Class 12

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Ex 8(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Exercise 8(a)

Question 1.
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules.
(i) s = 2t² + 3t + 1
Solution:
s = 2t² + 3t + 1

∴ Velocity is 11 units/sec and acceleration is 4 units/sec².

(ii) s = √t +1
Solution:
s = √t +1

(iii) s = \(\frac{3}{2 t+1}\)
Solution:

(iv) s = t³ – 6t² + 15t + 12
Solution:
s = t³ – 6t² + 15t + 12

∴ Velocity is 3 units/sec and acceleration is 0.

Question 2.
The sides of an equilateral triangle are increasing at the rate of √3 cm/sec. Find the rate at which the area of the triangle is increasing when the side is 4 cm long.
Solution:
Let x be the lenght of each side of an equilateral triangle.

∴ Area of the triangle is increasing at the rate of 6 cm²/sec

Question 3.
Find the rate at which the volume of a spherical balloon will increase when its radius is 2 metres if the rate of increase of its radius is 0.3 m/min.
Solution:
Let r be the radius of a spherical balloon.

∴ The volume increase at the rate of 4.8π m³/min.

Question 4.
The surface area of a cube is decreasing at the rate of 15 sq. cm/sec. Find the rate at which its edge is decreasing when the length of the edge is 5 cm.
Solution:
Let s be the surface area of a cube.
Let x be the length of each side of the cube.

∴ The edge is decreasing at the rate of 0.25 cm/sec

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(m) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(m)

Question 1.
Verify Rolle’s theorem for the function
f(x) = x (x – 2)², 0 ≤ x ≤ 2.
Solution:
f(x) = x (x – 2)², 0 ≤ x ≤ 2
Here a = 0, b = 2
f(x) is a polynomial function hence it is continuous and also differentiable.
∴ f is continuous on [0. 2]
f is differentiable on (0, 2)
f(0) = 0 = f(2)
Thus conditions of Rolle’s theorem are satisfied.
f'(x) = (x – 2)² + 2x (x – 2)
= (x – 2) (x – 2 + 2x)
= (x – 2) (3x – 2)
f'(x)= 0 ⇒ x = 2, x = \(\frac{2}{3}\)
But x = 2 ∉ (0, 2). Thus c = \(\frac{2}{3}\) such that f'(c) = 0
Thus Rolle’s theorem is verified.

Question 2.
Examine if Rolle’s theorem is applicable to the following functions:
(i) f(x) = |x| on [-1, 1]
Solution:
f(x) = |x| on [-1, 1]
As f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)
We have Rolle’s theorem is not applicable.

(ii) f(x) = [x] on [-1, 1]
Solution:
f(x) = [x] on [-1, 1]
f(x) = [x] is not continuous at 0 ∈ [-1, 1]
Rolle’s theorem is not applicable.

(iii) f(x) = sin x on [0, π]
Solution:
f(x) = sin x on [0, π]
f is a trigonometric function hence continuous and differentiable on its domain.
∴ f is continuous on [0, π]
f is differentiable on (0, π]
f(0) = f(π)
Thus Rolle’s theorem is applicable for f(x) = sin x on [0, π]

(iv) f(x) = cot x on [0, π]
Solution:
f(x) = cot x on [0, π]
Clearly cot (0) and cot (π] are not defined hence Rolle’s theorem is not applicable.

Question 3.
Verify Lagrange’s Mean-Value theorem for
F(x) = x³ – 2x² – x + 3 on [1, 2]
Solution:
f(x) = x³ – 2x² – x + 3 on [1, 2]
f is a polynomial function hence continuous as well as differentiable.
∴ f is continuous on [1, 2]
f is differentiable on (1, 2)
Thus Largange’s mean value theorem is applicable.
Now f(1) = 1 – 2 – 1 + 3 = 1
f(2) = 8 – 8 – 2 + 3 = 1
∴ f(x) = 2x² – 4x – 1

Thus Lagrange’s mean value theorem is verified.

Question 4.
Test if Lagrange’s mean value theorem holds for the functions given in question no. 2.
Solution:
(i) f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)
Thus Lagrange’s mean value theorem, does not hold.

(ii) f(x) = [x] is discontinuous at 0 ∈ [-1, 1]
Thus Lagrange’s mean value theorem is not applicable.

(iii) f(x) = sin x is a trigonometric function, which is continuous as well as differentiable in its domain.
∴ f is continuous on [0, π]
f is differentiable on (0, π)
Thus conditions of Lagrange’s mean value theorem are satisfied.
Hence mean value theorem is applicable.

(iv) f(x) = cot x
Which is undefined x = 0 and x = π
Thus Lagrange’s mean value theorem is not applicable.

Question 5.
(Not for examination) Verify Cauchy’s mean value theorem for the functions x² and x³
in [1, 2].
Solution:
Let f(x) = x², and g(x) = x³ on [1, 2]
Both f and g are polynomial functions, hence continuous and differentiable.
∴ f and g are continuous on [1, 2]
f and g are differentiable on (1, 2)
g'(x) = 3x² ≠ 0 ∀ x ∈ (1, 2)
Thus conditions of Cauchy’s mean value theorem are satisfied.
Now f(1) = 1, f(2) = 4, g(1) = 1, g(2) = 8
f'(x) = 2, and g'(x) = 3x²

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(i) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(i)

Differentiate.
Question 1.
√x w.r.t x².
Solution:

Question 2.
sin x. w.r.t. cot x.
Solution:
Let y = sin x and z = cot x

Question 3.
\(\frac{1-\cos x}{1+\cos x}\) w.r.t \(\frac{1-\sin x}{1+\sin x}\)
Solution:

Question 4.
tan-1 x w.r.t. tan^-1 \( \sqrt{1+x^2} \)
Solution:

Question 5.
sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\)
Solution:
Let y = sin^-1 \(\frac{2 x}{1+x^2}\) and z = cos^-1 \(\frac{1-x^2}{1+x^2}\)
Then y = 2 tan^-1 x and z = 2 tan^-1 x
So y = z
∴ \(\frac{d y}{d z}\) = 1.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(a)

Question 1.
Examine the continuity of the following functions at indicated points.
(i) f(x) = \(\left\{\begin{array}{cl}
\frac{x^2-a^2}{x-a} & \text { if } x \neq a \\
a & \text { if } x=a
\end{array}\right.\) at x = a
Solution:

(ii) f(x) = \(\left\{\begin{aligned}
\frac{\sin 2 x}{x} & \text { if } x \neq 0 \\
2 & \text { if } x=0
\end{aligned}\right.\) at x = 0
Solution:

(iii) f(x) = \(\begin{cases}(1+2 x)^{\frac{1}{x}} & \text { if } x \neq 0 \\ e^2 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(iv) f(x) = \(\left\{\begin{array}{l}
x \sin \frac{1}{x} \text { if } x \neq 0 \\
0
\end{array}\right.\) at x = 0
Solution:

(v) f(x) = \(\left\{\begin{array}{l}
\frac{x^2-1}{x-1} \text { if } x \neq 1 \\
2
\end{array}\right.\) at x = 1
Solution:

(vi) f(x) = \(\begin{cases}\sin \frac{1}{x} & \text { if } x \neq a \\ 0 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(vii) f(x) = [3x + 11] at x = –\(\frac{11}{3}\)
Solution:

(viii) f(x) = \(\left\{\begin{array}{l}
\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} \text { if } x \neq 0 \\
0
\end{array}\right.\) at x= 0
Solution:

(ix) f(x) = \(\left\{\begin{array}{l}
\frac{1}{x+[x]} \text { if } x<0 \\
-1 \quad \text { if } x \geq 0
\end{array}\right.\) at x = 0
Solution:

because [- h]is the greatest integer not exceeding – h
and so [- h ] = – 1
As L.H.L. = R.H.L. = f(0)
f(x) is cntinuous at x = 0.

(x) f(x) = \(\begin{cases}\frac{|x|}{x} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(xi) f(x) = \(\left\{\begin{array}{l}
2 x+1 \text { if } x \leq 0 \\
x \quad \text { if } 0<x<1 \\
2 x-1 \text { if } x \geq 1
\end{array}\right.\) at x = 0, 1
Solution:

(xii) f(x) = \(\left\{\begin{array}{l}
\frac{1}{e^{\frac{1}{x}}-1} \text { if } x>0 \\
0
\end{array} \text { if } x \leq 0\right.\) at x = 0
Solution:

(xiii) f(x) = sin\(\frac{\pi[x]}{2}\) at x = 0
Solution:

(xiv) f(x) = \(\frac{g(x)-g(1)}{x-1}\) at x = 1
Solution:
g(x) = |x – 1|
Then g(1) = |1 – 11| = 0
Now f(1) = \(\frac{g(1)-f(1)}{1-1}\) = 0/0
which we cannot determine.
Hence f(x) is discontinuous at x = 1.

Question 2.
If a function is continuous at x = a, then find
(i) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)+f(a-h)\}\)
(ii) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)-f(a-h)\}\)
Solution:

Question 3.
Find the value ofa such that the function f defined by \(\begin{cases}\frac{\sin a x}{\sin x} & \text { if } x \neq 0 \\ \frac{1}{a} & \text { if } x=0\end{cases}\)
is continuous at x = 0.
Solution:

Question 4.
If f(x) = \(\left\{\begin{array}{l}
a x^2+b \text { if } x<1 \\
1 \quad \text { if } x=1 \\
2 a x-b \text { if } x>1
\end{array}\right.\)
is continuous at x = 1, then find a and b.
Solution:
Let f(x) be continuous at x = 1
Then L.H.L. = R.H.L. = f(1)

Question 5.
Show that sin x is continuous for every real x.
Solution:
Let f(x) = sin x
Consider the point x = a, where ‘a’ is any real number.
Then f(a) = sin a

Thus L.H.L. = R.H.L. = f(a)
Hence f(x) = sin x is continuous for every real x.
(Proved)

Question 6.
Show that the function f defined by \(\left\{\begin{array}{l}
1 \text { if } x \text { is rational } \\
0 \text { if } x \text { is irrational }
\end{array}\right.\) is discontinuous ∀ ≠ 0 ∈ R.
Solution:
Consider any real point x = a
If a is rational then f(a) = 1.
Again lim_x→a+f(x) = lim_h→0f(a + h)
which does not exist because a + h may be rational or irrational
Similarly lim_x→a-f(x) does not exist.
Thus f(x) is discontinuous at any rational point. Similarly we can show that f(x) is discontinuous at any irrational point.
Hence f(x) is discontinuous for all x ∈ R
(Proved)

Question 7.
Show that the function f defined by f(x) = \(f(x)=\left\{\begin{array}{l}
x \text { if } x \text { is rational } \\
-x \text { if } x \text { is irrational }
\end{array}\right.\)
is continuous at x = 0 and discontinuous ∀ x ≠ 0 ∈ R.
Solution:
f(0) = 0
L.H.L. = lim_x→0f(x) = lim_h→0f(-h)
= \(\lim _{h \rightarrow 0} \begin{cases}-h & \text { if } h \text { is rational } \\ h & \text { if } h \text { is irrational }\end{cases}\) = 0
Similarly R.H.L. = 0
Thus L.H.L. = R.H.L. = f(0)
Hence f(x) is continuous at x = 0.
We can easily show that f(x) is discontinuous at all real points x ≠ 0.

Question 8.
Show that the function f defined by
f(x) = \(\left\{\begin{array}{l}
x \text { if } x \text { is rational } \\
0 \text { if } x \text { is irrational }
\end{array}\right.\)
is discontinuous everywhere except at x = 0.
Solution:
f(0) = 0
L.H.L. = lim_h→0f(-h)
= lim_h→0\(\begin{cases}-h & \text { if }-h \text { is rational } \\ 0 & \text { if }-h \text { is irrational }\end{cases}\)
Similarly R.H.L. = 0
Thus L.H.L. = R.H.L = f(0)
Hence f(x) is continuous at x = 0.
Let a be any real number except 0.
If a is rational then f(a) = a.
L.H.L. = lim_h→0f(a – h) which does not exist because a – h may be rational or may be irrational.
Similarly R.H.L. does not exist.
Thus f(x) is discontinuous at any rational point x – a ≠ 0.
Similarly f(x) is discontinuous at any irrational point.
Hence f(x) is discontinuous everywhere except at x = 0.
(Proved)

Question 9.
Show that f(x) = \(\begin{cases}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}\) is continuous at x = 0.
Solution:
Refer to No. 1(iv) of Exercise – 7(a).

Question 10.
Prove that e^x – 2 = 0 has a solution between 0 and 1. [Hints: Use continuity of e^x– 2 and fact – 2]
Solution:

∴ f(x) is continuous in [0, 1]
f(0). f(1) = (-1) (e – 2) < 0
∴ f(x) has a zero between 0 and 1
i.e. e^x – 2 = 0 has a solution between 0 and 1

Question 11.
So that x⁵ + x +1 = 0 for some value of x between -1 and 0.
Solution:
Let f(x) = x⁵ + x + 1 and any a ∈ (-1, 0)
f(a) = a⁵ + a + 1

= -1 = f(-1)
∴ f is continuous on [-1, 0]
But f(-1) f(0) = 1 × -1 < 0
∴ f has a zero between -1 and 0
⇒ x5 + x + 1 = 0 for some value of x between -1 and 0.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(c)

Question 1.
There are 3 bags B₁, B₂ and B₃ having respectively 4 white, 5 black; 3 white, 5 black and 5 white, 2 black balls. A bag is chosen at random and a ball is drawn from it. Find the probability that the ball is white.
Solution:
Let
E₁ = The selected bag is B₁,
E₂ = The selected bag is B₂,
E₃ = The selected bag is B₃.
A = The ball drawn is white.

Question 2.
There are 25 girls and 15 boys in class XI and 30 boys and 20 girls in class XII. If a student chosen from a class, selected at random, happens to be a boy, find the probability that he has been chosen from class XII.
Solution:
Let
E₁ = The student is choosen from class XI.
E₂ = The student is choosen from class XII.
A = The student is a boy.

Question 3.
Out of the adult population in a village 50% are farmers, 30% do business and 20% are service holders. It is known that 10% of the farmers, 20% of the business holders and 50% of service holders are above poverty line. What is the probability that a member chosen from any one of the adult population, selected at random, is above poverty line?
Solution:
Let
E₁ = The person is a farmer.
E₂ = The person is a businessman.
E₃ = The person is a service holder.
A = The person is above poverty line.

Question 4.
Take the data of question number 3. If a member from any one of the adult population of the village, chosen at random, happens to be above poverty line, then estimate the probability that he is a farmer.
Solution:
P (a farmer / he is above poverty line)
= P (E₁ | A)

Question 5.
From a survey conducted in a cancer hospital it is found that 10% of the patients were alcoholics, 30% chew gutka and 40% have no specific carcinogenic habits. If cancer strikes 80% of the smokers, 70% of alcoholics, 50% of gutka chewers and 10% of the nonspecific, then estimates the probability that a cancer patient chosen from any one of the above types, selected at random,
(i) is a smoker
(ii) is alcoholic
(iii) chews gutka
(iv) has no specific carcinogenic habits.
Solution:
The question should be modified as from a survey conducted in a cancer hospital, 10% patients are smokers, 20% alcoholics, 30% chew gutka and 40% have no specific carcinogenic habits. If cancer strikes 80% of the smokers, 70% of alcoholic, 50% of gutka chewers and 10% of non specific, then estimate the probability that a cancer patient chosen from any one of the following types selected at random
(i) is a smoker
(ii) is alcoholic
(iii) chews gutka
(iv) has no specific carcinogenic habits.
Let
E₁ = The person is a smoker.
E₂ = The person is alcoholic.
E₃ = The person chew Gutka.
E₄ = The person have no specific habits.
A = The person is a cancer patient.
According to the question

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(b)

Question 1.
A bag contains 5 white and 3 black marbles and a second bag contains 3 white and 4 black marbles. A bag is selected at random and a marble is drawn from it. Find the probability that it is white. Assume that either bag can be chosen with the same probability.
Solution:
A bag contains 5 white and 3 black marbles and a second bag contains 3 white and 4 black marbles. A bag is selected at random and a marble is drawn from it.
Let W₁ be the event that 1st bag is choosen and a white marble is drawn and let W₂ be the event that 2nd bag is choosen and a white marble is drawn and these two events are mutually exclusive.

