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Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 Random Variables and Probability Distributions to solve questions creatively.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Formulas

Random variable:
→ Suppose S is the sample space of a random experiment then any function X : S → R is called a random variable.

→ Let S be a sample space and X : S → R.be a random variable. The function F : R → R defined by F(x) = P(X ≤ x), is called probability distribution function of the random variable X.

→ A set ‘A’ is said to be countable if there exists a bijection from A onto a subset of N.

→ Let S be a sample space. A random variable X : S → R is said to be discrete or discontinuous if the range of X is countable.

→ If X : S → R is a discrete random variable with range {x₁, x₂, x₃, ……………. } then
\(\sum_{r=1}^{\infty}\) P(x = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁, x₂, x₃, …………….} If Σx_r P(X = x_r) exists, then Σx_r. P(X = x) is called the mean of the random variable X. It is denoted by µ or x . If Σ (x_r – µ)² P(X = X_r) exists, then Σ(x_r – µ)² P(X = X_r) is called variance of the random variable X. It is denoted by σ².

→ The positive square root of the variance is called the standard deviation of the Fandom variable x. It is denoted by σ.

Binomial distribution:
→ Let n be a positive integer and p be a real number-such that 0 ≤ p ≤ 1. A random variable x with range {0, 1, 2, 3, ……….. n} is said to follows (or have) binomial distribution or Bernoulli distribution with parameters n and p if P (X = r) = ⁿC_r p^r q^{n – r} for r = 0, 1, 2, ………. n where q = 1 – p. Its Mean µ = np and variance σ² = npq. n and p are called. parameters of the Binomial distribution.

Poisson distribution:
→ Let λ > 0 be a real, number. A, random variable x with range {0, 1, 2, ……… n} is said to follows (have) poisson distribution with parameter λ if P(X = r) = \(\frac{e^{-\lambda} \lambda^{r}}{r !}\) for r = 0, 1, 2, ……………. . Its Mean = λ and variance = λ. Its parameter is λ.

→ If X : S → R is a discrete random variable with range {x₁ x₂, x₃, …. } then \(\sum_{r=1}^{\infty}\) P (X = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁ x₂, x₃, …..} .If Σ x_r P(X = x_r) exists, then Σ x_r P(X = x_r) is called the mean of the random variable X. It is denoted by or x.

→ If Σ(x_r – μ)² P(X = x_r) exists, then Σ (x_r – μ)² P(X = x_r) is called variance of the random variable X. It is denoted by σ². The positive square root of the variance is called the standard deviation of the random variable X. It is denoted by σ

→ If the range of discrete random variable X is {x₁ x₂, x₃, …. x_n, ..} and P(X = x_n) = P_n for every Integer n is given then σ² + μ² = Σx_n²P_n

Binomial Distribution:
A random variable X which takes values 0, 1, 2, ., n is said to follow binomial distribution if its probability distribution function is given by
P(X = r) = ⁿc_rp^rq^n-r, r = 0,1,2, ……………. , n where p, q > 0 such that p + q = 1.

→ If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is p^r.

→ Mean and variance of the binomial distribution

• The mean of this distribution is \(\sum_{i=1}^{n}\) X_ip_i = latex]\sum_{X=1}^{n}[/latex] X. ⁿC_xq^n-xp^X = np,
• The variance of the Binomial distribution is σ² = npq and the standard deviation is σ = \(\sqrt{(n p q)}\).

→ The Poisson Distribution : Let X be a discrete random variable which can take on the values 0, 1, 2,… such that the probability function of X is given by
f(x) = P(X = x) = \(\frac{\lambda^{x} e^{-\lambda}}{x !}\), x = 0, 1, 2, ………….
where λ is a given positive constant. This distribution is called the Poisson distribution and a random variable having this distribution is said to be Poisson distributed.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 8 Measures of Dispersion to solve questions creatively.

Intermediate 2nd Year Maths 2A Measures of Dispersion Formulas

→ The arithmetic mean of ungrouped data = \(\frac{\Sigma x_{i}}{n}\) where n is the number of observations.

→ The median of ungrouped data
First expressing the n data points in the ascending order of magnitude.

→ If n is odd then \(\frac{n+1}{2}\)the data points in the median of the given ungrouped data.

→ If n is even then the average of \(\frac{n}{2}, \frac{n+2}{2}\) the data points in the median of the given ungrouped data.

• Range,
• Mean deviation,
• Variance and,
• Standard deviation are some measures of dispusion for ungrouped and grouped data.

→ Range is defined as the difference between the maximum value and the minimum value of the series of observations.

Mean Deviation for ungrouped distribution:
→ Mean deviation about the mean = \(\frac{\text { Sum of the absolutevalues of deviationsfrom } \bar{x}}{\text { Number of Observations }} .\)
= \(\frac{\sum\left|x_{i}-\bar{x}\right|}{n}\) where x̅ is the mean

→ Mean Deviation about the median = \(\frac{\sum\left|x_{i}-{median}\right|}{n}\)

Mean Deviation for grouped data:
→ Mean Deviation about mean = \(\frac{1}{N}\) Σf_i|x – x̅_i|
Where N = Σf_i and x̅ is the mean

→ Mean Deviation about median = \(\frac{1}{N}\) Σf_i|x_i – median|
When N = Σf_i

→ Ungrouped data variance σ² = \(\frac{1}{n}\) = Σ(x_i – x̅)²
Standard deviation σ = \(\sqrt{\frac{1}{n} \sum\left(x_{i}-\bar{x}\right)^{2}}\)

→ Discrete frequency distribution variance σ² = \(\frac{1}{n}\) = Σf_i(x_i – x̅)² where x̅ is the mean]
Standard deviation σ = \(\frac{1}{N}\) \(\sqrt{\sum f_{i}\left(x_{i}-\bar{x}\right)^{2}}\)

→ Continuous frequency distribution standard deviation σ = \(\frac{1}{N} \sqrt{N \sum f_{i} x_{i}{ }^{2}-\left(\sum f_{i} x_{i}\right)^{2}}\)
(or) σ = \(\frac{h}{N}\) \(\sqrt{N \Sigma f_{i} y_{i}^{2}-\left(\sum f_{i} y_{i}\right)^{2}}\)

→ Co-efficient of variation = \(\frac{\sigma}{x}\) × 100 (x̅ ≠ 0)

→ If each observation in a data is. multiplied by a constant K, then the variance of the resulting observations is K² times that or the variance of original observations.

→ If each of the observations x₁, x₂, ………, x_n is increased by K, where K is a positive or negative number then the variance remains unchanged.

Measure of Central Tendency:
1. Mathematical Average:
a) Arithmetic mean (A.M.)
b) Geometric mean (G.M.)
c) Harmonic mean (H.M.)

2. Averages of Position:
a) Median
b) Mode

Arithmetic Mean:
(1) Simple arithmetic mean in individual series
(i) Direct method: If the series in this case be x₁, x₂, …………. x_n then the arithmetic mean x̄ is given by
x̄ = \(\frac{\text { Sum of the series }}{\text { Number of terms }}\)
i.e x̄ = \(\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\frac{1}{n} \sum_{i=1}^{n} x_{i}\)

(2) Simple arithmetic mean in continuous series If the terms of the given series be x₁, x₂, …………. x_n and the corresponding frequencies be f₁, f₂, …………. f_n then the arithmetic mean x̄ is given by,
x̄ = \(\frac{f_{1} x_{1}+f_{2} x_{2}+\ldots+f_{n} x_{n}}{f_{1}+f_{2}+\ldots+f_{n}}=\frac{\sum_{i=1}^{n} f_{i} x_{i}}{\sum_{i=1}^{n} f_{i}}\)

Continuous Series:
If the series is continuous then x_ii’s are to be replaced by m_i’s where m_i’s are the mid values of the class intervals.

Mean of the Composite Series:
If x̄_i,(i = 1, 2, …………… k) are the means of k-component series of sizes n_i(i = 1,2,….,k) respectively, then the mean x̄ of the composite series obtained on combining the component series is given by the formula x̄ = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}+\ldots+n_{k} \bar{x}_{k}}{n_{1}+n_{2}+\ldots+n_{k}}=\frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum_{i=1}^{n} n_{i}}\)

Geometric Mean:
If x₁, x₂, …………. x_n are n values of a variate x, none of them being zero, thengeometric mean (G.M.) is given by G.M. (x₁, x₂, …………. x_n)^1/n

In case of frequency distribution, G.M. of n values x₁, x₂, …………. x_n of a variate x occurring with frequency f₁, f₂, …………. f_n is given by G.M = (x₁^f₁, x₂^f₂, …………. x_n^f_n)^1/N, where N = f₁ + f₂ + ……….. + f_n

Continuous Series:
If the series is continuous then xjj’s are to be replaced by m_ii’s where m_ii’s are the mid values of the class intervals.

Harmonic Mean:
The harmonic mean of n items x₁, x₂, …………. x_n is defined as H.M. = \(\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots .+\frac{1}{x_{n}}}\)
If the frequency distribution is f₁, f₂, …………. f_n respectively, then H.M = \(\frac{f_{1}+f_{2}+f_{3}+\ldots . .+f_{n}}{\left(\frac{f_{1}}{x_{1}}+\frac{f_{2}}{x_{2}}+\ldots .+\frac{f_{n}}{x_{n}}\right)}\)

Median:
The median is the central value of the set of observations provided all the observations are arranged in the ascending or descending orders. It is generally used, when effect of extreme items is to be kept out.

(1) Calculation of median
(i) Individual series: If the data is raw, arrange in ascending or descending order. Let n be the number of observations.
If n is odd, Median = value of \(\left(\frac{n+1}{2}\right)^{t h}\) item.
If n is even, Median = \(\frac{1}{2}\)[value of \(\left(\frac{n}{2}\right)^{\text {th }}\) item + value of \(\left(\frac{n}{2}+1\right)^{\text {th }}\) item]

(ii) Discrete series: In this case, we first find the cumulative frequencies of the variables arranged in ascending or descending order and the median is given by
Median = \(\left(\frac{n+1}{2}\right)^{t h}\) observations, where n is the cumulative frequency.

(iii) For grouped or continuous distributions: In this case, following formula can be used.
(a) For series in ascending order, Median = l + \(\frac{\left(\frac{N}{2}-C\right)}{f}\) × i
Where l = Lower limit of the median class
f = Frequency of the median class
N = The sum of all frequencies
I = The width of the median class
C = The cumulative frequency of the class preceding to median class.

