TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 10 Random Variables and Probability Distributions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 1.
A cubical die is thrown. Find the mean and variance of X, giving the number on the face that shows up.
Solution:
Let S be the sample space and X be the random variable associated with S, where P(X) is given by the following table.
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 1

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 2.
The probability distribution of a random variable X is given below:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 2
Find the value of k and the mean and variance of X.
Solution:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 3

Question 3.
If x is a random variable with probability distribution p(X=k)=\(\frac{(k+1) c}{2^k}\) k = 0,1,2 ………… then find c.
Solution:
Since p(X=k)=\(\frac{(k+1) c}{2^k}\) k = 0,1,2 ……….. is the probability distribution of x
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 4

Question 4.
Let X be a random variable such that
P(x=-2) = P(X = -1) = P(X=2)
P(X = 1) = \(\frac{1}{6}\) and P(X = 0) \(\frac{1}{3}\)
Find the mean and variance of X.
Solution:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 5

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 5.
Two dice are roiled at random. Find the probability distribution of the sum of the numbers on them. Find the mean of the random variable.
Solution:
When two dice are rolled, the sample space S consists of 6 x 6 = 36 sample points
S = ((1, 1), (1, 2), (1, 6), (2, 1), (2, 6),(6, 6)).
Let X denote the sum of the numbers on the two dice. Then the range of X = {(2, 3, 4 , 12)}
The probability distribution for X is given here under:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 6

Question 6.
8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.
Solution:
In the experiment of tossing a coin, the probability of getting a head \(\frac{1}{2}\) and the probability of getting a tail \(\frac{1}{2}\). The probability of getting r heads in a random throw of 8 coins is
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 7

Question 7.
The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X≥1).
Solution:
Here x = B(n,p) is specified by np = 4 = μ and npq = σ2 = 3
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 8

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 8.
The probability that a person chosen at random is left handed (in hand wilting) is 0.1. What is the probability that in a group of 10 people, there is one who is left handed.
Solution:
Heye n = 10, find p =\(\frac{1}{10}\) = 0.1.
Hence q = 0.9
We have to find P(X = 1); the probability that
exactly one out of 10 is left handed
P(X = 1) = 10C1 p1 q10-1
= 10 x 0.1 x (0.9)9 = (0.9)9

Question 9.
In a book of 450 pages, there are 400 typo graphical errors. Assuming that the number of errors per page follow the Poisson law, find the probability that a random sample of 5 pages will contain no typo graphical error.
Solution:
The average number of errors per page in the book is \(\lambda=\frac{400}{450}=\frac{8}{9}\)
The probability that there are r errors per page:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 9
Hence P(X=0) = e-8/9
The required probability that a random sample of 5 pages will contain no error is [P(X) = 0)]5 = (e-8/9)5

Question 10.
The deficiency of red cells In the blood cells is determined by examining a specimen of blood under a microscope. Suppose a small fixed volume contains on an average 20 red cells for normal persons. Using the Poisson distribution, find the probability that a specimen of blood taken from a normal person will contain less than 15 red cells.
Solution:
Here λ = 20.
Let P(X = r) denote the probability that a specimen taken from a normal person will contain r red cells.
Then we have P(X < 15)
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 10

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 11.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5).
Solution:
TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions 11

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

I.
Question 1.
Express x dy – y dx = \(\sqrt{x^2+y^2}\) dx in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x dy – y dx = \(\sqrt{x^2+y^2}\) dx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 1

Question 2.
Express (x – y tan-1 \(\frac{y}{x}\)) dx + x tan-1 \(\frac{y}{x}\) dy = 0 in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 2

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 3.
Express x \(\frac{d y}{d x}\) = y (log y – log x + 1) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x \(\frac{d y}{d x}\) = y (log y – log x + 1)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\left[\log \left(\frac{y}{x}\right)+1\right]=F\left(\frac{y}{x}\right)\)

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{x-y}{x+y}\)
Solution:
Let y = vx then \(\frac{d y}{d x}\) = v + x . \(\frac{d v}{d x}\)
∴ The given differential equation is

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 3

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 2.
(x2 + y2) dy = 2xy dx.
Solution:
The given differential equation is (x2 + y2) dy = 2xy dx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 4

∴ 1 + v2 = A (1 – v2) + Bv (1 + v) + Cv (1 – v)
take v = – 1, we get
2C = – 2
⇒ C = – 1
coefficient of v2 gives
⇒ – A + B – C = 1
⇒ – A + B = 0
coefficient of v gives B + C = 0
⇒ B = 1
∴ A = 1
∴ \(\frac{1+v^2}{v-v^3}=\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\)
∴ From (1)
\(\int\left(\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\right) \mathrm{d} v=\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
⇒ log v – log (1 – v – log (1 + v) – log x + log c
⇒ log v – log(1 – v) – log (1 + v) = log cx
⇒ log v – [Iog(1 – v2)] = log cx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 5

⇒ yx = cx (x2 – y2)
⇒ y = c (x2 – y2).

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 3.
\(\frac{d y}{d x}=\frac{-\left(x^2+3 y^2\right)}{3 x^2+y^2}\)
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 6

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 7

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 8

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 4.
y2 dx + (x2 – xy) dy = 0
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 9

⇒ u – log v = log x + log c
⇒ v = log (cx v)
⇒ \(\frac{y}{x}\) = log (cx \(\frac{y}{x}\)) = log (cy)
⇒ cy = ey/x is the solution.

Question 5.
\(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\)
Solution:
Let y = vx
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 10

2 tan-1 v = log x + log c
2 tan-1 (\(\frac{y}{x}\)) = log cx is the solution of the differential equation.

Question 6.
(x2 – y2) dx – xy dy = 0
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 11

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 7.
(x2y – 2xy2) dx = (x3 – 3x2y) dy
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 12

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 13

Question 8.
y2 dx + (x2 – xy + y2) dy = 0
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 14

∴ 1 + v + v2 = A (1 + v2) + (Bv + C) v
Comparing coefficient of v2,
A + B = 1
Also A = 1,
∴ B = 0.
Comparing coefficient of v,
C = – 1
∴ \(\frac{1-v+v^2}{v\left(1+v^2\right)}=\frac{1}{v}-\frac{1}{1+v^2}\)
∴ From (1)
\(\int \frac{1}{v} \mathrm{~d} v-\int \frac{1}{1+v^2} \mathrm{~d} v=-\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
log v – tan-1 v = – log x + log c

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 15

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 9.
(y2 – 2xy) dx + (2xy – x2) dy = 0
Solution:
The given equation is (y2 – 2xy) dx = – (2xy – x2) dy

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 16

2v – 1 = A (1 – v) + 3Bv
Put v = 1,
1 = 3B
⇒ B = \(\frac{1}{3}\)
Also, – A + 3B = 2
⇒ 3B = 2 + A
⇒ 1 = 2 + A
⇒A = – 1
∴ \(\frac{2 v-1}{3 v(1-v)}=-\frac{1}{3 v}+\frac{1}{3} \frac{1}{1-v}\)
∴ From (1)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 17

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 10.
\(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
Given \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 18

v – 2 = c2 x2 . v
\(\frac{y}{x}\) – 2 = c2x2 \(\frac{y}{x}\)
y – 2x = c2x3 \(\frac{y}{x}\)
= c2x2y
= kx2y
where c2 = k
∴ Solution of the given equation is y – 2x= kx2y.

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\) dx
Solution:
x dy = (y + \(\sqrt{x^2+y^2}\)) dx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 19

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 12.
(2x – y) dy = (2y – x) dx
Solution:
Given \(\frac{d y}{d x}=\frac{2 y-x}{2 x-y}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 20

⇒ (y – x)2 = c2 (y + x)2 (y2 – x2)
⇒ y – x = c2 (y + x)3
⇒ (x + y)3 = c (x – y) where c = \(-\frac{1}{c^2}\) (constant)
Solution of the given differential equation is (x + y)3 = c (x – y)

Question 13.
(x2 – y2) \(\frac{d y}{d x}\) = xy.
Solution:
The given equation is \(\frac{d y}{d x}\) = \(\frac{x y}{x^2-y^2}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
v + x \(\frac{d v}{d x}\) = \(\frac{\mathrm{x}(v \mathrm{x})}{\mathrm{x}^2-v^2 \mathrm{x}^2}=\frac{v}{1-v^2}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 21

∴ x2 = – 2y2 [log c + log y]
⇒ x2 + 2y2 (c + log y) = 0 is the solution of the given equation where log c = c.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 14.
2 \(\frac{d y}{d x}\) = \(\frac{y}{x}+\frac{y^2}{x^2}\)
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ 2 [v + x \(\frac{d v}{d x}\)] = v + v2
∴ 2v + 2x \(\frac{d v}{d x}\) = v + v2
⇒ 2x \(\frac{d v}{d x}\) = v + v2 – 2v = v2 – v
⇒ \(\frac{\mathrm{d} v}{v^2-v}=\frac{\mathrm{dx}}{2 \mathrm{x}}\) ………..(1)
writing in variable separable lorm
⇒ \(\frac{1}{v^2-v}=\frac{1}{v(v-1)}=\frac{\mathrm{A}}{v}+\frac{\mathrm{B}}{v-1}\)
∴ 1 = A (v – 1) + Bv
Put v = 1 then B = 1, and A + B = 0
⇒ A = – B = – 1
∴ \(\frac{1}{v^2-v}=-\frac{1}{v}+\frac{1}{v-1}\)
∴ From (1)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 22

⇒ (y – x)2 = y2xc2
⇒ c1 (x – y)2 = y2x
where c1 = \(\frac{1}{c^2}\)
∴ The solution of the given differential equation is y2x = c1 (x – y)2.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

III.
Question 1.
Solve: (1 + ex/y) dx + ex/y (1 – \(\frac{x}{y}\)) dy = 0
Solution:
The given equation is
(1 + ex/y) dx + ex/y (1 – \(\frac{x}{y}\)) dy = 0
⇒ (1 + ex/y) \(\frac{d x}{d y}\) + ex/y (1 – \(\frac{x}{y}\)) = 0
Let \(\frac{x}{y}\) = v then x = vy
∴ \(\frac{d x}{d y}\) = v + y \(\frac{d v}{d y}\)
∴ From (1)
(1 + ey) (v + y \(\frac{d v}{d y}\) ) + ev (1 – v) = 0

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 23

⇒ log (ev + v) = – log y + log c
⇒ ev + v = \(\frac{c}{y}\)
⇒ ex/y + \(\frac{x}{y}\) = \(\frac{c}{y}\)
⇒ yex/y + x = c is the solution of the given equation.

Question 2.
Solve x sin \(\frac{y}{x}\) . \(\frac{d y}{d x}\) = y sin \(\frac{4}{4}\) – x.
Solution:

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 24

⇒ – cos v = – log x + log c
⇒ cos v = log x + log c = log (cx)
⇒ cos (\(\frac{y}{x}\)) = log (cx)
∴ The solution of the given equation is cx = ecos(y/x).

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 3.
x dy = (y + x cos2 \(\frac{y}{x}\)) dx.
Solution:
Given x dy = (y + x cos2 \(\frac{y}{x}\)) dx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 25

⇒ ∫ sec2 v dv = ∫ \(\frac{\mathrm{dx}}{\mathrm{x}}\) + c
⇒ tan v = log x + c
⇒ tan (\(\frac{y}{x}\)) = log x + c is the solution.

Question 4.
Solve (x – y log y + y log x) dx + x (log y – log x) dy = 0.
Solution:
Given (x – y log y + y log x) dx + x (log y – log x) dy = 0
⇒ x (log y – log x) dy = – (x – y log y + log x) dx

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 26

Writing in variable separable form we get
log v dv = – \(\frac{d x}{x}\)
∴ ∫ log v dv = – ∫ \(\frac{d x}{x}\) + c
⇒ v log v – v = – log x + c
⇒ \(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
∴ Solution of the given differential equation is
\(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
⇒ \(\frac{y}{x}\) (log y – log x) – \(\frac{y}{x}\) = – log x + c
⇒ y log y – y log x – y – x log x + cx
⇒ y log y + log x [(x – y)] = y + Cx.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 5.
Solve (y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))
Solution:
Given equation is
(y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 27

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) 28

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

Question 6.
Find the equation of a curve whose gradient is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\) where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Given gradient of the curve as \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ v + x \(\frac{d v}{d x}\) = v – cos2 v
⇒ x \(\frac{d v}{d x}\) = – cos2 v
⇒ x \(\frac{d v}{d x}\) = – cos2 v
⇒ ∫ sec2 v dv = – ∫ \(\frac{d x}{x}\)
⇒ tan v = – log x + log c
⇒ \(\tan \left(\frac{y}{x}\right)=\log \left(\frac{\mathrm{c}}{\mathrm{x}}\right)\)
Given curve passes through (1, \(\frac{\pi}{4}\)) we have
tan (\(\frac{\pi}{4}\)) = log (c)
⇒ c = e
∴ Solution of the given differential equation is
\(\tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)\)
= log e – log x
∴ Equation of the required curve is
tan \(\frac{y}{x}\) = 1 – log x.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

I.
Question 1.
Find the general solution of \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0.
Solution:
Given equation is \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 1

⇒ sin-1 y = – sin-1 x + c
⇒ sin-1 x + sin-1 y + c is the general solution.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 2.
Find the general solution of \(\frac{d y}{d x}=\frac{2 y}{x}\).
?Solution:
The given equation can be written in variable separable form as \(\frac{d y}{d x}=\frac{2 y}{x}\).
⇒ \(\frac{\mathrm{dy}}{\mathrm{y}}=2\left(\frac{\mathrm{dx}}{x}\right)\)
⇒ log |y| = 2 log |x| + log c1
⇒ log y = log x2 + log c
⇒ log \(\left(\frac{y}{x^2}\right)\) = log c
⇒ y = cx2
⇒ x2 = \(\frac{1}{c}\) y
⇒ x2 = c1y where c1 is a constant is the general solution.

