Class 12

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.
Maximize Z = 5x₁+ 6x₂
Subject to: 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation, we get 2x₁ + 3x₂ = 6
Step – 2 Let us draw the graph

x1	3	0
x2	0	0

Step – 3 Clearly (0,0) statisfies 2x₁ + 3x₂ ≤ 6
The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).
Step – 4

Corner point	Z = 5x1+ 6x2
0(0.0)	0
A(3,0)	15 → maximum
B(0,2)	12

Z is maximum at A (3,0)
∴ The solution of LPP is x₁ = 3, x₂ = 0
Z_max = 15

Question 2.
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation we get x₁ + 2x₂ = 0
Step – 2 Let us draw the graph of x₁ + 2x₂ = 4

x1	0	4
x2	2	0

Step – 3 Clearly 0(0,0) does not satisfy
x₁ + 2x₂ > 4, x₁ > 0, x₂ > 0 is the first quadrant.
The feasible region is the shaded region with vertices A(4, 0), B(0, 2).
Step – 4 Z (4, 0) = 24
Z (0, 2) = 14 → minimum
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).
Let us draw the half-plane 6x₁ + 7x₂ < 14

x1	0	3.5
x2	2	-1

As this half-plane has no point common with the feasible region, we have Z is minimum for x₁= 0, x₂ = 2 and the minimum value of Z = 14.

Question 3.
Maximize Z = 20x₁+ 40x₂
Subject to: x₁ + x₂ ≤ 1
6x₁ + 2x₂ ≤ 3
x₁, x₂ ≥ 0.
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ = 1    …. (1)
6x₁ + 2x₂ = 3   …. (2)
x₁, x₂ ≥ 0
Step – 2 Let us draw the graph:
Table – 1

x1	0	1
x2	1	0

Table – 2

x1	0	0.5
x2	1.5	0

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.
Step – 4 Solving
x₁ + x₂ = 1
6x₁ + 2x₂ = 3
we have x₁ = ¼ x₂ = ¾
The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)
Now Z(O) = 0
Z(A) = 10
Z(B) = 40
Z(C) = 20 × ¼ + 40 × ¾ = 35
∴ Z attains maximum at B for x₁= 0, x₂ = 1
Z_max = 40

Question 4.
Minimize: Z = 30x₁ + 45x₂
Subject to: 2x₁ + 6x₂ ≥ 4
5x₁ + 2x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Consider the constraints as equations
2x₁ + 6x₂ = 4
5x₁ + 2x₂ = 5
Step – 2
Table – 1

x1	2	-1
x2	0	1

Table – 2

x1	1	0
x2	0	2.5

Step – 3 Clearly 0(0,0) does not satisfy 2x₁ + 6x₂ ≥ 4 and 5x₁ + 2x₂ ≥ 5.
Thus the shaded region is the feasible region.
Solving the equations we get
x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\).
∴ The vertices are A(2, 0)
B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).
Step – 4 Z(A) = 60
Z(B) = \(\frac{555}{13}\) → minimum
Z(C) = \(\frac{225}{2}\)
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))
Let us draw the half plane
30x₁ + 45x₂ < \(\frac{555}{13}\)

x1	\(\frac{11}{13}\)	0
x2	\(\frac{5}{13}\)	\(\frac{27}{39}\)

As this half plane and the feasible region has no point in common we have Z is minimum for x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\), and Z_min = \(\frac{555}{13}\)

Question 5.
Maximize: Z = 3x₁+ 2x₂
Subject to: -2x₁ + x₂ ≤ 1
x₁ ≤ 2
x₁+ x₂ ≤ 3
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-2x₁ + x₂ = 1        …..(1)
x₁ = 2                   …..(2)
x₁+ x₂ = 3            …..(3)
Step – 2 Let us draw the lines.
Table – 1

x1	0	-1
x2	1	-1

Table – 2

x1	2	2
x2	0	1

Table – 3

x1	0	3
x2	3	0

Step – 3 (0, 0) satisfies all the constraints and x₁, x₂ > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x₁ + x₂ = 1
x₁+ x₂ = 3
we have 3x₁ = 2
⇒ x₁ = \(\frac{2}{3}\), x₂ = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)
From x₁+ x₂ = 3 and x₁ = 2 we have x₁ = 2, x₂ = 1
∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)
Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2
Z is maximum at B.
∴ The solution of given LPP is x₁ = 2, x₂ = 1, Z_(max) = 8.

Question 6.
Maximize: Z = 50x₁+ 60x₂
Subject to: x₁ + x₂ ≤ 5
x₁+ 2x₂ ≤ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + x₂ = 5     ….(1)
x₁+ 2x₂ = 4    ….(2)
Step – 2 Let us draw the graph
Table – 1

x1	5	5
x2	0	0

Table – 2

x1	4	0
x2	0	2

Step – 3 0(0,0) satisfies x₁ + x₂ ≤ 5 and does not satisfy x₁+ 2x₂ ≤ 4
Thus the shaded region is the feasible region.
Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)
∴

Corner point	z = 50x1+ 60x2
A(4,0)	200
B (5,0)	250 → maximum
C(0,5)	300
D(0,2)	120

Z is maximum for x₁ = 0, x₂ = 5, Z_(max) = 300.

Question 7.
Maximize: Z = 5x₁+ 7x₂
Subject to: x₁ + x₂ ≤ 4
5x₁+ 8x₂ ≤ 30
10x₁+ 7x₂ ≤ 35
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get,
x₁ + x₂ = 4           …. (1)
5x₁+ 8x₂ = 30      …. (2)
10x₁+ 7x₂ = 35    …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	4	0
x2	0	4

Table – 2

x1	6	2
x2	0	2.5

Table – 3

x1	0	3.5
x2	5	0

Step – 3 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))
From (1) and (3) we get
x₁ = \(\frac{7}{3}\), x₁ = \(\frac{5}{3}\)
∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))
Step – 4

Corner point	z = 5x1+ 7x2
0(0,0)	0
A(\(\frac{7}{2}\), 0)	\(\frac{35}{2}\)
B(\(\frac{7}{3}\), \(\frac{5}{3}\))	\(\frac{70}{3}\)
C(\(\frac{2}{3}\), \(\frac{10}{3}\))	\(\frac{80}{3}\)
D(0, \(\frac{15}{4}\))	\(\frac{105}{4}\)

Z attains its maximum value \(\frac{80}{3}\) for x₁ = \(\frac{2}{3}\) and x₂ = \(\frac{10}{3}\).