(b) For series in descending order
Median = u – \(\left(\frac{\frac{N}{2}-C}{f}\right)\) × i, where u = upper limit of the median class, N = \(\sum_{i=1}^{n}\)f_i
As median divides a distribution into two equal parts, similarly the quartiles, quintiles, deciles and percentiles divide the distribution respectively into 4, 5, 10 and 100 equal parts. The th quartile is given by Q = l + \(\left(\frac{j \frac{N}{4}-C}{f}\right)\)i:j= 1,2,3. Q₁ is the lower quartile, Q₂ is the median and Q₃ is called the upper quartile.

(2) Lower qualities

• Discrete series: Q₁ = size of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) item
• Continuous series : Q₁ = l + \(\frac{\left(\frac{N}{4}-C\right)}{f}\) × i

(3) Upper qualities

• Discrete series : Q₃ = size of \(\left[\frac{3(n+1)}{4}\right]^{\text {th }}\) item
• Continuous series : Q₃ = l + \(\frac{\left(\frac{3N}{4}-C\right)}{f}\) × i

Mode:
The mode or model value of a distribution is that value of the variable for which the frequency is maximum. For continuous series, mode is calculated as,
Mode = l₁ + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × i
Where, l₁ = The lower limit of the model class
f₁ = The frequency of the model class
f₀ = The frequency of the class preceding the model class
f₂ = The frequency of the class succeeding the model class
i = The size of the model class.

Empirical relation :
Mean – Mode = 3(Mean – Median) ⇒ Mode = 3 Median – 2 Mean.

Measure of dispersion:
The degree to which numerical data tend to spread about an average value is called the dispersion of the data. The four measure of dispersion are

1. Range
2. Mean deviation
3. Standard deviation
4. Square deviation

(1) Range:
It is the difference between the values of extreme items in a series. Range = X_max – X_min

• The coefficient of range (scatter) = \(\frac{x_{\max }-x_{\min }}{x_{\max }+x_{\min }}\)
  Range is not the measure of central tendency. Range is widely used in statistical series relating to quality control in production.
• Range is commonly used measures of dispersion in case of changes in interest rates, exchange rate, share prices and like statistical information. It helps us to determine changes in the qualities of the goods produced in factories.

Quartile deviation or semi inter-quartile range:
It is one-half of the difference between the third quartile and first quartile i. e., Q.D. = \(\frac{Q_{3}-Q_{1}}{2}\) and coefficient of quartile deviation = \(\frac{Q_{3}-Q_{1}}{Q_{3}+Q_{1}}\), where Q₃ is the third or upper quartile and Q₁ is the lowest or first quartile.

(2) Mean Deviation:
The arithmetic average of the deviations (all taking positive) from the mean, median or mode is known as mean deviation.
Mean deviation is used for calculating dispersion of the series relating to economic and social inequalities. Dispersion in the distribution of income and wealth is measured in term of mean deviation.

• Mean deviation from ungrouped data (or individual series) Mean deviation = \(\frac{\sum|x-M|}{n}\) where |x – M| means the modulus of the deviation of the variate from the mean (mean, median or mode) and n is the number of terms.
• Mean deviation from continuous series: Here first of all we find the mean from which deviation is to be taken. Then we find the deviation dM =| x -M| of each variate from the mean M so obtained.
  • Next we multiply these deviations by the corresponding frequency and find the product f.dM and then the sum ΣfdM of these products.
  • Lastly we use the formula, mean deviation = \(\frac{\sum f|x-M|}{n}=\frac{\sum f d M}{n}\), where n = Σf.

(3) Standard Deviation:
Standard deviation (or S.D.) is the square root of the arithmetic mean of the square of deviations of various values from their arithmetic mean and is generally denoted by aread as sigma. It is used in statistical analysis.
(i) Coefficient of standard deviation: To compare the dispersion of two frequency distributions the relative measure of standard deviation is computed which is known as coefficient of standard deviation and is given by
Coefficient of S.D. = \(\frac{\sigma}{\bar{x}}\), where x̄ is the A.M.

(ii) Standard deviation from individual series σ = \(\sqrt{\frac{\sum(x-\bar{x})^{2}}{N}}\)
where, x̄ = The arithmetic mean of series
N = The total frequency.

(iii) Standard deviation from continuous series σ = \(\sqrt{\frac{\sum f_{i}\left(x_{i}-\bar{x}\right)^{2}}{N}}\)
where, x̄ = Arithmetic mean of series
x_i = Mid value of the class
f_i = Frequency of the corresponding
N = Σf = The total frequency

Short cut Method:
(i) σ = \(\sqrt{\frac{\sum f d^{2}}{N}-\left(\frac{\sum f d}{N}\right)^{2}}\)
(ii) σ = \(\sqrt{\frac{\sum d^{2}}{N}-\left(\frac{\sum d}{N}\right)^{2}}\)
where, d = x – A = Deviation from the assumed mean A
f = Frequency of the item
N = Σf = Sum of frequencies

(4) Square Deviation:
(i) Root mean square deviation
S = \(\sqrt{\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-A\right)^{2}}\)
where A is any arbitrary number and S is called mean square deviation.

(ii) Relation between S.D. and root mean square deviation : If σ be the standard deviation and S be the root mean square deviation.
Then, S² = σ² + d².
Obviously, S² will be least when d = 0 i.e., x̄ = A
Hence, mean square deviation and consequently root mean square deviation is least, if the deviations are taken from the mean.

Variance:
The square of standard deviation is called the variance. Coefficient of standard deviation and variance : The coefficient of standard deviation is the ratio of the S.D. to A.M. i.e., \(\frac{\sigma}{x}\).
Coefficient of variance = coefficient of S.D.× 100 = \(\frac{\sigma}{x}\) × 100 .

Variance of the combined series :
If n₁,n₂ are the sizes, x̄₁, x̄₂ the means and σ₁, σ₂ the standard deviation of two series, then σ² = \(\frac{1}{n_{1}+n}\)[n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]
where d₁ = x̄₁ – x̄, d₂ = x̄₂ – x̄ and x̄ = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 7 Partial Fractions to solve questions creatively.

Intermediate 2nd Year Maths 2A Partial Fractions Formulas

→ The quotient of two polynomials f(x) and Φ(x) where Φ(x) ≠ 0, is called a rational fraction.

→ If the degree of f(x) < the degree of Φ(x) in a rational fraction \(\frac{f(x)}{\phi(x)}\), then the rational fraction is called a proper fraction.

→ If the degree of f(x) ≥ the degree of Φ(x) in a rational fraction \(\frac{f(x)}{\phi(x)}\), then the rational fraction is called a proper fraction.

→ Let \(\frac{f(x)}{\phi(x)}\) be a proper fraction

→ When Φ(x) contains non-repeated linear factors only corresponding to every non-repeated linear factor (ax +b) of Φ(x) there exists a partial fraction of the form \(\frac{A}{a x+b}\) where A is a real number.

→ When Φ(x) contains repeated and non-repeated linear factors only corresponding to every repeated linear factor (ax + b)^p of Φ(x) there exist fractions of the form.
\(\frac{A_{1}}{a x+b}+\frac{A_{2}}{(a x+b)^{2}}+\ldots+\frac{A_{p}}{(a x+b)^{p}}\) Where A₁, A₂, A₃ ………… A_p are real numbers.

→ When Φ(x) contains non-repeated irreducible factors only. Corresponding to every non-repeated irreducible quadratic factor ax² + bx + c of Φ(x) of exists a partial fraction of the form \(\frac{A x+B}{a x^{2}+b x+C}\) where A and B are real numbers.

→ When Φ(x) contains repeated and non-repeated irreducible factors only. Corresponding to every repeated quadratic factor (ax² + bx + c0^p of Φ(x) there exists the partial fractions of the form
\(\frac{A_{1} x+B_{1}}{a x^{2}+b x+c}\) + \(\frac{A_{2} x+B_{2}}{\left(a x^{2}+b x+c\right)^{2}}\) + ……… + \(\frac{A_{p} x+\dot{B}_{p}}{\left(a x^{2}+b x+c\right)^{p}}\) where A₁, A₂, …….. A^p and B₁, B₂, ………. B_p are real numbers.

→ Let \(\frac{f(x)}{\phi(x)}\) be a improper fraction, then
\(\frac{f(x)}{\phi(x)}\) = Q(x) + \(\frac{R(x)}{\phi(x)}\) where Q(x) is quotient and R(x) is the remainder, and Degree R(x) < that of Φ(x).

→ Remainder obtained when f(x) is divided by x – a is f(a).
If degree of divisor is ‘n’, then the degree of remainder is (n – 1)
f(x), g(x) are two polynomials. If g(x) ≠ 0, then ∃ two polynomials q(x) , r(x) such that
\(\frac{f(x)}{g(x)}\) = q(x) + \(\frac{r(x)}{g(x)}\) if the degree of f(x) is > that of g(x)

II. Method of resolving proper fraction \(\frac{f(x)}{g(x)}\) into partial fractions.
Type 1 : When the denominator g(x) contains non-repeated factors i.e
g(x) = (x – a)(x – b)(x – c)
\(\frac{f(x)}{(x-a)(x-b)(x-c)}=\frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}}\)

Type 2 : When the denominator g(x) contains repeated and non repeated linear factors
i.e g(x) = (x – a)²(x – b)
\(\frac{f(x)}{(x-a)^{2}(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a)^{2}}+\frac{C}{(x-b)}\)

Type 3 : When the denominator g(x) contains non repeated irreducible quadratic factors
i.e g(x) = (ax² + bx + c)(x – d)

Type 4 : When the denominator g(x) contains repeated irreducible quadratic factors
i.e g(x) = (ax² + bx + c)²(x – d)
\(\frac{f(\mathrm{x})}{\left(\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}\right)^{2}(\mathrm{x}-\mathrm{d})}=\frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)^{2}}+\frac{\mathrm{E}}{\mathrm{x}-\mathrm{d}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 Permutations and Combinations to solve questions creatively.

Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas

→ Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.

→ If n is a non-negative integer then

• 0 ! = 1
• n ! = n (n – 1) ! if n > 0

→ The number of permutations of n dissimilar things taken ‘r at a time is denoted by ⁿP_r and ⁿP_r = \(\frac{n !}{(n-r) !}\) for 0 ≤ r ≤ n.