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
Solution:
The given equation can be written in variable seperable form as
\(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
∴ \(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
⇒ tan-1 y = tan-1 x + tan-1 c
⇒ tan-1 y = tan-1 x + tan-1 c is the solution of the given differential equation.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 2.
\(\frac{d y}{d x}\) = ey-x
Solution:
The given equation can be written in variable separable form as
\(\frac{d y}{d x}\) = ey-x
⇒ \(\frac{d y}{e^y}=\frac{d x}{e^x}\)
⇒ ∫ e-y dy = ∫ e-x dx
⇒ – e-y dy = – e-x + c
⇒ e-x – e-y = c is the solution 0f the given differential equation.

Question 3.
(ex + 1) y dy + (y + 1) dx = 0.
Solution:
The given differential equation can be written as (ex + 1)y dy = – (y + 1) dx
⇒ \(\frac{y d y}{y+1}=-\frac{d x}{e^x+1}\)
⇒ \(\left[\frac{(y+1)-1}{y+1}\right] d y=\frac{-e^{-x}}{1+e^{-x}} d x\)
∴ ∫ dy – ∫ \(\frac{d y}{y+1}\) dx = ∫ \(\frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}}\) dx
⇒ y – log (y + 1) = log (1 + e-x)
⇒ y = log (y + 1) + log (1 + e-x) + log c
= log [(y + 1) (e-x + 1) c]
∴ ey = c(y + 1) (e-x + 1)
∴ The solution of the given equation is
ey = c (y + 1) (1 + e-x).

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 4.
\(\frac{d y}{d x}\) = ex-y + x2e-y
Solution:
\(\frac{d y}{d x}\) = ex-y + x2e-y
= e-y (ex + x2)
Writing in variable separable form we get
\(\frac{d y}{d x}\) = \(\frac{1}{y}\) (ex + x2)
⇒ ∫ ey dy = ∫ (ex + x2) dx
⇒ ey = ex + \(\frac{x^3}{3}\) + c
The solution of the given equation is ey = ex + \(\frac{x^3}{3}\) + c.

Question 5.
tan y dx + tan x dy = 0.
Solution:
The given equation can be written as
\(\frac{d x}{\tan x}+\frac{d y}{\tan y}\) = 0
⇒ \(\int \frac{\mathrm{dx}}{\tan x}+\int \frac{\mathrm{dy}}{\tan y}\) = 0
⇒ ∫ cot x dx + ∫ cot y dy = 0
⇒ log (sin x) + log (sin y) = log c
⇒ sin x sin y = c is the solution of the given differential equation.

Question 6.
\(\sqrt{1+x^2} d x+\sqrt{1+y^2} d y\) = 0
Solution:
The given equation can be written in variable seperable form as

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 2

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 7.
y – x\(\frac{d y}{d x}\) = 5 (y2 + \(\frac{d y}{d x}\))
Solution:
The given differential equation is

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 3

∴ 1 = A (1 – 5y) + By
⇒ A = 1 and B – 5A = 0
⇒ B = 5
∴ \(\int \frac{1}{y-5 y^2} d y=\int \frac{1}{y} d y+\int \frac{5}{1-5 y} d y\)
= log |y| – log (1 – 5y)
∴ From (1)
log |x + 5| = log |y| – log (1 – 5y) + log c
⇒ x + 5 = \(\frac{c y}{1-5 y}\)
∴ Solution of the given dillerential equation is 5 + x = \(\frac{c y}{1-5 y}\)

Question 8.
\(\frac{d y}{d x}=\frac{y(x+1)}{x(y+1)}\)
Solution:
The given equation \(\frac{d y}{d x}=\frac{y(x+1)}{x(y+1)}\) writing in variable separable form
\(\frac{(y+1) d y}{y}=\frac{(x+1) d x}{x}\)
\(\int\left(\frac{y+1}{y}\right) d y=\int \frac{(x+1) d x}{x}\)
⇒ y + log |y| = x + log |x| + log c
⇒ y – x = |og |x| – log |y| + log c
= log \(\left(\frac{c x}{y}\right)\)
∴ y – x = log \(\left(\frac{c x}{y}\right)\) is the solution.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

III. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{1+y^2}{\left(1+x^2\right) x y}\)
Solution:
The given equation is \(\frac{d y}{d x}=\frac{1+y^2}{\left(1+x^2\right) x y}\) which can be written in variable separable form as

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 4

⇒ (1 + x2) (1 + y2) = c2x2
= cx2 where c2 = c is a constant
∴ The solution of the given differential equation is
(1 + x2) (1 + y2) = cx2

Question 2.
\(\frac{d y}{d x}\) + x2 = x2 e3y
Solution:
The given equation can be written in variable separable form as
\(\frac{d y}{d x}\) = x2 e3y – x2
= x2 (e3y – 1)
\(\frac{d y}{e^{3 y}-1}\) = xsup>2 dx
∴ ∫ \(\frac{d y}{e^{3 y}-1}\) = ∫ x2 dx
⇒ ∫ \(\left(\frac{e^{-3 y}}{1-e^{-3 y}}\right)\) dy = \(\frac{x^3}{3}\) + c
⇒ log (1 – e-3y) = x3 + c
⇒ 1 – e-3y = ex3 + ec
= k ex3
∴ The solution of the given differential equation is 1 – e-3y = k ex3.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 3.
(xy2 + x) dx + (yx2 + y) dy = 0
Solution:
The given dilferential equation can be written as
x (y2 + 1) dx + y(x2 + 1) dy = 0 which can be expressed in variable separable form as
\(\frac{1}{2} \int \frac{2 x d x}{x^2+1}+\frac{1}{2} \int \frac{2 y d y}{y^2+1}\) = 0
⇒ \(\frac{1}{2}\) log(x2 + 1) + log (y2 + 1) = log c
⇒ log \(\sqrt{\mathrm{x}^2+1}\) + log \(\sqrt{\mathrm{y}^2+1}\) = log c
⇒ (1 + x2) (1 + y2) = c2.

Question 4.
\(\frac{d y}{d x}\) = 2y tanh x
Solution:
The given equation is \(\frac{d y}{d x}\) = 2y tanh x.
\(\frac{d y}{y}\) = 2 tanh x (variable separable form)
∫ \(\frac{d y}{y}\) = 2 ∫ tanh x dx
⇒ log y = 2 log |cosh x| + log c
= log |cosh2 x| + log c
log y = log (c cosh2 x)
y = c .cosh2 x which is the solution of the given differential equation.

Question 5.
Sin-1 (\(\frac{d y}{d x}\)) = x + y
Solution:
The given equation is Sin-1 (\(\frac{d y}{d x}\)) = x + y
⇒ sin (x + y) = \(\frac{d y}{d x}\) …………..(1)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 5

⇒ ∫ sec2 dz – ∫ sec z tan x dx = ∫ dx + c
⇒ tan z – sec z = x + c
⇒ tan (x + y) – sec (x + y) = x + c is the solution of the given differential equation.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

Question 6.
\(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}\) = 0
Solution:
Given equation in variable separable form is
\(\frac{d y}{y^2+y+1}=-\frac{d x}{x^2+x+1}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) 6

Question 7.
\(\frac{d y}{d x}\) = tan2 (x + y)
Solution:
Let x + y = z then 1 + \(\frac{d y}{d x}\) = \(\frac{d z}{d x}\)
from the given equation
∴ 1 + \(\frac{d y}{d x}\) = 1 + tan2 (x + y)
⇒ \(\frac{d z}{d x}\) = sec2 z
⇒ ∫ \(\frac{\mathrm{d} z}{\sec ^2 z}\) = ∫ dx + c
⇒ ∫ cos2 z dz = x + c
⇒ ∫ \(\left(\frac{1+\cos 2 z}{2}\right)\) dz = x + c
⇒ \(\frac{1}{2}\) z + \(\frac{1}{4}\) sin 2z = x + c
⇒ 2z + sin 2z = 4x + 4c
⇒ 2 (x + y) + sin 2 (x + y) = 4x + 4c
⇒ sin 2 (x + y) = 2x – 2y + 4c
= 2x – 2y + c1
where c1 = 4c.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 5 Permutations and Combinations Important Questions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions Important Questions

Question 1.
If nP4= 1680, find n.
Solution:
We know that ‘P4 is the product of 4 consecutive integers of which n is the largest.
That is nP4 = n(n – 1) (n – 2) (n – 3) and 1680 = 8  x 7 x 6 x 5
on comparing the largest integers, we get n = 8.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 2.
If 12Pr = 1320, find r.
Solution:
1320 = 12 x 11 x 10= 12P3 .
Thus r = 3.

Question 3.
If (n+1)P5 : nP5 = 3 : 2, find n.
Solution:
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 1

Question 4.
If 56(r+6) : 54P(r+3) = 30800 : I, find r.
Solution:
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 2

Question 5.
In how many ways 9 mathematics papers can be arranged so that the best and the worst
(i) may come together
(ii) may not come together?
Solution:
(i) lf the best and worst papers are treated as one unit, then we have 9 – 2 + 1 = 7+ 1+ 8 papers.
Now these can be arranged in (7+1) ! ways and the best and worst papers between themselves can be permuted in 2 ! ways. Therefore the number of arrangements in which best and worst papers come together is 8 ! 2 !

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

(ii) Total number of ways of arranging 9 mathematics papers is 9! . The best and worst papers come together in 8! 2! ways. Therefore the number of ways they may not come together is 9! – 8! 2! = 8!(9-2)= 8 ! × 7.

Question 6.
Find the number of ways of arranging 6 boys and 6 girls In a row. In how many of these arrangements
(i) all the girls are together
(ii) no two girls are together
(iii) boys and girls come alternately?
Solution:
6 boys + 6 girls = 12 persons. They can be arranged in a row in (12) ! ways.
(i) Treat the 6 girls as one unit. Then we have 6 boys + 1 unit of girls. They can be arranged in 7! ways. Now, the 6 girls among themselves can be permuted in 6! ways. Hence, by the fundamental principle, the number of arrangements in which all 6 girls are together 7! x 6!.

(ii) First we arrange 6 boys in a row in 6! ways. The girls can be arranged in the 7 gaps between the boys. These gaps are shown below by the letter X.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 3
Now, the girls can be arranged In these 7 gaps in 7P6 ways. Hence, by the fundamental principle, the number of
arrangements in which no two girls come together is 6! x 7P6 = 6! x 7! = 7 x 6! x 6!.

(iii) Let us take 12 places. The row may begin with either a boy or a girl. That is, 2 ways. If it begins with a boy, then all odd places (1, 3, 5, 7, 9, Ii) will be occupied by boys and the even places (2, 4, 6, 8. 10, 12) by girls. The 6 boys can be arranged in the 6 odd places in 6! ways and the 6 girls can be arranged in the 6 even places in 6! ways. Thus the number of arrangements in which boys and girls come alternately is 2 x 6! x 6!.

Note: In the above, one may think that ques tions (ii) and (iii) are same. But they are not (as evident Irom the answers). In Question (ii), after arranging 6 boys, we found 7 gaps and 6 girls are arranged in these 7 gaps. Hence one place remains vacant. It can be any one of the 7 gaps. But in Question (iii), the vacant place should either be at the beginning or at the ending but not in between. Thus, only 2 choices for the vacant place.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 7.
Find the number of 4 letter words that can he formed using the letters of the word
MIRACLE. How many of them
(i) begin with an vowel
(ii) begin and end with vowels
(iii) end with a consonant?
Solution:
The word MIRACLE has 7 letters. Hence the number of 4 letter worlds that can be formed using these letters is
7P4 = 7 x 6 x 5 x 4 = 840. Let us take 4 blanks.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 4

(i) We can fill the first place with one of the 3 vowels (1, A, E) 3P1 days. Now, the remaining 3 places can be filled using the remaining 6 letters in 6P3 120 ways. Thus the number of 4 letter words that begin with an vowel is 3 x 120 360.

(ii) Fill the first and last places with 2 vowels in 6P2 6 ways. The remaining 2 places can be filled with the remaining 5 letters in 5P2 = 20 ways. Thus the number of 4 letter words that begin and end with vowels is 6 x 20= 120.