Question 2.
A bag contains 5 white and 3 black balls; a second bag contains 4 white and 5 black balls; a third bag contains 3 white and 6 black balls. A bag is selected at random and a ball is drawn. Find the probability that the ball is black.
(i) Do the problem assuming that the probability of choosing each bag is same.
(ii) Do the problem assuming that the probability of choosing the first bag is twice as much as choosing the second bag, which is twice as much as choosing the third bag.
Solution:
A bag contains 5 white and 3 black balls, a 2nd bag contains 4 white and 5 black balls, a 3rd bag contains 3 white and 6 black balls. A bag is selected at random and a ball is drawn.

(i) Let B₁, B₂, B₃ be the events that 1st bag is choosen and a black ball is draw. 2nd bag is choosen and a black ball is drawn, 3rd bag is drawn and a black ball is drawn. These events are mutually exclusive.
Probability of drawing a black ball
= P(B₁) + P(B₂) + P(B₃)
= \(\frac{1}{3}\) × \(\frac{3}{8}\) + \(\frac{1}{3}\) × \(\frac{5}{9}\) + \(\frac{1}{3}\) × \(\frac{6}{9}\) = \(\frac{115}{216}\)

(ii) Let the probability of choosing 1st bag be 4x. The probability of choosing the 2nd bag is 2x and that of 3rd bag is x.
It is obvious that probability of choosing 3 bags = 1
4x + 2x + x = 1 or, 7x = 1.
x = \(\frac{1}{7}\)
Probability of drawing a black ball
= \(\frac{4}{7}\) × \(\frac{3}{8}\) + \(\frac{2}{7}\) × \(\frac{5}{9}\) + \(\frac{1}{7}\) × \(\frac{6}{9}\)
= \(\frac{108+80+48}{8 \times 9 \times 7}\) = \(\frac{236}{8 \times 9 \times 7}\) = \(\frac{59}{126}\)

Question 3.
A and B play a game by alternately throwing a pair of dice. One who throws 8 wins the game. If A starts the game, find their chances of winning.
Solution:
A and B play a game by alternately throwing a pair of dice. One who throws 8 wins the game. A starts the game.
We can obtain 8 as follows:
{(6, 2), (5, 3), (4, 4) (3, 5), (6, 2)}.
∴ |S| = 6² = 36
∴ P(B) = \(\frac{5}{36}\)
⇒ P(not 8) = 1 – \(\frac{5}{36}\) = \(\frac{31}{36}\)
Since A starts the game, A can win the following situations.
(i) A throws 8
(ii) A does not throws, B does not throw 8, A throws 8,
(iii) A does not throw 8, B does not throw 8, A does not throw 8, B does not throw 8, A throws 8, etc.

Question 4.
A, B, C play a game by throwing a pair of dice in that order. One who gets 8 wins the game. If A starts the game, find their chances of winning.
Solution:
A, B, C play a game by throwing a pair of dice in that order. One who gets 8 wins the game and A starts the game.
P(B) = \(\frac{5}{36}\), P(not 8) = \(\frac{31}{36}\)

If A starts the game, then
(i) A throws 8.
(ii) A does not throw 8, B does not throw 8, C does not throw 8, A throws 8.
(iii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B does not throw 8, C does not throw 8, A throws 8, etc.

Similarly, If B wins the game, then
(i) A does not throw 8, B throw 8.
(ii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B throws 8.
(iii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B throws 8, etc.

Question 5.
There are 6 white and 4 black balls in a bag. If four are drawn successively (and not replaced), find the probability that they are alternately of different colour.
Solution:
There are 6 white and 4 black balls in a bag. Four balls are drawn without replacement. Let W and B denotes the white and black ball. There are two mutually exclusive cases WBWB and BWBW.

Question 6.
Five boys and four girls randomly stand in a line. Find the probability that no two girls come together.
Solution:
Five boys and 4 girls randomly stand in a line such that no two girls come together.
|S| = 9!

The 4 girls can stand in 6 positions in 6P4 ways. Further 5 boys can stand in 5! ways.
Probability that they will stand in a line such that no two girls come together.
= \(\frac{5 ! \times{ }^6 P_4}{9 !}\) = \(\frac{5}{42}\)

Question 7.
If you throw a pair of dice n times, find the probability of getting at least one doublet. [When you get identical members you call it a doublet. You can get a double in six ways: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6); thus the probability of getting a doublet is \(\frac{6}{36}\) = \(\frac{1}{6}\), so that the probability of not getting a doublet in one throw is \(\frac{5}{6}\)].
Solution:
A pair of dice is thrown n times. We get
the doublet as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Probability of getting a doublet in one throw
= \(\frac{6}{36}\) = \(\frac{1}{6}\)
Probability of not getting a doublet
= 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
If a pair of dice is thrown n-times, the probability of not getting a doublet
= \(\left(\frac{5}{6}\right)^n\)
Probability of getting atleast one doublet
= 1 – \(\left(\frac{5}{6}\right)^n\)

Question 8.
Suppose that the probability that your alarm goes off in the morning is 0.9. If the alarm goes off, the probability is 0.8 that you attend your 8 a.m. class. If the alarm does not go off, the probability that you make your 8 a.m. class is 0.5. Find the probability that you make your 8 a.m. class.
Solution:
Let A be the event that my alarm goes off and let B be the event that I make my 8 a. m. class.
Since S = a ∪ A’, B = (B ∩ A) ∪ (B ∩ A’)
Where B ∩ A and B ∩ A’ are mutually
exclusive events.
P(B) = P (B ∩ A) + P (B ∩ A’)
= P(A). P(\(\frac{B}{A}\)) + P(A’). P(\(\frac{\mathrm{B}}{\mathrm{A}^{\prime}}\))
= 0.9 × 0.8 + 0.1 × 0.5 = 0.77

Question 9.
If a fair coin is tossed 6 times, find the probability that you get just one head.
Solution:
A fair coin is tossed 6 times.
∴ |S| = 2⁶
The six mutually exclusive events are
HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH.
Probability of getting just one head = \(\frac{6}{2^6}\)

Question 10.
Can you generalize this situation? If a fair coin is tossed six times, find the probability of getting exactly 2 heads.
Solution:
A fair coin is tossed 6 times. Let A be the event of getting exactly 2 heads.
∴ |A| = ⁶C₂ = 15
∴ P(A) = \(\frac{15}{2^6}\)
Yes we can generalize the situation, i.e., if a fair coin is tossed n-times, then probability of getting exactly 2 heads
= \(\frac{{ }^n \mathrm{C}_2}{2^n}\) = \(\frac{{ }^6 C_2}{2^6}\)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(a)

Question 1.
Evaluate the following determinants.
(i) \(\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\) = 3 – 2 = 1

(ii) \(\left|\begin{array}{ll}
2 & -3 \\
1 & -4
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
2 & -3 \\
1 & -4
\end{array}\right|\) = -8 + 3 = -5

(iii) \(\left|\begin{array}{ll}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\) = sec² θ – tan² θ = 1

(iv) \(\left|\begin{array}{ll}
0 & x \\
2 & 0
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
0 & x \\
2 & 0
\end{array}\right|\) = 0 – 2x = -2x

(v) \(\left|\begin{array}{cc}
1 & \omega \\
-\omega & \omega
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
1 & \omega \\
-\omega & \omega
\end{array}\right|\) = ω + ω² = -1

(vi) \(\left|\begin{array}{cc}
4 & -1 \\
3 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
4 & -1 \\
3 & 2
\end{array}\right|\) = 8 + 3 = 11

(vii) \(\left|\begin{array}{ll}
\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\) = cos² θ – sin² θ = cos 2θ

(viii) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\) = 0
as the rows are identical.

(ix) \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = 1\(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1

(x) \(\left|\begin{array}{ccc}
2 & 3 & 1 \\
0 & 0 & 0 \\
-1 & 2 & 0
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
2 & 3 & 1 \\
0 & 0 & 0 \\
-1 & 2 & 0
\end{array}\right|\) = 0
as all the entries in the 2nd row are zero.

(xi) \(\left|\begin{array}{ccc}
1 & x & y \\
0 & \sin x & \sin y \\
0 & \cos x & \cos y
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & x & y \\
0 & \sin x & \sin y \\
0 & \cos x & \cos y
\end{array}\right|\) = 1\(\left|\begin{array}{cc}
\sin x & \sin y \\
\cos x & \cos y
\end{array}\right|\)
= sin x cos y – cos x sin y = sin (x – y)

(xii) \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\) = 0 (∵ R₁ = R₂)

(xiii) \(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\)
= 2\(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\) = 0 (∵ C₁ = C₂)

(xiv) \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\)

(xv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3
\end{array}\right|\) = 0 (∵ C₁ = C₂)

(xvi) \(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
= (-6) \(\left|\begin{array}{cc}
-5 & 7 \\
8 & 11
\end{array}\right|\) = = (-6) (- 55 – 56)
= (-6) (-111) = 666

(xvii) \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 5 \\
4 & 1 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 5 \\
4 & 1 & 3
\end{array}\right|\)
= 1 \(\left|\begin{array}{ll}
3 & 5 \\
1 & 3
\end{array}\right|\) = 9 – 5 = 4

(xviii) \(\left|\begin{array}{ccc}
-18 & 17 & 19 \\
3 & 0 & 0 \\
-14 & 5 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
-18 & 17 & 19 \\
3 & 0 & 0 \\
-14 & 5 & 2
\end{array}\right|\)
= -3 \(\left|\begin{array}{cc}
17 & 19 \\
5 & 2
\end{array}\right|\)
(Expanding along 2nd row)
= – 3 (34 – 95)
= (-3) (-61) = 183

Question 2.
State true or false.
(i) If the first and second rows of a determinant be interchanged then the sign of the determinant is changed.
Solution:
True

(ii) If first and third rows of a determinant be interchanged then the sign of the determinent does not change.
Solution:
False

(iii) If in a third order determinant first row be changed to second column. Second row to 1st column and third row to third column, then the value of the determinant does not change.
Solution:
False

(iv) A row and a column of a determinant can have two or more common elements.
Solution:
False

(v) The minor and the co-factor of the element a₃₂ of a determinant of third order are equal.
Solution:
False

(vi) \(\left|\begin{array}{lll}
3 & 1 & 3 \\
0 & 4 & 0 \\
1 & 3 & 1
\end{array}\right|\) = 0
Solution:
True

(vii) \(\left|\begin{array}{lll}
6 & 4 & 2 \\
4 & 0 & 7 \\
5 & 3 & 4
\end{array}\right|\) = \(\left|\begin{array}{lll}
6 & 4 & 5 \\
4 & 0 & 3 \\
2 & 7 & 3
\end{array}\right|\)
Solution:
True

(viii) \(\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 7 \\
1 & 2 & 3
\end{array}\right|\) = \(\left|\begin{array}{lll}
4 & 2 & 3 \\
7 & 5 & 6 \\
3 & 1 & 2
\end{array}\right|\)
Solution:
True

Question 3.
Fill in the blanks with appropriate answer from the brackes.
(i) The value of \(\left|\begin{array}{ccc}
0 & 8 & 0 \\
25 & 520 & 25 \\
1 & 410 & 0
\end{array}\right|\) = _______. (0, 25, 200, -250)
Solution:
200

(ii) If ω is the cube root of unity, then \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\) = _______. (1, 0, ω, ω²)
Solution:
0

(iii) The value of the determinant \(\left|\begin{array}{lll}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\) = _______. (a + b – c, (a + b + c)², 0, 1 + a + b + c)
Solution:
0

(iv) If \(\left|\begin{array}{lll}
a & b & c \\
b & a & b \\
x & b & c
\end{array}\right|\) = 0, then x = _______. (a, b, c, a + b + c)
Solution:
a

(v) \(\left|\begin{array}{lll}
a_1+a_2 & a_3+a_4 & a_5 \\
b_1+b_2 & b_3+b_4 & b_5 \\
c_1+c_2 & c_3+c_4 & c_5
\end{array}\right|\) can be expressed at the most as _______, different 3rd order determinants. (1, 2, 3, 4)
Solution:
4

(vi) Minimum value of \(\left|\begin{array}{cc}
\sin x & \cos x \\
-\cos x & 1+\sin x
\end{array}\right|\) is _______. (-1, 0, 1, 2)
Solution:
0

(vii) The determinant \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & 6
\end{array}\right|\) is not equal to _______. \(\left(\left|\begin{array}{lll}
2 & 1 & 1 \\
2 & 2 & 3 \\
2 & 3 & 6
\end{array}\right|,\left|\begin{array}{lll}
2 & 1 & 1 \\
3 & 2 & 3 \\
4 & 3 & 6
\end{array}\right|,\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 5 & 3 \\
1 & 9 & 6
\end{array}\right|,\left|\begin{array}{ccc}
3 & 1 & 1 \\
6 & 2 & 3 \\
10 & 3 & 6
\end{array}\right|\right)\)
Solution:
\(\left|\begin{array}{lll}
2 & 1 & 1 \\
2 & 2 & 3 \\
2 & 3 & 6
\end{array}\right|\)

(viii) With 4 different elements we can construct _______ number of different determinants of order 2. (1, 6, 8, 24)
Solution:
6

Question 4.
Solve the following:
(i) \(\left|\begin{array}{cc}
4 & x+1 \\
3 & x
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
4 & x+1 \\
3 & x
\end{array}\right|\) = 5
or, 4x – 3x – 3 = 5 or, x = 8

(ii) \(\left|\begin{array}{ccc}
\boldsymbol{x} & a & a \\
m & m & m \\
b & x & b
\end{array}\right|\) = 0
Solution:

(Replacing C₁ and C₂ by C₁ – C₃ and C₂ – C₃ respectively)
⇒ m |(x – a) (-x + b)| = 0
⇒ m (x – a) (b – x) – 0 ⇒ x = a, b

(iii) \(\left|\begin{array}{lll}
7 & 6 & x \\
2 & x & 2 \\
x & 3 & 7
\end{array}\right|\) = 0
Solution:

or, (x – 7) (7x + x² – 1 8) = 0
or, (x – 7) (x² + 7x – 18) = 0
or, (x – 7) (x + 9) (x – 2) = 0
∴ x = -9, 2, 7

(iv) \(\left|\begin{array}{ccc}
0 & x-a & x-b \\
x+a & 0 & x-c \\
x+b & x+c & 0
\end{array}\right|\) = 0
Solution:

or, – (x – a) {0 – (x + b) (x – c)} + (x – b) (x + a) (x + c) = 0
or, (x – a) (x + b) (x – c) + (x – b) (x + a) (x + c) = 0
or, (x² + bx – ax – ab) (x – c) + (x² + ax – bx – ab) (x + c) = 0
or, x³ – cx² + bx² – bcx – ax² + acx – abx + abc + x³ + cx² + ax² + acx- bx² – bcx – abx – abc = 0
or, 2x³ – 2abx – 2bcx + 2acx = 0
or, 2x (x² – ab – bc + ac) = 0
x = 0, x² = ab + bc – ca
∴ x = 0, x = \(\sqrt{a b+b c-c a}\)

(v) \(\left|\begin{array}{ccc}
\boldsymbol{x}+\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{x}+\boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{x}+\boldsymbol{b}
\end{array}\right|\) = 0
Solution:

⇒ (x + a + b + c) {x² + bx + cx + bc – a² – bx – b² + ca + ab – cx – c² = 0}
⇒ (x + a + b + c) {x² – a² – b² – c² + ab + bc + ca} = 0
⇒ x + a + b + c = 0
or, x² – a² – b² – c² + ab + bc + ca = 0
⇒ x = – (a + b + c)
∴ or x = \(\sqrt{a^2+b^2+c^2-a b-b c-c a}\)

(vi) \(\left|\begin{array}{ccc}
1+x & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+x
\end{array}\right|\) = 0
Solution:

(vii) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|\) = 0
Solution:

⇒ -30x² + 30 + 30x + 30 = 0
⇒ -30x² + 30x + 60 = 0
⇒ x² – x – 2 = 0
⇒ x² – 2x + x + 2 = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, -1

(viii) \(\left|\begin{array}{ccc}
x+1 & \omega & \omega^2 \\
\omega & x+\omega^2 & 1 \\
\omega^2 & 1 & x+\omega
\end{array}\right|\) = 0
Solution:

⇒ x(x² + xω – xω² + xω² + ω³ – ω⁴ – xω – ω² + ω³ – 1 + ω² + ω – ω³) = 0
⇒ x(x² + ω³ – ω⁴ – ω² + ω³ – 1 + ω² + ω – ω³) = 0
⇒ x(x² + ω³ – ω + ω – 1) = 0
⇒ x(x² + 1 – ω + ω – 1) = 0 (∵ ω³ = 1)
⇒ x³ = 0
⇒ x = 0

(ix) \(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0
Solution:

or, 10 + 10x – x² – 8x – x – 8 = 0
or, x² – 2x + x – 2 = 0
or, (x – 2) (x + 1) = 0
x = 2, x = -1

(x) \(\left|\begin{array}{lll}
x & 1 & 3 \\
1 & x & 1 \\
3 & 6 & 3
\end{array}\right|\) = 0
Solution:

or, x(3x – 6) – 0 + 3(6 – 3x) = 0
or, 3x² – 6x + 18 – 9x = 0
or, 3x² – 15x + 18 = 0
or, x² – 5x + 6 = 0
or, (x – 3) (x – 2) = 0
x = 3 or, x = 2

Question 5.
Evaluate the following
(i) \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -1 & 3 \\
4 & 1 & 10
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{lll}
\boldsymbol{x} & \mathbf{1} & 2 \\
\boldsymbol{y} & \mathbf{3} & 1 \\
z & 2 & 2
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{ccc}
x & 1 & -1 \\
2 & y & 1 \\
3 & -1 & z
\end{array}\right|\)
Solution:

= x (yz + z – z + 1) – (2z – 2 – 3y – 3)
= xyz + x – 2z + 3y + 5
= xyz + x + 3y – 2z + 5

(iv) \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)
Solution:

= a(bc – f²) – h (ch – fg) + g (hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(v) \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
\sin ^2 \theta & \cos ^2 \theta & 1 \\
\cos ^2 \theta & \sin ^2 \theta & 1 \\
-10 & 12 & 2
\end{array}\right|\)
Solution:

(vii) \(\left|\begin{array}{ccc}
-1 & 3 & 2 \\
1 & 3 & 2 \\
1 & -3 & -1
\end{array}\right|\)
Solution:

(viii) \(\left|\begin{array}{ccc}
11 & 23 & 31 \\
12 & 19 & 14 \\
6 & 9 & 7
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
37 & -3 & 11 \\
16 & 2 & 3 \\
5 & 3 & -2
\end{array}\right|\)
Solution:

(x) \(\left|\begin{array}{ccc}
2 & -3 & 4 \\
-4 & 2 & -3 \\
11 & -15 & 20
\end{array}\right|\)
Solution:

= 2(40 – 45) + 3(-80 + 33) + 4(60 – 22)
= -10 – 141 + 152 = -151 + 152 = 1

Question 6.
Show that x = 1 is a solution of \(\left|\begin{array}{ccc}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0
Solution:

or, (x + 9) {(x – 1)²} – 0
or, x = -9, 1
∴ x = 1 is a solution of the given equation.

Question 7.
Show that (a + 1) is a factor of \(\left|\begin{array}{ccc}
a+1 & 2 & 3 \\
1 & a+1 & 3 \\
3 & -6 & a+1
\end{array}\right|\) = 0
Solution:

= (a+ 1) {(a + 1)² + 18} – 2(a + 1 – 9) + 3(- 6 – 3a – 3)
= (a + 1) (a² + 2a + 1 + 18) – 2(a – 8) + 3(- 9 – 3a)
= (a + 1) (a² + 2a + 19) – 2a + 16 – 27 – 9a
= (a + 1) (a² + 2a + 19) – 11a – 11
= (a + 1) (a² + 2a + 19) – 11(a + 1)
= (a + 1) (a² + 2a + 19 – 11)
= (a + 1) (a² + 2a + 8)
∴ (a + 1) is a factor of the above determinant.

Question 8.
Show that \(\left|\begin{array}{ccc}
a_1 & b_1 & -c_1 \\
-a_2 & b_2 & c_2 \\
a_3 & b_3 & -c_3
\end{array}\right|=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\)
Solution:

Question 9.
Prove the following
(i) \(\left|\begin{array}{lll}
a & b & c \\
\boldsymbol{x} & y & z \\
\boldsymbol{p} & q & r
\end{array}\right|=\left|\begin{array}{lll}
\boldsymbol{y} & \boldsymbol{b} & \boldsymbol{q} \\
\boldsymbol{x} & \boldsymbol{a} & p \\
z & c & r
\end{array}\right|=\left|\begin{array}{lll}
\boldsymbol{x} & \boldsymbol{y} & z \\
\boldsymbol{p} & \boldsymbol{q} & r \\
a & b & c
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = abc \(\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Solution:

(iii) \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
q+r & r+p & p+q \\
y+z & z+x & x+y
\end{array}\right|=2\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
Solution:

(iv) \(\left|\begin{array}{lll}
(a+1)(a+2) & a+2 & 1 \\
(a+2)(a+3) & a+3 & 1 \\
(a+3)(a+4) & a+4 & 1
\end{array}\right|\) = -2
Solution:

(v) \(\left|\begin{array}{ccc}
a+d & a+d+k & a+d+c \\
c & c+b & c \\
d & d+k & d+c
\end{array}\right|\) = abc
Solution:

(vi) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c & c+a & c+a \\
b^2+c^2 & c^2+a^2 & a^2+b^2
\end{array}\right|\) = (b – c) (c – a) (a – b)
Solution:

(vii) \(\left|\begin{array}{lll}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a)
Solution:

(viii) \(\left|\begin{array}{ccc}
\boldsymbol{b}+\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{a} \\
\boldsymbol{b} & \boldsymbol{c}+\boldsymbol{a} & \boldsymbol{b} \\
\boldsymbol{c} & \boldsymbol{c} & \boldsymbol{a}+\boldsymbol{b}
\end{array}\right|\) = 4abc
Solution:

= (b + c – a) {(a + b) (c + a – b) – b (c – a – b)} + a (b – c – a) (c – a – b)
= (b + c – a)(ca + a² – ab + bc + ab – b² – bc + ab + b²) + a(bc – ab – b² – c² + ca + bc – ac + a² + ab)
= (b + c – a) (a² + ab + ca) + a (a² – b² – c² + 2bc)
= a²b + ab² + abc + ca² + abc + c²a – a³ – a²b – ca² + a³ – b²a – c²a + 2abc = 4abc

(ix) \(\left|\begin{array}{ccc}
b^2+c^2 & a b & a c \\
a b & c^2+a^2 & b c \\
c a & c b & a^2+b^2
\end{array}\right|\) = 4a²b²c²
Solution:

(x) \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|\) = (b – c) (c – a) (a – b) (bc + ca + ab)
Solution:

= (a – b) (b – c) (c – a) – (- ab + c²) + c (a + b + c)
= (a – b) (b – c) (c – a) (ab – c² + ca + bc + c²)
= (a – b) (b – c) (c – a) (ab + bc + ca)

(xi) \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) = (a + b+ c)³
Solution:

(xii) \(\left|\begin{array}{ccc}
(v+w)^2 & u^2 & u^2 \\
v^2 & (w+u)^2 & v^2 \\
w^2 & w^2 & (u+v)^2
\end{array}\right|\) = 2uvw (u + v + w)³
Solution:

Question 10.
Factorize the following
(i) \(\left|\begin{array}{ccc}
x+a & b & c \\
b & x+c & a \\
c & a & x+b
\end{array}\right|\)
Solution:

= (x + a + b + c) [(x + c – b) (x + b – a) – (a – c) (a – x – c)]
= (x + a + b + c) (x² + xb – ax + cx +bc – ca – bx – b² + ab – a² + ax + ac + ac – cx – c²)
= (x + a + b + c) (x² – a² – b² – c² + ab + bc + ca)

(ii) \(\left|\begin{array}{ccc}
a & b & c \\
b+c & c+a & a+b \\
a^2 & b^2 & c^2
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{ccc}
x & 2 & 3 \\
1 & x+1 & 3 \\
1 & 4 & x
\end{array}\right|\)
Solution:

Question 11.
Show that by eliminating α and from the equations.
a_i α + b_i β + c_i = 0, i = 1, 2, 3 we get \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_2 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
We have
a₁ α + b₁ β + c₁ = 0    …..(1)
a₂ α + b₂ β + c₂ = 0    …..(2)
a₃ α + b₃ β + c₃ = 0    …..(3)
Solving (2) and (3) by cross-multiplication method we have

Question 12.
Prove the following:
(i) \(\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & b(c+a) \\
1 & a b & c(a+b)
\end{array}\right|\) = 0
Solution:

(ii) \(\left|\begin{array}{ccc}
x+4 & 2 x & 2 x \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\) = (5x + 4) (4- x)²
Solution:

(iii) \(\left|\begin{array}{l}
\sin \alpha \cos \alpha \cos (\alpha+\delta) \\
\sin \beta \cos \beta \cos (\beta+\delta) \\
\sin \alpha \cos \gamma \cos (\gamma+\delta)
\end{array}\right|\) = 0
Solution:

(iv) \(\left|\begin{array}{ccc}
1 & x & x^2 \\
x^2 & 1 & x \\
x & x^2 & 1
\end{array}\right|\) = (1 -x³)²
Solution:

Question 13.
Prove that the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are collinear if \(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) = 0
Solution:
From geometry, we know that, if the points A, B, C, are collinear, then the area of the triangle ABC with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is zero.
\(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) = 0

Question 14.
If A + B + C = π, prove that \(\left|\begin{array}{lll}
\sin ^2 A & \cot A & 1 \\
\sin ^2 B & \cot B & 1 \\
\sin ^2 C & \cot C & 1
\end{array}\right|\) = 0
Solution:

Question 15.
Eliminate x, y, z from a = \(\frac{x}{y-z}\), b = \(\frac{y}{z-x}\), c = \(\frac{z}{x-y}\)
Solution:
We have
a = \(\frac{x}{y-z}\), b = \(\frac{y}{z-x}\), c = \(\frac{z}{x-y}\)
ay – az – x = 0, bz – bx – y = 0, cx – cy – z = 0
x – ay + az = 0
bx + y – bz = 0
cx – cy – z = 0
Now eliminating x, y, z from the above equations we have,

or, – 1 – bc + a(-b + bc) + a(-bc – c) = 0
or, – 1 – bc – ab + abc – abc – ac = 0
or, ab + bc + ca + 1 = 0

Question 16.
Given the equations
x = cy + bz, y = az + ex and z = bx + ay where x, y and z are not all zero, prove that a² + b² + c² + 2abc = 1 by determinant method.
Solution:
x = cy + bz, y = az + cx and z = bx + ay

or, 1 – a² + c(-c – ab) – b(ca + b) = 0
or, 1 – a² – c² – abc – abc – b² = 0
or, a² + b² + c² + 2abc = 1

Question 17.
If ax + hy + g = 0, hx + by +f = 0 and gx + fy + c = λ, find the value of λ, in the form of a determinant.
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
Write the value of cos^-1 cos(3π/2).
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Solution:
(b) \(\frac{\pi}{2}\)

Question 2.
Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively.
(a) m = 3, n = 2
(b) m = 2, n = 2
(c) m = 2, n = 3
(d) m = 3, n = 3
Solution:
(c) m = 2, n = 3

Question 3.
Write the principal value of
sin^-1 (\(-\frac{1}{2}\)) + cos^-1 cos(\( -\frac{\pi}{2}\))
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) π
Solution:
(b) \(\frac{\pi}{3}\)

Question 4.
Write the maximum value of x + y subject to: 2x + 3y < 6, x > 0, y > 0.
(a) 3
(b) 1
(c) 2
(d) 0
Solution:
(a) 3

Question 5.
Let A has 3 elements and B has m elements. Number of relations from A to B = 4096. Find the value of m.
(a) 2
(b) 4
(c) 1
(d) 3
Solution:
(b) 4

Question 6.
Let A is any non-empty set. Number of binary operations on A is 16. Find |A|.
(a) 2
(b) 1
(c) 3
(d) 4
Solution:
(a) 2

Question 7.
Give an example of a relation which is reflexive, transitive but not symmetric.
(a) x < y on Z
(b) x = y on Z
(c) x > y on Z
(d) None of the above
Solution:
(a) x < y on Z

Question 8.
Find the least positive integer r such that – 375 ∈ [r]₁₁
(a) r = 5
(b) r = 6
(c) r = 3
(d) r = 10
Solution:
(d) r = 10

Question 9.
Find three positive integers x_i, i = 1, 2, 3 satisfying 3x ≡ 2 (mod 7)
(a) x = 1, 3, 9…
(b) x = 2, 4, 6…
(c) x = 3, 10, 17…
(d) x = 2, 10, 18…
Solution:
(c) x = 3, 10, 17…

Question 10.
If the inversible function f is defined as f(x) = \(\frac{3 x-4}{5}\) write f^-1(x)
(a) \(\frac{5 x+4}{3}\)
(b) \(\frac{4 x+5}{3}\)
(c) \(\frac{5 x-4}{3}\)
(d) \(\frac{5 x+4}{2}\)
Solution:
(a) \(\frac{5 x+4}{3}\)

Question 11.
Let f : R → R and g : R → R defined as f(x) = |x|, g(x) = |5x – 2| then find fog.
(a) |5x + 2|
(b) |5x – 2|
(c) |2x – 2|
(d) |2x – 5|
Solution:
(b) |5x – 2|

Question 12.
Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.
(a) 20
(b) 12
(c) 30
(d) 36
Solution:
(c) 30

Question 13.
Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.
(a) -2
(b) -1
(c) 2
(d) 1
Solution:
(a) -2

Question 14.
Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.
(a) e = 1
(b) e = 5
(c) e = -5
(d) e = -1
Solution:
(b) e = 5

Question 15.
Find the number of binary, operations on the set {a, b}.
(a) 12
(b) 14
(c) 15
(d) 16
Solution:
(d) 16

Question 16.
Let ∗ is a binary operation on [0, ¥) defined as a ∗ b = \(\sqrt{a^2+b^2}\) find the identity element.
(a) e = 0
(b) e = 2
(c) e = 1
(d) e = 3
Solution:
(a) e = 0

Question 17.
Find least non-negative integer r such that 7 × 13 × 23 × 413 ≡ r (mod 11).
(a) r = 13
(b) r = 49
(c) r = 7
(d) r = 23
Solution:
(c) r = 7

Question 18.
Find least non-negative integer r such that 1237(mod 4) + 985 (mod 4) ≡ r (mod 4).
(a) r = 1
(b) r = 2
(c) r = -2
(d) r = -1
Solution:
(b) r = 2

Question 19.
Let ∗ is a binary operation on R – {0} defined as a ∗ b = \(\frac{a b}{5}\). If 2 ∗ (x ∗ 5) = 10, then find x:
(a) x = 25
(b) x = -5
(c) x = 5
(d) x = 1
Solution:
(a) x = 25

Question 20.
Find the principal value of cos^-1 (\( -\frac{1}{2}\)) + 2sin^-1 (\( \frac{1}{2}\)).
(a) \(\frac{5 \pi}{6}\)
(b) \(\frac{5 \pi}{2}\)
(c) \(\frac{5 \pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{5 \pi}{6}\)

Question 21.
Evaluate sin^-1 (\(\frac{1}{\sqrt{5}}\)) + cos^-1 (\(\frac{3}{\sqrt{10}}\))
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{4}\)
(d) π
Solution:
(c) \(\frac{\pi}{4}\)

Question 22.
Evaluate cos^-1 (\(\frac{1}{2}\)) + 2sin^-1 (\(\frac{1}{2}\)).
(a) \(\frac{2 \pi}{5}\)
(b) \(\frac{2 \pi}{3}\)
(c) π
(d) \(\frac{\pi}{3}\)
Solution:
(b) \(\frac{2 \pi}{3}\)

Question 23.
Find the value of tan^-1 √3 – sec^-1 (-2).
(a) –\(\frac{\pi}{3}\)
(b) \(\frac{2 \pi}{3}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{3 \pi}{2}\)
Solution:
(a) –\(\frac{\pi}{3}\)

Question 24.
Evaluate tan (2 tan^-1 \(\frac{1}{3}\))
(a) 2
(b) 1
(c) 0
(d) -1
Solution:
(a) 2

Question 25.
Evaluate : sin^-1 (sin \(\frac{3 \pi}{5}\)).
(a) \(-\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{2 \pi}{5}\)
(d) \(\frac{\pi}{5}\)
Solution:
(c) \(\frac{2 \pi}{5}\)

Question 26.
tan^-1 (2cos\(\frac{\pi}{3}\)) is ________.
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{3}\)
Solution:
(b) \(\frac{\pi}{4}\)

Question 27.
Evaluate : sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________.
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{3}\)
Solution:
(d) \(\frac{\pi}{3}\)

Question 28.
The value of sin(tan^-1 x + tan^-1 \( \frac{1}{x}\)), x > 0 = ________.
(a) -1
(b) 1
(c) 0
(d) 2
Solution:
(b) 1

Question 29.
2sin^-1 \( \frac{4}{5}\) + sin^-1 \( \frac{24}{25}\) = ________.
(a) 12
(b) 15
(c) 16
(d) 20
Solution:
(b) 15

Question 30.
Evaluate: tan^-1 1 = (2cos\(\frac{\pi}{3}\))
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{2 \pi}{3}\)
Solution:
(a) \(\frac{\pi}{4}\)

Question 31.
If sin^-1 (\( \frac{\pi}{5}\)) + cosec^-1 (\(\frac{5}{4}\)) = \(\frac{5}{2}\) then find the value of x.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 32.
Evaluate:
tan^-1 (\( \frac{-1}{\sqrt{3}}\)) + cot^-1 (\( \frac{1}{\sqrt{3}}\)) + tan^-1(sin(\( -\frac{\pi}{2}\))).
(a) \(\frac{- \pi}{12}\)
(b) \(\frac{2 \pi}{5}\)
(c) \(\frac{\pi}{12}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{- \pi}{12}\)

Question 33.
Evaluate sin^-1 (cos(\( \frac{33 \pi}{5}\)))
(a) \(\frac{\pi}{10}\)
(b) \(\frac{- \pi}{10}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{2}\)
Solution:
(b) \(\frac{- \pi}{10}\)

Question 34.
Express the value of the following in simplest form. tan (\( \frac{\pi}{4}\) + 2cot^-1 3)
(a) 7
(b) 12
(c) 3
(d) 6
Solution:
(a) 7

Question 35.
Express the value of the following in simplest form sin cos^-1 tan sec^-1 √2
(a) cos 0
(b) cot 0
(c) tan 0
(d) sin 0
Solution:
(d) sin 0

Question 36.
tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
(a) \(\frac{x-y}{1+x y}\)
(b) \(\frac{x+y}{1-x y}\)
(c) \(\frac{x-y}{1+y}\)
(d) \(\frac{x+y}{x y}\)
Solution:
(b) \(\frac{x+y}{1-x y}\)

Question 37.
The relation R on the set A = [1, 2, 3] given by R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} is:
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(a) Reflexive

Question 38.
Let f : R → R be defined as f(x) = 3x – 2. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
(a) f is one-one onto

Question 39.
Let R be a relation defined on Z as R = {(a, b) ; a² + b² = 25 }, the domain of R is:
(a) {3, 4, 5}
(b) {0, 3, 4, 5}
(c) {0, 3, 4, 5, -3, -4, -5}
(d) None of the above
Solution:
(c) {0, 3, 4, 5, -3, -4, -5}

Question 40.
let R be the relation in the set N given by R={(a, b) : a = b – 2, b > 6}. Choose the correct answer.
(a) (2, 4) • R
(b) (3, 8) • R
(c) (6, 8) • R
(d) (8, 10) • R
Solution:
(d) (8, 10) • R

Question 41.
Set A has 3 elements and set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is
(a) 144
(b) 12
(c) 24
(d) 64
Solution:
(c) 24

Question 42.
Let R be a relation on set of lines as L₁ R L₂ if L₁ is perpendicular to L₂. Then
(a) R is Reflexive
(b) R is transitive
(c) R is symmetric
(d) R is an equivalence relation
Solution:
(c) R is symmetric

Question 43.
A Relation from A to B is an arbitrary subset of:
(a) A × B
(b) B × B
(c) A × A
(d) B × B
Solution:
(a) A × B

Question 44.
Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is
(a) reflexive but not transitive
(b) transitive but not symmetric
(c) equivalence
(d) None of these
Solution:
(c) equivalence

Question 45.
The maximum number of equivalence relations on the set A = {1, 2, 3} are
(a) 1
(b) 2
(c) 3
(d) 5
Solution:
(d) 5

Question 46.
Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric nor transitive
Solution:
(a) reflexive but not symmetric

Question 47.
Which of the following functions from Z into Z are bijective?
f(x) = x³
f(x) = x + 2
f(x) = 2x + 1
f(x) = x² + 1
Solution:
f(x) = x + 2

Question 48.
Let R be a relation on the set N of natural numbers denoted by nRm <=> n is a factor of m (i.e. n | m). Then, R is
(a) Reflexive and symmetric
(b) Transitive and symmetric
(c) Equivalence
(d) Reflexive, transitive but not symmetric
Solution:
(c) Equivalence

Question 49.
Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows: (a, b) R (c, d) iff ad = cb. Then, R is
(a) reflexive only
(b) Symmetric only
(c) Transitive only
(d) Equivalence relation
Solution:
(d) Equivalence relation

Question 50.
Let X = {-1, 0, 1}, Y = {0, 2} and a function f : X → Y defined by y = 2x⁴, is
(a) one-one onto
(b) one-one into
(c) many-one onto
(d)many-one into
Solution:
(c) many-one onto

Question 51.
Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f(x) = (x-2)/(x-3). Then,
(a) f is bijective
(b) f is one-one but not onto
(c) f is onto but not one-one
(d) None of these
Solution:
(a) f is bijective

Question 52.
The function f : R → R given by f(x) = x³ – 1 is
(a) a one-one function
(b) an onto function
(c) a bijection
(d) neither one-one nor onto
Solution:
(c) a bijection

Question 53.
Let f : [0, ∞) → [0, 2] be defined by f(x) = 2x/1+x, then f is
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
Solution:
(a) one-one but not onto

Question 54.
If N be the set of all natural numbers, consider f : N → N such that f(x) = 2x, ∀ × ∈ N, then f is
(a) one-one onto
(b) one-one into
(c) many-one onto
(d) None of these
Solution:
(b) one-one into

Question 55.
Let f : R → R be a function defined by f(x) = x³ + 4, then f is
(a) injective
(b) surjective
(c) bijective
(d) none of these
Solution:
(c) bijective

Question 56.
Given set A = {a, b, c}. An identity relation in set A is
(a) R = {(a, b), (a, c)}
(b) R = {(a, a), (b, b), (c, c)}
(c) R = {(a, a), (b, b), (c, c), (a, c)}
(d) R= {(c, a), (b, a), (a, a)}
Solution:
(b) R = {(a, a), (b, b), (c, c)}

Question 57.
Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is
(a) 144
(b) 12
(c) 24
(d) 64
Solution:
(c) 24

Question 58.
sin(sec^-1 x + cosec^-1 x) =
(a) 1
(b) -1
(c) π/2
(d) π/3
Solution:
(a) 1

Question 59.
The principle value of sin^-1 (√3/2) is:
(a) 2π/3
(b) π/6
(C) π/4
(d) π/3
Solution:
(d) π/3

Question 60.
Simplified form of cos^-1 (4x³ – 3x)
(a) 3 sin^-1 x
(b) 3 cos^-1 x
(c) π – 3 sin^-1 x
(d) None of these
Solution:
(b) 3 cos^-1 x

Question 61.
tan^-1 √3 – sec^-1 (-2) is equal to
(a) π
(b) -π/3
(c) π/3,
(d) 2π/3
Solution:
(b) -π/3

Question 62.
If y = sec^-1 x then
(a) 0 ≤ y ≤ π
(b) 0 ≤ y ≤ π/2
(c) -π/2 ≤ y ≤ π/2
(d) None of these
Solution:
(d) None of these

Question 63.
If x + (1/x) = 2 then the principal value of sin^-1 x is
(a) π/4
(b) π/2
(c) π
(d) 3π/2
Solution:
(b) π/2

Question 64.
The principle value of sin^-1 (sin 2π/3) is :
(a) 2π/3
(b) π/3
(c) -π/6
(d) π/6
Solution:
(b) π/3

Question 65.
The value of cos^-1 (1/2) + 2sin^-1 (1/2) is equal to
(a) π/4
(b) π/6
(c) 2π/3
(d) 5π/6
Solution:
(b) π/6

Question 66.
Principal value of tan^-1 (-1) is
(a) π/4
(b) -π/2
(c) 5π/4
(d) -π/4
Solution:
(d) -π/4

Question 67.
A Linear function, which is minimized or maximized is called
(a) an objective function
(b) an optimal function
(c) A feasible function
(d) None of these
Solution:
(a) an objective function

Question 68.
The maximum value of Z = 3x + 4y subject to the constraints : x+ y ≤ 4, x ≥ 0, y ≥ 0 is:
(a) 0
(b) 12
(c) 16
(d) 18
Solution:
(c) 16

Question 69.
The maximum value of Z = 2x +3y subject to the constraints : x + y ≤ 1, 3x + y ≤ 4, x, y ≥ 0 is
(a) 2
(b) 4
(c) 5
(d) 3
Solution:
(c) 5

Question 70.
The point in the half plane 2x + 3y – 12 ≥ 0 is:
(a) (-7,8)
(c) (-7,-8)
(b) (7, -8)
(d) (7, 8)
Solution:
(d) (7, 8)

Question 71.
Any feasible solution which maximizes or minimizes the objective function is Called:
(a) A regional feasible solution
(b) An optimal feasible solution
(c) An objective feasible solution
(d) None of these
Solution:
(b) An optimal feasible solution

Question 72.
Objective function of a LPP is
(a) a constraint
(b) a function to be optimized
(c) a relation between the variables
(d) none of these
Solution:
(b) a function to be optimized

(B) Very Short Type Questions With Answers

Question 1.
If R is a relation on A such that R = R^-1, then write the type of the relation R.
Solution:
We know that (a, b) ∈ R ⇒ (b, a) ∈ R^-1
As R = R^-1, so R is symmetric. [2019(A)

Question 2.
Write the value of cos^-1 cos (\(\frac{3 \pi}{2}\)). [2019(A)
Solution:
cos^-1 cos (\(\frac{3 \pi}{2}\)) = cos^-1 (0) = \(\frac{\pi}{2}\)

Question 3.
Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively. [2018(A)
Solution:
|A| = m and |B| = n
Number of relations from A to B = 2^mn.
A.T.Q. 2^mn = 64 = 2⁶.
⇒ mn = 6, m < n with m ≠ 1.
∴m = 2, n = 3

Question 4.
Write the principal value of sin^-1 (\(\frac{- 1}{2}\)) + cos^-1 cos(\(\frac{- \pi}{2}\)) [2018(A)
Solution:
sin^-1 (\(\frac{- 1}{2}\)) + cos^-1 (cos\(\frac{- \pi}{2}\)) = \(\frac{- \pi}{6}\) + \(\frac{\pi}{2}\) = \(-\frac{\pi}{3}\)

Question 5.
Write the maximum value of x + y subject to: 2x + 3y ≤ 6, x ≥ 0, y ≥ 0. [2011(A)
Solution:
2x + 3y = 6 intersects the axes at (3, 0) and (0, 2)
∴ The maximum value of x + y = 3.

Question 6.
Define ‘feasible’ solution of an LPP. [2009(A)
Solution:
The solutions of LPP which satisfy all the constraints and non-negative restrictions are called feasible solutions.

Question 7.
Mention the quadrant in which the solution of an LPP with two decision variables lies when the graphical method is adopted. [2008(A)
Solution:
The solution lies in XOY or 1st quadrant.

Question 8.
Write the smallest equivalence relation on A = {1, 2, 3}.
Solution:
The relation R = {(1, 1), (2, 2), (3, 3)} is the smallest equivalence relation on set A.

Question 9.
Congruence modulo 3 relation partitions the set Z into how many equivalence classes?
Solution:
The relation congruence modulo 3 on the set Z partitions Z into three equivalence classes.

Question 10.
Give an example of a relation which is reflexive, symmetric but not transitive.
Solution:
The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.

Question 11.
Give an example of a relation which is reflexive, transitive but not symmetric.
Solution:
‘‘The relation x ≤ y on Z” is reflexive, transitive but not symmetric.

Question 12.
Give an example of a relation which is reflexive but neither symmetric nor transitive.
Solution:
The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.

Question 13.
Find three positive integers x_i, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)
Solution:
3x ≡ 2 mod 7
Least positive value of x = 3
Each member of [3] is a solution
∴ x = 3, 10, 17…

Question 14.
State the reason for relation R on {1, 2, 3} defined as {(1, 2), (2, 1)} is not transitive.
Solution:
(1, 2), (2, 1) ∈ R but (1, 1) ∉ R
∴ R is not transitive.

Question 15.
Give an example of a function which is injective but not surjective.
Solution:
f(x) = \(\frac{x}{2}\) from Z → R is injective but not surjective.

Question 16.
Let X = {1, 2, 3, 4}. Determine whether
f : X → X defined as given below have inverses. Find f^-1 if it exists:
f = {(1, 2), (2, 2), (3, 2), (4, 2)}
Solution:
f is not injective hence not invertible.

Question 17.
Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.
Solution:
4 ∗ 5 = 3 × 4 + 4 × 5 – 2
= 12 + 20 – 2
= 30

Question 18.
Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.
Solution:
3 ∗ 4 = 6 + 4 – 12 = -2

Question 19.
Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.
Solution:
Let e is the identity element.
⇒ a ∗ e = e ∗ a = a
⇒ a + e – 5 = a
⇒ e = 5

Question 20.
Find the number of binary operations on the set {a, b}.
Solution:
Number of binary operations on
{a, b} = 2²² = e⁴ =16.

Question 21.
Let * is a binary operation on [0, ¥) defined as a * b = \(\sqrt{\mathbf{a}^2+\mathbf{b}^2}\) find the identity element.
Solution:
Let e is the identity element
⇒ a * e = \(\sqrt{\mathbf{a}^2+\mathbf{e}^2}\) = a
⇒ a² + e² = a²
⇒ e = 0

Question 22.
Evaluate cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 (\(\frac{1}{2}\)).
Solution:
cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 (\(\frac{1}{2}\))
= \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\) = \(\frac{2 \pi}{3}\)

Question 23.
Find the value of tan^-1 √3 – sec^-1 (-2)
Solution:
tan^-1 √3 – sec^-1 (-2)
= \(\frac{\pi}{3}\) – \(\frac{2 \pi}{3}\) = – \(\frac{\pi}{3}\).

Question 24.
Evaluate tan (2  tan^-1 \(\frac{1}{3}\))
Solution:
tan (2  tan^-1 \(\frac{1}{3}\)) = tan tan^-1 \(\left(\frac{\frac{2}{3}}{1-\frac{2}{3}}\right)\)
= tan tan^-1 (2) = 2

Question 25.
Evaluate: sin^-1 (sin \(\frac{3 \pi}{5}\)).
Solution:
sin^-1 (sin \(\frac{3 \pi}{5}\)) = sin^-1 sin (\(\pi \frac{-2 \pi}{5}\))
= sin^-1 sin \(\frac{2 \pi}{5}\) = \(\frac{2 \pi}{5}\)

Question 26.
Evaluate tan^-1 1 = (2 cos \(\frac{\pi}{3}\))
Solution:
tan^-1 (2 cos \(\frac{\pi}{3}\))
= tan^-1 (2 × 1/2) = tan^-1 1 = \(\frac{\pi}{4}\)

Question 27.
Define the objective function.
Solution:
If C₁, C₂, C₃ …. Cn are constants and x₁, x₂, …… x_n are variables then the linear function z = C₁x₁ + C₂x₂ +…… C_nx_n which is to be optimized is called an objective function.

Question 28.
Define feasible solution.
Solution:
A set of values of the variables x₁, x₂, …… x_n is called a feasible solution of LPP if it satisfies the constraints and non-negative restrictions of the problem.

Question 29.
Define a convex set.
Solution:
A set is convex if every point on the line segment joining any two points lies on it.

Question 30.
State extreme point theorem.
Solution:
Let S is a convex polygon bounded by the straight lines. The linear function z = Ax + By attains its optimum value at the vertices of S.

(C) Short Type Questions With Answers

Question 1.
Construct the multiplication table X₇ on the set {1, 2, 3, 4, 5, 6}. Also find the inverse element of 4 if it exists. [2019(A)
Solution:
Given set A = { 1, 2, 3, 4, 5, 6} Binary operation ∗ defined on A is X₇.
i.e. a ∗ b = a × b mod 7
= The remainder on dividing a × b by 7
The composition table for this operation is:

∗	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	2	5	1	4
4	4	1	5	2	6	3
5	5	3	1	6	4	2
6	6	5	4	3	2	1

As the second row is identical to first row, we have ‘1’ as the identity element.
As 4 ∗ 2 = 2 ∗ 4 = 1 we have 4^-1 = 2

Question 2.
Let R be a relation on the set R of real numbers such that aRb iff a – b is an integer. Test whether R is an equivalence relation. If so, find the equivalence class of 1 and \(\frac{1}{2}\). [2019(A)
Solution:
The relation R on the set of real numbers is defined as
R = { (a, b) : a – b ∈ Z}
Reflexive:
∀ a ∈ R (set of real numbers)
a – a = 0 ∈ Z
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetric:
Let (a, b) ∈ R
⇒ a – b ∈ Z
⇒ b – a ∈ Z
⇒ (b, a) ∈ R
⇒ R is symmetric.
Transitive:
Let (a, b), (b, c) ∈ R
⇒ a – b and b – c ∈ Z
⇒ a – b + b – c ∈ Z
⇒ a – c ∈ Z
⇒ (a, c) ∈ R
⇒ R is transitive
Thus R is an equivalence relation
[1] = { x ∈ R : x – 1 ∈ Z} = Z
\(\frac{1}{2}\) = { x ∈ R : x – \(\frac{1}{2}\) ∈ Z}
= {x ∈ R : x = \(\frac{2 k+1}{2}\), k ∈ Z}

Question 3.
Two types of food X and Y are mixed to prepare a mixture in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. These vitamins are available in 1 kg of food as per the table given below: [2019(A)

Vitamin
Food	A	B	C
X	1	2	3
Y	2	2	1

1 kg of food X costs ₹16 and 1 kg of food Y costs ₹20. Formulate the LPP so as to determine the least cost of the mixture containing the required amount of vitamins.
Solution:
Let x kg of food X and Y kg of food y are to be mixed to prepare the mixture.
Total cost = 16x + 20y to be minimum.
According to the question
Total vitamin A = x + 2y ≥ 10 units
Total vitamin B = 2x + 2y ≥ 12 units
Total vitamin C = 3x + y ≥ 8 units.
∴ The required LPP is minimize
Z= 16x + 20y
Subject to : x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0

Question 4.
Let ~ be defined by (m, n) ~ (p, q) if mq = np, where m, n, p, q e Z – {0}. Show that it is an equivalence relation. [2018(A)
Solution:
Let A = z – {0}
~ is a relation on A x A defined as (m, n) ~ (p, q) ⇔ mq = np
Reflexive : For all (m, n) ∈ A × A
We have mn = nm
⇒ (m, n) ~ (m, n)
∴ ~ is reflexive.
Symmetric: Let {m, n), (p, q) ∈ A × A and (m, n) ~ (p, q)
⇒ mq = np
⇒ pn = qm
(p, q) ~ (m, n)
∴ ~ is symmetric.
Transitive: Let (m, n), (p, q), (x, y) ∈ A × A
and (m, n) ~ (p, q), (p, q) ~ (x, y)
⇒ mq = np and py = qx
⇒ mqpy = npqx
⇒ my = nx
⇒ (m, n) ~ (x, y)
∴ ~ is transitive.
Thus ~ is an equivalence relation.

Question 5.
Solve the following LPP graphically:
Minimize Z = 4x + 3y
subject to 2x + 5y ≥ 10
x, y ≥ 0. [2018(A)
Solution:
Given LPP is:
Minimize: Z = 4x + 3y
Subject to: 2x + 5y ≥ 10
x, y ≥ 0
Step – 1 Considering the constraints as equations we have 2x + 5y = 10

x	5	0
y	0	2

Let us draw the graph.

Step – 2 As 0(0, 0) does not satisfy 2x + 5y > 0 and x, y > 0 is the first quadrant, we have the shaded region is the feasible region whose vertices are A(5, 0) B(0, 2).
Step – 3 Z (5, 0) = 20
Z (0, 2) = 6 … Minimum
As the feasible region is unbounded. Let us draw the half plane.
4x + 3y < 6

x	0	\(\frac{3}{2}\)
y	2	0

Step – 4 As there is no point common to the feasible region and the half plane 4x + 3y < 6, we have Z is minimum for x = 0, y = 2 and Z(min) = 6

Question 6.
Find the feasible region of the system 2y – x > 0, 6y – 3x < 21, x > 0, y > 0. [2017 (A)
Solution:
Step – 1: Treating the constraints as equations we have
2y – x = 0
6y – 3x = 21
⇒ 2y – x = 0
2y – x = 7
Step – 2: Let us draw the lines.
Table – 1

x	0	2
y	0	1

x	1	37
y	4	5

Step – 3: Clearly A(1, 3) Satisfies both the constraints, x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 7.
Solve the following LPP graphically:
Maximize: Z = 20x + 30y
Subject to: 3x + 5y ≤ 15
x, y ≥ 0. [2014 (A), 2016 (A), 2017 (A)
Solution:
Given LPP is
Maximize: Z = 20x + 30y
Subject to: 3x + 5y ≤ 15
x, y ≥ 0
Step – 1 Treating the constraints as equations we get 3x + 5y = 15.
Step- 2 Let us draw the graph

x	5	0
y	0	3

Step – 3
As 0(0, 0) satisfies 3x + 5y ≤ 15 the shaded region is the feasible region.
Step – 4
The vertices ofthe feasible region are 0(0, 0), A(5, 0) and B(0, 3).
Z(0) = 0, Z(A) = 100, Z(B) = 90
Z attains maximum at A for x = 5 and y = 0.
The given LPP has a solution, x = 5, y = 0 and Z(max) = 100.

Question 8.
Find the feasible region of the following system:
2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0. [2016 (A)
Solution:
Given system of inequations are
2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0
Step- 1: Consider 2x + y = 6
x – y = 3
Step – 2: Let us draw the graph
Table- 1

x	3	0
y	0	6

Table- 2

x	3	0
y	0	-3

Step – 3: 0(0, 0) satisfies x – y < 3, does not satisfy 2x + y > 6 and x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 9.
Solve the following LPP graphically:
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 1
x₁, x₂ ≥ 0. [2015 (A)
Solution:
Given LPP is
Minimize: Z = 6x₁+ 7x₂
Subject to: x₁ + 2x₂ ≥ 2
x₁, x₂ ≥ 0
Let us draw the line x₁ + 2x₂ = 2

x1	0	2
x2	1	0

Clearly (0, 0) does not satisfy x₁ + 2x₂ ≥ 2 and x₁, x₂ ≥ 0 is the first quadrant.
The shaded region is the feasible region.
The coordinates of vertices are A(2, 0) and B(0, 1).

Point	Z = 6x1 + 7x2
A (2, 0)	12
B (0,1)	7 → Minimum

As there is no point common to the half plane 6x₁ + 7x₂ < 7 and the feasible region.
Z is minimum when x₁ = 0, y₁ =1 and the minimum value of z = 7

Question 10.
Find the feasible region of the following system:
2y – x ≥ 0, 6y – 3x ≤ 21, x ≥ 0, y ≥ 0. [2015 (A)
Solution:
Given system is
2y – x ≥ 0
6y – 3x ≤ 21
x, y ≥ 0.
Considering the constraints as equations we have
2y – x = 0
and 6y – 3x = 21

x	0	2
x	0	1

⇒ 3y – x = 7

x	-7	2
x	0	3

Let us draw the lines

Clearly (2, 0) does not satisfy 2y – x ≥ 0 and satisfies 6y – 3x ≤ 21
∴ The shaded region is the feasible region.

Question 11.
Find the maximum value of z = 50x₁ + 60x₂
subject to 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0. [2013 (A)
Solution:
Let us consider the constraints as equations.
2x₁ + 3x₂ = 6 … (1)
The table of some points on (1) is

x1	0	3
x2	2	0

Let us draw the line 2x₁ + 3x₂ = 6

As (0, 0) satisfies the inequality 2x₁ + 3x₂ ≤ 6 and x₁, x₂ ≥ 0 is the first quadrant, the shaded region is the feasible region with corner points O(0, 0), A(3, 0) and B(0, 2).

Corner point	z = 50x1 + 60x2
O(0, 0)	0
A(3, 0)	150
B(0, 2)	120

Thus Z_(max) = 150 for x₁ = 3, x₂ = 0

Question 12.
Shade the feasible region satisfying the inequations 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 in a rough sketch. [2011(A)
Solution:
Let us consider the line 2x + 3y = 6

x1	0	3
x2	2	0

Let us draw the line on the graph

The feasible region is shaded in the figure.

Question 13.
Show the feasible region for the following constraints in a graph:
2x + y ≤ 4, x ≥ 0, y ≥ 0. [2010(A)
Solution:
Let us draw the graph of 2x + y = 4.

The shaded region shows the feasible region.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
If \(\left|\begin{array}{ccc}
1+x & x & x^2 \\
x & 1+x & x^2 \\
x^2 & x & 1+x
\end{array}\right|\) = a + bx + cx² + dx³ + ex⁴ + fx⁵ then write the value of a.
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(c) 1

Question 2.
If every element of a third order determinant of value 8 is multiplied by 2, then write the value of the new determinant.
(a) 32
(b) 64
(c) 16
(d) 128
Answer:
(b) 64

Question 3.
If A is a 4 x 5 matrix and B is a matrix such that A^TB and BA^T both are defined, then write the order of B
(a) 4 x 5
(b) 1 x 5
(c) 5 x 4
(d) None of these
Answer:
(a) 4 x 5

Question 4.
If \(\left[\begin{array}{lll}
3 & 5 & 3 \\
2 & 4 & 2 \\
\lambda & 7 & 8
\end{array}\right]\) is a singular matrix, write die value of 1.
(a) λ = 2
(b) λ = 1
(c) λ = 4
(d) λ = 8
Answer:
(d) λ = 8

Question 5.
Determine the maximum value of \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b) 2

Question 6.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find x.
(a) x = 1
(b) x = 0
(c) x = 2
(d) x = -1
Answer:
(b) x = 0

Question 7.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find y.
(a) y = 1
(b) y = 3
(c) y = 2
(d) y = 0
Answer:
(a) y = 1

Question 8.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find t.
(a) t = 1
(b) t = 2
(c) t = 3
(d) t = 0
Answer:
(c) t = 3

Question 9.
Which matrix is a unit matrix?
(a) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)
(b) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)
(c) \(\left(\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right)\)
(d) \(\left(\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right)\)
Answer:
(b) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)

Question 10.
If \(\left(\begin{array}{cc}
\mathbf{x}_1 & \mathbf{x}_2 \\
\mathbf{y}_1 & \mathbf{y}_2
\end{array}\right)\) – \(\left(\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right)\) = \(\left(\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right)\) then find x₁, x₂, y₁, y₂.
(a) x₁ = 8, x₂ = 5, y₁ = 3, y₂ = 1
(b) x₁ = 1, x₂ = 8, y₁ = 5, y₂ = 3
(c) x₁ = 5, x₂ = 8, y₁ = 1, y₂ = 3
(d) x₁ = 3, x₂ = 1, y₁ = 8, y₂ = 5
Answer:
(c) x₁ = 5, x₂ = 8, y₁ = 1, y₂ = 3

Question 11.
If \(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0, what is the value of k?
(a) 3
(b) 4
(c) 2
(d) 6
Answer:
(a) 3

Question 12.
If \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{b}_1 \\
\mathbf{c}_1 & \mathbf{d}_1
\end{array}\right|\) = k \(\left|\begin{array}{ll}
a_1 & c_1 \\
b_1 & d_1
\end{array}\right|\) hen what is the value of k?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 13.
If A = \(\left(\begin{array}{lll}
1 & 0 & 2 \\
5 & 1 & x \\
1 & 1 & 1
\end{array}\right)\) is a singular matrix then what is the value of x?
(a) 6
(b) 7
(c) 8
(d) 9
Answer:
(d) 9

Question 14.
Evaluate \(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
(a) 66
(b) 666
(c) 6666
(d) 6
Answer:
(b) 666

Question 15.
Evaluate \(\left|\begin{array}{lll}
1 & 1 & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
(a) 0
(b) 1
(c) 11
(d) 2
Answer:
(a) 0

Question 16.
Evaluate \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
(a) 54
(b) 58
(c) -54
(d) 60
Answer:
(c) -54

Question 17.
If A and B are square matrices of order 3, such that |A| = -1, |B| = 3 then |3 AB| = –
(a) 1
(b) 11
(c) 9
(d) 81
Answer:
(d) 81

Question 18.
For what k
x + 2y – 3z = 2
(k + 3)z = 3
(2k + 1)y + z = 2 is inconsistent?
(a) -3
(b) -6
(c) 3
(d) 6
Answer:
(a) -3

Question 19.
The sum of two nonintegral roots of \(\left|\begin{array}{lll}
x & 2 & 5 \\
3 & x & 3 \\
5 & 4 & x
\end{array}\right|\) = 0 is ______.
(a) 5
(b) -5
(c) 3
(d) 15
Answer:
(b) -5

Question 20.
The value of \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 2 \\
8 & 14 & 20
\end{array}\right|\) is ______.
(a) 1
(b) 2
(c) 0
(d) 3
Answer:
(c) 0

Question 21.
If [x 1] \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) = 0, then x equals:
(a) 0
(b) -2
(c) -1
(d) 2
Answer:
(d) 2

Question 22.
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(a) 27
(b) 18
(c) 81
(d) 512
Answer:
(d) 512

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) , and A + A’ = I, then the value of α is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{3}\)
(c) π
(d) \(\frac{3 \pi}{2}\)
Answer:
(b) \(\frac{\pi}{3}\)

Question 24.
Matrix A and B will be inverse of each other only if
(a) AB = BA
(b) AB = BA = 0
(c) AB = 0, BA = I
(d) AB = BA = I
Answer:
(d) AB = BA = I

Question 25.
The matrix P = \(\left[\begin{array}{lll}
0 & 0 & 4 \\
0 & 4 & 0 \\
4 & 0 & 0
\end{array}\right]\) is a
(a) square matrix
(b) diagonal matrix
(c) unit matrix
(d) None of these
Answer:
(a) square matrix

Question 26.
If A and B are symmetric matrices of same order, then AB – BA is a
(a) Skew-symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity
Answer:
(a) Skew-symmetric matrix

Question 27.
If A is a square matrix of order 3, such that A(adj A) = 10I, then |adj A| is equal to
(a) 1
(b) 10
(c) 100
(d) 1000
Answer:
(c) 100

Question 28.
Let A be a square matrix of order 2 × 2, then |KA| is equal to
(a) K|A|
(b) K²|A|
(c) K³|A|
(d) 2K|A|
Answer:
(b) K²|A|

Question 29.
If A and B are invertible matrices then which of the following is not correct
(a) Adj A = |A|. A^-1
(b) det (A^-1) = (det A)^-1
(c) (AB)^-1 = B^-1A^-1
(d) (A + B)^-1 = A^-1 + B^-1
Answer:
(d) (A + B)^-1 = A^-1 + B^-1

Question 30.
If A is a skew-symmetric matrix of order 3, then the value of |A| is
(a) 3
(b) 0
(c) 9
(d) 27
Answer:
(b) 0

Question 31.
If A is a square matrix of order 3, such that A(adjA) = 10I, then ladj Al is equal to
(a) 1
(b) 10
(c) 100
(d) 1000
Answer:
(c) 100

Question 32.
Let A be a non-angular square matrix of order 3 x 3, then |A. adj Al is equal to
(a) |A|³
(b) |A|²
(c) |A|
(d) 3|A|
Answer:
(a) |A|³

Question 33.
Let A be a square matrix of order 3 × 3 and k a scalar, then |kA| is equal to
(a) k|A|
(b) |k||A|
(c) k³|A|
(d) none of these
Answer:
(c) k³|A|

Question 34.
If a, b, c are all distinct, and \(\left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 then the value of abc is
(a) 0
(b) -1
(c) 3
(d) -3
Answer:
(b) -1

Question 35.
If a, b, c are in AP, then the value of \(\left|\begin{array}{lll}
x+1 & x+2 & x+a \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|\) is:
(a) 4
(b) -3
(c) 0
(d) abc
Answer:
(c) 0

Question 36.
If A is a skew-symmetric matrix of order 3, then the value of |A| is
(a) 3
(b) 0
(c) 9
(d) 27
Answer:
(b) 0

Question 37.
A bag contains 3 white, 4 black and 2 red balls. If 2 balls are choosen at random (without replacement), then the probability that both the balls are white is:
(a) \(\frac{1}{18}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{24}\)
Answer:
(c) \(\frac{1}{12}\)

Question 38.
Three diece are thrown simultaneously. The probability of obtaining a total score of 5 is:
(a) \(\frac{5}{216}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{36}\)
(d) \(\frac{1}{49}\)
Answer:
(c) \(\frac{1}{36}\)

Question 39.
An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colour is:
(a) \(\frac{2}{5}\)
(b) \(\frac{1}{15}\)
(c) \(\frac{8}{15}\)
(d) \(\frac{4}{15}\)
Answer:
(d) \(\frac{4}{15}\)

Question 40.
The probability of obtaining an even prime number on each die when a pair of dice is rolled is:
(a) 0
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{36}\)
Answer:
(d) \(\frac{1}{36}\)

Question 41.
Two events A and B are said to be independent if:
(a) A and B are mutually exclusive
(b) P (A’B’) = [1 – P(A)][1 – P(B)]
(c) P(A) = P(B)
(d) P(A) + P(B) = 1
Answer:
(b) P (A’B’) = [1 – P(A)][1 – P(B)]

Question 42.
A die is. thrown once, then the probability of getting number greater than 3 is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) 6
(d) 0
Answer:
(a) \(\frac{1}{2}\)

Question 43.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A ∪ B) = \(\frac{7}{11}\), then P(A/B) is:
(a) \(\frac{2}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{4}{5}\)
(d) 1
Answer:
(c) \(\frac{4}{5}\)

Question 44.
Let the target be hit A and B: the target be hit by B and C: the target be hit by A and C. Then the probability that A, B and C all will hit, is:
(a) \(\frac{4}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{1}{5}\)
Answer:
(c) \(\frac{2}{5}\)

Question 45.
What is the probability that ‘none of them will hit the target’?
(a) \(\frac{1}{30}\)
(b) \(\frac{1}{60}\)
(c) \(\frac{1}{15}\)
(d) \(\frac{2}{15}\)
Answer:
(b) \(\frac{1}{60}\)

(B) Very Short Type Questions With Answers

Question 1.
If \(\left|\begin{array}{ccc}
1+\mathbf{x} & \mathbf{x} & \mathbf{x}^2 \\
\mathbf{x} & 1+\mathbf{x} & \mathbf{x}^2 \\
\mathbf{x}^2 & \mathbf{x} & 1+\mathbf{x}
\end{array}\right|\) = a + bx + cx² + dx³ + ex⁴ + fx⁵ then write the value of a.
Solution:
\(\left|\begin{array}{ccc}
1+\mathbf{x} & \mathbf{x} & \mathbf{x}^2 \\
\mathbf{x} & 1+\mathbf{x} & \mathbf{x}^2 \\
\mathbf{x}^2 & \mathbf{x} & 1+\mathbf{x}
\end{array}\right|\)
= a + bx + cx² + dx³ + ex⁴ + fx⁵
which is an identity
Putting x = 0 we get
a = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = 1

Question 2.
If every element of a third order determinant of value 8 is multiplied by 2, then write the value of the new determinant.
Solution:
According to the question
|A| = 8
Now |KA| = Kⁿ|A|
⇒ |2A| = 2³|A| = 8 × 8 = 64
Value of the new determinant is 64.

Question 3.
If I is an identity matrix of order n, then k being a natural number, write the matrix I^k_n.
Solution:
If I is an identity matrix of order n, then I^k_n = I_n

Question 4.
If A is a 4 × 5 matrix and B is a matrix such that A^TB and BA^T both are defined, then write the order of B.
Solution:
Order of A = 4 × 5
Order of A^T = 5 × 4
Let order of B = m × n
A^TB is well defined ⇒ m = 4
BA^T is well defined ⇒ n = 5
Order of B = 4 × 5

Question 5.
Write the matrix which when added to the matrix \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives the matrix \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:
Let the required matrix is A.
\(\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)\) + A = \(\left(\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right)\)
A = \(\left(\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right)\) – \(\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)\) = \(\left(\begin{array}{cc}
2 & 4 \\
7 & -5
\end{array}\right)\)

Question 6.
Determine the maximum value of \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
Solution:
Let f(x) = \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
= cos²x – cos x + sin²x = 1 – cos x
As – 1 < cos x ≤ 1
⇒ 1 >- cos x ≥ – 1
⇒ 2 > 1 – cos x ≥ 0
The maximum value of f(x) = 2.

Question 7.
Write the value of k if:
\(\left|\begin{array}{lll}
\mathbf{a a _ { 1 }} & \mathbf{a a}_2 & \mathbf{a} \mathbf{a}_3 \\
\mathbf{a b _ { 1 }} & \mathbf{a b}_2 & \mathbf{a b} \\
\mathbf{a c _ { 2 }} & \mathbf{a c}_2 & \mathbf{a c _ { 3 }}
\end{array}\right|\) = k\(\left|\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_3 & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
\mathbf{a a _ { 1 }} & \mathbf{a a}_2 & \mathbf{a} \mathbf{a}_3 \\
\mathbf{a b _ { 1 }} & \mathbf{a b}_2 & \mathbf{a b} \\
\mathbf{a c _ { 2 }} & \mathbf{a c}_2 & \mathbf{a c _ { 3 }}
\end{array}\right|\) = k\(\left|\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_3 & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right|\)
k = a³.

Question 8.
If A is a 3 × 3 matrix and |A| = 3, then write the matrix represented by A × adj A.
Solution:
|A| = 3 ⇒ A × Adj A = \(\left(\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right)\)

Question 9.
If ω is a complex cube root of 1, then for what value of λ the determinant
\(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \lambda & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\) = 0?
Solution:

⇒ for any value of ‘A,’ the given determinant is ‘0’

Question 10.
If [1 2 3] A = [0], then what is the der of the matrix A?
Solution:
If [1 2 3] A = [0]
A is a 3 × 1 matrix

Question 11.
What is A + B if A = \(\left(\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right)\), B = \(\left(\begin{array}{cc}
0 & -1 \\
-2 & 1
\end{array}\right)\)?
Solution:
For A = \(\left(\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right)\), B = \(\left(\begin{array}{cc}
0 & -1 \\
-2 & 1
\end{array}\right)\)
A + B = \(\left(\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right)\)

Question 12.
Give an example of a unit matrix.
Solution:
\(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\) is a unit matrix of 2nd order.

Question 13.
Construct a 2 × 3 matrix having elements defined by a_ij = i – j.
Solution:
a_ij = i – j
a₁₁ =0, a₁₂ = 1 – 2 = – 1, a₁₃ = 1 – 3 =- 2
a₂₁ = 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ = 2 – 3 = -1
∴ The required matrix is \(\left(\begin{array}{ccc}
0 & -1 & -2 \\
0 & 0 & -1
\end{array}\right)\).

Question 14.
Find x, y if A = A’ where A = \(\left(\begin{array}{ll}
5 & \mathbf{x} \\
\mathbf{y} & 0
\end{array}\right)\)
Solution:
A = A’
⇒ \(\left(\begin{array}{ll}
5 & \mathbf{x} \\
\mathbf{y} & 0
\end{array}\right)\) = \(\left(\begin{array}{ll}
5 & \mathbf{y} \\
\mathbf{x} & 0
\end{array}\right)\) ⇒ x = y
∴ x and y are any real number where x = y

Question 15.
Cana matrix be constructed by taking 29 elements?
Solution:
Only two matrices can be formed by taking 29 elements. They are of order 1 × 29 and 29 × 1.

Question 16.
If \(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0 , what is the value of k?
Solution:
\(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0 ⇒ 12 – 4k = 0 ⇒ k = 3

Question 17.
If \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{b}_1 \\
\mathbf{c}_{\mathbf{1}} & \mathbf{d}_1
\end{array}\right|\) = k = \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{c}_1 \\
\mathbf{b}_1 & \mathbf{d}_1
\end{array}\right|\) then what is the value of k?
Solution:
k = 1

Question 18.
If A and B are square matrices of order 3, such that |A| = -1, |B| = 3 then |3 AB| = ______.
Solution:
|3 AB| = 27 |A| |B| = 81

Question 19.
Solve: \(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0 => \(\left|\begin{array}{ccc}
0 & 2 & x \\
-1-x & x & 4 \\
0 & 1 & 1
\end{array}\right|\) = 0
⇒ – (- 1 – x) (2 – x) = 0 ⇒ x = -1, x = 2.

(C) Short Type Questions With Answers

Question 1.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\) then show that A³ – 23A – 40I = 0
Solution:

Question 2.
Solve: \(\left|\begin{array}{ccc}
\mathbf{x + 1} & \omega & \omega \\
\omega & x+\omega^2 & 1 \\
\omega^2 & 1 & x+\omega
\end{array}\right|\) = 0
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A + A’ is symmetric and A – A’ is skew symmetric.
Solution:

Question 4.
If A, B, C are matrices of order 2 × 2 each and 2A + B + C = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 0
\end{array}\right]\), A + B + C = \(\left[\begin{array}{ll}
0 & 1 \\
2 & 1
\end{array}\right]\) and A + B – C = \(\left[\begin{array}{ll}
1 & 2 \\
1 & 0
\end{array}\right]\), then find A, B and C.
Solution:

Question 5.
Find the inverse of the following matrix: \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Method – I
Let us find A^-1 by using elementary row transformation.
Let A = IA

Method – II
|A| = 1(1 – 4) – 1(0- 2) + 2(0- 1)
= 1(-3) – 1(-2) + 2(-1)
= -3 ≠ 0
∴ A^-1 exists.
A₁₁ = -3, A₁₂ = 2, A₁₃ = -1

Question 6.
Show that \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) = (a+b +c)³
Solution:

= (a + b + c)³ (1 – 0) = (a + b + c)³

Question 7.
Find the inverse of the following matrix: \(\left[\begin{array}{lll}
0 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 0
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
0 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 0
\end{array}\right]\)
|A| = 2(- 4) = – 8 ≠ 0
∴ A^-1 exists.
A₁₁ = 0, A₁₂ = 0, A₁₃ = – 4
A₂₁ = 0, A₂₂ = – 4, A₂₃ = 0
A₃₁ = 0,A₃₂ = 0, A₃₃ = – 4
∴ The matrix of cofactors

Question 8.
If the matrix A is such that \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)A = \(\left[\begin{array}{cc}
-4 & 1 \\
7 & 7
\end{array}\right]\), find A.
Solution:

Question 9.
Show that (a + 1) is a factor of \(\left|\begin{array}{ccc}
a+1 & 2 & 3 \\
1 & a+1 & 3 \\
3 & -6 & a+1
\end{array}\right|\).
Solution:

= (a + 1) (a² + 2a + 1 + 18) – 2(a + 1 – 9) + 3(-6 – 3a – 3)
= (a + 1)(a² + 2a + 19) – 2a + 16 – 27 – 9a
= (a + 1) (a² + 2a + 19) – 11a – 11
= (a + 1) (a² + 2a + 19) – 11(a + 1)
= (a + 1) (a² + 2a + 8)
⇒ (a + 1) is a factor of the given determinant.

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 0 \\
5 & 1
\end{array}\right]\) show that for no values of α, A² = B.
Solution:

⇒ α2 = 1 and α + 1 = 5
⇒ α = ± 1 and α = 4
Which is not possible.
There is no α for which A² = B

Question 12.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then show that A^k = \(\left[\begin{array}{cc}
1+2 \mathrm{k} & -4 \mathrm{k} \\
\mathrm{k} & 1-2 \mathrm{k}
\end{array}\right]\), k ∈ N.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 2 \\
3 & 1 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
1 & 2 \\
3 & -1
\end{array}\right]\), verify that (AB)^T = B^TA^T.

Question 14.
Show that for each real value of λ the system of equations
(λ + 3) + λy = 0
x + (2λ + 5)y = 0 has a unique solution.
Solution:
Given system of equations is a
homogeneous system of linear
equations.
Now
Δ = \(\left|\begin{array}{cc}
\lambda+3 & \lambda \\
1 & 2 \lambda+5
\end{array}\right|\)
= (λ + 3)(2λ + 5) – λ
= 2λ² + 11λ + 15 – λ
= 2λ² + 10λ + 15
As for 2λ² + 10A + 15, D = 100 – 120 < 0
the polynomial 2λ² + 10λ + 15 has no roots i.e. Δ ≠ 0.
Thus the system has a unique trivial solution for every real value of λ.

Question 15.
If A and B are square matrices of same order then show by means of an example that AB ≠ BA in general.
Solution:

∴ AB ≠ BA we have AB ≠ BA in general.

Question 16.
If A = \(\left|\begin{array}{cc}
0 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 0
\end{array}\right|\), then prove that det{(I + A)(I – A)-1} = 1
Solution:

Question 17.
Solve for x: \(\left|\begin{array}{ccc}
15-2 x & 11 & 10 \\
11-3 x & 17 & 16 \\
7-x & 14 & 13
\end{array}\right|\) = 0
Solution:

Question 18.
If A = \(\left[\begin{array}{ccc}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\) find A³ – A².
Solution:

A² = A ⇒ A²A = A²
⇒ A³ – A² = 0

Question 19.
Prove that: A = \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -2 & -3 \\
3 & 1 & -8
\end{array}\right|\) ⇒ A² – 5A + 71 = 0.
Solution:

Question 20.
Test whether the following system of equations have non-zero solution.
Write the solution set:
2x + 3y + 4z = 0,
x – 2y – 3z = 0,
3x + y – 8z = 0.
Solution:
Given equations are
2x + 3y + 4z = 0
x – 2y – 3z = 0
3x + y – 8z = 0
Now \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -2 & -3 \\
3 & 1 & -8
\end{array}\right|\)
= 2(19) – 3(1) + 4(7) 0
∴ The system has no non-zero solution.
The solution set is x = 0; y = 0, z = 0.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(a)

Question 1.
Two balls are drawn from a bag containing 5 white and 7 black balls. Find the probability of selecting 2 white balls if
Solution:
Two balls are drawn from a bag containing 5 white and 7 black balls.
∴ |S| = 12.

(i) the first ball is replaced before drawing the second.
Solution:
The 1st ball is replaced before the 2nd ball is drawn. We are to select 2 white balls. So in both the draws we will get white balls. Drawing a white ball in 1st draw and in 2nd draw are independent events.
Probability of getting 2 white balls = \(\frac{5}{12}\) × \(\frac{5}{12}\) = \(\frac{25}{144}\)

(ii) the first ball is not replaced before drawing the second.
Solution:
Here the 1st ball is not replaced before the 2nd ball is drawn. Since we are to get 2 white balls in each draw, we must get a white ball.
Now probability of getting a white ball in 1st draw = \(\frac{5}{12}\).
Probability of getting a white ball in 2nd = \(\frac{4}{11}\).
Since the two draws are independent, we have the probability of getting 2 white balls
= \(\frac{5}{12}\) × \(\frac{4}{11}\) = \(\frac{20}{132}\).

Question 2.
Two cards are drawn from a pack of 52 cards; find the probability that
(i) they are of different suits.
(ii) they are of different denominations.
Solution:
Two cards are drawn from a pack of 52 cards. The cards are drawn one after another. Each suit has 13 cards.
|S| = ⁵²C₂
(i) As the two cards are of different suits, their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\)
(ii) Each denomination contains 4 cards. As the two cards drawn are of different denominations, their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\)

Question 3.
Do both parts of problem 2 if 3 cards drawn at random.
Solution:
(i) 3 cards are drawn one after another. As they are of different suits, we have their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\) × \(\frac{26}{50}\).
(ii) As the 3 cards are of different denominations, we have their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\) × \(\frac{44}{50}\).

Question 4.
Do both parts of problem 2 if 4 cards are drawn at random.
Solution:
(i) 4 cards are drawn one after another. As they are of different suits, we have their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\) × \(\frac{26}{50}\) × \(\frac{13}{49}\).
(ii) As the cards are of different denominations, we have their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\) × \(\frac{44}{50}\) × \(\frac{40}{49}\).

Question 5.
A lot contains 15 items of which 5 are defective. If three items are drawn at random, find the probability that
(i) all three are defective
(ii) none of the three is defective.
Do this problem directly.
Solution:
(i) A lot contains 15 items of which 5 are defective. Three items are drawn at random. As the items are drawn one after another.
Their probability = \(\frac{5}{15}\) × \(\frac{4}{14}\) × \(\frac{3}{13}\)
(ii) As none of the 3 items are defective, we have to draw 3 non-defective items one after another.
Their probability = \(\frac{10}{15}\) × \(\frac{9}{14}\) × \(\frac{8}{13}\)

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of at least 9 if 5 appears on at least one of the dice.
Solution:
A pair of dice is thrown. Let A be the event of getting at least 9 points and B, the event that 5 appears on at least one of the dice.
∴ B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ A ∩ B = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ P (A | B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\) = \(\frac{\frac{5}{36}}{\frac{31}{36}}\) = \(\frac{5}{11}\)

Question 7.
A pair of dice is thrown. If the two numbers appearing are different, find the probability that
(i) the sum of points is 8.
(ii) the sum of points exceeds 8.
(iii) 6 appears on one die.
Solution:
A pair of dice is thrown as two numbers are different
We have |S| = 30
(i) Let A be the. event that the sum of points on the dice is 8, where the numbers on the dice are different.
A = {(2, 6), (3, 5), (5, 3), (6, 2)}
P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}\) = \(\frac{4}{30}\)
(ii) Let B be the event that sum of the points exceeds 8.
B = {(3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5), (4, 6), (6, 4)}
P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}\) = \(\frac{8}{30}\)
(iii) Let C be the event that 6 appears on one die.
C = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
P(C) = \(\frac{|\mathrm{C}|}{|\mathrm{S}|}\) = \(\frac{10}{30}\) = \(\frac{1}{3}\)

Question 8.
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. A student is selected at random.
Solution:
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. Let A be the event that a student fails in Mathematics and B be the events that he fails in English.
P(A) = \(\frac{30}{100}\), P(B) = \(\frac{20}{100}\)
Where |S| = 100, P (A ∩ B) = \(\frac{10}{100}\)

(i) If he has failed in English, what is the probability that he has failed in Mathematics?
Solution:
If he has failed in English, then the probability that he has failed in Mathematics.
i.e., P\(\left(\frac{A}{B}\right)\) = \(\frac{P(A \cap B)}{P(B)}\) = \(\frac{\frac{10}{100}}{\frac{20}{100}}\) = \(\frac{1}{2}\)

(ii) If he has failed in Mathematics, what is the probability that he has failed in English?
Solution:
If he has failed in Mathematics, then the probability that he has failed in English
i.e., P\(\left(\frac{B}{A}\right)\) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{\frac{10}{100}}{\frac{30}{100}}\) = \(\frac{1}{3}\)

(iii) What is the probability that he has failed in both?
Solution:
Probability that he has failed in both
i.e., P (A ∩ B) = \(\frac{10}{100}\) = \(\frac{1}{10}\)

Question 9.
IfA, B are two events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
Find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | B^c)
(iv) P (B | A^c)
Solution:
A and B are two set events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
We have
P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
or, 0.6 = 0.3 + 0.4 – P (A ∩ B)
or, P (A ∩ B) = 0.7 – 0.6 = 0.1

Question 10.
If A, B are events such that P(A) = 0.6, P(B) = 0.4 and P (A ∩ B) = 0.2, then find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | B^c)
(iv) P (B | A^c)
Solution:
A and B are events such that
P(A) = 0.6, P(B) = 0.4, P (A ∩ B) = 0.2

Question 11.
If A and B are independent events, show that
(i) A^c and B^c are independent,
(ii) A and B^c are independent,
(iii) A^c and B are independent.
Solution:
A and B are independent events.
(i) We have P (A ∩ B) = P(A). P(B)
P (A’ ∩ B’) = P (A ∪ B)’ = 1 – P (A ∪ B)
= 1 – [P(A) + P(B) – P (A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= 1 [1 – P(A)] – P(B) [1 – P(A)]
= [1 – P(A)] [1 – P(B)] = P(A’) P(B’)
∴ A’ and B’ are independent events.

(ii) P (A ∩ B^c) = P (A – B)
= P(A) – P (A ∩ B)
= P(A) – P(A) P(B)
= P(A) [1 – P(B)]
= P(A). P(B^c).
∴ A and B^c are independent events.

(iii) P (A^c ∪ B) = P (B – A)
= P(B) – P (A ∩ B)
= P(B) – P(A) P(B)
= P(B) [1 – P(A)] = P(B) P(A^c)
∴ A^c and B are independent events.

Question 12.
Two different digits are selected at random from the digits 1 through 9.
(i) If the sum is even, what is the probability that 3 is one of the digits selected?
(ii) If the sum is odd, what is the probability that 3 is one of the digits selected?
(iii) If 3 is one of the digits selected, what is the probability that the sum is odd?
(iv) If 3 is one of the digits selected, what is the probability that the sum is even?
Solution:
Two different digits are selected at random from the digits 1 through 9.
(i) Let A be the event that the sum is even and B be the event that 3 is one of the number selected.
We have to find P (B | A).
There are 4 even digits and 5 odd digits.
∴ The sum is even if both the numbers are odd or both are even.
∴ |A| = ⁴C₂ + ⁵C₂ = 6 + 10 = 16
∴ P(A) = \(\frac{16}{{ }^9 \mathrm{C}_2}\) = \(\frac{16}{36}\)
Also B = {(1, 3), (5, 3), (7, 3), (9, 3), (3, 2), (3, 4), (3, 8), (3, 6)}
∴ A ∩ B = {(1, 3), (5, 3), (7, 3), (9, 3)}
P(B) = \(\frac{8}{36}\) P (A ∩ B) = \(\frac{4}{36}\)
P(\(\frac{B}{A}\)) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{\frac{4}{36}}{\frac{16}{36}}\) = \(\frac{1}{4}\)

(ii) Let A be the event that the sum is odd. The sum is odd if one of the numbers selected is odd and other is even.
∴ P(A) = \(\frac{5}{9}\) × \(\frac{4}{8}\) + \(\frac{4}{9}\) × \(\frac{5}{8}\) = \(\frac{20}{36}\)
Let B be the event that one of the numbers selected is 3.
∴ B = {(1, 3), (2, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3)}
∴ A ∩ B = {(2, 3), (4, 3), (6, 3), (8, 3)}

Question 13.
If P(A) = 0.4, P (B | A) = 0.3 and P (B^c | A^c) = 0.2. find
(i) P (A | B)
(ii) P (B | A^c)
(iii) P(B)
(iv) P(A^c)
(v) P (A ∪ B)
Solution:
P(A) = 0.4, P (B | A) = 0.3 and P (B^c | A^c) = 0.2.


(iii) P(B) = 0.6
= \( \frac{6}{10} \)
= \( \frac{3}{5} \)

(iv) P(A^c) = 1 – P(A)
= 1 – 0.4 = 0.6
= \( \frac{6}{10} \) = \( \frac{3}{5} \)

(v) P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
= 0.4 + 0.6 – 0.12
= 1.0 – 0.12 = 0.88

Question 14.
If P(A) = 0.6, P (B | A) = 0.5, find P (A ∪ B) if A, B are independent.
Solution:
P(A) = 0.6, P (B | A) = 0.5
We have P (B | A) = \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 0.5
or, P (B ∩ A) = 0.5 × P(A)
= 0.5 × 0.6 = 0.3
As A and B are independent events, we have
P (B ∩ A) = P(B) P(A) = 0.3
or, P(B) = \(\frac{0.3}{\mathrm{P}(\mathrm{A})}\)
= \(\frac{0.3}{0.6}\)
= \(\frac{1}{2}\) = 0.5
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.5 – 0.3 = 0.8

Question 15.
Two cards are drawn in succession from a deck of 52 cards. What is the probability that both cards are of denomination greater than 2 and less than 9?
Solution:
Two cards are drawn in succession from a deck of 52 cards.
There are 6 denominations which are greater than 2 and less than 9. So there are 24 cards whose denominations are greater than 2 and less than 9.
∴ Their probability = \(\frac{24}{52}\) × \(\frac{23}{51}\).

Question 16.
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession. Find the probability that
(i) all three are of the same colour.
(ii) each colour is represented.
Solution:
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession.
(i) The 3 balls drawn are of same colour.
∴ Probability of drawing 3 balls of black colour
= \(\frac{5}{12}\) × \(\frac{4}{11}\) × \(\frac{3}{10}\) = \(\frac{1}{22}\)
Probability of drawing 3 white balls
= \(\frac{7}{12}\) × \(\frac{6}{11}\) × \(\frac{5}{10}\) = \(\frac{7}{44}\)
∴ Probability of drawing 3 balls of same colour
= \(\frac{5}{12}\) × \(\frac{4}{11}\) × \(\frac{3}{10}\) + \(\frac{7}{12}\) × \(\frac{6}{11}\) × \(\frac{5}{10}\) = \(\frac{9}{44}\)

(ii) Balls of both colour will be drawn. If B represents black ball and W represents the white ball.
∴ The possible draws are WWB, WBW, BWW.

Question 17.
A die is rolled until a 6 is obtained. What is the probability that
(i) you end up in the second roll
(ii) you end up in the third roll.
Solution:
A die is rolled until a 6 is obtained
(i) We are to end up in the 2nd roll i.e., we get 6 in the 2nd roll. Let A be the event of getting a 6 in one roll of a die.
∴ P(A) = \(\frac{1}{6}\) ⇒ P(A’) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∴ Probability of getting a 6 in the 2nd roll
= \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{5}{36}\)
(ii) Probability of getting a 6 in the 3rd roll
= \(\frac{5}{6}\) × \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{25}{216}\)

Question 18.
A person takes 3 tests in succession. The probability of his (her) passing the first test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test. Find the probability of his (her) passing at least 2 tests.
Solution:
A person takes 3 tests in succession. The probability of his passing the 1st test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test.
Let S denotes the success (passing) in a test and F denotes the failure in a test.
∴ P(S) = 0.8
∴ P(F) = 1 – P(S) = 1 – 0.8 = 0.2
We have the following mutually exclusive cases:

Event			Probability
S	S	S	0.8 × 0.8 × 0.8 = 0.512
S	S	F	0.8 × 0.8 × 0.2 = 0.128
S	F	S	0.8 × 0.2 × 0.5 = 0.080
F	S	S	0.2 × 0.5 x 0.8 = 0.080

∴ Probability of atleast 2 successes
= 0.512 + 0.128 + 0.080 + 0.080
= 0.8 = \(\frac{4}{5}\)

Question 19.
A person takes 4 tests in succession. The probability of his passing the first test is p, that of his passing each succeeding test is p or y depending on his passing or failing the preceding test. Find the probability of his passing
(i) at least three test
(ii) just three tests.
Solution:
A person takes 4 tests in succession. The probability of his passing the 1st test is P, that of his passing each succeeding test is P or P/2 depending bn his passing or failing the preceding test. Let S and F denotes the success and failure in the test.
∴ P(S) = P, P(F) = 1 – P
We have the following mutually exclusive tests:

Question 20.
Given that all three faces are different in a throw of three dice, find the probability that
(i) at least one is a six
(ii) the sum is 9.
Solution:
Three dice are thrown once showing different faces in a throw.
|S| = 6³ = 216
Let A be the event that atleast one is a six.
Let B be the event that all three faces are different.
|B| = ⁶6₃
(i) Now A^c is the event that there is no six. A^c ∩ B is the event that all 3 faces are different and 6 does not occur.
|A^c ∩ B| = ⁵C₃
P (A^c | B) = \(\frac{P\left(A^C \cap B\right)}{P(B)}\)
= \(\frac{{ }^5 \mathrm{C}_3 / 216}{{ }^6 \mathrm{C}_3 / 216}\) = \(\frac{1}{2}\)

(ii) Let A be the event that the sum is 9.
A ∩ B = {(1,3, 5), (1,5, 3), (3, 5,1), (3, 1, 5), (5, 1, 3), (5, 3, 1), (1, 2, 6), (1, 6, 2), (2, 1, 6), (2, 6, 1), (6, 2, 1), (6, 1, 2), (2, 3, 4), (2, 4, 3), (3, 2, 4), (2, 3, 4), (3, 4, 2), (4, 3, 2)}
|A ∩ B| = 18
P (A | B) = \(\frac{P(A \cap B)}{P(B)}\)
= \(\frac{18 / 216}{20 / 216}\) = \(\frac{9}{10}\)

Question 21.
From the set of all families having three children, a family is picked at random.
(i) If the eldest child happens to be a girl, find the probability that she has two brothers.
(ii) If one child of the family is a son, find the probability that he has two sisters.
Solution:
A family is picked up at random from a set of families having 3 children.
(i) The eldest child happens to be a girl. We have to find the probability that she has two brothers. Let G denotes a girl and B denotes a boy.
∴ P(B) = \(\frac{1}{2}\), P(G) = \(\frac{1}{2}\)
P(BB | G) = P(B) × P(G) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

(ii) The one child of the family is a son. We have to find the probability that he has two sisters. We have the following mutually exclusive events:
BGG, GBG, GGB.
∴ The required probability
= P(B) × P(G) × P(G) + P(G) × P(B) + P(G) + P(G) × P(G) × P(B)
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\)

Question 22.
Three persons hit a target with probability \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. If each one shoots at the target once,
(i) find the probability that exactly one of them hits the target
(ii) if only one of them hits the target what is the probability that it was the first person?
Solution:
Three persons hit a target with probability
\(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\)
Let A, B, C be the events that the 1st person, 2nd person, 3rd person hit the target respectively.

(i) As the events are independent, the probability that exactly one of them hit the target
= P(AB’C’) + P(A’BC’) + P(A’B’C)
= P(A) P(B’) P(C’) + P(A’) P(B) P(C’) + P(A’) P(B’) P(C)

(ii) Let E₁ be the event that exactly one person hits the target.
∴ P(E₁) = \(\frac{11}{24}\)
Let E₂ be the event that 1st person hits the target
∴ P(E₂) = P(A) = \(\frac{1}{2}\)
∴ E₁ ∩ E₂ = AB’C’
⇒ P(E₁ ∩ E₂)
= P(A) × P(B’) P(C’) = \(\frac{6}{24}\)
∴ P(E₂ | E₁) = \(\frac{6 / 24}{11 / 24}\) = \(\frac{6}{11}\)

Odisha State Board CHSE Odisha Class 12 Psychology Solutions Unit 4 Long Answer Questions Part-3.

CHSE Odisha 12th Class Psychology Unit 4 Long Answer Questions Part-3

Long Questions With Answers

Question 1.
What are the different types of psychotherapy? On what basis are they classified?
Answer:
Different types of psychotherapy are:

• Psychodynamic therapy
• Behaviour therapy
• Humanistic-existential therapy
• Biomedical therapy

Also, there are many alternative therapies such as yoga, meditation, acupuncture, herbal remedies etc.
Basis of classification of psychotherapy:

On the cause which has led to the problem:
Psychodynamic therapy is of the view that intrapsychic conflicts, i.e. the conflicts that are present within the psyche of the person, are the source of psychological problems.

On how did the cause come into existence:
The psychodynamic therapy, unfulfilled desires of childhood and unresolved childhood fears lead to intrapsychic conflicts.

What is the chief method of treatment?
Psychodynamic therapy uses the methods of free association and reporting of dreams to elicit the thoughts and feelings of the client.

What is the nature of the therapeutic relationship between the client and the therapist?
Psychodynamic therapy assumes that the therapist understands the client’s intrapsychic conflicts better than the client and hence it is the therapist who interprets the thoughts and feelings of the client to her/him so that s/he gains an understanding of the same.

What is the chief benefit to the client?
Psychodynamic therapy values emotional insight as the important benefit that the client derives from the treatment. Emotional insight is present when the client understands her/his conflicts intellectually; is able to accept the same emotionally, and is able to change her/his emotions towards the conflicts.

On the duration of treatment:
Hie duration of classical psychoanalysis may continue for several years. However, several recent versions of psychodynamic therapies are completed in 10—15 sessions.

Question 2.
A therapist asks the client to reveal all her/his thoughts including early childhood experiences. Describe the technique and type of therapy being used.
Answer:
In this case psychodynamic, therapy is used in the treatment of the client. Since the psychoanalytic approach views intrapsychic conflicts to be the cause of the psychological disorder. The first step in the treatment is to elicit this intrapsychic conflict. Psychoanalysis has invented free association and dream interpretation as two important methods for eliciting intrapsychic conflicts.

The free association method is the main method for understanding the client’s problems. Once a therapeutic relationship is established, and the client feels comfortable, the therapist makes her/him lie down on the couch, close her/his eyes and asks her/him to speak whatever comes to mind without censoring it in any way. The client is encouraged to freely associate one thought with another, and this method is called the method of free association.

The censoring superego and the watchful ego are kept in abeyance as the client speaks whatever comes to mind in an atmosphere that is relaxed and trusting. As the therapist does not interrupt, the free flow of ideas, desires and conflicts of the unconscious, which had been suppressed by the ego, emerges into the conscious mind. This free uncensored verbal narrative of the client is a window into the client’s unconscious to which the therapist gains access.

Along with this technique, the client is asked to write down her/his dreams upon waking up. Psychoanalysts look upon dreams as symbols of the unfulfilled desires present in the unconscious. The images of dead dreams are symbols which signify intrapsychic forces. Dreams use symbols because they are indirect expressions and hence would not alert the ego.

If the unfulfilled desires are expressed directly, the ever-vigilant ego would suppress them and that would leads to anxiety. These symbols are interpreted according to an accepted convention of translation as indicators of unfulfilled desires and conflicts.

Question 3.
Discuss the various techniques used in behaviour therapy.
Answer:
Various techniques used in behaviour therapy:
A range of techniques is available for changing behaviour. The principles of these techniques are to reduce the arousal level of the client, alter behaviour through classical conditioning or operant conditioning with different contingencies of reinforcements, as well as to use vicarious learning procedures, if necessary. Negative reinforcement and aversive conditioning are the two major techniques of behaviour modification.

Negative reinforcement refers to following an undesired response with an outcome that is gainful or not liked. For example, one learns to put on woollen clothes, bum firewood or use electric heaters to avoid the unpleasant cold weather. One learns to move away from dangerous stimuli because they provide negative reinforcement.

Aversive conditioning refers to the repeated association of an undesired response with an aversive consequence. For example, an alcoholic is given a mild electric shock and asked to smell the alcohol. With repeated pairings, the smell of alcohol is aversive as the pain of the shock is associated with it and the person will give up alcohol.

Positive reinforcement is given to increase the deficit if adaptive behaviour occurs rarely. For example, if a child does not do homework regularly, positive reinforcement may be used by the child’s mother by preparing the child’s favourite dish whenever s/he does homework at the appointed time. The positive reinforcement of food will increase the behaviour of doing homework at the appointed time.

The token economy in which persons with behavioural problems can be given a token as a reward every time a wanted behaviour occurs. The tokens are collected and exchanged for a reward such as an outing for the patient or a treat for the child. Unwanted behaviour can be reduced and waited behaviour can be increased simultaneously through differential reinforcement.

Positive reinforcement for the wanted behaviour and negative reinforcement for the unwanted behaviour attempted together may be one such method. The other method is to positively reinforce the wanted behaviour and ignore the unwanted behaviour. The latter method is less painful and equally effective. For example, let us consider the case of a girl who sulks and cries when she is not taken to the cinema when she asks.

The parent is instructed to take her to the cinema if she does not cry and sulk but not to take her if she does. Further, the parent is instructed to ignore the girl when she cries and sulks. The wanted behaviour of politely asking to be taken to the cinema increases and the unwanted behaviour of crying and sulking decreases.

Question 4.
Explain with the help of an example how cognitive distortions take place.
Answer:
Cognitive distortions are ways of thinking which are general in nature but which distort reality in a negative manner. These patterns of thought are called dysfunctional cognitive structures. They lead to errors of cognition about social reality. Aaron Beck’s theory of psychological distress states that childhood experiences provided by the family and society develop core, schemas or systems, which include beliefs and action patterns in the individual.

Thus, a client, who was neglected by the parents as a child, develops the core schema of “I am not wanted”. During the course of their life, a critical incident occurs in her/his life. S/he is publicly ridiculed by the teacher in school. This critical incident triggers the core schema of “I am not wanted” leading to the development of negative automatic thoughts. Negative thoughts are persistent irrational thoughts such as “nobody loves me”, “I am ugly”, “l am stupid”, “I will not succeed”, etc.

Such negative automatic thoughts are characterised by cognitive distortions. Repeated occurrence of these thoughts leads to the development of feelings of anxiety and depression. The therapist uses questioning, which is a gentle, non-threatening disputation of the client’s beliefs and thoughts. Examples of such questions would be, “Why should everyone love you?”, “What does it mean to you to succeed?” etc.

Question 5.
Which therapy encourages the client to seek personal growth and actualise their potential? Write about the therapies which are based on this principle.
Answer:
Humanistic-existential therapy encourages the client to seek personal growth and actualise their potential. It states that psychological distress arises from feelings of loneliness, alienation, and an inability to find meaning and genuine fulfilment in life.
The therapies which are based on this principle are:

Existential therapy:
There is a spiritual unconscious, which is the storehouse of love, aesthetic awareness, and values of life. Neurotic anxieties arise when the problems of life are attached t6 the physical, psychological or spiritual aspects of one’s existence. Frankl emphasised the role of spiritual anxieties in leading to meaninglessness and hence it may be called existential anxiety, i.e. neurotic anxiety of spiritual origin.

Client-centred therapy:
Client-centred therapy was given by Carl Rogers. He combined scientific rigour with the individualised practice of client-centred psychotherapy. Rogers brought into psychotherapy the concept of self, with freedom and choice as the core of one’s being. The therapy provides a warm relationship in which the client can reconnect with her/his disintegrated feelings. The therapist shows empathy, i.e. understanding the client’s experience as if it were her/his own, is warm and has unconditional positive regard, i.e. total acceptance of the client as s/he is. Empathy sets up an emotional resonance between the therapist and the client.

Gestalt therapy:
The German word gestalt means ‘whole’. This therapy was given by Frederick (Fritz) Peris together with his wife Laura Peris. The goal of gestalt therapy is to increase an individual’s self-awareness and self-acceptance. The client is taught to recognise the bodily processes and die emotions that are being blocked out from awareness. The therapist does this by encouraging the client to act out fantasies about feelings and conflicts. This therapy can also be used in group settings.

Question 6.
What are the factors that contribute to healing in psychotherapy? Enumerate some of the alternative therapies.
Answer:
Factors Contributing to Healing in Psychotherapy are:

A major factor in healing is the techniques adopted by the therapist and the implementation of the same with the patient/client. If the behavioural system and the CBT school are adopted to heal an anxious client, the relaxation procedures and the cognitive restructuring largely contribute to the healing.

The therapeutic alliance, which is formed between the therapist and the patient/ client, has healing properties, because of the regular availability of the therapist and the warmth and empathy provided by the therapist.

At the outset of therapy, while the patient/client is being interviewed in the initial sessions to understand the nature of the problem, s/he unburdens the emotional problems being faced. This process of emotional unburdening is known as catharsis and it has healing properties.

There are several non-specific factors associated with psychotherapy. Some of these factors are attributed to the patient/client and some to the therapist. These factors are called non-specific because they occur across different systems of psychotherapy and across .different clients/patients and different therapists. Non-specific factors attributable to the client/patient are the motivation for change, the expectation of improvement due to the treatment, etc.

These are called patient variables. Non-specific factors attributable to the therapist are positive nature, absence of unresolved emotional conflicts, presence of good mental health, etc. These are called therapist variables. Some of the alternative therapies are Yoga, meditation, acupuncture, herbal remedies etc.

Question 7.
What are the techniques used in the rehabilitation of the mentally ill?
Answer:
The treatment of psychological disorders has two components, i.e. reduction of symptoms, and improving the level of functioning or quality of life. In the case of milder disorders such as generalised anxiety, reactive depression or phobia, reduction of symptoms is associated with an improvement in the quality of life. However, in the case of severe mental disorders such as schizophrenia, reduction of symptoms may not be associated with an improvement in the quality of life.

Many patients suffer from negative symptoms such as disinterest and lack of motivation to do work or interact with people. The aim of rehabilitation is to empower the patient to become a productive member of society to the extent possible. In rehabilitation, the patients are given occupational therapy, social skills training, and vocational therapy. In occupational therapy, the patients are taught skills such as candle making, paper bag making and weaving to help them to form a work discipline.

Social skills. training helps the patients to develop interpersonal skills through role play, imitation and instruction. The objective is to teach the patient to function in a Social group. Cognitive retraining is given to improve the basic cognitive functions of attention, memory and executive functions. After the patient improves sufficiently, vocational training is given wherein the patient is helped to gain the skills necessary to undertake productive employment.

Question 8.
How would a social learning theorist account for a phobic fear of lizards/ cockroaches? How would a psychoanalyst account for the same phobia?
Answer:
Systematic desensitisation is a technique introduced by Wolpe for treating phobias or irrational fears. The client is interviewed to elicit fear-provoking situations and together with the client, the therapist prepares a hierarchy of anxiety-provoking stimuli with the least anxiety-provoking stimuli at the bottom of the hierarchy. The therapist relaxes the client and asks the client to think about the least anxiety-provoking situation.

The client is asked to stop thinking of the fearful situation if the slightest tension is felt. Over sessions, the client is able to imagine more severe fear-provoking situations while maintaining relaxation. The client gets systematically desensitised to the fear.

Question 9.
Should Electroconvulsive Therapy (ECT) be used in the treatment of mental disorders?
Answer:
Yes, Electro-convulsive Therapy (ECT) can be used in the treatment of mental disorders. Electroconvulsive Therapy (ECT) is another form of biomedical therapy. Mild electric shock is given via electrodes to the brain of the patient to induce convulsions. The shock is given by the psychiatrist only when it is necessary for the improvement of the patient. ECT is not a routine treatment and is given only when drugs are not effective in controlling, the symptoms of the patient.

Question 10.
What kind of problems is cognitive behaviour therapy best suited for?
Answer:
Cognitive behaviour treatment best suited for a wide range of psychological disorders such as anxiety, depression, panic attacks, borderline personality, etc. adopts a bio-CBT psychosocial approach to the delineation of psychopathology. It combines cognitive therapy with behavioural techniques.

Question 11.
What is the nature and process of therapeutic approaches?
Answer:
Psychotherapy is a voluntary relationship between the one seeking treatment or the client and the one who treats the therapist. The purpose of the relationship is to help the client to solve the psychological problems faces by her or him. The relationship is conducive to building the trust of the client so that problems may be freely discussed.

Psychotherapies aim at changing maladaptive behaviours, decreasing the sense of personal distress and helping the client to adapt better to her/his environment. The inadequate marital, occupational and social adjustment also requires that major changes be made in an individual’s personal environment. All psychotherapeutic approaches have the following characteristics:

• there is the systematic application of principles underlying the different theories of therapy.
• persons who have received practical training under expert supervision can practice psychotherapy and not everybody. An untrained person may unintentionally cause more harm than good.
• the therapeutic situation involves a therapist and a client who seeks and receives help for her/his emotional problems (this person is the focus of attention in the therapeutic process).
• the interaction of these two persons — the therapist and the client— results in the consolidation/formation of the therapeutic relationship. This is a confidential, interpersonal and dynamic relationship.

This human relationship is central to any sort of psychological therapy and is the vehicle for change. All psychotherapies aim at a few or all of the following goals :

• Reinforcing the client’s resolve for betterment.
• Lessening emotional pressure.
• Unfolding the potential for positive growth.
• Modifying habits.
• Changing thinking patterns
• Increasing self-awareness.
• Improving interpersonal relations and communication.
• Facilitating decision-making.
• Becoming aware of one’s choices in life.
• Relating to one’s social environment in a more creative and self-aware manner.

Question 12.
What is the relationship between the client and therapist?
Answer:
Therapeutic Relationship :
The special relationship between the client and the therapist is known as the therapeutic relationship or alliance. It is neither a passing acquaintance nor a permanent and lasting relationship. There are two major components of a therapeutic alliance. The first component is the contractual nature Of the relationship in which two willing individuals, the client and the therapist, enter into a partnership which aims at helping the client overcome her/his problems.

The second component of the therapeutic alliance is the limited duration of the therapy. This alliance lasts until the client becomes able to deal with her/his problems and take control of her/ his life. This relationship has several unique properties. It is a trusting and confiding relationship. The high level of trust enables the client to unburden herself/himself to the therapist and confide her/his psychological and personal problems to the latter.

The therapist encourages this by being accepting, empathic, genuine and warm to the client. The therapist conveys by her/his words and behaviours that s/he is not judging the client and will continue to show the same positive feelings towards the client even if the client is rude or confides in all the ‘wrong’ things that s/he may have done or thought about. This is the unconditional positive regard that the therapist has for the client. The therapist has empathy for the client.

Empathy:
Empathy is different from sympathy and intellectual understanding of another person’s situation. Iii sympathy, one has compassion and pity towards, the .suffering of another but is not able to feel like the other person. Intellectual understanding is cold in the sense that the person is unable to feel like the other person and does not feel sympathy either. On the other hand, empathy is present when one is able to understand the plight of another person and feel like the other person.

It means understanding things from the other person’s perspective, i.e. putting oneself in the other person’s shoes. Empathy enriches the therapeutic relationship and transforms it into a healing relationship. The therapeutic alliance also requires that the therapist must keep strict confidentiality of the experiences, events, feelings or thoughts disclosed by the client. The therapist must not exploit the trust and confidence of the client in any way. Finally, it is a professional relationship and must remain so.

Question 13.
Write the types of therapies.
Answer:
Though all psychotherapies aim at removing human distress and fostering effective behaviour, they differ greatly in concepts, methods, and techniques. Psychotherapies may be classified into three broad groups, viz. the psychodynamic, behaviour sad existential psychotherapies. In terms of chronological order, psychodynamic therapy emerged first followed by behaviour therapy while existential therapies which are also called the third force, emerged last. The classification of psychotherapies is based on the following parameters:

What is the cause, which has led to the problem?
Psychodynamic therapy is of the view that intrapsychic conflicts, i.e. the conflicts that are present within the psyche of the person, are the source of psychological problems. According to behaviour therapies, psychological problems arise due to faulty learning of behaviours and cognitions. Existential therapies postulate that questions about the meaning of one’s life and existence are the cause of psychological problems.

How did the cause come into existence?
In psychodynamic therapy, unfulfilled desires of childhood and unresolved childhood fears lead to intrapsychic conflicts. Behaviour therapy postulates that faulty conditioning patterns, faulty learning, and faulty thinking and beliefs lead to maladaptive behaviours that, in turn, lead to psychological problems. Existential therapy places importance on the present. It is the current feelings of loneliness, alienation, a sense of the futility of one’s existence, etc., which cause psychological problems.

What is the chief method of treatment?
Psychodynamic therapy uses the methods of free association and reporting of dreams to elicit the thoughts and feelings of the client. This material is interpreted to the client to help her/him to confront and resolve the conflicts and thus overcome problems. Behaviour therapy identifies faulty conditioning patterns and sets up alternate behavioural contingencies to improve behaviour.

The cognitive methods employed in this type of therapy challenge the faulty thinking patterns of the client to help her/him overcome psychological distress. Existential therapy provides a therapeutic environment which is positive, accepting and non-judgmental. The client is able to talk about the problems and the therapist acts as a facilitator. The client arrives at the solutions through a process of personal growth.

What is the nature of the therapeutic relationship between the client and the therapist?
Psychodynamic therapy assumes that the therapist understands the client’s intrapsychic conflicts better than the client and hence it is the therapist who interprets the. thoughts and feelings of the client to her/him so that s/he gains an understanding of the same. Behaviour therapy assumes that the therapist is able to discern the faulty behaviour and thought patterns of the client.

It further assumes that the therapist is capable of finding out the correct behaviour and thought patterns, which would be adaptive for the client. Both psychodynamic and behaviour therapies assume that the therapist is capable of arriving at solutions to the client’s problems. In contrast to these therapies, existential therapies emphasise that the therapist merely provides a warm, empathic relationship in . which the client feels secure to explore the nature and causes of her/his problems by herself/ himself.

What is the chief benefit to the client?
Psychodynamic therapy values emotional insight as the important benefit that the client derives from the treatment. Emotional insight is present when the client understands her/his conflicts intellectually; is able to accept the same emotionally and is able to change her/his emotions towards the conflicts. The client’s symptoms and distresses reduce as a consequence of this emotional insight.

Behaviour therapy considers changing faulty behaviour and thought patterns to adaptive ones as the chief benefit of the treatment. Instituting adaptive or healthy behaviour and thought patterns ensures the reduction of distress and the removal of symptoms. Humanistic therapy values personal growth as the chief benefit. Personal growth is the process of gaining an increasing understanding of oneself and one’s aspirations, emotions and motives.

What is the duration of treatment?
The duration of classical psychoanalysis may continue for several years. However, several recent versions of psychodynamic therapies are completed in 10-15 sessions. Behaviour and cognitive behaviour therapies as well as existential therapies are shorter and are completed in a few months. Thus, different types of psychotherapies differ on multiple parameters.

However, they all share the common method of providing treatment for psychological distress’ through psychological means. The therapist, the therapeutic relationship, and the process of therapy become the agents of change in the client leading to the alleviation of psychological distress. The process of psychotherapy begins by formulating the client’s problem.