→ If n, r positive integers and r ≤ n, then

• ⁿP_r = n. ^{n – 1}P_{r – 1} if r ≥ 1
• ⁿP_r = n.(n – 1) ^{n – 2}P_{r – 2} if r ≥ 2
• ⁿP_r = ^{n – 1}P_r + r. ^{(n – 1)}P_{(r – 1)}

→ The number of injections that can be defined from set A into set B is ^n(B)p_n(A) n(A) ≤ n(B)

→ The number of bijections that can be defined from set A into set B having same number of elements with A is n(A)!

→ The number of permutations of1n’ dissimilar things taken ‘r’ at a time.

• Containing a particular thing is (r) ^{n – 1}P_{r – 1}
• Not containing a particular thing is ^{n – 1}P_r
• Containing a particular thing in a particular place is ^{n – 1}P_{r – i}.

→ The number of functions that can be defined from set A into set B in [n(B)]^n(A)

→ The sum of the r – digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is ^{(n – 1)}P_{(r – 1)} × sum of all n digits × 111 …… 1 (r times)

→ In the above, if ‘0’ is one among the given n digits, then the sum is ^{(n – 1)}P_{(r – 1)} × sum of the digits × 111 … 1 (r times) ^{(n – 2)}P_{(r – 2)} × sum of the digits × 111 … 1 [(r – 1) times]

→ The number of permutations of n dissimilar things taken r at a time when repetitions are allowed any number of times is n^r.

→ The number of circular permutations of n dissimilar things is (n – 1) !

→ In case of hanging type circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is \(\frac{1}{2}\) [(n – 1) !]

→ If in the given n things, p alike things are of one kind, q alike things are of the second kind, r alike things are of the third kind and the rest are dissimilar, then the number of permutations (of these n things) is \(\frac{n !}{p ! q ! r !}\)

→ The number of combinations of n things taken r at a time is denoted by ⁿC_r and ⁿC_r = \(\frac{n !}{(n-r) ! r !}\) for 0 ≤ r ≤ n and \(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\) and ⁿP_r = r! ⁿC_r

→ If n, r are integers and 0 ≤ r ≤ n then ⁿc_r = ⁿC_{r – 1}

→ Let n, r, s are integers and 0 ≤ r ≤ n, 0 ≤ s ≤ n. If ⁿC_r = ⁿC_s then r = s or r + s = n.

→ The number of ways of dividing ‘m + n’ things (m ≠ n) into two groups containing m, n things is ^{(m + n)}C_m = ^{(m + n)}C_m = \(\frac{(m+n) !}{m ! n !}\)

→ The number of ways of dividing (m + n + p) things (m, n, p are distinct) into 3 groups of m, n, p things is \(\frac{(m+n+p) !}{m ! n ! p !}\).

→ The number of ways of dividing mn things into m equal groups is \(\frac{(m n) !}{(n !)^{m} m !}\)

→ The number of ways if distributing mn things equally to m persons is \(\frac{(m n) !}{(n !)^{m}}\)

→ If p alike things are of one kind, q alike things are of the second kind and r alike things are of the third kind, then the number of ways of selecting one or more things out of them is (p + 1) (q + 1) (r +1) – 1.

→ If m is a positive integer and m = p₁^α₁, p₁^α₂ …… p_k^α_k where p₁, p₂ …… p_k are distinct primes and α₁, α₂, …. α_k are non-negative integers, then the number of divisors of m is (α₁ + 1) (α₂ + 1) ……… (α_k + 1). [This includes 1 and m].

→ The total number of combinations of n different things taken any number of times is 2ⁿ.

→ The total number of combinations of n different things taken one or more at a timers 2ⁿ – 1.

→ The number of diagonals in a regular polygon of n sides is \(\frac{n(n-3)}{2}\)

→ Permutations are arrangements of things taken some or all at a time.

→ In a permutation, order of the things is taken into consideration.

→ ⁿp_r represents the number of permutations (without repetitions) of n dissimilar things taken r at a time.

→ Fundamental Principle: If an event is done in ‘m’ ways and another event is done in ‘n’ ways, then the two events can be together done in mn ways provided the events are independent.

→ ⁿp_r = \(\frac{n !}{(n-r) !}\) = n(n – 1)(n – 2) …………. (n – r + 1)

→ ⁿp₁ = n, ⁿp₂ = n(n-1), ⁿp₃ = n(n-1)(n-2).

→ ⁿp_r = n!

→ \(\frac{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}-1}}\)ⁿp_n = n!

→ ⁿ⁺¹p_r = r. ⁿp_r-1 + ⁿp_r

→ ⁿp_r = r. ^n-1p_r-1 + ⁿp_r

→ ⁿp_r = r.^n-1p_r-1 + ^n-1p_r

→ The number of permutations of n things taken r at a time containing a particular thing is r × ^n-1p_r-1

→ The number of permutations of n things taken r at a time not containing a particular thing is ^n-1p_r

• The number of permutations of n things taken r at a time allowing repetitions is n^r.
• The number of permutations of n things taken not more than r at a time allowing repetitions is \(\frac{n\left(n^{\mathrm{r}}-1\right)}{(n-1)}\)

→ The number of permutations of n things of which p things are of one kind and q things are of another kind etc., is \(\frac{n !}{p ! q ! \ldots \ldots \cdots}\)

→ The sum of all possible numbers formed out of all the ‘n’ digits without zero is (n-1)! (sum of all the digits) (111 ……….. n times).

→ The sum of all possible numbers formed out of all the ‘n’ digits which includes zero is [(n -1)! (sum of all the digits) (111 …………… n times)] – [( n – 2)! (sum of all the digits)(111 ………….. (n -1) times]

→ The sum of all possible numbers formed by taking r digits from the given n digits which do not include zero is ^n-1p_r-1 (sum of all the digits)(111 …………… r times).

→ The sum of all possible numbers formed by taking r digits from the given n digits which include zero is ^n-1p_r-1(sum of all the digits)(111 ……………….. r times) – ^n-2p_r-2(sum of all the digits)(111 ……………….. (r-1) times).

• The number of permutations of n things when arranged round a circle is (n -1)!
• In case of necklace or garland number of circular permutations is \(\frac{(n-1) !}{2}\)

→ Number of permutations of n things taken r at a time in which there is at least one repetition is n^r – ⁿp_r.

→ The number of circular permutations of ‘n’ different things taken ‘r’ at a time is \(\frac{{ }^{n} \mathrm{P}_{\mathrm{r}}}{\mathrm{r}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 4 Theory of Equations to solve questions creatively.

Intermediate 2nd Year Maths 2A Theory of Equations Formulas

→ If n is a non-negative integer and a₀, a₁, a₂, ……….. a_n are real or complex numbers and a₀ ≠ 0, then the expression f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n is called a polynomial in x of degree n.

→ f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n = 0 is called a polynomial equation in x of degree n (a₀ ≠ 0). Every non-constant polynomial equation has atleast one root.

→ If f(α) = 0 then α is called a root of the equation f(x) = 0.

→ If f(α) = 0 then (x – α) is a factor of f(x).

Relation between roots and coefficients of an equation:
→ If α β γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 then sum of the roots s₁ = α + β + γ = – p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = p₂.
Product of all the roots, s₃ = αβγ = – p₃.

→ If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then sum of the roots s₁ = α + β + γ + δ = – p₁.
Sum of the products of roots taken two at a time
s₂ = αβ + αγ + αδ + βγ + βδ + γδ = p₂.
Sum of the products of roots taken three at a time .
s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
Product of the roots, s₄ = αβγδ = p₄.

→ For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d.
• In G.P., then they are taken as \(\frac{a}{r}\), a, ar.
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a^{\prime}}, \frac{1}{a+d}\).

→ For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a – d, a + d, a + 3d.
• In C.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\).

→ In an equation with real coefficients, imaginary roots occur in conjugate pairs.

→ In an equation with rational coefficients, irrational roots occur in pairs of conjugate surds.

→ The equation whose roots are those of the equation f(x) = 0 with contrary signs is f(- x) = 0.

→ The equation whose roots are multiplied by kfa 0) of those of the proposed equation f(x) = 0 is f \(\left(\frac{x}{k}\right)\) = 0.

→ The equation whose roots are reciprocals of the roots of f(x) = 0 is f \(\left(\frac{1}{x}\right)\) = 0.

→ The equation whose roots are exceed by h than those of f(x) = 0 is f(x – h) = 0.

→ The equation whose roots are diminished by h than those of f(x) 0 is f(x + h) = 0.

→ The equation whose roots are the square of the roots of f(x) = 0 is obtained by eliminating square root from f(√x) = 0.

→ If f(x) = p₀xⁿ + p₁x^{n – 1} + p₂x^{n – 2} + ……. + p_n = 0 then to eliminate the second term,
f(x) = 0 can be transformed to f(x + h) = 0 where h = \(\frac{-p_{1}}{n \cdot p_{0}}\).

→ If an equation is unaltered by changing x into \(\frac{1}{x}\) then it is a reciprocal equation.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of first class p_i = p_{n – i} for all i.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of second class if p_i = – p_{n – i} for all i.

→ For an odd degree reciprocal equation of class one, – 1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.

→ For an even degree reciprocal equation of class two, 1 and – 1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate r^th term, f(x) = 0 can be transformed to f(x + h) = 0 where h is a constant such that f^{(n – r + 1)}(h) = 0 i.e.,(n – r + 1)^th derivative of f(h) is zero.

→ Every n^th degree equation has exactly n roots real or imaginary.

→ Relation between, roots and coefficients of an equation.

(i) If α, β, γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 the sum of the roots s₁ = α + β + γ = -p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = -p₂
Product of all the roots, s₃ = αβγ= – p₃.

(ii) If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then

• Sum of the roots s₁ = a+P+y+S = -p₁.
  s₂ = αβ + αγ + αδ + βα + βδ + γδ = p₂.
• Sum of the products of roots taken three at a time
  s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
• Product of the roots, s₄ = αβγδ = p₄

→ For the equation xⁿ + p₁x^n-1 + p₂x^n-2 + ……… + p_n = 0

• Σ α₂ = p₁² – 2p₂
• Σ α₃ = -p₁³ + 3p₁p₂ – 3p₃
• Σ α₄ =p₁⁴ – 4p₁²p² + 2p₂² + 4p₁p₃ – 4p₄
• Σ α₂β = 3p₃ – p₁p₂
• Σ α₂βγ = p₁p₃ – 4p₄

Note: For the equation x³ + p₁x² + p₂x + p₃ = 0 Σα₂β₂ — p₂ -2p₁p₃

→ To remove the second term from a nth degree equation, the roots must be diminished by \(\frac{-\mathrm{a}_{1}}{\mathrm{na}_{0}}\) and the resultant equation will not contain the term with x^n-1.