Question 8.
Maximize: Z = 14x₁ – 4x₂
Subject to: x₁ + 12x₂ ≤ 65
7x₁ – 2x₂ ≤ 25
2x₁+ 3x₂ ≤ 10
x₁, x₂ ≥ 0
Also find two other points which maximize Z.
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + 12x₂ = 65   …. (1)
7x₁ – 2x₂ = 25    …. (2)
2x₁ + 3x₂ = 10   …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	65	5
x2	0	5

Table – 2

x1	5	10
x2	5	22.5

Table – 3

x1	5	2
x2	0	2

Step – 3 Clearly 0(0,0) satisfies x₁ + 12x₂ ≤ 65 and 7x₁ – 2x₂ ≤ 25 but does not satisfy 2x₁+ 3x₂ ≤ 10. Thus shaded region is the feasible region.
Equation (1) and (2) meet at (5, 5).
From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).
Step – 4

Corner point	z = 14x1 – 4x2
A(0, \(\frac{10}{3}\))	\(\frac{-40}{3}\)
B(\(\frac{19}{5}\), \(\frac{4}{5}\))	50 → maximum
C(5, 5)	50 → maximum
D(0, \(\frac{65}{12}\))	\(\frac{65}{3}\)

Z is maximum for x₁ = \(\frac{19}{5}\), x₂ = \(\frac{4}{5}\) or x₁ = 5, x₂ = 5 and Z_max = 50
There is no other point that maximizes Z.

Question 9.
Maximize: Z = 10x₁ + 12x₂ + 8x₃
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁ + x₂ + x₃ = 20
x₁, x₂ ≥ 0
[Hints: Eliminate x₃ from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]
Solution:
Eliminating x₃ this LPP can be written as Maximize Z = 2x₁ + 4x₂ + 160
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁, x₂ ≥ 0
Step – 1 Treating the consraints as equations we get
x₁ + 2x₂ = 30    …..(1)
5x₁ – 7x₃ = 12   …..(2)
Step – 2 Let us draw the graph
Table – 1

x1	30	0
x2	0	15

Table – 2

x1	8	1
x2	8	20

Step – 3 Clearly 0(0,0) satisfies x₁ + 2x₂ ≤ 30 and does not satisfy 12x₁ + 7x₂ ≤ 152
∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x₁ = 30, x₂ = 0 and Z_max = 220

Question 10.
Maximize: Z = 20x₁ + 10x₂
Subject to: x₁ + 2x₂ ≤ 40
3x₁ + x₂ ≥ 30
4x₁+ 3x₂ ≥ 60
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equalities we have:
x₁ + 2x₂ = 40   ….(1)
3x₁ + x₂ = 30   ….(2)
4x₁+ 3x₂ = 60  ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + 2x₂ ≤ 40 and does not satisfy 3x₁ + x₂ ≥ 30 and 4x₁+ 3x₂ ≥ 60, x₁, x₂ ≥ 0 is the first quadrant.
∴ The shaded region is the feasible region.
Step – 4 x₁ + 2x₂ = 40 and 3x₁ + x₂ = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)
Z(A) = 300, Z(B) = 800
Z (C) = 20 x 4 + 10 x 18 = 260
Z (D) = 120 + 120 = 240
Z attains minimum at D(6 ,12).
∴ The required solution x₁ = 6, x₂ =12 and Z_min = 240

Question 11.
Maximize: Z = 4x₁ + 3x₂
Subject to: x₁ + x₂ ≤ 50
x₁ + 2x₂ ≥ 80
2x₁+ x₂ ≥ 20
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ ≤ 50    ….(1)
x₁ + 2x₂ ≥ 80  ….(2)
2x₁+ x₂ ≥ 20   ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + x₂ < 50, x₁ + 2x₂ < 80 but does not satisfy
2x₁ + x₂ > 20, x₁ > 0, x₂ > 0 is the 1st quadrant.
Hence the shaded region is the feasible region.
Step – 4 x₁ + x₂ = 50
x₁ + 2x₂ = 80
=> x₂ = 30, x₁ = 20
The vertices of feasible region are
A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point	Z = 4x1 + 3x2
A(10,0)	40
5(50,0)	200
C(20,30)	170
D(0,40)	120
E(0,120)	60

Question 12.
Optimize: Z = 5x₁ + 25x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x₁ + x₂ ≤ 2, but does not satisfy x₁ + x₂ ≥ 2 and -x₁+ 5x₂ ≥ 5, x₁ > 0, x₂ > 0 is the 1st quadrant.
The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).
Step – 4 Z can be made arbitrarily large.
∴ Problem has no maximum.
But Z(A) = \(\frac{100}{3}\), Z(B) = 50
Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).
But the feasible region is unbounded.
Hence we cannot immediately decide, Z is minimum at A.
Let us draw the half plane
5x₁ + 25x₂ < \(\frac{100}{3}\)
⇒ 3x₁ + 15x₂ < 20
As there is no point common to this half plane and the feasible region.
we have Z is minimum for x₁ = \(\frac{5}{6}\), x₂ = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.
Optimize: Z = 5x₁ + 2x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.
Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.
Let us draw the half plane 5x₁ + 2x₂ < 4

x1	0	4/5
x2	2	0

As there is no point common to this half plane and the feasible region,
we have Z is minimum for x₁ = 0, x₂ = 2 and the minimum value of Z = 4.