(iii) We can fill the last place with one of the 4 consonants (M, R, C, L) in 4P1 = ways. The remaining 3 places can be filled with the letters in 6P3 ways. Thus the number of 4 letter words that end with an vowel is 4 x 6P3 = 4 x 120 = 480.

Question 8.
Find the number of ways of permuting the letters of the word PICTURE so that
(i) all vowels come together
(ii) no two vowels come together.
(iii) the relative positions of vowels and consonants are not disturbed.
Solution:
The word PICTURE has 3 vowels (I, U, E) and 4 consonants (P, C, T, R).
(i) Treat the 3 vowels as one unit. Then we can arrange 4 consonants + 1 unit of vowels in 5! ways. Now the 3 vowels among themselves can be permuted in 3! ways. Hence the number of permutations in which the 3 vowels come together is 5! x 3! 720.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

(ii) First arrange the 4 consonants in 4! ways. Then in between the vowels, in the beginning and in the ending, there are 5 gaps as shown below by the letter X.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 5
In these 5 places we can arrange the 3 vowels 5P3 ways. Thus the number of words in which rio two vowels come
to gether is 4! x 5P3 = 24 x 60 = 1440.

(iii)The three vowels can be arranged in thier relative positions in 3’ ways and the 4 consonants can be arranged in their relative positions in 4 ways. The required number of arrangements is 3! . 4! = 144.

Note: In the above problem, from (i) we get that the number of permutations in which the vowels do not come together is = Total number of permutations – number of permutations in which 3 vowels come together.
7! – 5!. 3! = 5040 – 720 = 4320.

But the number of permutations in which no two vowels come together is only 1440. In the remaining 2880 permutations, two vowels come together and third appears away from these.

Question 9.
If the letters of the word PRISON are permuted in all possible ways and the words thus formed are arranged In dictionary order, find the rank of the word PRISON.
Solution:
The letters of the given word in dictionary order is
N  O  P  R  S
In the dictionary order, first we write all words that begin with I. If we fill the first place with I, the remaining 5 places can be filled with the remaining 5 letters in 5! ways. That is, there are 5! words that begin with I. Proceeding like this, after writing all words that begin with I, N, O, we have to write the words begin with P. Among them first come the words with first two letters P, I. As above there are 4! such words. On proceeding like this, we get
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 7
Hence the rank of the word PRISON Is
3×5! + 3×4! + 2×2! + 1! + 1
= 360+72+4+1 + 1= 438

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8 (without repetition). How many of them are divisible by
(i) 2 (ii) 3 (iii) 4 (iv) 5 (v)25
Solution:
The number of 4 digit numbers that can be formed using the 5 digit 2, 3, 5, 6, 8 is = 120.
(i) Divisible by 2: For a number to be divisible by 2, the units place should be filled with an even digit. This can be done in 3 ways (2 or 6 or 8).
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 8
Now, the remaining 3 places can be filled with the remaining 4 digits in = 24 ways. Hence, the number of 4-digit numbers divisible by 2 is 3 x 24 = 72.

(ii) Divisible by 3: A number is divisible by 3 ii the sum of the digits in it is a multiple of 3. Since the sum of the given 5 digits is 24, we have to leave either 3 or 6 and use the digits 2, 5, 6, 8 or 2, 3, 5, 8. In each cae, we can permute them In 4! ways. Thus the number of 4 – digit numbers divisible by 3 is 2 x 4! = 48.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

(iii) Divisible by 4 : A number is divisible by 4 if the number formed by the digit in the last two places (tens and units places) is a multiple of 4.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 9
Thus we fill the last two places (as shown in the figure) with one of 28,32,36,52,56,68 That is done in 6 ways. After filling the last two places, we can fill the remaining two places with the remaining 3 digits in 3P2 ways.
Thus, the number of 4 – digit numbers divisible by 4 is 6 ×6=36.

(v) Divisible by 5 : After filling the units place with 5 (one way), the remaining 3 places can be filled with the remaining 4 digits in 4P3 = 24 ways. Hence the number of 4 digit numbers divisible by 5 is 24.

(vi) Divisible by 25 : Here also we have to fill. the last two places (that is, units and tens place) with 25 (one way) as shown below.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 10
Now the remaining 2 places can be filled with the remaining 3 digits in 3P2 = 6 ways. Hence the number of 4 digit numbers divisible by 25 is 6.

Question 11.
Find the sum of all 4-digit numbers that can be fonned using the digits 1, 3, 5, 7, 9.
Solution:
We know that the number of 4-digit numbers that can be formed using the given 5P4 digits is = 120. Now we find their sum. We first find the sum of the digits in the unit place of all these 120 numbers. If we fill the units place with 1 as shown below
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 11
then the remaining 3 places can be filled with the remaining 4 digits in 4P3 ways. This means, the number of 4 digit numbers having 1 in units place is 4P3 . Similarly, each of the digits 3, 5, 7, 9 appear 24 times in units place. By adding aB these digits we get the sum of the digits in units place of all 120 numbers as
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 12
Similarly, we get the sum of the digits in tens place as 4P3 x 25.
Since it is in 10’s place, its value is = 4P3 x 25 x 10.
Similarly, the value of the sum of the digits in 100s place and 1000s place are 4P3 x 25 x 100 and 4P3 x 25 x 1000
respectively. Hence the sum of the 4 digit numbers formed by using the digits 1, 3,5, 7, 9 is.
4P3 x 25 x 1+4P3 x 25 x 10 + 4P3 x 25 x 100
= 4P3 x 25 x 1000
= 4P3 x 25 x 1111 ……………………. (*)
= 24 x 25 x  1111 = 6,66,600

Note:
1. From (*) in the above example, we can derive that the sum of all r-digit numbers that can be formed using the given ‘n non-zero digits (1 ≤ r ≤ n ≤ 9) is
(n-1)P(r-1) x sum of the given digits x 111 …. 1 (r times)

2. In the above, if ‘0’ is one digit among the given n digits, then we get the sum of the r – digit numbers that can be formed using thegiven n digits (including ‘0’)
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 13

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 12.
How many four digited numbers can be formed using the digits 1, 2, 5,7, 8, 9? How many of them begin with 9 and end with 2?
Solution:
The number of four digited numbers that can be formed using the given digits 1, 2, 5, 7, 8, 9 is 6P4 = 360. Now, the first place and last place can be filled with 9 and 2 in one way.

The remaining 2 places can be filled by the remaining 4 digits 1, 5, 7, 8. Therefore these two places can be filled in 4P2 ways. Hence, the required number of ways = 1 . 4P2 = 12.

Question 13.
Find the number of injections of a set A with 5 elements to a set B with 7 elements.
Solution:
The first element of A can be mapped to any one of the 7 elements in 7 ways. The second element of A can be mapped to any one of the remaining 6 elements in 6 ways. Proceeding like this we get the number of injections from
A to B as 7P5 = 2520.

Note : If a set A has m elements and set B has n elements, then the number of injections from A into B is nPm if m≤n and 0 if m > n.

Question 14.
Find the number of ways in which 4 letters can be put in 4 addressed envelopes so that no letter goes into the envelope meant for it.
Solution:
Required number of ways is
\(4 !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)=12-4+1=9\)

Note : If there are n things is a row, a permutation of these n things such that none of them occupies its original position is called a derangement of n things. The number of derangements of n distinct things is
\(\mathrm{n} !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots . .+(-1)^{\mathrm{n}} \frac{1}{\mathrm{n} !}\right)=9\)

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 15.
Find the number of 5-letter words that can be formed using the letters of the word ‘ MIXTURE which begin with an vowel when repetitions are allowed.
Solution:
We have to fill up 5 blanks using the letters of the word MIXTURE having 7 letters among which there are 3 vowels. Fill the first place with one of the vowels (I or U or E) in 3 ways as shown below.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 14
Each of the remaining 4 places can be filled in 7 ways (since we can use all 7 letters each time). Thus the number of 5 letter words is 3 x 7 x 7 x 7 x 7 3 x 74.

Question 16.
Find the number of functions from a set A with in elements to a set B with n elements.
Solution:
Let A {a1,a2, ……………….. am} and B {b1, b2,…., bn} To define the image of a1 we have n choices (any element of B). Then we can define the image of a2 again in n ways (since a1, a2 can have same image). Thus we can define the image of each of the m elements in n ways. Therefore the number of functions from A to B is n x n x …………x n (m times) = nm.

Question 17.
Find the number of surjections from a set A with n elements to a set B with 2 elements when n > i.
Solution:
Let A {a1,a2, ……………….. an}and B = {x, y}. From the above problem. the total number of functions from A onto B is 2. For a function to be a surjection its range should contain both x, y. Observe that the number of functions which are not surjections that is, the functions which contain x or y alone in the range is 2. Hence the number of surjections from A to B is 2n – 2.

Note: In the above problem. even if B has more than 2 elements also we can derive a formula to find the number of surjections from A to B. But this result is beyond the scope of this book and hence it is not included here.

Question 18.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that ate divisible by
(i) 2
(ii) 3 when repetition is allowed.
Solution:
(i) Numbers divisible by 2:
Take 4 blanks. For a number to be divisible by 2, the units place should be filled with an even digit. This can be done in 3
ways (2 or 4 or 6).
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 15
Now, each of the remaining 3 places can be filled in 6 ways. Hence the number of 4 digit numbers that are divisible by 2 is 3 x 63 = 3 x 216 = 648.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

(ii) Numbers divisible by 3:
Fill the first 3 places with the given 6 digits in 63 ways.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 16
Now, after filling up the first 3 places with three digits, if we fill up the units place in 6 ways, we get 6 consecutive positive integers. Out of any six consecutive integers exactly two are divisible by 3. Hence the units place can be filled in 2 ways. hence the number of 4 digit numbers divisible by 3 is 2 x 216 = 432.

Question 19.
Find the number of 5- letter words that can be formed using the letters of the word Explain that begin and end with a vowel when repetitions are allowed.
Solution:
We can fill the first and last places with vowels each in 3 ways (E or A or I).
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 17
Now each of the remaining 3 places can be filled in 7 ways (using any letter of given 7 letters). Hence the number of 5 letter words which begin and end with vowels is 32 x 73 = 9 x 343 = 3087.

Question 20.
Find the number of ways of arranging 8 men and 4 women around a circular table. In how many of them
(i) all the women come together
(ii) no two women come together.
Solution:
Total number of persons = 12 (8 men + 4 women)
Therefore, the number of circular permutations is (1 1)

(i) Treat the 4 women as single unit. Then we have 8 men. 1 unit of women = 9 entities. They can be arranged around a circular table In 8! ways. Now the 4 women among themselves can be arranged in 4! ways. Hence by the Fundamental principle, the required number of arrangements is 8! x 4!.’

(ii)
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 18
First we arrange 8 men around the circular table in 7! ways. There are 8 places In between them as shown In figure by the symbol x. (one place in between any two consecutive men).

Now we can arrange the 4 women in these 8 places in 8P4 ways. Thus, the number of circular permutations in which no
two women come together is 7! x 8P4.

Question 21.
Find the number of ways of seating 5 indians, 4 Americans and 3 Russians at a round table so that
i) all Indians sit together
ii) no two Russians sit together
iii) persons of same nationality sit together.
Solution:
(i) Treat the 5 indians as one unit. Then we have 4 Americans + 3 Russians + 1 unit of Indians = 8 entities.
They can be arranged at a round table in (8 – 1)! = 7! ways. Now, the 5 Indians among themselves can be arranged in 5! ways. Hence, the required number of arrangements is 7! x 5!.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

(ii) First we arrange the 5 Indians + 4 Americans around the table in (9 – 1)! = 8! ways. Now, there are 9 gaps in between these 9 persons (one gap between any two consecutive persons). The 3 Russians can be arranged in these 9 gaps in 9P3 ways. Hence, the required number of arrangments is 8! x 9P3.

(iii)Treat the 5 Indinas as one unit, the 4 Americans as one unit and the 3 Russians as one unit. These 3 units can be
arranged at round table in (3 – 1)! = 2! ways.

Now, the 5 Indians among themselves can be permuted in 5! ways. Similarly, the 4 Americans in 4! ways and 3 Russians in 3! ways. Hence, the required number of arrangments is 2! x 5! x 4! x 3!.

Question 22.
Find the number of different chains that can be prepared using 7 different coloured beads.
Solution:
We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{1}{2}\) ((n -1)!). Hence the number of different ways of preparing the chains = \(\frac{1}{2}\{(7-1) !\}=\frac{6 !}{2}=360\)

Question 23.
Find the number of different ways of preparing a garland using 7 distinct red roses and 4 distinct yellow roses such that no two yellow roses come together.
Solution:
First we arrange 7 red roses in a circular form (garland form) in (7 – 1)! = 6! ways. Now, there are 7 gaps in between the red roses and we can arrange the 4 yellow roses in these 7 gaps 7P4 ways. Thus the total number of circular permutations is 6! x 7P4. But, this being the case of garlands, clock wise and anti-clock-wise arrangements look a like. Hence the required number of ways is \(\frac{1}{2}\) (6! x 7P4)

Question 24.
Find the number of ways of arranging the letters of the word SINGING so that
i) they begin and end with l
ii) the two G’s come together
iii) relative positions of vowels and consonants are not disturbed.
Solution:
(i) First we fill the first and last places with
I’s in \(\frac{2 !}{2 !}\) = 1 way as shown below
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 19
Now, we fill the remaining 5 places with the remaining 5 letters S, N, G, N, G in
\(\frac{5 !}{2 ! 2 !}\)  = 30 ways.
Hence, the number of required permutations is 30.