→ If α₁ , α₂ ………………. , α_n are the roots of f(x) = 0, the equation

• Whose roots are \(\) is f\(\left(\frac{1}{x}\right)\) = 0
• Whose roots are kα₁, kα₂ …,kα_n is f\(\left(\frac{x}{h}\right)\) = 0
• Whose roots are α₁ – h, α₂ – h, …. α_n – h is f(x + h) = 0.
• Whose roots are α₁ + h, α₂ + h, ………….. α_n + h is f(x – h) = 0
• Whose roots are α₁², α₂²…. α₁² is f (f√y) = 0

→ In any equation with rational coefficients, irrational roots occur in conjugate pairs.

→ In any equation with real coefficients, complex roots occur in conjugate pairs.

→ If α is r – multiple root of f(x) = 0, then a is a (r – 1) – multiple root of f¹(x) = 0 and (r-2) – Multiple root of f ¹¹(x) = 0 and non multiple root of f^r-1(x) =0.

→ If f(x) = xⁿ + p₁x^n-1 + …………. + p_n-1x + p_n and f(a) and f(b) are of opposite sign, then at least
one real root of f(x) =0 lies between a and b.

(a) For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d
• In G.P., then are taken as \(\frac{a}{r}\), a, ar
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}\)

(b) For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a + d, a + 3d
• In G.P., then they are \(\frac{a}{d^{3}}, \frac{a}{d}\), ad, ad³
• In H.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\)

→ It an equation is unaltered by changing x into \(\frac{1}{x}\), then it is a reciprocal equation.

• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 is said to be a reciprocal equation of first class p_i = p_n-i for all i.
• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 = 0 is said to be a reciprocal equation of second class p_i = p_n-i for all i.
• For an odd degree reciprocal equation of class one, -1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.
• For an even degree reciprocal equation of class two, 1 and -1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate rth term, .f(x) = 0 can be transformed to f(x+h) = 0 where h is a constant such that f^(n-r+1)(h) =0 i.e., (n – r + 1)^th derivative of f(h) is zero.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 3 Quadratic Expressions to solve questions creatively.

Intermediate 2nd Year Maths 2A Quadratic Expressions Formulas

→ If a, b, c are real or complex numbers and a ≠ 0, then the expression ax² + bx + c is called a quadratic expression in the variable x.
Eg: 4x² – 2x + 3

→ If a, b, c are real or complex numbers and a ≠ 0, then ax² + bx + c = 0 is called a quadratic equation in x.
Eg: 2x² – 5x + 6 = 0

→ A complex number α is said to be a root or solution of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ If α, β are roots of ax² + bx + c = 0, then α + β = \(\frac{-b}{a}\) and αβ = \(\frac{C}{a}\)

→ The equation of whose roots are α, β is x² – (α + β)x + αβ = 0

→ Nature of the roots: ∆ = b² – 4ac is called the discriminant of the quadratic equation ax² + bx + c = 0. Let α, β be the roots of the quadratic equation ax² + bx + c = 0
Case 1: If a, b, c are real numbers, then

• ∆ = 0 ⇔ α = β = \(\frac{-b}{2 a}\) (a repeated root or double root)
• ∆ > 0 ⇔ α and β are real and distinct.
• ∆ < 0, ⇔ α and β are non- real complex numbers conjugate to each other.

Case 2: If a, b, c are rational numbers, then

• ∆ = 0 ⇔ α and β are rational and equal (ei) α = \(\frac{-b}{2 a}\), a double root or a repeated root.
• ∆ > 0 and is a square of a rational number ⇔ α and β are rational and distinct.
• ∆ > 0 but not a square of a rational number ⇔ α and β are conjugate surds.
• ∆ < 0, ⇔ α and β are non- real ⇔ α and β are non-real con conjugate complex numbers.

→ Let a, b and c are rational numbers, α and β be the roots of the equations ax² + bx + c = 0. Then

• α, β are equal rational numbers if ∆ = 0.
• α, β are distinct rational numbers if ∆ is the square of a non zero rational numbers.
• α, β are conjugate surds if ∆ > 0 and ∆ is not the square of a nonzero square of a rational number.

→ If a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have two same roots, then \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ If α, β are roots of ax² + bx + c = 0,

• the equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is f \(\left(\frac{1}{x}\right)\) = 0. If c ≠ 0 (ie) αβ ≠ 0
• the equation whose roots are α + k, β + k is f(x – k) = 0
• the equation whose roots are kα and kβ is f\(\left(\frac{x}{k}\right)\) = 0
• the equation whose roots are equal but opposite in sign is f(-x) = 0
  (ie) the equation whose roots are – α, – β is f(-x) = 0.

→ If the roots of ax² + bx + c = 0 are complex roots then for x ∈ R, ax² + bx + c and ‘a’ have the same sign.

→ If α and β (α < β) are the roots of ax² + bx + c = 0 then

• ax² + bx + c and ‘a’ are of opposite sign when α < x < β
• ax² + bx + c and ‘a’ are of the same sign if x < α or x > β.

→ Let f(x) = ax² + bx + c be a quadratic function

• If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value is given by \(\frac{4 a c-b^{2}}{4 a}\)
• If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value is given by \(\frac{4 a c-b^{2}}{4 a}\)

→ A necessary and sufficient condition for the quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).

→ If a₁b₂ – a₂b₁ = 0 then common root of a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\).

→ The standard form of a quadratic ax² + bx + c = 0 where a, b, c ∈ R and a ≠ 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ For the equation ax² + bx + c = 0, sum of the roots = \(-\frac{b}{a}\), product of the roots = \(\frac{c}{a}\).

→ If the roots of a quadratic are known, the equation is x² – (sum of the roots)x +(product of the roots)= 0

→ “Irrational roots” of a quadratic equation with “rational coefficients” occur in conjugate pairs. If p + √q is a root of ax² + bx + c = 0, then p – √q is also a root of the equation.

→ “Imaginary” or “Complex Roots” of a quadratic equation with “real coefficients” occur in conjugate pairs. If p + iq is a root of ax² + bx + c = 0. Then p – iq is also a root of the equation.

→ Nature of the roots of ax² + bx + c = 0

→ Two equations a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have exactly the same roots if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ The equations a₁x² + b₁x + c₁ = 0, a2x² + b₂x + c₂ = 0 have a common root, if (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁) and the common root is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\) if a₁b₂ ≠ a₂b₁

→ If f(x) = 0 is a quadratic equation, then the equation whose roots are

• The reciprocals of the roots of f(x) = 0 is f\(\left(\frac{1}{x}\right)\) = 0
• The roots of f(x) = 0, each ‘increased’ by k is f(x – k) = 0
• The roots of f(x) = 0, each ‘diminished’ by k is f(x + k) = 0
• The roots of f(x) = 0 with sign changed is f(-x) = 0
• The roots of f(x) = 0 each multiplied by k(≠0) is f\(\left(\frac{x}{k}\right)\) = 0

→ Sign of the expression ax² + bx + c = 0

• The sign of the expression ax² + bx + c is same as that of ‘a’ for all values of x if b² – 4ac ≤ 0 i.e. if the roots of ax² + bx + c = 0 are imaginary or equal.
• If the roots of the equation ax² + bx + c = 0 are real and different i.e b² – 4ac > 0, the sign of the expression is same as that of ‘a’ if x does not lie between the two roots of the equation and opposite to that of ‘a’ if x lies between the roots of the equation.

→ The expression ax² + bx + c is positive for all real values of x if b² – 4ac < 0 and a > 0.

→ The expression ax² + bx + c has a maximum value when ‘a’ is negative and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Maximum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

→ The expression ax² + bx + c has a maximum value when ‘a’ is positive and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Minimum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

Theorem 1:
If the roots of ax² + bx + c = 0 are imaginary, then for x ∈ R , ax² + bx + c and a have the same sign.
Proof:
The root are imaginary
b² – 4ac < 0 4ac – b² > 0
\(\frac{a x^{2}+b x+c}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
∴ For x ∈ R, ax² + bx + c = 0 and a have the same sign.

Theorem 2.
If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\), then α ≠ x ∈ R ax² + bx + c and a will have same sign.
Proof:
The roots of ax² + bx + c = 0 are real and equal
⇒ b² = 4ac ⇒ 4ac – b² = 0
\(\frac{a x^{2}+b x+c}{a}\) = x + \(\frac{b}{a}\)x + \(\frac{c}{a}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}\) > 0 for x ≠ \(\frac{-b}{2 a}\) = α
For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Theorem 3.
Let be the real roots of ax² + bx + c = 0 and α < β. Then
(i) x ∈ R, α < x< β ax² + bx + c and a have the opposite signs
(ii) x ∈ R, x < α or x> β ax² + bx + c and a have the same sign.
Proof:
α, β are the roots of ax² + bx + c = 0
Therefore, ax² + bx + c = a(x – α)(x – β)
\(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

(i) Suppose x ∈ R, α < x < β
⇒ x < α < β then x – α < 0, x – β < 0 ⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have a same sign

(ii) Suppose x ∈ R, x > β, x > β > α then x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have same sign
∴ x ∈ R, x < α or x > β ⇒ ax² + bx + c and a have the same sign.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 2 De Moivre’s Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Formulas

Statement:
→ If ‘n’ is an integer, then (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
If n’ is a rational number, then one of the values of
(cos θ + i sin θ)ⁿ is cos nθ + i sin nθ

n^th roots of unity:
→ n^th roots of unity are {1, ω, ω² …….. ω^{n – 1}}.
Where ω = \(\left[\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right]\) k = 0, 1, 2 ……. (n – 1).
If ω is a n^th root of unity, then

• ωⁿ = 1
• 1 + ω + ω² + ………… + ω^{n – 1} = 0

Cube roots of unity:
→ 1, ω, ω² are cube roots of unity when

• ω³ = 1
• 1 + ω + ω² = 0
• ω = \(\frac{-1+i \sqrt{3}}{2}\), ω² = \(\frac{-1-i \sqrt{3}}{2}\)
• Fourth roots of unity roots are 1, – 1, i, – i

→ If Z₀ = r₀ cis θ₀ ≠ 0, then the n^th roots of Z₀ are α_k = (r₀)^1/n cis\(\left(\frac{2 k \pi+\theta_{0}}{n}\right)\) where k = 0, 1, 2, ……… (n – 1)

→ If n is any integer, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

→ If n is any fraction, one of the values of (cosθ + i sinθ)ⁿ is cos nθ + i sin nθ.