Question 14.
Optimize: Z = -10x₁ + 2x₂
Subject to: -x₁ + x₂ ≥ -1
x₁ + x₂ ≤ 6
x₂ ≤ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-x₁ + x₂ = -1     ….(1)
x₁ + x₂ = 6        ….(2)
x₂ = 5                ….(3)
Step – 2 Let us draw the graph
Table – 1

x1	1	0
x2	0	-1

Table – 2

x1	6	0
x2	0	1

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)
Step – 4 Z(O) = 0
Z(A) = -10
Z(B) = – 30
Z(C) = 0
Z(D) = 10
∴ Z is maximum for x₁= 0, x, = 5 and Z_(max) = 10
Z is minimum for x₁ = \(\frac{7}{2}\)  x₂ = \(\frac{5}{2}\) and Z_(min) = -30

Question 15.
Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.
(1) Maximise: Z = 1500x + 2000y
Subject to: x + y < 20
x + 2y < 24
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 20
x + 2y = 24
Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get
y = 14
x = 16
With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).
Step – 4 Z(0) = 0
Z(A) = 30,000
Z(B) = 32,000 → Maximum
Z(C) = 24000
Z is maximum for x = 16, y = 4 with Z = 32000
To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.
Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y
Subject: x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
2x +3y = 600
2a + y = 480
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)
Step – 4 Z(0) = 6
Z(A) = 3600
Z(B) = 3150 + 600
= 3750 → maximum
Z(C) = 2000
Thus Z is maximum for x = 210 and y = 60
and Z_(max) = 3750

(3) Maximize: Z = 20x + 30y
Subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10       …(1)
x + y = 6           …(2)
x = 4
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
Solving (1) and (2) we get x = 2, y = 4.
The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).
Step – 4 Z(0) =0
Z(A) = 80
Z (B) =140
Z(C) = 1 60 → maximum
Z (D) = 150
∴ Z is Maximum when x = 2, y = 4 and Z_(max) = 160

(4) Maximize: Z = 15x + 17y
Subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y > 300
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
4x + 7y = 150      ….(1)
x + y = 30            ….(2)
15x + 17y = 300  ….(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.
∴ The shaded region is the feasible region.
From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_(max) = 470

(5) Maximize: Z = 2x + 4y
Subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
3x + 2y = 10  …(1)
2x + 5y = 15  …(2)
5x + 6y = 21  …(3)
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y
Subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 5    ….(1)
2x + y = 9  ….(2)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
∴ Thus the shaded region is the feasible region.
From (1) and (2) we get x = 4, y = 1.
∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).
Step – 4 Z(0) =0
Z (A) = 4500
Z (B) = 4800 → Maximum
Z (C) = 4000
Z is maximum for x = 4 and y = 1, Z_(max) = 4800

(7) Minimize: Z = 4960 – 70x – 130y
Subject to: x + y ≤ 12
x + y ≥ 6
x ≤ 8
y ≤ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 12   ….(1)
x + y = 6     ….(2)
x = 8           ….(3)
y = 4           ….(4)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.
The shaded region is the feasible region.
The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).
Step – 4 Z (A) = 4540
Z (B) = 4400
Z (C) = 3880
Z (D) = 3640 → Minimum
Z (E) = 3920
Z (F) = 4180
∴ Z is maximum for x = 4 and y = 8 and Z_(min) = 3640.

(8) Minimize: Z = 16x + 20y
Subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10  ….(1)
x + y = 6      …(2)
3x + y = 8    …(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.
From (1) and (2) we get y = 4, x = 2.
From (2) and (3) we get x = 1, y = 5.
The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).
Step – 4 Z (A) = 160
Z (B) = 112 → Minimum
Z (C) =116
Z (D) = 160
As the region is unbounded, let us draw the half plane Z < Z_(min)
⇒ 16x + 20y < 112
⇒ 4x + 5y < 28

x1	7	0
x2	0	5.6

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).
∴ Z is minimum for x = 2, y = 4 and Z_(min) = 112.

(9) Minimize: Z = (512.5)x + 800y
Subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0
Solution:
Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3

x1	8	0
x2	0	10

Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3
Step – 2 The line segment AB is the feasible region.
Step – 3 Z (A) = 3587.5 + 1000 = 4587.5
Z (B) = 2870 + 2400 = 5270
Clearly Z is minimum for
x = 7, y = \(\frac{5}{4}\) and Z_(min) = 4587.5

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 2 Inverse Trigonometric Functions Ex 2 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. (\(\frac{2 x}{1-x^2}\), \(\frac{2 x}{\sqrt{1-x^2}}\), \(\frac{2 x}{1+x^2}\))
Solution:
\(\frac{2 x}{1+x^2}\)

(ii) If the value of sin^-1 x = \(\frac{\pi}{5}\) for some x ∈ (-1, 1) then the value of cos^-1 x is ________. (\(\frac{3 \pi}{10}\), \(\frac{5 \pi}{10}\),\(\frac{3 \pi}{10}\))
Solution:
\(\frac{3 \pi}{10}\)

(iii) The value of tan^-1 x (2cos\(\frac{\pi}{3}\)) is ________. (1, \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. (\(\frac{3 \pi}{4}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{2}\)

(v) The value of cot^-1 2 + tan^-1 \(\frac{1}{3}\) = ________. (\(\frac{\pi}{4}\), 1, \(\frac{\pi}{2}\))
Solution:
\(\frac{\pi}{4}\)

(vi) The principal value of sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________. (\(\frac{2 \pi}{3}\), \(\frac{\pi}{3}\), \(\frac{4 \pi}{3}\))
Solution:
\(\frac{\pi}{3}\)

(vii) If sin^-1 \(\frac{x}{5}\) + cosec^-1 \(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 \(\frac{1}{x}\)), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 \(\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\) = ________. (2π – \(\frac{x}{2}\), \(\frac{x}{2}\), π – \(\frac{x}{2}\))
Solution:
π – \(\frac{x}{2}\)

(x) 2sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{24}{25}\) = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [(\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), [0, \(\frac{\pi}{2}\)), (0, \(\frac{\pi}{2}\)])
Solution:
(0, \(\frac{\pi}{2}\)]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 \(\frac{1}{x}\) cosec^-1 x = 1
Solution:
False

(ii) cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{2}{3}\) = tan^-1 \(\frac{17}{6}\)
Solution:
True