(ii) Treat the two G’s as one unit. Then we have 6 letters In which there are 2I’s and 2N’s.
Hence they can be arranged in
\(\frac{5 !}{2 ! 2 !}\) = 180 ways
Now, the two G’s among themselves can be arranged in \(\frac{2 !}{2 !}\) = 1 way. Hence the number of required permutations is 180.

(iii) In the word SINGING, there are 2 vowels which are alike i.e., 1, and there are 5 consonants of which 2Ns and 2Gs are
alike and one S is different.
C   V   C   C  V  C  C
The two vowels can be interchanged among themselves in \(\frac{2 !}{2 !}\) = 1 way. Now, the 5 consonants can be arranged in the remaining 5 places in \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
∴ Number of required arrangements = 1 x 30 = 30.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 25.
Find the number of ways of arranging the letters of the word a4b3c5 in its expanded form.
Solution:
The expanded form of a4b3c5 is
aaaa  bbb  ccccc
This word has 12 letters in which there are 4 a’s, 4 b’s and 5c’s. By Theorem 5.5.2, they can be arranged in ways.
\(\frac{12 !}{4 ! 3 ! 5 !}\) ways.

Question 26.
Find the number of 5 – digit numbers that can be formed using the digit 1, 1, 2, 2, 3. How many of them are even?
Solution:
In the given 5 digits, there are two l’s and two 2’s. Hence they can be arranged in 5!
\(\frac{5 !}{2 ! 2 !}\) = 30 ways.

Now, to find even numbers fill the units place by 2. Now the remaining 4 places can be filled using the remaining digits 1, 1, 2, 3, in
\(\frac{4 !}{2 !}\) = 12 ways.
Thus the number of 5 – digit even numbers that can be formed using the digits 1, 1, 2, 2, 3 is 12.

Question 27.
There are 4 copies (alike) each of 3 different books. Find the number of ways of arranging these 12 books in a shelf in a single row.
Solution:
We have 12 books in which 4 books are alike of one kind, 4 books are alike of second kind and 4 books are alike of third kind. Hence, by Therorem 5.5.2., they can be arranged in a shelf in a row in \(\frac{12 !}{4 ! 4 ! 4 !}\) ways.

In problem 9 of solved problems 5.2.12, we have calculated the rank of the word PRISION. In the following problem we find the rank of a word when it contains repreated letters.

Question 28.
If the letters of the word EAMCET are permuted in ail possible ways and If the words thus formed are arranged in the dictionary order, find the rank of the word EAMCET.
Solution:
The dictionary order of the letters of given word is A C E E M T
In the dictionary order the words which begin with the letter A come first. If we fill the first place with A, remaining 5 letters can be arranged \(\frac{5 !}{2 !}\) ways (since there are two Es).

On proceeding like this (as in problem 9 or 5.2.12) we get
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 20
Question 29.
Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls.
Solution:
4 boys can be selected from the given 8 boys in 5C4 ways and 3 girls can be selected from the given 5 girls in 5C3 ways. Hence, by the Fundamental principle, the number of required selections is
8C4 x 5C3 = 70 x 10 = 700

Question 30.
Find the number of ways of selecting
4 English, 3 Telugu and 2 Hindi books out of 7 English, 6 Telugu and 5 Hindi books.
Solution:
The number of ways of selecting
4 English books out of 7 books = 7C4
3 Telugu books out of 6 books = 6C3
2 HIndi books out of 5 books = 5C2
Hence, the number of required ways
7C4 x 6C3 x 5C2  = 35 x 20 x 10 = 7000

Question 31.
Find the number of ways of forming a committee of 4 members out of 6 boys and 4 girls such that there is least one girl in the committee.
Solution:
The number of ways of forming a committee of 4 members out of 10 members (6 boys + 4 girls) is 10C4 . Out of these, the number of ways of forming the committee having no girl is 6C4 (we select all 4 members from boys). Therefore, the number of ways of forming the committees having atleast one girl is 10C4– 6C4 = 210 – 15 = 195.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 32.
Find the number of ways of selecting 11 member cricket team from 7 batsmen, 6 bowlers and 2 wicket-keepers so taht the team contaIns 2 wicket-keepers and atleast 4 bowlers.
Solution:
The required cricket team can have the following compositions.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 21
Therefore, the number of ways of selecting the required cricket team = 315 + 210 + 35 = 560

Question 33.
If a set of rn’ parallel lines intersect another set of ‘n’ parallel lines (not parallel to the lines in the first set), then find the number of parallelograms formed In this lattice structure.
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 22
Solution:
Whenever we select 2 lines from the first set of m lines ad 2 lines from the second set of n lines, one parallelogram is formed as shown in the figure. Thus, the number of parallelogram formed mC2 x mC2

Question 34.
There are rn’ points in a plane out of which ‘p’ points are colinear and no three of the points are collinear unless all the three are from these p points. Find the number of different
(i) straight lines passing through pairs of distinct points.
(ii) triangles formed by joining these points (by line segments).
Solution:
(i) From the given ‘m’ points, by drawing straight lines passing through 2 distinct points at a time, we are supposed to get mC2 number of lines. But, since p’ out of these ‘m’ points are coil mear, by forming lines passing through these p points 2 at a time we get only one line instead of getting pC2. Therefore, the number of different lines as required is
mC2pC2 + 1.

(ii) From the given m points, by joining 3 at a time, we are supposed to get mC3 number of triangles. Since p points out of these m point are collinear, by joining these p points 3 at a time we do not get any triangle (we get only a Line) when we are supposed to get number of triangles. Hence the number of triangles formed by joining the given m points is
mC3 – pC3

Note : The number of diagonals in an n-sided polygon = \({ }^n C_2-n=\frac{n(n-3)}{2}\)

Question 35.
A teacher wants to take 10 students to a park. He can take exactly 3 students at a time and will not take the same group of 3 students more than once. Find the number of times (i) each student can go to the park (ii) the teacher can go to the part.
Solution:
i) To find the number of times a student can go to the park, we have to select 2 more students from the remaining 9
students. This can be done in pC2 ways. Hence, each student can go to park = 36 times.

ii) The number of times the teacher can go to park = The number of different ways of selecting 3 students out of 10
= 10C3 = 120

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions

Question 36.
A double decker minibus has 8 seats in the lower deck and 10 seats On the upper deck. Find the number of ways of arranging 18 persons in the bus If 3 children want to go to the upper deck and 4 old people can not go to the upper deck.
Solution:
Allowing 3 children, to the upper deck and 4 old people to the lower deck we are left with li people and 11 seats (7 in the upper deck and 4 in the lower deck). We can select 7 people for the upper deck out of the 11 people in 11C7 ways. The remaining 4 persons go to lower deck. Now we can arrange 10 persons (3 children and 7 others) in the upper deck and 8 persons (4 old people and 4 others) in the lower deck in 10! and 8! ways respectively. Hence, the required number arrangements = 11C7 x 10! 8!

Question 37.
Prove that
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 23
Solution:
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 24

Question 38.
(i) If \({ }^{12} C_{s+1)}={ }^{12} C_{(2 s-5)}\), find s
(ii) If \({ }^n C_{21}={ }^n C_{27} \text {, find }{ }^{50} C_n\)
Solution:
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 25
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 26

Question 39.
14 persons are seated at a round table. Find the number of ways of selecting two persons out of them who are not seated adjacent to each other.
Solution:
TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions 27
Let the seating arrangement of given 14 persons at the round table be as shown in figure.
Number of ways of selecting 2 persons out of 14 persons 14C2 = 91.
In the above arrangement two persons sitting adjacent to each other can be selected in 14 ways
(they are a1, a2, a3, a13,a14, a15 a1).
Therefore, the required number of ways = 91 – 14 = 77

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

I.
Question 1.
Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = cex+ be-x + x2.
Solution:
Given equation is xy = cex + be-x + x2 ………….(1)
Differentiating (1) w.r.t x, we get
xy1 + y = cex – be-x + x2
Again differentiating w.r.t x, we get
xy2 + y1 + y1 = cex + be-x + 2
= (xy – x2) + 2
∴ xy2 + 2y1 – (xy – x2) – 2 = 0 ………………(2)
Arbitrary constants a and b are eliminated in the differential equation (2).
The order of the differential equation (2) is 2.

Question 2.
Find the order of the differential equation of the family of all circles with their cen¬tres at the origin.
Solution:
The equation of circle with centre 0 is given by x2 + y2 = a2 where a is any constant.
Differentiating w.r.t x we get 2x + 2yy1 = 0
⇒ x + yy1 = 0
Which is the required differential equation of family of all circles with their centres at origin.
The order of the above differential equation is 1.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

II.
Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets.
(i) y = c (x – c)2; (c)
Given y = c(x – c)2 ……………..(1)
Differentiating w.r.t ‘x’ we have
y1 = 2c(x – c) ………….(2)

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) 1

∴ y . y13 = (xy1 – 2y) 4y2
⇒ y13 = (xy1 – 2y) 4y
⇒ y13 = 4xyy1 – 8y2
⇒ y13 – 4xyy1 + 8y2 = 0
⇒ \(\left(\frac{d y}{d x}\right)^3\) – 4xy \(\frac{d y}{d x}\) + 8y2 = 0
This the differential equation in which c is eliminated.

ii) xy = aex + be-x ; (a, b)
Solution:
Given xy = aex + be-x and ………….(1)
Differentiating (1) w.r.t x
xy1 + y = aex – be-x
Again differentiating w.r.t x,
xy2 + y1 + y1 = aex + be-x = xy
= xy2 + 2y1 – xy = 0
which is the required equation obtained on the elimination of a and b.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

(iii) y = (a + bx) ekx; (a, b)
Solution:
Given y = (a + bx) ekx …………(1)
and Differentiating (1) w.r.t x, we get
y1 = (a + bx) kekx + ekx . b
= ky + ekx . b
∴ y1 – ky = bekx ………….(2)
Again differentiating w.r.t x,
y2 – ky1 = kbekx
= k(y1 – ky)
⇒ y2 – 2ky1 + k2y = 0
⇒ \(\frac{d^2 y}{d x^2}-2 \mathrm{k} \frac{d y}{d x}\) + k2y = 0
is the required equation obtained on the elimination of a, b.

v) y = a cos (nx + b); (a, b)
Solution:
Given equation is y = a cos (nx + b)
∴ y1 = – an sin (nx + b)
= – an2 cos (nx + b)
= – n2y
∴ y1 + n2y = 0
⇒ \(\frac{d^2 \mathrm{y}}{d x^2}\) + n2y = 0
is the required differential equation obtained on elimination of a and b.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
(i) The rectangular hyperbolas which have the coordinate axes as asymptotes.
Solution:
Equation of rectangular hyperbolas which have the coordinate axes as asymptotes is
xy = c2.
Differentiating w.r.t x,
xy1 + y = 0
⇒ x\(\frac{d y}{d x}\) + y = 0 is the required equation.

(ii) The ellipses with centres at the origin and having coordinate axes as axes.
Solution:
Equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Differentiating w.r.t x’ we get,

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) 2

⇒ \(\frac{2}{\mathrm{~b}^2}\) [yy1 – xyy2 – xy12] = 0
⇒ yy1 – xyy2 – xy12 = 0
⇒ xyy2 + xy1 – yy1 = 0
⇒ \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\) is the required differential equation.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

III.
Question 1.
Form the differential equations of the following family of curves whose parameters are given in brackets.
(i) y = ae3x + be4x; (a, b)
Solution:
Given y = ae3x + be4x
Differentiating w.r.t. ‘X’
y1 = 3ae3x + 4be4x
⇒ y1 – 3ae3x = 4be4x
= 4 [y – ae3x]
⇒ y1 – 4y = ae3x ………….(1)
Again differentiating w.r.t. x,
y2 – 4y1 = – 3ae3x
⇒ y2 – 4y1 = 3 (y1 – 4y)
⇒ y2 – 7y1 + 12y = 0 is the required differential equation.

(ii) y = ax2 + bx, (a, b)
Solution:
Given equation is
y = ax2 + bx …………..(1)
and dill erentiating w.r.t. x
y1 = 2ax + b …………(2)
Again differentiating w.r.t. x,
y2 = 2a
⇒ x2y2 = 2ax2 …………..(3)
Also from (2)
– 2xy1 = – 4x2a – 2bx …………..(4)
From (1)
2y = 2ax2 + 2bx …………(5)
Adding (3), (4), (5) we get
x2y2 – 2xy1 + 2y = 2ax2 – 4ax2 – 2bx + 2ax2 + 2bx = 0
∴ \(x^2 \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=0\) is the required differential equation in which a, b are eliminated.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

(iii) ax2 + by2 = 1; (a, b)
Solution:
Given equation of the curve is
ax2 + by2 = 1 ……………(1)
Differentiating (1) w.r.t. ‘x we get
2ax + 2by \(\frac{d y}{d x}\) = 0 and
by2 = 1 – ax2 from (1)
⇒ 2ax + 2byy1 = 0 ……………(2)
⇒ b(2yy1) = – 2ax …………….(3)
From (3) + (2) we get

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) 3

⇒ – xay = y1 (1 – ax2)
⇒ – axy = y1 – ax2y1
⇒ y1 = ax (xy1 – y)
⇒ a = \(\frac{y_1}{x\left(x y_1-y\right)}\)
Differentiating w.r.t x,
0 = \(\frac{d}{d x}\left[\frac{y_1}{x\left(x y_1-y\right)}\right]\)
= \(\frac{y_2\left(x^2 y_1-x y\right)-y_1\left(\frac{d}{d x}\left(x^2 y_1-x y\right)\right)}{x^2\left(x y_1-y\right)^2}\)
⇒ (x2y1 – xy)y2 – y1(x2y2 + 2xy1 – xy1 – y) = 0
⇒ x2y1y2 – xyy1 – x2y1y2 – 2xy12 + xy12 + yy1 = 0
⇒ xyy2 + xy12 – yy1 = 0
⇒ \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)\) = 0 is the required differential equation obtained on elimination of constants a and b.

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

(iv) xy = ax2 + \(\frac{b}{x}\); (a, b)
Solution:
Given equation is x2y = ax3 + b ………….(1)
Differentiating (1) w.r.t. ‘x’
2xy + x2y1 = 3ax2 ………….(2)
Again differentiating w.r.t x,
x2y2 + 2xy1 + 2xy1 + 2y = 6ax
⇒ x2y2 + 4xy1 + 2y = 6ax
⇒ x3y2 + 4x2y1 + 2xy = 6ax2
= 2(3ax2)
= 2 [2xy + x2y1]
= 2x2y1 + 4xy
⇒ x3y2 + 2x2y1 – 2xy = 0
⇒ x2y2 + 2xy1 – 2y = 0
⇒ \(x^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}\) – 2y = 0 which is the required differential equation on elimination of constants a and b from (1).

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
(i) The circles which touch the Y – axis at the origin.
Solution:
The cquation of circle which touch the Y-axis at the origin is x2 + y2 + 2gx = 0 ………..(1)
Differentiating wr.t. x we get
2x + 2yy1 + 2g = 0
⇒ g = – (x + yy1)
Hence from (1)
x2 + y2 + 2x [- (x + yy1)] = 0]
x2 + y2 – 2x2 – 2xyy1 = 0
⇒ – y2 – x2 = 2xy . \(\frac{d y}{d x}\) which is the required differential equation obtained on elimination of ‘g’ from (1).

TS Board Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

(ii) The parabola each of which has a laws rectum 4a and whose axis are parallel to X- axis.
Solution:
Equation of parabola which has latus rectum 4a and whose axes are parallel to X-axis is
(y – k)2 = 4a(x – h) ………….(1)
Differentiating w.r.t ‘x’
2 (y – k) y1 = 4a
⇒ (y – k) y1 = 2a ……………(2)
Differentiating again w.r.t ‘x’
(y – k) y2 + y12 = 0
From (2)
y – k = \(\frac{2 a}{y_1}\)
∴ From (3)
\(\frac{2 a}{y_1}\) y2 + y12 = 0
⇒ 2ay2 + y13 = 0
⇒ 2a \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\) = 0 which is the required differential equation obtained on elimination of constants h, k from (1).

(iii) The parabolas having their focli at the origin and axis along the X-axis.
Solution:
Equation of parabola having focii at origin and axis is along X-axis is given by
y2 = 4a(x + a) ……….(1)
Differentiating w.r.t x
2yy1 = 4a
a = \(\frac{\mathrm{yy}_1}{2}\)
∴ From (1)
y2 = 4a(x + a)
= 4 \(\frac{\mathrm{yy}_1}{2}\) (x + \(\frac{\mathrm{yy}_1}{2}\))
= 2yy1 (x + \(\frac{\mathrm{yy}_1}{2}\))
= 2xyy1 + y2y12
⇒ y = 2xy1 + yy12
⇒ yy12 + 2xy1 – y = 0
⇒ \(y\left(\frac{d y}{d x}\right)^2+2 x\left(\frac{d y}{d x}\right)\) – y = 0
which is the required differential equation obtained on elimination of ‘a’ from (1).

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

I.
Question 1.
In the experiment of tossing a coin n times, If the variable X denotes the number of heads and P (X = 4), P (X = 5), p (X = 6) are in arithmetic progression then find n.
Solution:
X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
P (X = 4), P (X = 5), P (X = 6) are in A.P.
∴ 2P (X = 5) = P (X = 4) + P (X = 6)

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 1

⇒ n2 – 21n + 98 = 0
⇒ (n – 7) (n – 14) = 0
⇒ n = 7 or 14.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 2.
Find the minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is 0.8.
Solution:
When a coin is tossed we have p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
P(X = x) = nCx pxqn-x
P(X = 0) = nC0 p0qn-0
Given probability 0f getting atleast one head is atleast 0.8
P(X ≥ 1 ) ≥ 0.8
⇒ 1 – P (X = 0) ≥ 0.8
⇒ 1 – \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) ≥ 0.8
⇒ – \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) ≥ – 0.2
⇒ \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) < (0.2). Which is true for n = 3.

Question 3.
The probability of a bomb hitting a bridge is 1/2 and three direct hits (not necessarily consecutive) are needed to destroy it. Find the mmiinuni number of bombs required so that the probability of bridge being destroyed is greater than 0.9.
Solution:
Given probability of a bomb hitting a bridge is \(\frac{1}{2}\).
∴ p = \(\frac{1}{2}\);
q + p = 1 ⇒ q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let X represents the minimum number of bombs to be dropped so that the bridge can be destroyed. Given that P (X ≥ 3) > 0.9
⇒ 1 – P(X < 3) > 0.9
⇒ 1 – [ P(X = 0) + P(X = 1) + P (X = 2)] > 0.9
⇒ 1 – [\({ }^n \mathrm{C}_0\left(\frac{1}{2}\right)^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_1\left(\frac{1}{2}\right)^{\mathrm{n}-1}\left(\frac{1}{2}\right)\) + \({ }^{\mathrm{n}} C_2\left(\frac{1}{2}\right)^{\mathrm{n}-2}\left(\frac{1}{2}\right)^2\)] > 0.9

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 2

By substituting the values for ‘n’ we see that n = 9, 10, 11. satisfies the above in equation.
∴ The minimum number of bombs to be dropped so that the bridge Is to be destroyed is n = 9.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 4.
If the difference between the mean and the variance of a binomial variate is 5/9 then find the probability for the event of 2 successes when the experiment is conducted 5 times.
Solution:
Given n = 5 and
given that mean – variance of a Binomial variate is \(\frac{5}{9}\).
∴ np – npq = \(\frac{5}{9}\)
np(1 – q) = \(\frac{5}{9}\)
⇒ npq = \(\frac{5}{9}\)
⇒ 5p (1 – q) = \(\frac{5}{9}\)
⇒ 5p (p) = \(\frac{5}{9}\)
⇒ p2 = \(\frac{5}{9}\)
⇒ p = \(\frac{1}{3}\)
∴ q = \(\frac{2}{3}\)
∴ P(X = 2) = 5C2 \(\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{5-2}\)
= 10 × \(\frac{1}{9} \times \frac{8}{27}=\frac{80}{243}\)

Question 5.
One in 9 ships is likely to be wrecked, when they are set on sail. When 6 ships set on sail, find the probability for
i) Atleast one will arrive safely
ii) Exactly three will arrive safely.
Solution:
Let q = Probability of one in 9 ships to be wrecked = \(\frac{1}{9}\)
p = 1 – q
= 1 – \(\frac{1}{9}\) = \(\frac{8}{9}\)
Here n = 6.

i) Probability ¡or atleast one will arrive safely
P (X ≥ 1) = 1 – P(X = 0)
= 1 – \({ }^6 C_0\left(\frac{8}{9}\right)^0\left(\frac{1}{9}\right)^{6-0}\)
= 1 – \(\frac{1}{9^6}\)

ii) Exactly, three will arrive safely
P (X = 3) = 6C3 \(\left(\frac{8}{9}\right)^3\left(\frac{1}{9}\right)^{6-3}\)
= \({ }^6 C_3 \frac{8^3}{9^3} \frac{1}{9^3}={ }^6 C_3 \frac{8^3}{9^6}\)
= \(20\left(\frac{8^3}{9^6}\right)\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 6.
If the mean and variance of a binomial variate X are 2.4 and 1.44 respectively, find P (1 < X ≤ 4).
Solution:
Mean = np = 2.4;
Variance = npq = 1.44
∴ \(\frac{\mathrm{npq}}{\mathrm{np}}=\frac{1.44}{2.4}\) = 0.6
⇒ q = 0.6
and p = 1 – 0.6 = 0.4
∴ np = 2.4
⇒ n (0.4) = 2.4
⇒ n = 6
∴ P(1 < X ≤ 4) = P (X = 2) + P(X = 3) + P (X = 4)

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 3

Question 7.
If is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.
Solution:
Probability for an electric bulb to be defective, p = \(\frac{10}{100}\) = 0.1
∴ Probability for a non defective bulb
q = 1 – p = 1 – 0.1 = 0.9
Probability to have more than 2 are defectives
⇒ P(X > 2) = 1 – P(X ≤ 2)
1 – [P (X = 0) + P(X = 1) + P(X = 2)]
We have
P(X = x) = nCx pxqn-x

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 4

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 8.
On an average, rain falls on 12 days in every 30 days, find the probability that, rain will fall on just 3 days of a given week.
Solution:
Given the probability for the day to be rainy = \(\frac{12}{30}=\frac{2}{5}\)
∴ p = \(\frac{2}{5}\) and
q = 1 – p
= 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Also, n = 7
∴ Probability for the rain to fall on just 3 days of a given week
P(X = 3) = 7C3 \(\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^{7-3}\)
= 7C3 \(\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^4\)
= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Given mean = np = 6
and variance = npq = 2
∴ q = \(\frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{6}=\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
∴ np = 6
⇒ \(\frac{2 n}{3}\) = 6
⇒ n = 9
We have P (X = x)
= nCx px qn-x

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 5

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 10.
In a city 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.
Solution:
Here λ = \(\frac{10}{50}=\frac{1}{5}\)
∴ The probability that there will be 3 or more accidents in a day using Poisson variate.
P (X = x) = \(\frac{c^{-\lambda} \lambda^x}{x !}\)
P (X ≥ 3) = 1 – [P (X = 0) + P(X = 1) + P(X = 2)]

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 6

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

II.
Question 1.
Five coins are tossed 320 times. Find the frequencies of the distribution of the number of heads and tabulate the result.
Solution:
When a coin is tossed we have p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\).
Here n = 5;
The frequencies of distribution of the number of heads is given using binomial distribution given by
fx = N. nCx px qn-x,
x = 0,1, 2, 3, 4, 5, (Number of heads)
when x = 0

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 7

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 2.
Find the probability of guessing atleast 6 out of 10 of answers in
(i) True or false type examination
(ii) Multiple choice with 4 possible answers.
Solution:
i) Probability of guessing atleast 6 out of 10 answers in
(i) True or false type examination is
P(X ≥ 6) = P(X = 6) P(X = 7) + P (X = 8) + P (X = 9)
Here P (X = x) = nCx px qn-x and
probability for an answer to be true or false,
p = \(\frac{1}{2}\), q = \(\frac{1}{2}\), n = 10
p – q = n = 10

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 8

ii) Probability for a question to guess in multiple choice type with 4 answers is
p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)
∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P (X = 9) + P(X = 10)
where P (X = x) = nCx px qn-x
= 10Cx \(\left(\frac{1}{4}\right)^x\left(\frac{3}{4}\right)^{10-x}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) 9

In the above two cases (i) and (ii).

i) Probability to guess 6 out of 10 questions in true or false examination is = 10C6 \(\left(\frac{1}{2}\right)^{10}\)

ii) Probability to guess 6 out of 10 questions in Multiple choice type with 4 answers = 10C6 . \(\frac{3^4}{4^{10}}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

Question 3.
The number of persons joining a cinema ticket counter in a minute has poisson distribution with parameter 6. Find the probability that
(i) no one joins the queue in a particular minute
(ii) two or more persons join in the queue in a minute.
Solution:
Given λ = 6, and poisson distribution function is given by
P (X = x) = \(\frac{\lambda^x e^{-\lambda}}{x !}\), x = 0, 1, 2, 3, ………..

i) Probability that no one joins the queue in a particular minute is
P (X = 0) = \(\frac{\lambda^0 \mathrm{e}^{-6}}{0 !}\) = e-6.

ii) Probability for two or more persons join in the queue in a minute is
P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)]
= 1 – \(\left[\mathrm{e}^{-6}+\frac{\lambda \mathrm{e}^{-\lambda}}{1 !}\right]\)
= 1 – e-6 – 6 e-6.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 4 Theory of Equations to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 1.
Form the monic polynomial eduation of degree 3 whose roots are 2, 3 and 6. Given roots of required polynomial are
2, 3 and 6.
Solution:
We know that monic polynomial those roots of α, β and γ is (x -α) (x -β) (x -γ) = 0.
∴ Equation of required monk polynomial is
(x – 2)(x – 3)(x – 6) = 0
on simplification, it becomes
x3 – 11x2 + 36x – 36 = 0.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 2.
Find the relations between the roots and the coefficients of the cubic equation
3x2 – 10x2 + 7x + 10 = 6.
Solution:
Given cubic equation is
3x2 – 10x2 + 7x + 10 = 0
On dividing the equation by ‘3’ we get,
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 1

Question 3.
Write down the relations between the roots and the coefficients of the biquadratic equation :
x4– 2x3 + 4x2 + 6x – 21 = 0.
Solution:
Given equation is x4– 2x3 + 4x2 + 6x – 21 = O
On comparing (1) with
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 2

Question 4.
If 1,2,3 and 4 are the roots of x4+ax3+bx2+ cx + d = 0, then find the values of a, b, c and d.
Solution:
Given roots of the given polynomal equation are 1, 2, 3 and 4.
∴ x4+ax3+bx2+ cx + d ≡ (x-1)(x-2) (x – 3) (x – 4)
= x4 – 10x3 35x2 – 50x + 24
On comparing coellicients of like powers of we get,
a = – 10, b = 35, c = – 50, d = 24.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 5.
If a,b and c are the roots of x3-px2+qx – r = 0 and r ≠0 then find \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) in terms of p,q,r.
Solution:
Given a ,b and c are the roots of
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 3

Question 6.
Find the sum of the squares and the sum of the cubes of the roots of the equation
x3 – px2 + qx – r = 0 in terms of p, q, r.
Solution:
Let α, β and γ be roots of the given equation x3 – px2 + qx – r = 0 ……………..(1)
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 4

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 7.
Obtain the monic cubic equation, whose roots are the squares of the roots of the equation x3 + p1x3 + p2x + p3 = 0.
Solution:
Let α, β and γ be the roots of the given equation x3 + p1x3 + p2x + p3 = 0 ………………… (1)
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 7
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 6

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 8.
Let α, β and γ be the roots of x3+px2+qx+r = 0. Then find
(i) Σα2
(ii) \(\Sigma \frac{1}{\alpha}\) ,If α, β,γ are non-zero
(iii) Σα3
(iv) Σβ2γ2
(v) (α+β)(β+γ)(γ+α)
Solution:
Let α, β, and γ be the roots of equation  x3+px2+qx+r = 0 ……………….(1)
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 8

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 9.
If α, β, γ are the roots of x3 + ax2 + bx + c = 0, then find ∑α2β + ∑αβ2.
Solution:
Given α, β and γ  are the roots of
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 9

Question 10.
If α, β, γ are the roots of x3 + px2 + qx + r = 0, then form the monic cubic equation whose rools are
α(β+ γ),β(y+α), y(α+ β).
Solution:
Given a, and y are the roots of
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 10
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 11

Question 11.
Solve x3 – 3x2 – 16x + 48 = 0.
Solution:
Let f(x) = x3 – 3x2 – 16x + 48
By inspection, we see that
f(3)= 27- 27 – 48+48 = 0
Hence 3 is a root of f(x) = 0
Now we divide 1(x) by (x-3), using synthetic division.
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 12
Thus the quotient is (x2 – 16) and the remainder is 0.
Therefore f(x) (x – 3) (x2 – 16)
= (x – 3) (x – 4) (x 4)
Hence 3, – 4, 4 are the roots of the given equation.

Question 12.
Find the roots of x4 – 16x3 + 86x2 – 176x + 105 = 0.
Solution:
Let f(x) = x4 – 16x3 + 86x2 – 176x + 105
By inspection we see that,
f(1)= 1- 16+86 – 176+ 105=0
Hence 1 is a root of 1(x) = 0
Now we divide f(x) by (x – 1), using synthetic division.
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 13
Therefore
x4 – 16x3 + 86x2 – 176x + 105
(x – 1)(x3-15x2 + 71x – 105) ……………… (1)
Let g(x) = x3 – 15x2 + 71x – 105
By inspection g(3) = 0.
Hence 3 is a root of g(x) = 0.
Now we divide g(x) by (x -3),
using synthetic division.
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 14
Therefore g(x) (x – 3) (x2 – 12x + 35) ……………. (2)
From (1) and (2),
f(x) (x – 1) (x – 3) (x2 – 12x + 35)
= (x – 1) (x – 3) (x – 5) (x – 7)
Hence 1, 3, 5 and 7 are the roots of the given equation.
Now we solve equations when a relation between some of the roots is given.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 13.
Solve x3 – 7x2 +36 = 0 given one root being twice the other.
Solution:
Let α, β, γ be the three roots of the given equation and β = 2a.
Now we have α+β+γ = 7
αβ + βγ + γα = 0
αβγ = – 36
On substituting β =2α in the above equations, we obtain
3α + γ = 7 …………………… (1)
2+3αγ = 0 ……………………. (2)
2 γ = – 36 ……………………. (3)
On eliminating γ from (1) and (2), we have
2 + 3α (7 – 3α) = 0
i.e., α2 -3α = 0 or α(α -3) = 0
Therefore α = 0 or α = 3
Since α = 0 does not satisfy the given equation, we ignore this value.
Therefore α = 3 is a root of the given equation.
So, β = 6 (since β = 2α) and γ = – 2
Hence 3, 6, – 2 are the roots of the given equation.

Question 14.
Given that 2 is a root of x3 – 6x2+3x+10 = 0, find the other roots.
Solution:
Let f(x) x3 – 6x2+3x+10
Since 2 is a root of f(x) 0,
we divide f(x) by (x – 2)
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 15
Therefore
x3 – 6x2 – 3x + 10 = (x – 2) (x2 – 4x – 5)
= (x – 2) (x+1) (x-5)
Thus – 1, 2 and 5 are the roots of the given equation.

Question 15.
Given that two roots of 4x3 + 20x2– 23x + 6 = 0 are equal, find all the roots of the given equation.
Solution:
Let α, β, γ, and δ are the roots of
4x3 + 20x2– 23x + 6 = 0 ………………… (1)
Given that two roots of (1) are equal. Let β = α
Since α, β, γ are the roots of (1), we have
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 16
Therefore
x3 – 6x2 – 3x + 10 = (x – 2) (x2 – 4x – 5)
=(x – 2) (x+ 1) (x – 5)
Thus – 1, 2 and 5 are the roots of the given equation.
Therefore α = \(\frac{1}{2}\) or α = \(-\frac{23}{6}\)
On verification we get that α = \(\frac{1}{2}\) is a root of (1)
On substituting this value in (2), we get γ = – 6
Therefore \(\frac{1}{2}, \frac{1}{2},-6\) are the roots of (1).

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 16.
Given that the sum of two roots of x4 – 2x3+ 4x2 + 6x – 21 = 0 is zero. Find the roots of the equation.
Solution:
Let α, β, γ, and δ be the roots of (1), and α + β = 0 ……………. (1)
From the relation between the coefficients and the roots, we have
α + β + γ +δ = 2 so γ + δ = 2 ……………… (2)
Therefore the quadratic equation having roots α and β is x2 – 0, x + αβ=0
The quadratic equation having roots γ + δ is x2-2x +q =0
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 17

Question 17.
Solve 4x3 – 24x2+ 23x+ 18=0 given that the roots of this equation are in arithmetic progression.
Solution:
Let a – d, a, a + d be the roots of the given equation. These are in A.P. From the relation between the coefficients and the roots, we have
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 18

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 18.
Solve x3-7x2+14x-8=0 given that the roots are in geometric progression.
Solution:
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 19
Hence the roots of the given equation are 1, 2 and 4.

Question 19.
Solve x4-5x3+5x2+ 5x – 6 = 0 given that the product of two of its roots is 3.
Solution:
Let α , β, γ be the roots of the given equation.
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 20
we get γ = 3 and δ = 1 or γ = 1 and δ = 3
Hence the roots of the given equation are 2,- 1, 3 and 1.

Question 20.
Solve x4+4x3– 2x4– 12x+ 9 = 0 given that it has two pairs of equal roots.
Solution:
Let the roots of the given equation be
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 21
Therefore 1, 1, -3, -3 are the roots of the given equation.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 21.
Prove that the sum of any two of the roots of the equation x4 +px3 + qx2 + rx + s = 0 is equal to the sum of the remaining two roots of the equation if p3 – 4pq + Sr = 0.
Solution:
Suppose that the sum of two of the roots of the given equation is equal to the sum of the remaining two roots.
Let α, β, γ, δ be roots of the given equation such that α + β = γ + δ
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 24
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 22
Take \(b=\frac{p}{2}\)
Then equations (1), (2) and (4) are satisfied.
In view of (5), equation (3) is also satisfied.
Hence (x2 + bx + c) (x2 + bx + d)
=x4+ 2bx3+(b2+c+d)x2+h(c +d)x+cd
= x4 +px3 +qx2 +rx+s
Hence the roots of the given equation are α1, β11 and δ1, where α1 and β1 are the roots of the equations
x2 + bx + c = 0 and γ1 and δ1 are those of the equation x2 + bx+ d = 0
We have α1 + β1 = -b = γ1 + δ1

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 22.
Form the monic polynomial equation of degree 4 whose roots are \(4+\sqrt{3}, 4-\sqrt{3}\)
Solution:
The required equation is
\(\{\mathrm{x}-(4+\sqrt{3})\}\{\mathrm{x}-(4-\sqrt{3})\}\)
{x – (2+i)} {x – (2 – i)) =0
i.e., (x2 – 8x + 13) (x2 – 4x + 5) = 0
i.e., x2 – 12x3 + 50x – 92x + 65 = 0

Question 23.
Solve 6x4 -13x2 – 35x2 – x+3 = 0 given that one of its roots is \(2+\sqrt{3}\)
Solution:
Since \(2+\sqrt{3}\) is a root of the giver equation, by Theorem 4.3.9, \(2+\sqrt{3}\) is also a root of it. The quadratic factor corresponding to these two roots is x2 – 4x + 1.
On dividing 6x4 -13x1 – 35x2 –  x + 3 by x2 – 4x + 1 (by synthetic division) weet the quotient 6x2 + 11x + 3.
Therefore 6x – 13x3 – 35x2 – x + 3 =(x2– 4x+ 1) (6x2+ 11x+3)
Hence the other roots are obtained from 6x2 + 11x + 3 = 0
On solving this equation, we get
\(\mathrm{x}=-\frac{1}{3} \text { or }-\frac{3}{2}\)
Thus the roots of the given equation are
\(-\frac{1}{3},-\frac{3}{2}, 2 \pm \sqrt{3}\)

Question 24.
Find the polynomial equation of degree 4 whose roots are the negatives of the roots
of x4 – 6x3 +7x2 – 2x + 1 = 0.
Solution:
Let f(x) = x4 – 6x3 +7x2 – 2x + 1
By Theorem 4.4.1, the equation (- x) = 0 has the desired property.
We have
f(-x) = (-x)4 – 6(-x)3. 7(-x)2 – 2(-x) + 1
= x4+6x3+7x2+2x+ 1
Hence x4 – 6x3 +7x2 – 2x + 1 + 6x3 + 7x2 + 2x+ 1 = 0 is the desired equation.

Question 25.
Find an algebraic equation of degree 4 whose roots are 3 times the roots of equation 6x4 – 7x3 + 8x2 – 7x + 2 = 0.
Solution:
Let f(x) = 6x4 – 7x3 + 8x2 – 7x + 2
By Theorem 4.4.3, the equation
\(f\left(\frac{x}{3}\right)=0\) has the desired properties. We have
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 25
Hence 6x4– 21x3+72x2-189x +162 = 0 is the desired equation.

Question 26.
Form the equation whose roots are m times the roots of the equation
\(x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{72}=0\) and deduce the case when m = 12.
Solution:
From the note 4.4.2 and note 4.4.4., it follows that
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 26
is a polynomial equation of degree 3, whose roots are m times those of the given equation. On taking m = 12, equation (1) reduces to
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 27

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 27.
Find the polynomial equation of degree 5 whose roots are the translates of the roots of x5+4x3 -x2+11 = 0 by – 3.
Solution:
By Theorem 4.4.6, the equation
(x+3)5+4(x+3)3_(x+3)2 +11 = 0 has the desired properties.
On simplifying the above equation, we get
x5 + 15x4 + 94x3 + 305x2 + 507x + 353 = 0.
The transformed equation can also be object by synthetic division
Let f(x) =x5+4x3-x2+ 11
Suppose that f(x + 3) A0x5 + A1x4 + A2x3 + A3x2 + A4x + A5
Then by note 4.4.8, the coefficients
A0, A1 , …………………………. A5 can be obtained as follows
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 28
Therefore the roots of the equation
x5 + 15x4 + 94x3 + 305x2 + 507x + 353 = 0 are the translates of the roots of the given equation by – 3.

Question 28.
Find the algebraic equation of degree 4 whose roots are the translates of the roots of 4x4 + 32x3 + 83x2 + 76x + 21 = O by 2.
Solution:
By Theorem 4.4.6, the equation
4 (x – 2)4 + 32 (x – 2)3 + 83 (x – 2)4+76(x-2) + 21 =0 has desired properties.
On simplifying the above equation, we get 4x4– 13x2+90
Other mehod: The equation A0x4 . A1x3 +………… + A4 = 0 whose coefficients are obtained by synthetic division as given below, has the desired properties
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 29
Hence the equation 4x4 – 13x2 + 9 = 0 has the desired properties.

Question 29.
Find the polynomial equation whose roots are the reciprocals of the roots of the equation x4 + 3x3 – 6x2 + 2x – 4 = 0.
Solution:
Let f(x) = x4 + 3x3 – 6x2 + 2x – 4
By theorem 4.4.9, the equation
\(x^4 f\left(\frac{1}{x}\right)=0\) has the desired properties.
Therefore
\(x^4\left[\frac{1}{x^4}+3 \frac{1}{x^3}-6 \frac{1}{x^2}+\frac{2}{x}-4\right]=0\)
i.e., 4x4 – 2x3 + 6x2 – 3x – 1 = 0 is the required equation.

Question 30.
Find the polynomial equation whose roots are the squares of the roots of x3– x2 + 8x – 6 = 0.
Solution:
Let f(x) = x3– x2 + 8x – 6 = 0.
Then as per the notation introduced in Note 4.413, we have
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 30
The equation x3+ 15x2 + 52x -36 = 0 has the desired properties.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 31.
Show that 2x3+5x2+5x+2=0 is are reciprocal equation of class one.
Solution:
Let f(x) = 2x3+5x2+5x+2=0
Then 2 is the leading coefficient.
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 31
Therefore the given equation is a reciprocal equation of class one.

Question 32.
Solve the equation 4x3 -13x2 – 13x + 4 = 0.
Solution:
The given equation is an odd degree reciprocal equation of class one.
By Note 4.4.24(1), – 1 is a root of this equation.
Therefore (x + 1) is a factor of
4x3 -13x2 – 13x + 4
Hence on dividing this expression by(x + 1), we get
4x3 -13x2 – 13x + 4 (x + 1)(4x2 – 17x+ 4)
Now the roots of the equation
4x2– 17x+4=0 are \(\frac{1}{4}\) and 4.
Therefore -1,\(\frac{1}{4}\),4 are the roots of the given equation.

Question 33.
Solve the equation
– 5x4 + 9x3 – 9x2 + 5x – 1 = 0.
Solution:
We observe that the given equation is an odd degree reciprocal equation of class two.
By Note 4.4.24(1), 1 is a root of this equation.
Therefore (x – 1) is a factor of x5 – 5x4 + 9x3– 9x2 + 5x – 1.
On dividing this expression by (x – 1), we get
x4 – 4x3 + 5x2 – 4x + 1 as the quotient.
Now we have to solve the equation
x4 – 4x + 5x2 – 4x + 1 = 0
On dividing this equation by x2, we get
x2– 4x+5 –\(\frac{4}{x}+\frac{1}{x^2}\) = 0
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 32

Question 34.
Solve the equation 6x4-35x3+62x2-35x+6=0.
Solution:
We observe that the given equation is an even degree reciprocal equation of class one.
On dividing both sides of the given equation by x2, we get
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 33
Then the above equation reduces to
6 (y2-2) – 35y + 62 = 0
i.e., 6y2 – 35y + 50 = 0
i.e., (2y – 5) (3y – 10) = 0
Hence the roots of
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 34

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 35.
Solve the equation
6x6 – 25x5 + 31x4 -31x2 + 25x – 6 = 0.
Solution:
We observe that the given equation is an even degree reciprocal equation of class two. By Note 4.4.24(2), + 1 and – 1 are the roots of this equation. Hence (x + 1) and (x – 1) are the factors of the given equation.
Let f(x) = 6x6 – 25x5 + 31x4 -31x2 + 25x – 6 = 0
On dividing this expression by (x + 1) and then by (x – 1), we get
f(x) = (x2 -1) (6x4 – 25x3 + 37x2 – 25x + 6)
Now we have to solve the equation
6x4 – 25x3 + 37x2 -25x + 6 = 0
On dividing both sides of this equation by x2, we obtain
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 35
TS Inter 2nd Year Maths 2A Theory of Equations Important Questions 36

 

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 9 Probability Ex 9(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(a)

I.
Question 1.
In the experiment of throwing a die, consider the following events A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3}. Are these events equally likely?
Solution:
If the die is thrown there is a possibility of getting 1 or 2 or 3 or 4 or 5 or 6 on any face.
Hence the events A = {1, 3, 5}, B = {2, 4, 6} and C = {1, 2, 3} are equiprobable since there is no reason to expect one in preference to others.
Hence the events A, B, C are equally likely.

Question 2.
In the experiment of throwing a die, consider the following events A = {1, 3, 5}, B = {2, 4}, C = {6} . Are these events mutually exclusive?
Solution:
The three events A, B, C are mutually exclusive since the occurrence of one of the events prevents the happening of any one of the remaining events.
Since A ∩ B ∩ C = {1, 3, 5} ∩ {2, 4} ∩ {6}
We say that the events are A, B, C are mutually exclusive.

Question 3.
In the experiment of throwing a die, consider the events A = {2, 4, 6}, B = {(3, 6}, C = {1, 5, 6}. Are these events exhaustive?
Solution:
The three events A, B, C are exhaustive if A ∪ B ∪ C = S
A ∪ B ∪ C = {2, 4, 6} ∪ {3, 6} ∪ {15 6}
= {1, 2, 3, 4, 5, 6} = S.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(a)

II.
Question 1.
Give two examples of mutually exclusive and exhaustive events.
Solution:
In tossing a coin there are two exhaustive events Head (H) and Tail (T).
In throwing a die there are six exhaustive events of getting I or 2 or 3 or 4 or 5 or 6.
In tossing a coin either heads comes up or tail but both cannot happen at the same time. These two events are mutually exclusive because happening of one event prevents the happening of the other.
In a well shuffled pack of cards if a card is drawn from 52 cards then getting an ace and getting a king are mutually exclusive events.

Question 2.
Give examples of two events that are neither mutually exclusive nor exhaustive.
Solution:
If a coin is tossed twice or two coins are tossed a time, then the events of getting head or tail are not mutually exclusive nor exhaustive.
Since we get {HH, HT, TH, TT} as events.
From a well shuffled pack of cards if two cards are drawn one after other with replacement, then getting aces on two attempts are not mutually exclusive nor exhaustive.

Question 3.
Give two examples of events that are neither equally likely nor exhaustive.
Solution:
If a die is thrown then the event of getting ‘1’ and the event of getting a prime number are neither equally likely events nor exhaustive events.
In the experiment of throwing a pair of dice then the events
E1 = A sum 7 ( of the numbers that appear on the uppermost faces of the dice ) and
E3 = A sum > 7 ( of the number that appear on the uppermost faces of the dice ) are neither equally likely nor mutually exclusive.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

I.
Question 1.
Find the mean deviation about the mean for the following data.
i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
ii) 3, 6, 10, 4, 9, 10
Solution:
i) Mean of the given data is \(\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
= \(\frac{500}{10}\) = 50
The absolute values of the deviations are \(\left|x_i-\bar{x}\right|\) =12, 20, 2, 10, 8, 5, 13, 4, 4, 6
∴ Mean Deviation about the Mean = \(\frac{\sum_{\mathrm{i}=1}^{10}\left|\mathbf{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{n}\)
= \(\frac{12+20+2+10+8+5+13+4+4+6}{10}\)
= \(\frac{84}{10}\) = 8.4.

ii) Mean of the given data (\(\bar{x}\)) = \(\frac{\sum_{\mathbf{i}=1}^6 x_i}{n}\)
∴ \(\bar{x}\) = \(\frac{3+6+10+4+9+10}{6}=\frac{42}{6}\) = 7
The absolute values of the deviations are |xi – \(\bar{x}\)| = 4, 1, 3, 3, 2, 3
Mean Deviation about the Mean = \(\frac{\sum_{i=1}^6\left|x_i-\bar{x}\right|}{n}\)
= \(\frac{4+1+3+3+2+3}{6}=\frac{16}{6}\) = 2.67.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 2.
Find the mean deviation about the median for the following data.
i) 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
ii) 4, 6, 9, 3, 10, 13, 2
Solution:
i) Expressing the given data in the ascending order, we get 10, 11, 11,12,13, 13, 16, 16, 17, 17, 18
Median (M) of these 11 observations is 13.
The absolute values of deviations are |xi – M| = \(\frac{3+2+2+1+0+0+3+3+4+4+5}{11}\)
= \(\frac{27}{11}\) = 2.45.

ii) Expressing the given data in the ascending order, we get 2, 3, 4, 6, 9, 10, 13.
Median (M) of given data = 6
The absolute values of deviations are |xi – M | = 4, 3, 2, 0, 3, 4, 7
∴ Mean Deviation about the Median = \(\frac{\sum_{\mathrm{i}=1}^7\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{n}}=\frac{4+3+2+0+3+4+7}{7}\)
= \(\frac{23}{7}\) = 3.29.

Question 3.
Find the mean deviation about the mean for the following distribution.
i) TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 1

ii) TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 2
Solution:
i)

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 3

∴ Mean (\(\bar{x}\)) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{534}{45}\) = 11.87

∴ Mean Deviation about the Mean = \(\frac{\sum_{i=1}^4 f_i\left|x_i-\bar{x}\right|}{N}=\frac{31.95}{45}\) = 0.71.

ii)

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 4

∴ Mean (\(\bar{x}\)) = \(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{4000}{80}\) = 50

∴ Mean Deviation about the Mean = \(\frac{\sum_{i=1}^5 f_i\left|x_i-\bar{x}\right|}{N}=\frac{1280}{80}\) = 16.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 4.
Find the mean deviation about the median for the following frequency distribution.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 5

Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 6

Hence N = 26 and \(\frac{N}{2}\) = 13
Median (M) = 7
Median Deviation about Median = \(\frac{\sum_{\mathrm{i}=1}^6 \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{84}{26}\) = 3023.

Note:
We shall identify the observation whose cumulative frequency is equal to or just greater than N/2. This is the median of the data. Here median is “7”.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

II.
Question 1.
Find the mean deviation about the median for the following continuous distribution.
i) TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 7

ii) TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 8
Solution:

i) TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 9

Hence L = 20, \(\frac{N}{2}\) = 25, f1 = 14, f = 14, h = 10
Median (M) = L + \(\left[\frac{\left[\frac{\mathrm{N}}{2}-\mathrm{f}_1\right]}{\mathrm{f}}\right]\) h
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + \(\frac{110}{14}\)
= 20 + 7.86 = 27.86.
∴ Mean Deviation about Median = \(\frac{\sum_{\mathrm{i}=1}^6 \mathrm{t}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}\)
= \(\frac{517.16}{50}\) = 10.34.

ii)

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 10

Here N = 100, \(\frac{N}{2}\) = 50, L = 40, f1 = 32, f = 28, h = 10
Median (M) = L + \(\left\{\frac{\left[\frac{\mathrm{N}}{2}-\mathrm{f}_1\right]}{\mathrm{f}}\right\}\) h
= 40 + \(\frac{50-32}{28}\) × 10
= 40 + \(\frac{180}{28}\)
= 40 + 6.43 = 46.43.
∴ Mean Deviation about Median = \(\frac{\sum_{i=1}^8 f_i\left|x_i-M\right|}{N}=\frac{1428.6}{100}\) = 14.29.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 2.
Find the mean deviation about the mean for the following continuous distribution.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 11

Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 12

Mean (\(\bar{x}\)) = A + \(\frac{\Sigma f_i \mathrm{~d}_{\mathrm{i}}}{\mathrm{N}}\) . h
= 130 + \(\left(\frac{-47}{100}\right)\) . 10
= 130 – 1.7 = 125.3.

∴ Mean Deviation about Mean = \(\frac{\sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|}{N}\)
= \(\frac{1128.8}{100}\) = 11.29.

Question 3.
Find the variance for the discrete data given below.
i) 6, 7, 10, 12, 13, 4, 8, 12
ii) 350, 361, 370, 373, 376, 379, 385, 387, 394, 395.
Solution:
i) Mean (\(\bar{x}\)) = \(\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 13

Variance (σ2) = \(\frac{\sum_{\mathrm{i}=1}^8\left(x_i-\bar{x}\right)^2}{n}=\frac{74}{8}\) = 9.25.

ii)
TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 14

Mean (\(\bar{x}\)) = \(\frac{350+361+370+373+376+379+385+387+394+395}{10}\)
= \(\frac{3770}{10}\) = 377.
Variance (σ2) = \(\frac{\sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2}{n}=\frac{1832}{10}\) = 183.2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 4.
Find the variance and standard deviation of the following frequency distribution.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 15

Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 16

Mean (x) = \(\frac{760}{40}\) = 19
Variance (σ2) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{~N}}=\frac{1736}{40}\) = 43.4
Standard Deviation (σ) = \(\sqrt{43.4}\) = 6.59.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

III.
Question 1.
Find the mean and variance using the step deviation method, of the following tabular data, giving the age distribution of 542 members.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 17

Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 18

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 19

Question 2.
The coefficent of variation of two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Solution:
C.V = \(\frac{\sigma}{\overline{\bar{X}}}\) × 100

i) 60 = \(\frac{21}{\overline{\mathrm{X}}}\) × 100
\(\overline{\mathrm{X}}\) = \(\frac{21 \times 100}{60}\) = 35

ii) 70 = \(\frac{16}{\overline{\mathrm{Y}}}\) × 100
\(\overline{\mathrm{Y}}\) = \(\frac{16 \times 100}{70}\) = 22.857.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 3.
From the prices of shares X and Y given below, for 10 days of trading, find out which share is more stable?

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 20

Solution:
Variance is independent ol change of origin.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 21

V(X) = \(\frac{\Sigma x_i^2}{n}-(\bar{x})^2\)
= \(\frac{360}{10}-\left(\frac{10}{10}\right)^2\)
= 36 – 1 = 35.

V(Y) = \(\frac{\Sigma Y_i^2}{n}-(\bar{Y})^2\)
= \(\frac{290}{10}-\left(\frac{50}{10}\right)^2\)
= 29 – 25 = 4.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 4.
The mean of 5 observations is 4.4. Their variance is 8.24. If three of the observations are 1, 2 and 6. Find the other
two observations.
Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 22

S.D = \(\sqrt{\frac{\Sigma \mathrm{m}^2}{\mathrm{n}}-(\overline{\mathrm{x}})^2}\)
\(\bar{x}\) = 4.4
⇒ 4.4 = \(\frac{1+2+6+x+y}{5}\)
⇒ 9 + x + y = 22
⇒ x + y = 13 …………..(1)
S.D2 = \(\frac{1+4+36+x^2+y^2}{5}\) – (4.4)2
= \(\frac{41+x^2+y^2}{5}\) – 19.36
S.D2 = Variance
Variance = \(\frac{41+x^2+y^2}{5}\) – 19.36
8.24 + 19.36 = \(\frac{41+x^2+y^2}{5}\)
41 + x2 + y2 = 5×27.6
x2 + y2 = 138 – 41
x2 + y2 = 97 …………..(2)
From (1) and (2),
x2 + (13 – x)2 = 97
x2 + 169 + x2 – 26x = 97
2x2 – 26x + 72 = 0
x2 – 13x + 36 = 0
x2 – 9x – 4x + 36 = 0
x (x – 9) – 4 (x – 9) = 0
(x – 9) (x – 4) = 0
x = 4, 9
Put x = 4 in (1)
y = 13 – 4= 9
Put x = 9 in (1)
y = 13 – 9 = 4
∴ If x = 4, then y = 9.
If x = 9, then y = 4.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Question 5.
The arithmetic mean and standard deviation of a set of 9 items are 43 and 5 respectively. If an item of value 63 is added to that set, find the new mean and standard deviation of 10 items set given.
Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 23

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) 24

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 2 De Moivre’s Theorem to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 1.
Simplify \(\frac{(\cos \alpha+i \sin \alpha)^4}{(\sin \beta+i \cos \beta)^8}\)
Solution:
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 1

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 2.
If m,n are integers and x = cos α + i sin α, y = cos β + i sin β then prove that
xm yn + \(\frac{1}{x^m y^n}\) = cos (mα +nβ) and
xm yn – \(\frac{1}{x^m y^n}\) = 2i sin (mα +nβ)
Solution:
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 2
Question 3.
If n is a positive Integer, show that \((1+i)^n+(1-i)^n=2^{\frac{n+2}{2}} \cos \left(\frac{n \pi}{4}\right)\)
Solution:
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 3
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 4

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 4.
If n is an Integer then show that
(1 + cos θ + i sin θ)n + (1 + cos θ – i sin θ)n \(=2^{n+1} \cos ^n\left(\frac{\theta}{2}\right) \cos \left(\frac{n \theta}{2}\right)\)
Solution:
L.H.S
(1 + cos θ + i sin θ)n + (1 + cos θ – i sin θ)n
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 5

Question 5.
If cos α+cos β + cos γ = 0 = sin α + sin β + sin γ, Prove that cos2 α +cos2 β +cos γ = \(\frac{3}{2}\) sin2 α + sin2 β + sin2 γ.
Solution:
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 6
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 7

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 6.
Find all the values of \((\sqrt{3}+i)^{1 / 4}\)
Solution:
The modulus amplitude form of
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 8
Question 7.
Find all the roots of the equation
x11 – x7 + x4 -1 = 0
Solution:
x11 – x7 + x4 -1  = x7(x4-1) +1 (x4– 1) = (x4-1)(x7. 1)
Therefore the roots of the given equations are precisely the roots of unity and 7th roots of – 1.
They are cis = \(\frac{2 \mathrm{k} \pi}{4} \) = cis \(\frac{\mathrm{k} \pi}{4}\) k∈{0,1,2,3} and
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 9

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 8.
If 1, ω, ω2 are the cube roots of unity, prove that
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 13
Solution:
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 10
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 11

TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 9.
If α, β are the roots of the equation x2 + x + 1 = 0 then prove that α4 + β4 + α-1 = β-1
Solution:
Since α, β are the complex cube roots of unity,
we may take α = ω, β = ω2
TS Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions 12

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

Resolve the following fractions into partial fractions.

I.
Question 1.
\(\frac{2 x+3}{(x+1)(x-3)}\)
Solution:
Let \(\frac{2 x+3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}\)
⇒ A (x – 3) + B (x + 1) = 2x – 3 …………..(1)
Substituting x = 3 in (1),
weget 4B = 9 .
⇒ B = \(\frac{9}{4}\)
Substituting x = – 1 in (1),
we get – 4A = 1
⇒ A = \(\frac{-1}{4}\)
∴ \(\frac{2 x+3}{(x+1)(x-3)}=\frac{9}{4(x-3)}-\frac{1}{4(x+1)}\).

Question 2.
\(\frac{5 x+6}{(2+x)(1-x)}\)
Solution:
Let \(\frac{5 x+6}{(2+x)(1-x)}=\frac{A}{2+x}+\frac{B}{1-x}\)
⇒ A (1 – x) + B (2 + x) = 5x + 6 ……………..(1)
Substituting x = 1 in (I),
weget 3B = 11
⇒ B = \(\frac{11}{3}\)
Substituting x = – 2 in (1),
we get 3A = – 4
⇒ A = \(\frac{-4}{3}\)
∴ \(\frac{5 x+6}{(2+x)(1-x)}=\frac{11}{3(1-x)}-\frac{4}{3(2+x)}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

II.
Question 1.
\(\frac{3 x+7}{x^2-3 x+2}\)
Solution:
We know that
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-2)(x-1)}\)
Let \(\frac{3 x+7}{(x-2)(x-1)}=\frac{A}{x-2}+\frac{B}{x-1}\)
⇒ A (x – 1) + B(x – 2) = 3x + 7 …………..(1)
SubstitutIng x = 2 in (1)
we get A = 13
Substituting x = 1 in (1)
we get – B = 10 i.e., B = – 10
∴ \(\frac{3 x+7}{x^2-3 x+2}=\frac{13}{x-2}-\frac{10}{x-1}\)

Question 2.
\(\frac{x+4}{\left(x^2-4\right)(x+1)}\)
Solution:
We know that
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{x+4}{(x-2)(x+2)(x+1)}\)
Let \(\frac{x+4}{(x-2)(x+2)(x+1)}\) = \(\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{x+1}\)
A (x + 2) (x + 1) + B (x – 2) (x + 1) + C (x – 2) (x + 2) = x + 4 …………..(1)
Substituting x = 2 in (1), we have
12A = 6
A = \(\frac{1}{2}\)
Substituting x = – 2 in (1), we have
4B = 2
⇒ B = \(\frac{1}{2}\)
Substituting x = – 1 in (1), we have
– 3C = 3
⇒ C = – 1
∴ \(\frac{x+4}{\left(x^2-4\right)(x+1)}\) = \(\frac{1}{2(x-2)}+\frac{1}{2(x+2)}-\frac{1}{x+1}\)

Question 3.
\(\frac{2 x^2+2 x+1}{x^3+x^2}\)
Solution:
We know that
\(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}\)
Let \(\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
⇒ Ax (x + 1) + B (x + 1) + Cx2 = 2x + 2x + 1
Substituting x = 0 in (1), we have B = 1
Substituting x = – 1 in (1), we have C = 1
Equating coefficient of x2 on both sides in (1), we have
A + C = 2
⇒ A = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

Question 4.
\(\frac{2 x+3}{(x-1)^3}\)
Solution:
Let \(\frac{2 x+3}{(x-1)^3}\) = \(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}\)
⇒ A(x – 1)2 + B(x – 1) + C = 2x + 3 ……………..(1)
Substituting x = 1 in (1).
we get C = 5
Equating coefficient of x2 on both sides in (1)
We get A = 0
Equating coefficient of x on both sides in (1)
We get – 2A + B = 2
⇒ B = 2.

Alternate method:
Let x – 1 = y

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) 1

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

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) 2

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

III.
Question 1.
\(\frac{x^2-x+1}{(x+1)(x-1)^2}\)
Solution:
Let \(\frac{x^2-x+1}{(x+1)(x-1)^2}\) = \(\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
A (x – 1)2 + B (x + 1) (x – 1) + C (x + 1) = x2 – x + 1 ………..(1)
Substituting x = 1 in (1), we get
2C = 1
⇒ C = \(\frac{1}{2}\)
Substituting x = – 1 in (1), we get
4A = 3
⇒ A = \(\frac{3}{4}\)
Equating coefficient of x2 on both sides in (1)
We get A + B = 1
\(\frac{3}{4}\) + B = 1
⇒ B = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}\) = \(\frac{3}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1}{2(x-1)^2}\)

Question 2.
\(\frac{9}{(x-1)(x+2)^2}\)
Solution:
Let \(\frac{9}{(x-1)(x+2)^2}\) = \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
⇒ A (x + 2)2 + B (x – 1) (x + 2) + C (x – 1) = 9 …………….(1)
Substituting x = 1 in (1), we get
9A = 9
⇒ A = 1
Substituting x = – 2 in (1), we get
– 3C = 9
⇒ C = – 3
Equating coefficient of x2 on both sides in (1),
we get A + B = 0
⇒ B = – 1
∴ \(\frac{9}{(x-1)(x+2)^2}\) = \(\frac{1}{x-1}-\frac{1}{(x+2)}-\frac{3}{(x+2)^2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

Question 3.
\(\frac{1}{(1-2 x)^2(1-3 x)}\)
Solution:
Let \(\frac{1}{(1-2 x)^2(1-3 x)}\) = \(\frac{A}{(1-3 x)}+\frac{B}{(1-2 x)}+\frac{C}{(1-2 x)^2}\)
⇒ A (1 – 2x)2 + B (1 – 3x) (1 – 2x) + C (1 – 3x) = 1 …………..(1)
Substituting x = \(\frac{1}{2}\) in (1),
we get \(\frac{-C}{2}\) = 1
⇒ C = – 2
Substituting x = \(\frac{1}{3}\) in (1),
we get \(\frac{\mathrm{A}}{9}\) = 1
⇒ A = 9
Substituting x = 0 in (1),
We get A + B + C = 1
⇒ 9 + B – 2 = 1
⇒ B = – 6
∴ \(\frac{1}{(1-2 x)^2(1-3 x)}\) = \(\frac{9}{1-3 x}-\frac{6}{1-2 x}-\frac{2}{(1-2 x)^2}\)

Question 4.
\(\frac{1}{x^3(x+a)}\)
Sol.
Let \(\frac{1}{x^3(x+a)}=\frac{A}{x+a}+\frac{B}{x}+\frac{C}{x^2}+\frac{D}{x^3}\)
⇒ Ax3 + Bx2 (x + a) + Cx (x + a) + D (x + a) = 1 …………(1)
Substituting x = – a in (1),
we get – a3A = 1
⇒ A = \(\frac{-1}{a^3}\)
Equating coefficient of x3 on both sides,
we get A + B = 0
⇒ B = \(\frac{-1}{a^3}\)
Substituting x = 0 in (1),
we get aD = 1
Equating coefficient of x on both sides,
we get aC + D = 0
⇒ aC + \(\frac{1}{a}\) = 0
⇒ C = \(\frac{-1}{a^2}\)
∴ \(\frac{1}{x^3(x+a)}\) = \(\frac{-1}{a^3(x+a)}+\frac{1}{a^3 x}-\frac{1}{a^2 x^2}+\frac{1}{a x^3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a)

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

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) 3

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

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(a) 4