→ (sinθ + i cosθ)ⁿ = cos(\(\frac{n \pi}{2}\) – nθ) + i sin(\(\frac{n \pi}{2}\) – nθ)

→ If x = cosθ + i sinθ, then x + \(\frac{1}{x}\) = 2 cosθ, x – \(\frac{1}{x}\) = 2i sinθ

→ xⁿ + \(\frac{1}{x^{n}}\) = 2cos nθ, xⁿ – \(\frac{1}{x^{n}}\) = 2i sin nθ

→ The nth roots of a complex number form a G.P. with common ratio cis\(\frac{2 \pi}{n}\) which is denoted by ω.

→ The points representing n^th roots of a complex number in the Argand diagram are concyclic.

→ The points representing n^th roots of a complex number in the Argand diagram form a regular polygon of n sides.

→ The points representing the cube roots of a complex number in the Argand diagram form an equilateral triangle.

→ The points representing the fourth roots of complex number in the Argand diagram form a square.

→ The nth roots of unity are 1, w, w²,………. , w^n-1 where w = cis\(\frac{2 \pi}{n}\)

→ The sum of the nth roots of unity is zero (or) the sum of the nth roots of any complex number is zero.

→ The cube roots of unity are 1, ω, ω² where ω = cis\(\frac{2 \pi}{3}\), ω² = cis\(\frac{4 \pi}{3}\) or
ω = \(\frac{-1+i \sqrt{3}}{2}\)
ω² = \(\frac{-1-i \sqrt{3}}{2}\)
1 + ω + ω² = 0
ω³ = 1

→ The product of the n^th roots of unity is (-1)^n-1 .

→ The product of the n^th roots of a complex number Z is Z(-1)^n-1 .

→ ω, ω² are the roots of the equation x² + x + 1 = 0

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 1 Complex Numbers to solve questions creatively.

Intermediate 2nd Year Maths 2A Complex Numbers Formulas

Definition of a complex number:
→ A number of the type z = x + yi, where x, y ∈ R and i = √- 1 i.e., i² = – 1 is called a complex number ‘x’ is called real part of z, and ‘y’ is called imaginary part of z. We write x = Re(z) and y = Im(z). A number z = x + yi is said to be purely real iff y = 0 (x ≠ 0) and purely imaginary iff x = 0 (y ≠ 0)
A complex number a + ib is an ordered pair of real numbers. It is denoted by (a, b), a ∈ R, b ∈ R.

→ Two complex numbers z₁ = (a, b), z₂ = (c, d) are said to be equal iff a = c and b = d.

→ If z₁ = (a, b), z₂ = (c, d) then

• z₁ + z₁ = (a + c, b + d)
• z₁ – z₂ = (a – c, b – d)
• z₁. Z₂ = (ac – bd, ad + bc) and
• \(\frac{z_{1}}{z_{2}}=\left(\frac{a c+b d}{c^{2}+d^{2}}, \frac{b c-a d}{c^{2}+d^{2}}\right)\)

Modulus and Amplitude:
→ The modulus of a complex number z = x + iy is defined as a non-negative real number r = \(\sqrt{x^{2}+y^{2}}\). It is denoted by |z|.

→ Any real number θ satisfying the equation cos θ = \(\frac{x}{r}\), sin θ = \(\frac{y}{r}\) is called an amplitude or argument of z. The unique argument θ of z satisfying – π < θ ≤ π is called the principal argument of z and is denoted by Arg z.

→ The polar form or modulus amplitude form of the complex number
z = x + iy is r(cos θ + i sin θ)

Conjugate of a complex number:
→ If z = x + iy i.e., x, y ∈ R, then the complex number x – iy is called conjugate of z and is written as z̅. The sum and product of a complex number and its Conjugate is always purely real.

Some properties of modulus, amplitude and conjugate:

• (z̅) = z
• z + z̅ = 2 Re (z) and z – z̅ = 2 Im (z)
• \(\overline{Z_{1} Z_{2}}\) = \(\overline{Z_{1}}\) \(\overline{Z_{2}}\)
• \(\) and \(\) (z₂ ≠ 0) and |z₁z₂| = |z₁| |z₂|
• zz̅ = |z|²
• |z| = |z̅|;
• |z₁ + z₂| ≤ |z₁| + |z₂|, |z₁ – z₂| ≤ |z₁| + |z₂|
• |z₁ – z₂| > | |z₁| – |z₂| |
  In (vii) and (viii) equality holds iff amp (z₁) – amp (z₂) is an integral multiple of 2π.
• amp (z₁ z₂) = amp (z₁) + amp (z₂) + nπ, for some n ∈ {- 1, 0, 1}
• amp \(\left(\frac{Z_{1}}{Z_{2}}\right)\) = amp (z₁) – amp (z₂) + nπ, for some n ∈ {- 1, 0, 1}
• \(\frac{1}{{cis} \alpha}\) = cis(- α)
• cis α cis β = cis (α + β)
• \(\frac{{cis} \alpha}{{cis} \beta}\) = cis (α – β)

De-Moivre’s theorem:

• If n is any integer, then (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
• If n = \(\frac{p}{q}\), where p and q are integers having no common factor and q > 1, then cos nθ + i sin nθ is one of the q values of (cos θ + i sin θ)ⁿ
• If z₀ = r₀ cis θ₀ ≠ 0, then the n^th roots of z₀ are α_k = r₀^1/n cis \(\left(\frac{2 k \pi+\theta_{0}}{n}\right)\),
  k = 0, 1, 2, 3 … (n-1)

Cube roots of unity:

• The cube roots of unity are 1, ω = \(\frac{-1+\sqrt{3 i}}{2}\) and ω² = \(\frac{-1-\sqrt{3 i}}{2}\)
• 1 + ω + ω² = 0 and w³ = 1; 1 + ω = – ω ², 1 + ω² = – ω, ω + ω² = – 1
• Either of the two non-real cube roots of unity is square of the other.
• For either of the two non – real cube roots a.and of unity α + β = – 1, αβ = – 1, α² = β, β² = α and α³ = β³ = 1
• (-1)^1/3 = – 1, – ω – ω²
• The n^th roots of unity are cis\(\left(\frac{2 k \pi}{n}\right)\), k = 0, 1, 2, ….. (n – 1)

Formulae:

→ Modulus of Z = \(\sqrt{x^{2}+y^{2}}\)

→ If \(\sqrt{a+i b}\) = (x + iy), then x = \(\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}\) and y = \(\sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\)

→ Conjugate of a + ib = a – ib

→ Conjugate of a – ib = a + ib

→ Any number of the form x + iy where x, y ∈ R and i² = -1 is called a Complex Number.

→ In the complex number x + iy, x is called the real part and y is called the imaginary part of the complex number.

→ A complex number is said to be purely imaginary if its real part is zero and is said to be purely real if its imaginary part is zero.
(a) Two complex numbers are said to be equal if their real parts are equal and their imaginary parts are equal.
(b) In the set of complex numbers, there is no meaning to the phrase one complex is greater than or less than another i.e. If two complex numbers are not equal, we say they are unequal.
(c) a+ ib > c + id is meaningful only when b = 0, d = 0.

→ Two complex numbers are conjugate if their sum and product are both real. They are of the form a + ib, a – ib.

→ cisθ₁ cisθ₂ = cis(θ₁ + θ₂), \(\frac{{cis} \theta_{1}}{{cis} \theta_{2}}\) = cis(θ₁ – θ₂), \(\frac{1}{\cos \theta+i \sin \theta}\) = cosθ – isinθ

→ \(\frac{a_{1}+i b_{1}}{a_{2}+i b_{2}}=\frac{\left(a_{1} a_{2}+b_{1} b_{2}\right)+i\left(a_{2} b_{1}-a_{1} b_{2}\right)}{a_{2}^{2}+b_{2}^{2}}\)

→ \(\frac{1+i}{1-i}\) = i, \(\frac{1-i}{1+i}\) = i

→ \(\sqrt{x^{2}+y^{2}}\) is called the modulus of the complex number x + iy and is denoted by r or |x + iy|

→ Any value of 0 obtained from the equations cos θ = \(\frac{x}{r}\), sin θ = \(\frac{y}{r}\) is called an amplitude of the complex number.

→ The amplitude lying between -π and π is called the principal amplitude of the complex number. Rule for choosing the principal amplitude.

→ If θ is the principal amplitude, then -π < 0 < π

→If α is the principle amplitude of a complex number, general amplitude = 2nπ + α where n ∈ Z.

• Amp (Z₁ Z₂) = Amp Z₁ + AmpZ₂
• Amp z + Amp z̅ = 2π (when z is a negative real number) = 0 (otherwise)

→ r(cos θ + i sin θ) is the modulus amplitude form of x + iy.

→ If the amplitude of a complex number is \(\frac{\pi}{2}\), its real part is zero.

→ If the amplitude of a complex number is \(\frac{\pi}{4}\), its real part is equal to its imaginary part.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(c)

I.

Question 1.
Solve the following inequations by the algebraic method.
(i) 15x² + 4x – 4 ≤ 0
Solution:
15x² + 4x – 4 ≤ 0
⇒ 15x² – 6x + 10x – 4 ≤ 0
⇒ 3x(5x – 2) + 2(5x – 2) ≤ 0
⇒ (3x + 2) (5x – 2) ≤ 0
Co-efficient of x² = 15 > 0,
Given Expression is ≤ 0
⇒ x lies between \(\frac{-2}{3}\) and \(\frac{2}{5}\)
i.e., \(\frac{-2}{3} \leq x \leq \frac{2}{5}\)

(ii) x² – 2x + 1 < 0
Solution:
x² – 2x + 1 < 0
⇒ (x – 1)² < 0
There is no real value of ‘x’ satisfying this inequality
Solution set = Φ (or) Solution does not exist.

(iii) 2 – 3x – 2x² ≥ 0
Solution:
-(2x² + 3x – 2) ≥ 0
⇒ -(2x² + 4x – x – 2) ≥ 0
⇒ -[2x(x + 2) – 1(x + 2)] ≥ 0
⇒ -(2x – 1) (x + 2) ≥ 0
Co-efficient of x² = -2 < 0,
The given expression is ≥ 0
⇒ x lies between -2 and \(\frac{1}{2}\)
i.e., -2 ≤ x ≤ \(\frac{1}{2}\)

(iv) x² – 4x – 21 ≥ 0
Solution:
x² – 4x – 21 ≥ 0
⇒ x² – 7x + 3x – 21 ≥ 0
⇒ x(x – 7) + 3(x – 7) ≥ 0
⇒ (x + 3) (x – 7) ≥ 0
Co-efficient of x² = 1 > 0,
The given expression is ≥ 0
x does not lie between -3 and 7
i.e., {x/x ∈ (-∞, -3] ∪ [7, ∞)}

II.

Question 1.
Solve the following inequations by graphical method.
(i) x² – 7x + 6 > 0
Solution:
f(x) = x² – 7x + 6

f(x) > 0 ⇒ y > 0
Solutions are given by x < 1 and x > 6

(ii) 4 – x² > 0
Solution:
Let f(x) = 4 – x²

f(x) > 0 ⇒ y > 0
Solution set = {x/-2 < x < 2}

(iii) 15x² + 4x – 4 < 0
Solution:
Let f(x) = 15x² + 4x – 4

f(x) ≤ 0 ⇒ y ≤ 0
Solution set = \(\left\{x / \frac{-2}{3} \leq x \leq \frac{2}{5}\right\}\)

(iv) x² – 4x – 21 ≥ 0
Solution:
Let f(x) = x² – 4x – 21

f(x) ≥ 0 ⇒ y ≥ 0
Solution set = {x/x ∈ (-∞, -3] ∪ [7, ∞)}

Question 2.
Solve the following inequations.
(i) \(\sqrt{3 x-8}\) < -2
Solution:
L.H.S. is positive and R.H.S. is negative.
∴ The given inequality holds for no real x.
Solution set = Φ (or) Solution does not exist.

(ii) \(\sqrt{-x^{2}+6 x-5}\) > 8 – 2x
Solution:
\(\sqrt{-x^{2}+6 x-5}\) > 8 – 2x
⇔ -x² + 6x – 5 > 0
and (i) 8 – 2x < 0 (or) (ii) 8 – 2x ≥ 0
We have -x² + 6x – 5 = -(x² – 6x + 5) = -(x – 1) (x – 5)
Hence -x² + 6x – 5 ≥ 0 ⇔ x ∈ [1, 5]
(i) -x² + 6x – 5 ≥ 0 and 8 – 2x < 0
⇔ x ∈ [1, 5] and x > 4
⇔ x ∈ [4, 5] ………(1)
(ii) -x² + 6x – 5 ≥ 0 and 8 – 2x ≥ 0
∵ \(\sqrt{\left(-x^{2}+6 x-5\right)}\) > 8 – 2x
⇔ -x² + 6x – 5 > (8 – 2x)² and 8 – 2x ≥ 0
⇔ -x² + 6x – 5 > 64 + 4x² – 32x and x ≤ 4
⇔ -5x² + 38x – 69 > 0 and x ≤ 4
⇔ 5x² – 38x + 69 < 0 and x ≤ 4
⇔ 5x² – 15x – 23x + 69 < 0 and x ≤ 4
⇔ (5x – 23)(x – 3) < 0 and x ≤ 4
⇔ x ∈ (3, \(\frac{23}{5}\)) and x ≤ 4
⇔ x ∈ (3, \(\frac{23}{5}\)) ∩ (-∞, 4)
⇔ x ∈ (3, 4)
Hence the solution set of the given equation is x ∈ (4, 5) ∪ (3, 4)
⇒ x ∈ (3, 5) (or) 3 < x ≤ 5.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(a)

I.

Question 1.
Find the roots of the following equations.
(i) x² – 7x + 12 = 0
Solution:
a = 1, b = -7, c = 12

∴ The roots are 4, 3

(ii) -x² + x + 2 = 0
Solution:
a = -1, b = 1, c = 2

∴ The roots are 2, -1

(iii) 2x² + 3x + 2 = 0
Solution:
a = 2, b = 3, c = 2

(iv) √3x² + 10x – 8√3 = 0
Solution:
a = √3 , b = 10, c = -8√3

∴ The roots are \(\frac{2}{\sqrt{3}}\), -4√3

(v) 6√5x² – 9x – 3√5 = 0
Solution:
a = 6√5 , b = -9, c = -3√5

∴ The roots are \(\frac{\sqrt{5}}{2},-\frac{1}{\sqrt{5}}\)

Question 2.
Form a quadratic equation whose roots are:
(i) 2, 5
Solution:
α + β = 2 + 5 = 7
αβ = 2 × 5 = 10
The required equation is x² – (α + β)x + αβ = 0
⇒ x² – 7x + 10 = 0

(ii) \(\frac{m}{n},-\frac{n}{m}\), (m ≠ 0, n ≠ 0)
Solution:

(iii) \(\frac{p-q}{p+q}, \frac{-p+q}{p-q}\), (p ≠ ±q)
Solution:

(iv) 7 ± 2√5
Solution:
α + β = 7 + 2√5 + 7 – 2√5 = 14
αβ = (7 + 2√5) (7 – 2√5)
= 49 – 20
= 29
The required equation is x² – (α + β)x + αβ = 0
⇒ x² – 14x + 29 = 0

(v) -3 ± 5i
Solution:
α + β = -3 + 5i – 3 – 5i = -6
αβ = (-3 + 5i) (-3 – 5i)
= 9 + 25
= 34
The required equation is x² – (α + β)x + αβ = 0
⇒ x² + 6x + 34 = 0

Question 3.
Find the nature of the roots of the following equation, without finding the roots.
(i) 2x² – 8x + 3 = 0
Solution:
a = 2, b = -8, c = 3
b² – 4ac = 64 – 24 = 40 > 0
∴ The roots are real and distinct.

(ii) 9x² – 30x + 25 = 0
Solution:
a = 9, b = -30, c = 25
b² – 4ac = 900 – 900 = 0
∴ The roots are rational and equal.

(iii) x² – 12x + 32 = 0
Solution:
a = 1, b = -12, c = 32
b² – 4ac = 144 – 128
= 16
= (4)²
= perfect square
∴ The roots are rational and distinct.

(iv) 2x² – 7x + 10 = 0
Solution:
a = 2, b = -7, c = 10
b² – 4ac = 49 – 80 = -31 < 0
∴ The roots are complex conjugate numbers.

Question 4.
If α, β are the roots of the equation ax² + bx + c = 0, find the values of the following expressions in terms of a, b, c.
(i) \(\frac{1}{\alpha}+\frac{1}{\beta}\)
(ii) \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\)
(iii) α⁴β⁷ + α⁷β⁴
(iv) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^{2}\), if c ≠ 0
(v) \(\frac{\alpha^{2}+\beta^{2}}{\alpha^{-2}+\beta^{-2}}\)
Solution:
α, β are the roots of the equation ax² + bx + c = 0
α + β = \(\frac{-b}{a}\), αβ = \(\frac{c}{a}\)

Question 5.
Find the values of m for which the following equations have equal roots.
(i) x² – 15 – m(2x – 8) = 0
Solution:
Given equation is x² – 15 – m(2x – 8) = 0
⇒ x² – 2mx + 8m – 15 = 0
a = 1, b = -2m, c = 8m – 15
b² – 4ac = (-2m)² – 4(1) (8m – 15)
= 4m² – 32m + 60
= 4(m² – 8m + 15)
= 4(m – 3) (m – 5)
If the equation ax² + bx + c = 0 has equal roots then its discriminant is zero.
∴ The roots are equal
⇒ b² – 4ac = 0
⇒ 4(m – 3) (m – 5) = 0
⇒ m – 3 = 0 or m – 5 = 0
⇒ m = 3 or 5

(ii) (m + 1)x² + 2(m + 3)x + (m + 8) = 0
Solution:
Given equation is (m + 1 )x² + 2(m + 3)x + (m + 8) = 0
a = m + 1, b = 2(m + 3), c = m + 8
b² – 4ac = [2(m + 3)]² – 4(m + 1) (m + 8)
= 4(m² + 6m + 9) – 4(m² + 8m + m + 8)
= 4m² + 24m + 36 – 4m² – 36m – 32
= -12m + 4
= -4(3m – 1)
∴ The roots are equal
⇒ b² – 4ac = 0
⇒ -4(3m – 1) = 0
⇒ 3m – 1 = 0
⇒ 3m = 1
⇒ m = \(\frac{1}{3}\)

(iii) x² + (m + 3)x + m + 6 = 0
Solution:
Given equation is x² + (m + 3)x + m + 6 = 0
a = 1, b = m + 3, c = m + 6
∴ The roots are equal
⇒ b² – 4ac = 0
⇒ (m + 3)² – 4(1) (m + 6) = 0
⇒ m² + 6m + 9 – 4m – 24 = 0
⇒ m² + 2m – 15 – 0
⇒ m² + 5m – 3m – 15 = 0
⇒ m(m + 5) – 3(m + 5) = 0
⇒ (m + 5) (m – 3) = 0
⇒ m = -5, 3

(iv) (3m + 1)x² + 2(m + 1)x + m = 0
Solution:
Given equation is (3m + 1)x² + 2(m + 1)x + m = 0
a = 3m + 1, b = 2(m + 1), c = m
b² – 4ac = 4(m + 1)² – 4m(3m + 1)
= 4[(m + 1)² – m(3m + 1)]
= 4(m² + 2m + 1 – 3m² – m)
= 4(-2m² + m + 1)
= -4(2m² – m – 1)
= -4(m – 1) (2m + 1)
∴ The roots are equal
⇒ discriminant = 0
⇒ b² – 4ac = 0
⇒ -4(m – 1) (2m + 1) = 0
⇒ m – 1 = 0 or 2m + 1 = 0
⇒ m = 1 or m = \(\frac{-1}{2}\)