(iii) tan^-1 \(\frac{4}{3}\) + cot^-1 (\(\frac{-3}{4}\)) = π
Solution:
True

(iv) sec^-1 \(\frac{1}{2}\) + cosec^-1 \(\frac{1}{2}\) = \(\frac{\pi}{2}\)
Solution:
False

(v) sec^-1 (-\(\frac{7}{5}\)) = π – cos^-1 \(\frac{5}{7}\)
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan \(\frac{3 \pi}{4}\)) is \(\frac{3 \pi}{4}\)
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 \(\frac{9}{13}\)
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – \(\frac{\pi}{4}\)
Solution:
False

(xi) 2 sin^-1 \(\frac{4}{5}\) = sin^-1 \(\frac{24}{25}\)
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the foilowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot^-1 3)
Solution:

(iii) cos (2 sin^-1 x)
Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 (\(\frac{x}{y}\)) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:

(vi) cosec (cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\))
Solution:

(vii) sin^-1 \(\frac{1}{\sqrt{5}}\) + cos^-1 \(\frac{3}{\sqrt{10}}\)
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec \(\frac{\pi}{4}\)
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 \(\sqrt{\frac{1-x}{1+x}}\))
Solution:

(x) tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
Solution:

(ii) sin^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution:

(iii) tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
Solution:
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

(iv) tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) tan ( 2tan^-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:

(ii) sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

(iii) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{70}\) + tan^-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:

(iv) 2 tan^-1 \(\frac{1}{5}\) + sec^-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) cos^-1 \(\frac{12}{13}\) + 2 cos^-1 \(\sqrt{\frac{64}{65}}\) + cos^-1 \(\sqrt{\frac{49}{50}}\) = cos^-1 \(\frac{1}{\sqrt{2}}\)
Solution:


(vi) tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\) = 6
Solution:
tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\)
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

(iii) tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\)) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\))
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 \(\frac{p q+1}{p-q}\) + cot^-1 \(\frac{q r+1}{q-r}\) + cot^-1 \(\frac{r p+1}{r-p}\) = 0
Solution:

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 \(\frac{2 a-b}{b \sqrt{3}}\) + tan^-1 \(\frac{2 b-a}{a \sqrt{3}}\) = \(\frac{\pi}{3}\)
Solution:

(ii) tan^-1 \(\frac{1}{x+y}\) + tan^-1 \(\frac{y}{x^2+x y+1}\) = tan^-1 \(\frac{1}{x}\)
Solution:

(iii) sin^-1 \(\sqrt{\frac{x-q}{p-q}}\) = cos^-1 \(\sqrt{\frac{p-x}{p-q}}\) = cot^-1 \(\sqrt{\frac{p-x}{x-q}}\)
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x\(\sqrt{1-x^2}\) + x\(\sqrt{1-y^2}\) + x\(\sqrt{1-z^2}\) = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

(iii) tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 \(\frac{y z}{x r}\) + tan^-1 \(\frac{z x}{y r}\) + tan^-1 \(\frac{x y}{z r}\) = \(\frac{\pi}{2}\)
Solution:

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\) = \(\frac{\pi}{4}\). where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\)

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
Solution:
sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
or, sin^-1 (1 – x) = \(\frac{\pi}{2}\) – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 \(\sqrt{1-x^2}\)
or, 1 – x = \(\sqrt{1-x^2}\)
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x  = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x      ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:

(v) tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:

(vi) tan^-1 \(\frac{1}{2 x+1}\) + tan^-1 \(\frac{1}{4 x+1}\) = tan^-1 \(\frac{2}{x^2}\)
Solution:

(vii) 3 sin^-1 \(\frac{2 x}{1+x^2}\) – 4 cos^-1 \(\frac{1-x^2}{1+x^2}\) + 2 tan^-1 \(\frac{2 x}{1-x^2}\) = \(\frac{\pi}{3}\)
Solution:

(viii) cot^-1 \(\frac{1}{x-1}\) + cot^-1 \(\frac{1}{x}\) + cot^-1 \(\frac{1}{x+1}\) = cot^-1 \(\frac{1}{3x}\)
Solution:

(ix) cot^-1 \(\frac{1-x^2}{2 x}\) =  cosec^-1 \(\frac{1+a^2}{2 a}\) – sec^-1 \(\frac{1+b^2}{1-b^2}\)
Solution:

(x) sin^-1 \(\left(\frac{2 a}{1+a^2}\right)\) + sin^-1 \(\left(\frac{2 b}{1+b^2}\right)\) = 2 tan^-1 x
Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1 \(\frac{\sqrt{3}}{2}\)
Solution:

(xii) sin^-1 2x + sin^-1 x = \(\frac{\pi}{3}\)
Solution:

Question 10.
Rectify the error ifany in the following:
sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{33}{65}\)
Solution:

Question 11.
Prove that:
(i) cos^-1 \(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) = 2 tan^-1 \(\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)\)
Solution:

(ii) tan \(\left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\) + tan \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\)
Solution:

(iii) tan^-1 \(\sqrt{\frac{x r}{y z}}\) + tan^-1 \(\sqrt{\frac{y r}{y x}}\) + tan^-1 \(\sqrt{\frac{z r}{x y}}\) = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 (\(\frac{x}{a}\)) + cos^-1 (\(\frac{y}{b}\)) = Θ, prove that \(\frac{x^2}{a^2}\) – \(\frac{2 x}{a b}\) cos Θ + \(\frac{y^2}{b^2}\) = sin² Θ.
Solution:

(ii) If cos^-1 (\(\frac{x}{y}\)) + cos^-1 (\(\frac{y}{3}\)) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = sin^-1 (\(\frac{c^2}{a b}\)) prove that b²x² + 2xy \(\sqrt{a^2 b^2-c^4}\) a²y² = c²
Solution:

(iv) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = α prove that \(\frac{x^2}{a^2}\) + \(\frac{2 x y}{a b}\) cos α + \(\frac{y^2}{b^2}\) = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 \(\frac{x-1}{x+1}\) + tan^-1 \(\frac{2 x-1}{2 x+1}\) = tan^-1 \(\frac{23}{36}\)
Solution:

(ii) tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 x = \(\frac{\pi}{4}\)
Solution:

(iii) cos^-1 \(\left(x+\frac{1}{2}\right)\) + cos^-1 x+ cos^-1 \(\left(x-\frac{1}{2}\right)\) = \(\frac{3 \pi}{2}\)
Solution:

(iv) 3tan^-1 \(\frac{1}{2+\sqrt{3}}\) – tan^-1 \(\frac{1}{x}\) = tan^-1 \(\frac{1}{3}\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.
State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:
(i) Symmetric
(ii) Neither Symmetric nor skew-symmetric
(iii) Symmetric
(iv) Skew symmetric
(v) Both
(vi) Neither symmetric nor skew-symmetric
(vii) Skew symmetric

Question 2.
State ‘True’ or ‘False’:
(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.
Solution:
True

(ii) For any square matrix A, AA’ is symmetric.
Solution:
True

(iii) If A is any skew-symmetric matrix, then A² is also skew-symmetric.
Solution:
False

(iv) If A is symmetric, then A², A³, …, Aⁿ are all symmetric.
Solution:
True

(v) If A is symmetric then A – A¹ is both symmetric and skew-symmetric.
Solution:
False

(vi) For any square matrix (A – A¹)² is skew-symmetric.
Solution:
True

(vii) A matrix which is not symmetric is skew-symmetric.
Solution:
False

Question 3.
(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.
Solution:
A and B are symmetric matrices;
Thus A’ = A and B’ = B
Now (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB = – (AB – BA)
∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.
Solution:
We know that zero matrix is both symmetric as well as skew-symmetric.
Let A is symmetric.
∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.
A and B are square matrices of the same order, prove that
(i) If A, B and AB are all symmetric, then AB – BA = 0
Solution:
Let A, B and AB are all symmetric.
∴A’ = A, B’ = B and (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB
⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0
Solution:
Let A, B and AB are all skew symmetric matrices
∴ A’ = -A, B’ = -B and (AB)’ = -AB
Now (AB)’ = -AB
⇒ B’A’ = -AB
⇒ (-B) (-A) = -AB
⇒ BA = -AB
⇒ AB + BA = 0

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
2 & 1 & 5 \\
0 & 3 & 3
\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.
Prove that a unit matrix is its own inverse. Is the converse true?
IfA = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\) show that A² = I and hence A= A^-1.
Solution:
No the converse is not true for example:

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.
What is the inverse of

Question 11.
Find inverse of the following matrices by elementary row/column operation (transformations):
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find the inverse of the following matrices using elementary transformation:
(i) \(\left[\begin{array}{lll}
\mathbf{0} & \mathbf{0} & 2 \\
\mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{2} & \mathbf{0} & \mathbf{0}
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

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

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

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

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

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

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

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

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

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

Question 1.
State which of the following is the probability distribution of a random variable X with reasons to your answer:
(a)

X = x	0	1	2	3	4
p(x)	0.1	0.2	0.3	0.4	0.1

(b)

X = x	0	1	2	3
p(x)	0.15	0.35	0.25	0.2

(c)

X = x	0	1	2	3	4	5
p(x)	0.4	R	0.6	R2	0.7	0.3

Solution:
(a) As ∑p_i = 1.1 > 1, the given distribution is not a probability distribution.
(b) As ∑p_i = 0.95 < 1, the given distribution is not a probability distribution.
(c) As the value of R is not known, the given distribution is not a probability distribution.

Question 2.
Find the probability distribution of number of doublets in four throws of a pair of dice. Find also the mean and the variance of the number of doublets.
Solution:
Let the random variable X represents the number of doublets in 4 throws of a pair of dice.
X can take values 0, 1, 2, 3, 4
In a single throw of two dice
X can take values 0, 1, 2, 3, 4
In a single throw of two dice.
P(doublet) = \(\frac{6}{36}\)
P(non-doublet) = 1 – P (doublet) = \(\frac{5}{36}\)
Clearly the given experiment is a binomial experiment with n = 4,
p = \(\frac{1}{6}\), q = \(\frac{5}{6}\)

N.B. We can use the definition to find mean and variance.

Question 3.
Four cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. Calculate the mean and variance of the number of aces.
Solution:
Let the random variable X denotes the number of aces.
Thus X can take values 0, 1, 2, 3, 4.
Clearly the given experiment is a binomial experiment with n = 4

N.B. We can use the definition to find mean and variance.

Question 4.
Find the probability distribution of
(a) number of heads in three tosses of a coin.
Solution:
Let the random variable X denotes the number of heads in three tosses of a coin.
X can take values 0, 1, 2, 3
In one toss p(H) = \(\frac{1}{2}\), p(T) = \(\frac{1}{2}\)
This experiment is a binomial experiment with n = 3, p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

(b) number of heads in simultaneous tosses of four coins.
Solution:
When 4 coins tossed simultaneously, let the random variable X denotes the number of heads.
X can take values 0, 1. 2, 3, 4
This experiment is a binomial experiment with n = 4, p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

Question 5.
A biased coin where the head is twice as likely to occur as the tail is, tossed thrice. Find the probability distribution of number of heads.
Solution:
Let in a toss P(T) = x
According to the question P(H) = 2x
Now 2x + x = 1
⇒ x = \(\frac{1}{3}\)
Thus P(T) = \(\frac{1}{3}\), P(H) = \(\frac{3}{3}\)
Clearly the given experiment is a binomial experiment with n = 3, p = P(H) = \(\frac{2}{3}\), q = P(T) = \(\frac{1}{3}\)
Let the random variable X denotes the number of heads in three throws of that coin.
X can take values 0, 1, 2, 3

Question 6.
Find the probability distribution of the number of aces in question no. 3 if the cards are drawn successively without replacement.
Solution:
Total number of cards = 52
Number of aces = 4
Number of cards drawn = 4 (one by one without replacement)
Let X = the random variable of number of aces drawn.
X can take values 0, 1, 2, 3 or 4

Question 7.
From a box containing 32 bulbs out of which 8 are defective 4 bulbs are drawn at random successively one after another with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Total number of bulbs = 32
Number of defective bulbs = 8
∴ Number of nondefective bulbs = 24
Number of bulbs drawn = 4 (one after another with replacement)
Clearly the experiment is a binomial experiment with n = 4
p = p ((drawing a defective bulb) = \(\frac{8}{32}\) = \(\frac{1}{4}\)
q = p (drawing a non defective bulb) = \(\frac{3}{4}\)
Let the random variable X denotes the number of defective bulbs.
∴ X can take values 0, 1, 2, 3, 4.

