint: The number of bijections from set A with n elements to set B with the same number of elements in A is n!
Let A = [a₁, a₂, a₃, a₄, a₅, a₆, a₇]
For a bijection, the element a₁ has 7 choices
for its image, the element a₂ has 6 choices
for its image, the element a₃ has 5 choices
for its image, the element a₄ has 4 choices
for its image, the element a₅ has 3 choices
for its image, the element a₆ has 2 choices
for its image and the element, a₇ has 1 choice for its image.
∴ The number of bijections from set A with 7 elements onto itself = 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040

Question 4.
Find the number of ways of arranging ‘r’ things in a line using the given ‘n’ different things in which atleast one thing is repeated.
Solution:
The number of ways of arranging, r things in a line using the given n different things
(i) when repetition is allowed is n^r
(ii) when repetition is not allowed is ⁿP_r
∴ The number of ways of arranging ‘r’ things in a line using the ‘n1 different things in which atleast one thing is repeated = n^r – ⁿP_r

Question 5.
Find the number of 5-letter words that can be formed using the letters of the word NATURE that begin with N when repetition is allowed.
Solution:
First, we can fill up the first place with N in one way.

The remaining 4 places can be filled with any one of the 6 letters in 6 × 6 × 6 × 6 = 6⁴ ways.
∴ The number of 5 letter words that can be formed using the letters of the word NATURE that begin with N when repetition is allowed = 1 × 6⁴ = 1296

Question 6.
Find the number of 5-digit numbers divisible by 5 that can be formed using the digits 0, 1, 2, 3, 4, 5 when repetition is allowed.
Solution:
The unit place of 5 digited numbers which can be divisible by 5 using the given digits can be filled by either 0 or 5 in two ways.
The first place can be filled in any one of the given digits except ‘0’ in 5 ways.
The remaining 3 places can be filled by any one of the given digits in 6 × 6 × 6 ways (∵ repetition is allowed)

∴ The number of 5 digited numbers divisible by 5 that can be formed using the given digits when repetition is allowed = 2 × 5 × 6 × 6 × 6 = 2160 ways.

Question 7.
Find the number of numbers less than 2000 that can be formed using the digits, 1, 2, 3, 4 if repetition is allowed.
Solution:
All the single digited numbers, two digited numbers, three digited numbers and the four digited numbers started with 1 are the numbers less than 2000 using the digits 1, 2, 3, 4.
The number of single digited numbers formed using the given digits = 4

The number of two digited numbers formed using the given digits when repetition is allowed = 4 × 4 = 16

The number of three digited numbers formed using the given digits = 4 × 4 × 4 = 64

The number of 4 digited numbers started with 1 formed using the given digits = 4 × 4 × 4 = 64

∴ The total number of numbers less than 2000 that can be formed using the digits 1, 2, 3, 4 if repetition is allowed = 4 + 16 + 64 + 64 = 148

III.

Question 1.
9 different letters of an alphabet are given. Find the number of 4 letter words that can be formed using these 9 letters which have
(i) no letter is repeated
(ii) At atleast one letter is repeated
Solution:
The number of 4 letter words can be formed using the 9 different letters of an alphabet when repetition is allowed = 9⁴
(i) The number of 4 letter words can be formed using the 9 different letters of an alphabet in which no letter is repeated = ⁹P₄
= 9 × 8 × 7 × 6
= 3024
(ii) The number of 4 letter words can be formed using the 9 different letters of an alphabet in which atleast one letter is repeated = 9⁴ – ⁹P₄
= 6561 – 3024
= 3537

Question 2.
Find the number of 4-digit numbers which can be formed using the digits 0, 2, 5, 7, 8 that are divisible by (i) 2 (ii) 4 when repetition is allowed.
Solution:
(i) First place can be filled by either 2 or 5 or 7 or 8 in 4 ways.

Second place can be filled by any one of the given digits in 5 ways.
Third place can be filled by any one of the given digits in 5 ways.
Last place (or units place) can be filled by either 0 or 2 or 8 in 3 ways.
∴ The number of 4 digited divisible by 2 numbers that can be formed using the digits 0, 2, 5, 7, 8 when repetition is allowed = 4 × 5 × 5 × 3 = 300
(ii) Since a number is divisible by 4, the last two places should be filled with one of the 00, 08, 20, 28, 52, 72, 80, 88 in 8 ways.
The first place can be filled 4 ways except 0. Second place can be filled in 5 ways.

Question 3.
Find the number of 4-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 which are divisible by 6 when repetition of the digits is allowed.
Solution:
Since a number is divisible by 4, the last two places should be filled with one of the 00, 04, 12, 20, 24, 32, 40, 44 = 8 ways

The first place can be filled in any one of the given digits except ‘0’ in 4 ways. The remaining 2 places can be filled by anyone the given digits in 5 × 5 ways.
∴ The number of 5 digited numbers that can be formed using the digits 0, 1, 2, 3, 4 that are divisible by 4 when repetition is allowed = 4 × 5² × 8 = 800

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(d)

I.

Question 1.
Find the algebraic equation whose roots are 3 times the roots of x³ + 2x² – 4x + 1 = 0
Solution:
Given equation is f(x) = x³ + 2x² – 4x + 1 = 0
We require an equation whose roots are 3 times the roots of f(x) = 0
i,e., Required equation is f(\(\frac{x}{3}\)) = 0
⇒ \(\left(\frac{x}{3}\right)^{3}+2\left(\frac{x}{3}\right)^{2}-\frac{4 x}{3}+1=0\)
⇒ \(\frac{x^{3}}{27}+\frac{2}{9} x^{2}-\frac{4}{3} x+1=0\)
Multiplying with 27, required equation is x³ + 6x² – 36x + 27 = 0

Question 2.
Find the algebraic equation whose roots are 2 times the roots of x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3 = 0
Solution:
Given equation is f(x) = x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3 = 0
We require an equation whose roots are 2 times the roots of f(x) = 0
Required equation is f(\(\frac{x}{2}\)) = 0
⇒ \(\left(\frac{x}{2}\right)^{5}-2\left(\frac{x}{2}\right)^{4}+3\left(\frac{x}{2}\right)^{3}-2\left(\frac{x}{2}\right)^{2}+4\left(\frac{x}{2}\right)\) + 3 = 0
⇒ \(\frac{x^{5}}{32}-2 \cdot \frac{x^{4}}{16}+3 \cdot \frac{x^{3}}{8}-2 \cdot \frac{x^{2}}{4}+4 \cdot \frac{x}{2}+3=0\)
Multiplying with 32, the required equation is
⇒ x⁵ – 4x⁴ + 12x³ – 16x² + 64x + 96 = 0

Question 3.
Find the transformed equation whose roots are the negative of the roots of x⁴ + 5x³ + 11x + 3 = 0
Solution:
Given f(x) = x⁴ + 5x³ + 11x + 3 = 0
We want an equation whose roots are -α₁, -α₂, -α₃, -α₄
Required equation f(-x) = 0
⇒ (-x)⁴ + 5(-x)³ + 11(-x) + 3 = 0
⇒ x⁴ – 5x³ – 11x + 3 = 0

Question 4.
Find the transformed equation whose roots are the negatives of the roots of x⁷ + 3x⁵ + x³ – x² + 7x + 2 = 0
Solution:
Given f(x) = x⁷ + 3x⁵ + x³ – x² + 7x + 2 = 0
We want an equation whose roots are -α₁, -α₂, ………., -α_n
Required equation is f(-x) = 0
⇒ (-x)⁷ + 3(-x)⁵ + (-x)³ – (-x)² + 7(-x) + 2 = 0
⇒ -x⁷ – 3x⁵ – x³ – x² – 7x + 2 = 0
⇒ x⁷ + 3x⁵ + x³ + x² + 7x – 2 = 0

Question 5.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁴ – 3x³ + 7x² + 5x – 2 = 0
Solution:
Given equation is f(x) = x⁴ – 3x³ + 7x² + 5x – 2 = 0
Required equation is f(\(\frac{1}{x}\)) = 0
i.e., \(\frac{1}{x^{4}}-\frac{3}{x^{3}}+\frac{7}{x^{2}}+\frac{5}{x}-2=0\)
Multiplying with x⁴
⇒ 1 – 3x + 7x² + 5x³ – 2x⁴ = 0
⇒ 2x⁴ – 5x³ – 7x² + 3x – 1 = 0

Question 6.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0
Solution:
Given equation is f(x) = x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0
Required equation is f(\(\frac{1}{x}\)) = 0
\(\frac{1}{x^5}+\frac{11}{x^4}+\frac{1}{x^3}+\frac{4}{x^2}-\frac{13}{x}+6=0\)
Multiplying by x⁵
⇒ 1 + 11x + x² + 4x³ – 13x⁴ + 6x⁵ = 0
⇒ 6x⁵ – 13x⁴ + 4x³ + x² + 11x + 1 = 0

II.

Question 1.
Find the polynomial equation whose roots are the squares of the roots of x⁴ + x³ + 2x² + x + 1 = 0
Solution:
Given equation is f(x) = x⁴ + x³ + 2x² + x + 1 = 0
Required equation f(√x) = 0
⇒ x² + x√x + 2x + √x + 1 = 0
⇒ √x(x + 1) = -(x² + 2x + 1)
Squaring both sides,
⇒ x(x + 1)² = (x² + 2x + 1)²
⇒ x(x² + 2x + 1) = x⁴ + 4x² + 1 + 4x³ + 4x + 2x²
⇒ x³ + 2x² + x = x⁴ + 4x³ + 6x² + 4x + 1
⇒ x⁴ + 3x³ + 4x² + 3x + 1 = 0

Question 2.
Form the polynomial equation whose roots are the squares of the roots of x³ + 3x² – 7x + 6 = 0
Solution:
Given equation is f(x) = x³ + 3x² – 7x + 6 = 0
Required equation is f(√x) = 0
⇒ x√x + 3x – 7√x + 6 = 0
⇒ √x(x – 7) = -(3x + 6)
Squaring on both sides,
⇒ x(x – 7)² = (3x + 6)²
⇒ x(x² – 14x + 49) = 9x² + 36 + 36x
⇒ x³ – 14x² + 49x – 9x² – 36x – 36 = 0
⇒ x³ – 23x² + 13x – 36 = 0

Question 3.
Form the polynomial equation whose roots are cubes of the roots of x³ + 3x² + 2 = 0
Solution:
Given equation is x³ + 3x² + 2 = 0
Put y = x³ so that x = y^1/3
∴ y + 3y^2/3 + 2 = 0
∴ 3y^2/3 = -(y + 2)
Cubing on both sides,
27y² = -(y + 2)³ = -(y³ + 6y² + 12y + 8)
∴ y³ + 6y² + 12y + 8 = 0
⇒ y³ + 33y² + 12y + 8 = 0
Required equation is x³ + 33x² + 12x + 8 = 0

III.

Question 1.
Find the polynomial equation whose roots are the translates of those of the equation x⁴ – 5x³ + 7x² – 17x + 11 = 0 by -2.
Solution:
Given equation is f(x) = x⁴ – 5x³ + 7x² – 17x + 11 = 0
The required equation is f(x + 2) = 0
(x + 2)⁴ – 5(x + 2)³ + 7(x + 2)² – 17(x + 2) + 11 = 0

Required equation is x⁴ + 3x³ + x² – 17x – 19 = 0

Question 2.
Find the polynomial equation whose roots are the translates of those of x⁵ – 4x⁴ + 3x² – 4x + 6 = 0 by -3.
Solution:
Given equation is f(x) = x⁵ – 4x⁴ + 3x² – 4x + 6 = 0
Required equation is f(x + 3) = 0
(x + 3)⁵ – 4(x + 3)⁴ + 3(x + 3)²

Required equation is x⁵ + 11x⁴ + 42x³ + 57x² – 13x – 60 = 0

Question 3.
Find the polynomial equation whose roots are the translates of the roots of the equation x⁴ – x³ – 10x² + 4x + 24 = 0 by 2.
Solution:
Given f(x) = x⁴ – x³ – 10x² + 4x + 24 = 0
Required equation is f(x – 2) = 0
(x – 2)⁴ – (x – 2)³ – 10(x – 2)² + 4(x – 2) + 24 = 0

Required equation is x⁴ – 9x³ + 20x² = 0

Question 4.
Find the polynomial equation whose roots are the translates of the equation 3x⁵ – 5x³ + 7 = 0 by 4.
Solution:
Given f(x) = 3x⁵ – 5x³ + 7 = 0
Required equation is f(x – 4) = 0
3(x – 4)⁵ – 5(x – 4)³ + 7 = 0

Required equation is x⁵ – 60x⁴ + 475x³ – 1860x² + 3600x – 2745 = 0

Question 5.
Transform each of the following equations into ones in which of the coefficients of the second highest power of x is zero and also find their transformed equations.
(i) x³ – 6x² + 10x – 3 = 0
Solution:
Given equation is x³ – 6x² + 10x – 3 = 0
To remove the second term diminish the roots by \(-\frac{a_1}{n a_0}=\frac{6}{3}=2\)

Required equation is x³ – 2x + 1 = 0

(ii) x⁴ + 4x³ + 2x² – 4x – 2 = 0
Solution:
Given equation is x⁴ + 4x³ + 2x² – 4x – 2 = 0
Diminishing the roots by \(-\frac{a_1}{n a_0}=\frac{-4}{4}=-1\)

Required equation is x⁴ – 4x² + 1 = 0

(iii) x³ – 6x² + 4x – 7 = 0
Solution:
Given equation is x³ – 6x² + 4x – 7 = 0
Diminishing the roots by \(-\frac{a_1}{n a_0}=\frac{6}{3}=2\)

Required equation is x³ – 8x – 15 = 0

(iv) x³ + 6x² + 4x + 4 = 0
Solution:
Given equation is x³ + 6x² + 4x + 4 = 0
To remove the second term diminish the roots by \(\frac{-a_1}{n a_0}=-\frac{6}{3}=-2\)

Required equation is x³ – 8x + 12 = 0

Question 6.
Transform each of the following equations into ones in which the coefficients of the third highest power of x are zero.
Hint: To remove the r^th term in an equation f(x) = 0 of degree n diminish the roots by ‘h’ such that \(f^{(n-r+1)}(h)=0\)
(i) x⁴ + 2x³ – 12x² + 2x – 1 = 0
Solution:
Let f(x) = x⁴ + 2x³ – 12x² + 2x – 1
To remove the 3^rd term, diminish the roots by h such that f”(h) = 0
f'(x) = 4x³ + 6x² – 24x + 2
f”(x) = 12x² + 12x – 24
f”(h) = 0
⇒ 12h² + 12h – 24 = 0
⇒ h² + h – 2 = 0
⇒ (h + 2) (h – 1) = 0
⇒ h = -2 or 1
Case (i):

Transformed equation is x⁴ – 6x³ + 42x – 53 = 0
Case (ii):

Transformed equation is x⁴ – 6x³ – 12x – 8 = 0
∴ The required equation is x⁴ – 6x³ + 42x – 53 = 0
or x⁴ + 6x³ – 12x – 8 = 0

(ii) x³ + 2x² + x + 1 = 0
Solution:
Let f(x) = x³ + 2x² + x + 1
To remove the 3^rd term, diminish the roots by h such that f'(h) = 0, f'(x) = 3x² + 4x + 1
f'(h) = 0
⇒ 3h² + 4h + 1 = 0
⇒ (3h + 1) (h + 1)
⇒ h = -1, \(-\frac{1}{3}\)
Case (i):

Transformed equation is x³ – x² + 1 = 0
Case (ii):

Transformed equation is x³ + x² + \(\frac{23}{27}\) = 0
⇒ 27x³ + 27x² + 23 = 0
∴ The required equation is x³ – x² + 1 = 0 or 27x³ + 27x² + 23 = 0

Question 7.
Solve the following equations.
(i) x⁴ – 10x³ + 26x² – 10x + 1 = 0
Solution:
This is a standard reciprocal equation.
Dividing with x²
\(x^2-10 x+26-\frac{10}{x}+\frac{1}{x^2}=0\)
\(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)+26=0\) …….(1)
put a = x + \(\frac{1}{x}\)
\(x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2\) = a² – 2
Substituting in (1)
a² – 2 – 10a + 26 = 0
⇒ a² – 10a + 24 = 0
⇒ (a – 4)(a – 6) = 0
⇒ a = 4 or 6
Case (i): a = 4
x + \(\frac{1}{x}\) = 4
⇒ x² + 1 = 4x
⇒ x² – 4x + 1 = 0
⇒ x = \(\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm 2 \sqrt{3}}{2}\)
⇒ x = 2 ± √3
Case (ii): a = 6
x + \(\frac{1}{x}\) = 6
⇒ x² + 1 = 6x
⇒ x² – 6x + 1 = 0
⇒ x = \(\frac{6 \pm \sqrt{36-4}}{2}\)
⇒ x = \(\frac{2(3 \pm 2 \sqrt{2)}}{2}\)
⇒ x = 3 ± 2√2
∴ The roots are 3 ± 2√2, 2 ± √3

(ii) 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
Solution:
Given f(x) = 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
This is an odd-degree reciprocal equation of the first type.
∴ -1 is a root.
Dividing f(x) with x + 1

Dividing f(x) by (x + 1), we get
2x⁴ – x³ – 11x² – x + 2 = 0
Dividing by x²
\(2 x^2-x-11-\frac{1}{x}+\frac{2}{x^2}=0\)
\(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)-11=0\) ……..(1)
Put a = x + \(\frac{1}{x}\) so that
\(x^2+\frac{1}{x^2}=a^2-2\)
Substituting in (1), the required equation is
⇒ 2(a² – 2) – a – 11 = 0
⇒ 2a² – 4 – a – 11 = 0
⇒ 2a² – a – 15 = 0
⇒ (a – 3) (2a + 5) = 0
⇒ a = 3 or \(-\frac{5}{2}\)
Case (i): a = 3
x + \(\frac{1}{x}\) = 3
⇒ x² + 1 = 3x
⇒ x² – 3x + 1 = 0
⇒ x = \(\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\)
Case (ii): a = \(-\frac{5}{2}\)
⇒ \(x+\frac{1}{x}=-\frac{5}{2}\)
⇒ \(\frac{x^2+1}{x}=-\frac{5}{2}\)
⇒ 2x² + 2 = -5x
⇒ 2x² + 5x + 2 = 0
⇒ (2x + 1) (x + 2) = 0
⇒ x = \(-\frac{1}{2}\), -2
∴ The roots are -1, \(-\frac{1}{2}\), -2, \(\frac{3 \pm \sqrt{5}}{2}\)