(v) (2m + 1)x² + 2(m + 3)x + (m + 5) = 0
Solution:
Given equation is (2m + 1)x² + 2(m + 3)x + m + 5 = 0
a = 2m + 1, b = 2(m + 3), c = m + 5
∴ The roots are equal
⇒ b² – 4ac = 0
⇒ 4(m + 3)² – 4(2m + 1) (m + 5) = 0
⇒ 4(m² + 6m + 9 – 2m² – 10m – m – 5) = 0
⇒ -m² – 5m + 4 = 0
⇒ m² + 5m – 4 = 0
⇒ m = \(\frac{-5 \pm \sqrt{25+16}}{2}\)
⇒ m = \(\frac{-5 \pm \sqrt{41}}{2}\)

Question 6.
If α and β are the roots of x² + px + q = 0, form a quadratic equation whose roots are (α – β)² and (α + β)².
Solution:
∵ α, β are the roots of the equation x² + px + q = 0
α + β = -p, αβ = q
(α – β)² + (α + β)² = 2(α² + β²)
= 2[(α + β)² – 2αβ]
= 2[p² – 2q]
(α – β)² (α + β)² = [(α + β)² – 4αβ)] (α + β)²
= (p² – 4q) (p²)
∴ The required equation is x² – (sum of the roots) x + product of the roots = 0
∴ x² – 2(p² – 2q)x + p²(p² – 4q) = 0

Question 7.
If x² + bx + c = 0, x² + cx + b = 0 (b ≠ c) have a common root, then show that b + c + 1 = 0.
Solution:
If α is a common root of x² + bx + c = 0, x² + cx + b = 0 then
α² + bα + c = 0 ………..(1)
α² + cα + b = 0 ……….(2)
(1) – (2) ⇒ (b – c)α + (c – b) = 0
⇒ α – 1 = 0
⇒ α = 1
From (1), 1 + b + c = 0
Hence b + c + 1 = 0

Question 8.
Prove that the roots of (x – a) (x – b) = h² are always real.
Solution:
Given equation is (x – a) (x – b) = h²
x² – (a + b)x + (ab – h²) = 0
Discriminant = (a + b)² – 4(ab – h²)
= (a + b)² – 4ab + 4h²
= (a – b)² + (2h)² > 0
∴ The roots are real.

Question 9.
Find the condition that one root of the quadratic equation ax² + bx + c = 0 shall be n times the other, where n is a positive integer.
Solution:
Let the roots of the equation ax² + bx + c = 0 be α, nα

Question 10.
Find two consecutive positive even integers, the sum of whose squares is 340.
Solution:
Let the two consecutive positive even integers be 2λ, 2λ + 2
Sum of squares = 340
⇒ (2λ)² + (2λ + 2)² = 340
⇒ λ² + (λ + 1)² = 85
⇒ λ² + λ² + 2λ + 1 – 85 = 0
⇒ 2λ² + 2λ – 84 = 0
⇒ λ² + λ – 42 = 0
⇒ (λ + 7) (λ – 6) = 0
⇒ λ = 6, λ = -7
∵ Given numbers are positive λ = 6
∴ The two consecutive positive even integers are
2λ = 2(6) = 12 and 2λ + 2 = 12 + 2 = 14

II.

Question 1.
If x₁, x₂ are the roots of the quadratic equation ax² + bx + c = 0 and c ≠ 0, find the value of (ax₁ + b)^-2 + (ax₂ + b)^-2 terms of a, b, c.
Solution:
x₁, x₂ are the roots of the equation ax² + bx + c = 0

Question 2.
If α, β are the roots of the quadratic equation ax² + bx + c = 0, form a quadratic equation whose roots are α² + β² and α^-2 + β^-2.
Solution:
α, β are the roots of ax² + bx + c = 0, then

Question 3.
Solve the following equation:
2x⁴ + x³ – 11x² + x + 2 = 0
Solution:
2x⁴ + x³ – 11x² + x + 2 = 0
Dividing by x²

Substituting in (1)
2(a² – 2) + a – 11 = 0
⇒ 2a² – 4 + a – 11 = 0
⇒ 2a² + a – 15 = 0
⇒ (a + 3) (2a – 5) = 0
⇒ a + 3 = 0 or 2a – 5 = 0
⇒ a = -3 or 2a = 5
⇒ a = -3 or a = \(\frac{5}{2}\)

⇒ 2x² + 2 = 5x
⇒ 2x² – 5x + 2 = 0
⇒ (2x – 1) (x – 2) = 0
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = \(\frac{1}{2}\), 2
∴ The roots are \(\frac{1}{2}\), 2, \(\frac{-3 \pm \sqrt{5}}{2}\)

Question 4.
Solve 3^1+x + 3^1-x = 10
Solution:
3^1+x + 3^1-x = 10
\(\text { 3. } 3^{x}+\frac{3}{3^{x}}=10\)
Put a = 3^x so that 3a + \(\frac{3}{a}\) = 10
⇒ 3a² + 3 = 10a
⇒ 3a² – 10a + 3 = 0
⇒ (a – 3) (3a – 1) = 0
⇒ a – 3 = 0 or 3a – 1 = 0
⇒ a = 3 or a = \(\frac{1}{3}\)
Case (i): a = 3
⇒ 3^x = 3¹
⇒ x = 1
Case (ii): a = \(\frac{1}{3}\)
⇒ 3^x = 3^-1
⇒ x = -1
∴ The roots are 1, -1

Question 5.
Solve 4^x-1 – 3 . 2^x-1 + 2 = 0
Solution:
4^x-1 – 3 . 2^x-1 + 2 = 0
Put a = 2^x-1 so that a² = (2^x-1)² = 4^x-1
∴ a² – 3a + 2 = 0
⇒ (a – 2) (a – 1) = 0
⇒ a – 2 = 0 or a – 1 = 0
⇒ a = 2 or 1
Case (i): a = 2
2^x-1 = 2¹
⇒ x – 1 = 1
⇒ x = 2
Case (ii): a = 1
2^x-1 = 2⁰
⇒ x – 1 = 0
⇒ x = 1
∴ The roots are 1, 2

Question 6.
\(\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}\), when x ≠ 0 and x ≠ 3
Solution:
\(\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}\)
Put a = \(\sqrt{\frac{x}{x-3}}\)
\(a+\frac{1}{a}=\frac{5}{2}\)
⇒ \(\frac{a^{2}+1}{a}=\frac{5}{2}\)
⇒ 2a² + 2 = 5a
⇒ 2a² – 5a + 2 = 0
⇒ (2a – 1) (a – 2) = 0
⇒ 2a – 1 = 0 or a – 2 = 0
⇒ a = \(\frac{1}{2}\) or 2
Case (i): a = 2
\(\sqrt{\frac{x}{x-3}}\) = 2
⇒ \(\frac{x}{x-3}\) = 4
⇒ x = 4x – 12
⇒ 3x = 12
⇒ x = 4
Case (ii): a = \(\frac{1}{2}\)
\(\sqrt{\frac{x}{x-3}}=\frac{1}{2}\)
⇒ \(\frac{x}{x-3}=\frac{1}{4}\)
⇒ 4x = x – 3
⇒ 3x = -3
⇒ x = -1
∴ The roots are -1, 4.

Question 7.
\(\sqrt{\frac{3 x}{x+1}}+\sqrt{\frac{x+1}{3 x}}=2\), when x ≠ 0 and x ≠ -1
Solution:
Put a = \(\sqrt{\frac{3 x}{x+1}}\)
\(a+\frac{1}{a}=2\)
⇒ \(\frac{a^{2}+1}{a}\) = 2
⇒ a² + 1 = 2a
⇒ a² – 2a + 1 = 0
⇒ (a – 1)² = 0
⇒ a – 1 = 0
⇒ a = 1, 1
∴ \(\sqrt{\frac{3 x}{x+1}}\) = 1
⇒ \(\frac{3 x}{x+1}\)
⇒ 3x = x + 1
⇒ 2x = 1
⇒ x = \(\frac{1}{2}\)
∴ The root is \(\frac{1}{2}\)

Question 8.
Solve \(2\left(x+\frac{1}{x}\right)^{2}-7\left(x+\frac{1}{x}\right)+5=0\), when x ≠ 0.
Solution:
\(2\left(x+\frac{1}{x}\right)^{2}-7\left(x+\frac{1}{x}\right)+5=0\)
Put a = x + \(\frac{1}{x}\)
⇒ 2a² – 7a + 5 = 0
⇒ (2a – 5)(a – 1) = 0
⇒ 2a – 5 = 0 or a -1 = 0
⇒ a = \(\frac{5}{2}\) or 1
Case (i): 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
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = \(\frac{1}{2}\) or 2
Case (ii): a = 1
\(x+\frac{1}{x}=1\)
⇒ \(\frac{x^{2}+1}{x}\)
⇒ x² + 1 = x
⇒ x² – x + 1 = 0
⇒ x = \(\frac{1 \pm \sqrt{1-4}}{2}=\frac{1 \pm \sqrt{3 i}}{2}\)
∴ The roots are \(\frac{1 \pm \sqrt{3 i}}{2}, \frac{1}{2}\), 2

Question 9.
Solve \(\left(x^{2}+\frac{1}{x^{2}}\right)-5\left(x+\frac{1}{x}\right)+6=0\), when x ≠ 0
Solution:
Put a = x + \(\frac{1}{x}\)
⇒ a² = \(\left(x+\frac{1}{x}\right)^{2}\)
⇒ a² = \(x^{2}+\frac{1}{x^{2}}+2\)
⇒ \(x^{2}+\frac{1}{x^{2}}\) = a² – 2
∴ a² – 2 – 5a + 6 = 0
⇒ a² – 5a + 4 = 0
⇒ (a – 1) (a – 4) = 0
⇒ a = 1 or 4
Case (i): a = 1
x + \(\frac{1}{x}\) = 1
⇒ \(\frac{x^{2}+1}{x}\) = 1
⇒ x² + 1 = x
⇒ x² – x + 1 = 0
⇒ x = \(\frac{1 \pm \sqrt{1-4}}{2}=\frac{1 \pm \sqrt{3 i}}{2}\)
Case (ii): a = 4
x + \(\frac{1}{x}\) = 4
⇒ \(\frac{x^{2}+1}{x}\) = 4
⇒ x² + 1 = 4x
⇒ x² – 4x + 1 = 0
⇒ x = \(\frac{4 \pm \sqrt{16-4}}{2}\)
⇒ x = \(\frac{4 \pm 2 \sqrt{3}}{2}\)
⇒ x = 2 ± √3
∴ The roots are 2 ± √3, \(\frac{1 \pm \sqrt{3 i}}{2}\)

Question 10.
Find a quadratic equation for which the sum of the roots is 7 and the sum of the squares of the roots is 25.
Solution:
Let α, β be the roots of quadratic equation
α + β = 7, α² + β² = 25
⇒ (α + β)² – 2αβ = 25
⇒ 49 – 25 = 2αβ
⇒ αβ = 12
The required equation is x² – (α + β)x + αβ = 0
⇒ x² – 7x + 12 = 0

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d)

I.

Question 1.
(i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i in the Argand diagram.
Solution:
A(7, 7); B(7, -7) represents the given complex numbers in the Argand diagram.

O is the mid-point of AB.
Co-ordinates of O are \(\left(\frac{7+7}{2}, \frac{7-7}{2}\right)\) = (7, 0)
Slope of \(\overleftrightarrow{\mathrm{AB}}=\frac{7+7}{7-7}=\frac{14}{0}\) = ∞
AB is parallel to Y-axis
PQ is perpendicular to AB
PQ is parallel to X-axis
Slope of PQ = 0
Equation of PQ is y – 0 = 0(x – 7)
i.e., y = 0

(ii) Find the equation of the straight line joining the point -9 + 6i, 11 – 4i in the Argand plane.
Solution:
Given points are -9 + 6i, 11 – 4i
Let A = (-9, 6); B = (11, -4)
Equation of the straight line AB is y – 6 = \(\frac{-4-6}{11+9}\) (x + 9)
⇒ y – 6 = \(\frac{-1}{2}\) (x + 9)
⇒ 2y – 12 = -x – 9
⇒ x + 2y – 3 = 0

Question 2.
If Z = x + iy and if the point P in the Argand plane represents Z, then describe geometrically the locus of z satisfying the equation.
(i) |z – 2 – 3i| = 5
(ii) 2|z – 2| = |z – 1|
(iii) Img z² = 4
(iv) \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
Solution:
(i) z = x + iy and |z – 2 – 3i| = 5
|z – 2 – 3i| = 5
⇒ |x + iy – 2 – 3i| = 5
⇒ |(x – 2) + i(y – 3)| = 5
⇒ \(\sqrt{(x-2)^{2}+(y-3)^{2}}\) = 5
⇒ (x – 2)² + (y – 3)² = 25
⇒ x² – 4x + 4 + y² – 6y + 9 = 25
∴ Locus of P is x² + y² – 4x – 6y – 12 = 0

(ii) 2|z – 2| = |z – 1|
2|x + iy – 2| = |x + iy – 1|
⇒ 2|(x – 2) + iy| = |(x – 1) + iy|
⇒ \(2 \sqrt{(x-2)^{2}+y^{2}}=\sqrt{(x-1)^{2}+y^{2}}\)
Squaring both sides
⇒ 4[(x – 2)² + y²] = (x – 1)² + y²
⇒ 4(x² – 4x + 4 + y²) = x² – 2x + 1 + y²
⇒ 4x² + 4y² – 16x + 16 = x² + y² – 2x + 1
∴ Locus of P is 3x² + 3y² – 14x + 15 = 0

(iii) Img z² = 4
∵ z = x + iy
⇒ z² = (x + iy)²
⇒ z² = x² + i²y² + 2ixy
⇒ z² = (x² – y²) + i(2xy)
∴ Img (z²) = 2xy = 4
∴ The locus of P is xy = 2

(iv) \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)

Since \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
∴ \(\frac{2 y}{x^{2}+y^{2}-1}=\tan \frac{\pi}{4}\)
⇒ 2y = x² + y² – 1
⇒ x² + y² – 2y – 1 = 0
∴ The locus of z is x² + y² – 2y – 1 = 0.

Question 3.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2√3 + 2√3 i are the vertices of an equilateral triangle.
Solution:
A(2, 2), B(-2, -2), C(-2√3, 2√3) represent the given complex number in the Argand diagram.

AB² = (2 + 2)² + (2 + 2)²
= 16 + 16
= 32
BC² = (-2 + 2√3)² + (-2 – 2√3)²
= 4 + 12 – 8√3 + 4 + 12 + 8√3
= 32
AC² = (-2√3 – 2)² + (2√3 – 2)²
= 12 + 4 + 8√3 + 12 + 4 – 8√3
= 32
AB² = BC² = AC²
⇒ AB = BC = CA
∴ ∆ABC is an Equilateral triangle.

Question 4.
Find the eccentricity of the ellipse whose equation is |z – 4| + |z – \(\frac{12}{5}\)| = 10
Solution:
SP + S’P = 2a
S = (4, 0), S’ = (\(\frac{12}{5}\), 0)
2a = 10 ⇒ a = 5
SS’ = 2ae
⇒ 4 – \(\frac{12}{5}\) = 2 × 5e
⇒ \(\frac{8}{5}\) = 10e
⇒ e = \(\frac{4}{25}\)

II.

Question 1.
If \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is a real number, show that the points represented by the complex numbers z₁, z₂, z₃ are collinear.
Solution:
Let z₁ = x₁ + iy₁; z₂ = x₂ + iy₂; z₃ = x₃ + iy₃

Given \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is a real number.
Imaginary part = 0
⇒ (y₃ – y₁) (x₂ – x₁) – (x₃ – x₁) (y₂ – y₁) = 0
⇒ (y₃ – y₁) (x₂ – x₁) = (x₃ – x₁) (y₂ – y₁)
⇒ \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
The points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) represents the complex numbers z₁, z₂, z₃ respectively.
Slope of \(\stackrel{\leftrightarrow}{\mathrm{AC}}\) = Slope of \(\stackrel{\leftrightarrow}{\mathrm{AB}}\)
∴ A, B, C are collinear.

Question 2.
Show that the points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are the vertices of a square.
Solution:
A(2, 1), B(4, 3) C(2, 5), D(0, 3) represent the given complex number in the Argand plane

AB² = (2 – 4)² + (1 – 3)² = 4 + 4 = 8
BC² = (4 – 2)² + (3 – 5)² = 4 + 4 = 8
CD² = (2 – 0)² + (5 – 3)² = 4 + 4 = 8
DA² = (0 – 2)² + (3 – 1)² = 4 + 4 = 8
AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA ………(1)
AC² = (2 – 2)² + (1 – 5)² = 0 + 16 = 16
BD² = (4 – 0)² + (3 – 3)² = 16 + 0 = 16
AC² = BD²
⇒ AC = BD …….(2)
By (1), (2)
A, B, C, D are the vertices of a square.

Question 3.
Show that the points in the Argand plane represented by the complex numbers -2 + 7i, \(-\frac{3}{2}+\frac{1}{2} i\), 4 – 3i, \(\frac{7}{2}\)(1 + i) are the vertices of a rhombus.
Solution:
A(-2, 7), B(\(-\frac{3}{2}\), \(\frac{1}{2}\)), C(4, -3), D(\(\frac{7}{2}\), \(\frac{7}{2}\)) represents the given complex numbers in the Argand diagram.

∴ AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA ……….(1)
AC² = (-2 – 4)² + (7 + 3)²
= 36 + 100
= 136
BD² = \(\left(-\frac{3}{2}-\frac{7}{2}\right)^{2}+\left(\frac{1}{2}-\frac{7}{2}\right)^{2}\)
= 25 + 9
= 34
AC ≠ BD ……..(2)
∴ A, B, C, D are the vertices of a Rhombus.

Question 4.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃ are collinear, if and only if there exist three real numbers p, q, r not all zero, satisfying pz₁ + qz₂ + rz₃ = 0 and p + q + r = 0.
Solution:
pz₁ + qz₂ + rz₃ = 0
⇔ rz₃ = -pz₁ – qz₂
⇔ z₃ = \(\frac{-p z_{1}-q z_{2}}{r}\) [∵ r ≠ 0]
∵ p + q + r = 0
⇔ r = -p – q
⇔ z₃ = \(-\frac{\left(p z_{1}+q z_{2}\right)}{-(p+q)}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ divides the line segment joining z₁, z₂ in the ratio q : p
⇔ z₁, z₂, z₃ are collinear

Question 5.
The points P, Q denotes the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If \(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}=0\), show that POQ = 90°.
Solution:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
Then P(x₁, y₁), Q(x₂, y₂), O(0, 0)
\(\overline{\mathbf{z}}_{1}\) = x1 – iy1, \(\overline{\mathbf{z}}_{2}\) = x₂ – iy₂
\(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}\) = (x₁ + iy₁) (x₂ – iy₂) + (x₂ + iy₂) (x₁ – iy₁) = 0
⇒ x₁x₂ + y₁y₂ – ix₁y₂ + ix₂y₁ + x₁x₂ + y₁y₂ – ix₂y₁ + ix₁y₂ = 0
⇒ 2(x₁x₂ + y₁y₂) = 0
⇒ x₁x₂ + y₁y₂ = 0
⇒ x₁x₂ = -y₁y₂
⇒ \(\left(\frac{-x_{1}}{y_{1}}\right)\left(-\frac{x_{2}}{y_{2}}\right)=-1\)
Slope of OP × Slope of OQ = -1
∴ OP, OQ are perpendicular
⇒ ∠POQ = 90°

Question 6.
The complex number z has argument θ, 0 < θ < \(\frac{\pi}{2}\) and satisfy the equation |z – 3i| = 3. Then prove that (cot θ – \(\frac{6}{z}\)) = i
Solution:
Let z = cos θ + i sin θ
Given |z – 3i| = 3.
⇒ |(cos θ + i sin θ) – 3i| = 3
⇒ |cos θ + i(sin θ – 3)| = 3
⇒ \(\sqrt{\cos ^{2} \theta+(\sin \theta-3)^{2}}\) = 3
⇒ cos²θ + sin²θ – 6 sin θ + 9 = 9
⇒ 1 – 6 sin θ = 0
⇒ 6 sin θ = 1
⇒ sin θ = \(\frac{1}{6}\)
since 0 < θ < \(\frac{\pi}{2}\)

∴ cos θ = \(\frac{\sqrt{35}}{6}\)
cot θ = √35
∴ cot θ – \(\frac{6}{z}\) = cot θ – \(\frac{6}{\cos \theta+i \sin \theta}\)
= cot θ – 6(cos θ – i sin θ)
= √35 – 6\(\left[\frac{\sqrt{35}}{6}-i \cdot \frac{1}{6}\right]\)
= √35 – √35 + i
= i
Hence cot θ – \(\frac{6}{z}\) = i