Question 8.
A random variable X has the following probability distribution.

X = x	0	1	2	3	4	5
p(x)	0	R	2R	3R	3R	R

Determine
(a) R
(b) P(X < 4)
(d) P(2 ≤ X ≤ 5)
(c) P(X ≥ 2)
Solution:
(a) Clearly ∑P_i = 1
⇒ R + 2R + 3R + 3R + R = 1

Question 9.
Find the mean and the variance of the number obtained on a throw of an unbiased coin.
Solution:
When an unbiased coin is tossed we donot get any number.
p(getting a number) = 0
Thus mean = variance = 0
If instead of coin it would be an unbiased die
Then let X = The number obtained in the throw.
X can take values 1, 2, 3, 4, 5 or 6.
P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4)

Question 10.
A pair of coins is tossed 7 times. Find the probability of getting
(i) exactly five tails
(ii) at least five tails
(iii) at most five tails
Solution:
Clearly the given experiment is a binomial experiment with

Question 11.
If a pair of dice is thrown 5 times then find the probability of getting three doublets.
Solution:
In a single through of a pair of dice
p (a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
p (a non doublet) = 1 – p (a doublet) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Clearly the given experiment is a binomial experiment with n = 5
p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)
p(3 doublets in 5 throw) = ⁵C₃ p³q²
= 10. \(\left(\frac{1}{6}\right)^3\) \(\left(\frac{5}{6}\right)^2\) = \(\frac{250}{6^5}\) = \(\frac{125}{3888}\)

Question 12.
Four cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that:
(i) all the four cards are diamonds
(ii) only two cards are diamonds
(iii) none of the cards is a diamond.
Solution:
Out of 52 cards there are 13 diamonds. 4 cards are drawn one by one with replacement.
When we draw a card

Question 13.
In an examination, there are twenty multiple-choice questions each of which is supplied with four possible answers. What is the probability that a candidate would score 80% or more in the answers to these questions?
Solution:
Total number of questions = 20 for each question
p (correct answer) = \(\frac{1}{4}\)
p (wrong answer) = \(\frac{3}{4}\)
p (the score is > 80%)
= p (no. of correct answer > 16)
= p (16 correct answers)
+ p(17 correct answers)
+ p (18 correct answers)
+ p (19 correct answers)
+ p (20 correct answers)

Question 14.
A bag contains 7 balls of different colours. If five balls are drawn successively with replacement then what is the probability that none of the balls drawn is white?
Solution:
Total number of balls = 7 (The colours are different)
Number of balls drawn = 5 (one by one with replacement)
Case – 1
(If a white coloured ball is not present)
p(non is white) = 1
Case – 2
(If one ball is white).
In one draw p(white ball) = \(\frac{1}{7}\)
p (non white ball) = \(\frac{6}{7}\)
When 5 balls are drawn
p (none of the balls drawn is white) = ⁵C₀ \(\left(\frac{1}{7}\right)^0\) \(\left(\frac{6}{5}\right)^5\) = \(\left(\frac{6}{7}\right)^5\)

Question 15.
Find the probability ofthrowing at least 3 sixes in 5 throws of a die.
Solution:
In one throw of a die

Question 16.
The probability that a student securing first division in an examination is \(\). What is the probability that out of 100 students twenty pass in first division?
Solution:
Clearly the given experiment is a binomial experiment with

Question 17.
Sita and Gita throw a die alternatively till one of them gets a 6 to win the game. Find their respective probability of winning if Sita starts first.
Solution:
In one throw of a die
p (getting a 6) = \(\frac{1}{6}\)
p (not getting a 6) = \(\frac{5}{6}\)
If Sita begins the game she may win in first round or second round or third round, etc.
Thus p(Sita wins)

Question 18.
If a random variable X has a binomial distribution B(8, \(\frac{1}{2}\)) then find X for which the outcome is the most likely. [Hint: Find X=x for which P(X = x) is the maximum, x = 0, 1, 2, 3,…….. 8.]
Solution:
Given binomial distribution is B(8, \(\frac{1}{2}\))
Thus n = 8, p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
We shall find X which is most likely i.e. for which p(x = r) is maximum where r = 0, 1, 2, ….., 8
As p = q, the probability is maximum when ⁸Cr is maximum
But ⁸Cr, r = 0, 1, 2, ….. ,8 is maximum when r = 4.
Thus the most likely outcome is x = 4.

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

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

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution:

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

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

Find derivatives of the following functions.
Question 1.
x^x
Solution:
Let y = x^x
Then In y = x . In x
⇒ \(\frac{d}{d x}\)(In y) = \(\frac{d}{d x}\)(x . In x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = In x + x . \(\frac{1}{x}\) = In x + 1 = 1 + In x
⇒ \(\frac{d y}{d x}\) = y (1 + In x) = x^x (1 + In x) = x^x In (ex)

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

Question 3.
x^{sin x}
Solution:
y = x^{sin x}
⇒ In y = sin x . In x
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = cos x In x + sin x . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = x^{sin x} (cos x . In x + \(\frac{\sin x}{x}\))

Question 4.
(log x)^{tan x}
Solution:
y = (log x)^{tan x}
⇒ log y = tan x . log (log x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = sec² x log (log x) + tan x . \(\frac{1}{\log x}\) . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = (log x)^{tan x} {sec² x . log log x + \(\frac{\tan x}{x \log x}\)}

Question 5.
\(2^{\left(2^x\right)}\)
Solution:
y = \(2^{\left(2^x\right)}\)
⇒ In y = 2^x . In 2
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = 2^x . In 2 . In 2
⇒ \(\frac{d y}{d x}\) = \(2^{\left(2^x\right)}\) . 2^x . (In 2)²

Question 6.
\((1+\sqrt{x})^{x^2}\)
Solution:

Question 7.
\(\left(\sin ^{-1} x\right)^{\sqrt{1-x^2}}\)
Solution:

Question 8.
\((\tan x)^{\log x^3}\)
Solution:

Question 9.
x^1/x + (sin x)^x
Solution:

Question 10.
(cos x)^x + x^{cos x}
Solution:

Question 11.
(x² + 1)^2/3 (3x + 1)^1/4 √x
Solution:

Question 12.
\(\frac{(x+1)(x+2)^2(x+3)^3}{(x-1)(x-2)^2(x-3)^3}\)
Solution:

Question 13.
(sin x)^x \(\sqrt{\sin x}\left(1+x^2\right)^{\frac{1}{2}+x}\)
Solution:

Question 14.
(sec x + tan x)^{cot x}
Solution:
y = (sec x + tan x)^{cot x}
⇒ In y = cot x . In (sec x + tan x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = -cosec² x . In (sec x + tan x) + cot x . \(\frac{1}{\sec x+\tan x}\) × (sec x . tan x + sec² x)
= -cosec² x . In (sec x + tan x) + cot x . sec x
∴ \(\frac{d y}{d x}\) = y {-cosec² x . In (sec x + tan x) + cosec x}
= (sec x + tan x)^{cot x} {cosec x – cosec² x . In (sec x + tan x)}

Question 15.
(2^√x)^1+√x
Solution:

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

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

Differentiate the following functions by proper substitution.
Question 1.
sin^-1 2x\( \sqrt{1-x^2} \)
Solution:
y = sin^-1 2x\( \sqrt{1-x^2} \)   [Put x = sin θ
= sin^-1 (2 sin θ . cos θ)
= sin^-1 sin 2θ = 2θ = 2 sin^-1 x.
\(\frac{d y}{d x}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 2.
tan^-1 \(\frac{2 x}{1-x^2}\)
Solution:
y = tan^-1 \(\frac{2 x}{1-x^2}\)
= tan^-1 \(\frac{2 \tan \theta}{1-\tan ^2 \theta}\) = tan^-1 (tan 2θ)
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1-x^2}\)

Question 3.
tan^-1 \(\sqrt{\frac{1-t}{1+t}}\)
Solution:

Question 4.
\(\left[\left(\frac{1+t^2}{1-t^2}\right)^2-1\right]^{\frac{1}{2}}\)
Solution:

Question 5.
tan^-1 \(\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{x a}}\right)\)
Solution:

Question 6.
sin^-1 (\(\frac{2 x}{1+x^2}\))
Solution:
y = sin^-1 \(\frac{2 x}{1+x^2}\)  [Put x = tan θ
= sin^-1 \(\frac{2 \tan \theta}{1+\tan ^2 \theta}\) = sin^-1 sin θ
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1+x^2}\)

Question 7.
sec^-1 \(\left(\frac{\sqrt{a^2+x^2}}{a}\right)\)
Solution:

Question 8.
sin^-1 \(\left(\frac{2 \sqrt{t^2-1}}{t^2}\right)\)
Solution:

Question 9.
cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\)
Solution:
y = cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\) [Put t = tan θ
= cos^-1 \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)
= cos^-1 cos 2θ = 2 tan^-1 t
∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+t^2}\)

Question 10.
cos^-1 (2t² – 1)
Solution:
y = cos^-1 (2t² – 1) [Put t = tan θ
= cos^-1 (2 cos² θ – 1)
= cos^-1 cos 2θ = 2θ = 2 cos^-1 t
∴ \(\frac{d y}{d x}\) = – \(\frac{2}{\sqrt{1-t^2}}\)

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

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

Question 1.
Prove the formulae (4) to (7).
Solution:
(4) \(\frac{d}{d x}\)(cos-1 x) = \(\frac{-1}{\sqrt{1-x^2}}\)
Let y = cos^-1 x
⇒ x = cos y
⇒ \(\frac{d}{d x}\) = \(\frac{1}{\left(\frac{d x}{d y}\right)}\) = \(\frac{1}{-\sin y}\)
But sin y ≥ 0 when
y ∈ [0, π] ( ∵ [0, π] is the principal value branch for cos^-1 x)
∴ \(\frac{d y}{d x}\) = \(\frac{-1}{\sqrt{1-\cos ^2 y}}\) = \(\frac{-1}{\sqrt{1-x^2}}\)

(5) Let y = tan^-1 x
⇒ x = tan y

(6) Let y = cot^-1 x
⇒ x = cot y

(7) Let y = cosec^-1 x
⇒ x = cosec y

Question 2.
Find derivatives of the following functions.
sin^-1 2x
Solution:
y = sin^-1 2x

Question 3.
cot^-1 √x
Solution:
cot^-1 √x

Question 4.
sec^-1 (2x + 1)
Solution:
y = sec^-1 (2x + 1)

Question 5.
cos^-1 \(\sqrt{\frac{1+x}{2}}\)
Solution:

Question 6.
cos^-1 \(\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\)
Solution:

Question 7.
tan^-1 (cos √x)
Solution:
y = tan^-1 (cos √x)

Question 8.
x² cosec^-1 \(\left(\frac{1}{\ln x}\right)\)
Solution:

Question 9.
cot^-1 \( \frac{\sqrt{1-x^2}}{x} \)
Solution:

Question 10.
(x sin^-1 x)¹⁵
Solution:
y = (x sin^-1 x)¹⁵

Question 11.
sin^-1 \( \sqrt{\frac{1-x}{1+x}} \)
Solution:

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

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

Find derivatives of the following functions.
Question 1.
(x² +5)⁸
Solution:
y = (x² +5)⁸
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x² +5)⁸
= 8(x² +5)⁷ × \(\frac{d}{d x}\)(x² +5) by chain rule
= 8(x² +5)⁷ . 2x
= 16x (x² +5)⁷

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

Question 3.
In (√x+1)
Solution:

Question 4.
sin 5x + cos 7x
Solution:
sin 5x + cos 7x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin 5x) + \(\frac{d}{d x}\)(cos 7x)
= cos 5x . \(\frac{d}{d x}\)(5x) – sin 7x . \(\frac{d}{d x}\)(7x)
= 5 cos 5x – 7 sin 7x

Question 5.
e^{sin t}
Solution:
y = e^{sin t}
\(\frac{d y}{d x}\) = \(\frac{d}{d t}\)(e^{sin t}) = e^{sin t} . \(\frac{d}{d t}\)(sin t)
= e^{sin t} . cos t

Question 6.
\(\sqrt{a x^2+b x+c}\)
Solution:
y = \(\sqrt{a x^2+b x+c}\)

Question 7.
\(\left(\frac{x+1}{x^2+3}\right)^{-3}\)
Solution:

Question 8.
sec (tan θ)
Solution:
y = sec (tan θ)
\(\frac{d y}{d θ}\) = \(\frac{d}{d θ}\) {sec (tan θ)}
= sec (tan θ) . tan (tan θ) . \(\frac{d}{d θ}\)(tan θ)
= sec (tan θ) . tan (tan θ) . sec² θ

Question 9.
sin \(\left(\frac{1-x^2}{1+x^2}\right)\)
Solution:

Question 10.
\(\sqrt{\tan (3 z)}\)
Solution:

Question 11.
tan³ x
Solution:
y = tan³x = (tan x)³
\(\frac{d y}{d x}\) = 3(tan x)². \(\frac{d}{d x}\)(tan x)
= 3tan² x . sec² x

Question 12.
sin⁴ x
Solution:
y = sin⁴x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin⁴ x)
= 4 (sin x)³ . \(\frac{d}{d x}\)(sin x)
= 4 sin³ x . cos x

Question 13.
sin² x cos² x
Solution:
y = sin² x cos² x = \(\frac{1}{4}\)sin² 2x
\(\frac{d y}{d x}\) = \(\frac{1}{4}\) \(\frac{d}{d x}\)(sin 2x)²
= \(\frac{1}{4}\). 2sin 2x . \(\frac{d}{d x}\)(sin 2x)
= \(\frac{1}{2}\) sin 2x . cos 2x . \(\frac{d}{d x}\)(2x)
= \(\frac{1}{2}\) sin 2x cos 2x . 2 = sin 2x . cos 2x

Question 14.
sin 5x cos 7x
Solution:
y = sin 5x cos 7x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin 5x) . cos 7x + \(\frac{d}{d x}\)sin 5x . (cos 7x)
= cos 5x . \(\frac{d}{d x}\)(5x) . cos 7x + sin 5x . (-sin 7x) . \(\frac{d}{d x}\)(7x)
= 5 cos 5x . cos 7x – 7 sin 5x . sin 7x

Question 15.
tan x cot 2x
Solution:
y = tan x. cot 2x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(tan x) . cot 2x + tan x . \(\frac{d}{d x}\)(cot 2x)
= sec² x / cot 2x + tan x . (-cosec² x) \(\frac{d}{d x}\)(2x)
= sec² x . cot 2x – 2 tan x . cosec² x

Question 16.
\(\sqrt{\sin \sqrt{x}}\)
Solution:

Question 17.
\(\sqrt{\sec (2 x+1)}\)
Solution:

Question 18.
cosec (ax + b)²
Solution:
y = cosec (ax + b)²
\(\frac{d y}{d x}\) = – cosec (ax + b)² cot (ax + b)² . \(\frac{d}{d x}\)(ax + b)²
[ ∵ \(\frac{d}{d x}\)(cosec u) = -cosec u . cot u . \(\frac{d u}{d x}\)
= – cosec (ax + b)² . cot (ax + b)² . 2(ax + b) . \(\frac{d}{d x}\)(ax + b)
[ ∵ \(\frac{d}{d x}\)(u²) = 2u \(\frac{d u}{d x}\)
= – cosec (ax + b)² . cot (ax + b)² . 2(ax + b) . a
= – 2a (ax + b) cosec (ax + b)² cot (ax + b)²

Question 19.
a^{In x}
Solution:
y = a^{In x} . In a . \(\frac{d}{d x}\)(In x)
[ ∵ \(\frac{d}{d x}\)(a^u) = a^u . In a . \(\frac{d u}{d x}\)
= a^{In x} . In a . \(\frac{1}{x}\) = \(\frac{a^{\ln x} \ln a}{x}\)

Question 20.
\(a^{x^2} b^{x^3}\)
Solution:

Question 21.
In tan x
Solution:

Question 22.
\(5^{\sin x^2}\)
Solution:

Question 23.
In tan\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)
Solution:

Question 24.
\(\sqrt{\left(a^{\sqrt{x}}\right)}\)
Solution:

Question 25.
In (e^nx + e^-nx)
Solution:
y = In (e^nx + e^-nx)
\(\frac{d y}{d x}\) = \(\frac{1}{e^{n x}+e^{-n x}}\) . \(\frac{d}{d x}\) (e^nx + e^-nx)
= \(\frac{n\left(e^{n x}-e^{-n x}\right)}{e^{n x}+e^{-n x}}\)

Question 26.
\(e^{\sqrt{a x}}\)
Solution:

Question 27.
\(\sqrt{\log x}\)
Solution:
y = \(\sqrt{\log x}\)
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{\log x}}\) . \(\frac{d}{d x}\)(log x)
= \(\frac{1}{2 \sqrt{\log x}}\) . \(\frac{1}{x}\)

Question 28.
e^{sin x} – a^{cos x}
Solution:
y = e^{sin x} – a^{cos x}
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(e^{sin x}) – \(\frac{d}{d x}\)(a^{cos x})
= e^{sin x} . \(\frac{d}{d x}\)(sin x) – a^{cos x} . In a . \(\frac{d}{d x}\)(cos x)
= e^{sin x} . cos x + a^{cos x} . In a . sin x

Question 29.
\(\frac{e^{3 x^2}}{\ln \sin x}\)
Solution:

Question 30.
Prove that
\(\frac{d}{d x}\left[\frac{1-\tan x}{1+\tan x}\right]^{\frac{1}{2}}\) = 1 / \(\sqrt{\cos 2 x}\) (cos x + sin x)
Solution: