Class 12

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(b)

Integrate the following:
(In some cases suggestions have been given for substitution.)
(i) ∫sin 3x dx
Solution:

(ii) ∫cos ax dx
Solution:

(iii) ∫cos (2 – 7x) dx
Solution:

(iv) ∫sin \(\frac{x}{2}\) dx
Solution:

(v) ∫sec² 4x dx
Solution:

(vi) ∫cosec² \(\frac{x}{3}\) dx
Solution:

(vii) ∫sec (x + 2) tan (x + 2) dx
Solution:
∫sec (x + 2) tan (x + 2) dx
[Put x + 2 = θ
Then dx = dθ]
= ∫sec θ . tan θ dθ
= sec θ + C = sec (x + 2) + C

(viii) ∫cosec (x + \(\frac{\pi}{4}\)) cot (x + \(\frac{\pi}{4}\)) dx (x + \(\frac{\pi}{4}\) = z)
Solution:

(ix) ∫x² cos x³ dx (x³ = z)
Solution:

(x) ∫e^x . sec e^x tan e^x dx (e^x = z)
Solution:
∫e^x . sec e^x . tan e^x dx
[Put e^x = z
Then e^x dx = dz]
= ∫ sec z . tan z dz
= sec z + C = sec e^x + C

(xi) ∫\(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
Solution:

Question 2.
(i) ∫sin x cos x dx (sin x = v)
Solution:
∫sin x cos x dx
[Put sin x = v
Then cos x dx = dv]
= ∫v dv = \(\frac{1}{2}\)v² + C
= \(\frac{1}{2}\)sin² x + C

(ii) ∫tan³ x sec² x dx (tan x = t)
Solution:
∫tan³ x sec² x dx
[Put tan x = t
Then sec² x dx = dt]
= ∫t³ dt = \(\frac{1}{4}\)t⁴ + C
= \(\frac{1}{4}\)tan⁴ x + C

(iii) ∫\(\frac{{cosec}^2 x}{1+\cot x}\) dx
Solution:
∫\(\frac{{cosec}^2 x}{1+\cot x}\) dx
[Put 1 + cot x = t
Then -cosec² x dx = dt
or, cosec² x dx = – dt]
= ∫\(\) = -∫\(\) = -ln |t| + C
= -ln |1 + cot x| + C

(iv) ∫\(\frac{\sin x}{\cos ^3 x}\) dx
Solution:

(v) ∫\(\frac{\cos x}{\sin ^5 x}\) dx
Solution:

(vi) ∫\(\frac{{cosec}^2 x(\ln x)}{x}\) dx (In x = z)
Solution:
∫\(\frac{{cosec}^2 x(\ln x)}{x}\) dx
[Put ln x = z
Then \(\frac{d x}{x}\) = dz]
= ∫cosec² z dz
= -cot z + C = -cot (ln x) + C

(vii) ∫\(\sqrt{1-\sin x}\) cos x dx
Solution:

Question 3.
(i) ∫x\(\sqrt{x^2+3}\) dx (x² + 3 = v)
Solution:

(ii) ∫\(\frac{7 d x}{2-3 x}\) dx
Solution:

(iii) ∫\(\frac{x}{\sqrt{x^2-a^2}}\) dx
Solution:

(iv) ∫\(\frac{x^2+1}{\left(x^3+3 x+7\right)^3}\) dx
Solution:

(v) ∫(x⁴ – 3x² + 1)⁴ (2x³ – 3x) dx
Solution:

Question 4.
(i) ∫e^3x dx
Solution:
∫e^3x dx
[Put 3x = t
Then 3 dx = dt
or, dx = \(\frac{1}{3}\)dt]
= ∫e^t . \(\frac{1}{3}\)dt = \(\frac{1}{3}\)e^t + C
= \(\frac{1}{3}\)e^3x + C

(ii) ∫e^2x+7 dx
Solution:
∫e^2x+7 dx
[Put 2x + 7 = t
Then 2 dx = dt
or, dx = \(\frac{1}{2}\)dt]
= ∫e^t . \(\frac{1}{2}\)dt = \(\frac{1}{2}\)e^t + C
= \(\frac{1}{2}\)e^2x+7 + C

(iii) ∫e^x/3 dx
Solution:
[Put \(\frac{x}{3}\) = t
Then dx = 3 dt
= ∫e^t . 3dt = 3e^t + C = 3e^x/3 + C

(iv) ∫\(e^{x^3}\) x² dx
Solution:

(v) ∫a^2x dx
Solution:

(vi) ∫2\(a^{x^2}\) x dx
Solution:

(vii) ∫e^2tanx sec² x dx
Solution:

(viii) ∫\(\frac{e^x}{\left(e^x-2\right)^2}\) dx
Solution:

(ix) ∫\(e^{\cos ^2 x}\) sin 2x dx
Solution:
∫\(e^{\cos ^2 x}\) sin 2x dx
[Put cos2 x = t
Then 2cos x (-sin x)dx – dt
or, -sin 2x dx = dt
or, sin 2x dx = -dt]
= ∫e^t (-dt) = e^t + C = –\(e^{\cos ^2 x}\) + C

Question 5.
(i)∫\(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
Solution:

(ii)∫\(\frac{d x}{\sqrt{1-(x-1)^2}}\)
Solution:

(iii)∫\(\frac{\left(\sec ^{-1} x\right)^2 d x}{x \sqrt{x^2-1}}\)
Solution:

(iv)∫\(\frac{d x}{x\left[1+(\ln x)^2\right]}\)
Solution:

(v)∫\(\frac{d x}{x^2+2 x+2}\) (Integrate \(\frac{d x}{(x+1)^2+1}\))
Solution:

Question 6.
(i) ∫tan 3x dx
Solution:

(ii) ∫cos \(\frac{x}{3}\) dx
Solution:

(iii) ∫sec (2x + 1) dx
Solution:
∫sec (2x + 1) dx
[Put 2x + 1 = t
Then 2 dx = dt
or, dx = \(\frac{1}{2}\)dt]
= ∫sec t . \(\frac{1}{2}\)dt = \(\frac{1}{2}\) ∫sec t dt
= \(\frac{1}{2}\) ln |sec t + tan t| + C
= \(\frac{1}{2}\) ln |sec (2x + 1) + tan (2x + 1)| + C

(iv) ∫cosec 7x dx
Solution:
∫cosec 7x dx
[Put 7x = t
Then 7 dx = dt
or, dx = \(\frac{1}{7}\)dt]
= ∫cosec t . \(\frac{1}{7}\)dt
= \(\frac{1}{7}\) ln |cosec t + cot t| + C
= \(\frac{1}{7}\) ln |cosec 7x + cot x| + C

(v) ∫2x cot (x² + 3) dx    (x² + 3 = z)
Solution:
∫2x cot (x² + 3) dx
[Put x² + 3 = z
Then 2x dx = dz]
= ∫cot dz
= In |sin z| + C
= In |sin (x² + 3)| + C

(vi) ∫e^x tan e^x dx
Solution:
∫e^x tan e^x dx
[Put e^x = t
Then e^x dx = dt]
= ∫tan t dt
= In |sec t| + C
= In |sec e^x| + C

(vii) ∫(sec 2x – 3)² dx
Solution:
∫(sec 2x – 3)² dx
∫sec² 2x dx – 6∫sec 2x dx + 9∫dx
= \(\frac{1}{2}\) tan 2x – 3 ln |sec 2x + tan 2x| + 9x + C

Question 7.
(i) ∫\(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
Solution:
∫\(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
[Put (e^x – e^-x) = t
Then (e^x + e^-x)dx = dt]
= ∫\(\frac{d t}{t}\) = ln |t| + C = ln |(e^x – e^-x| + C

(ii) ∫3^x e^2x dx
Solution:

(iii) ∫\(\frac{(x+1) \ln \left(x^2+2 x+2\right)}{x^2+2 x+2}\) dx
Solution:

Question 8.
(i) ∫\(\frac{\sin x}{\sin (x+\alpha)}\) dx
Solution:

= ∫(cos α – sin α . cot t) dt
= cos α ∫dt – sin α ∫cot dt
= cos α . t – sin α . ln |sin t| + C
= (x + α) cos α – sin α  ln |sin (x + α)| + C
= x cos α – sin α ln |sin (x + α)| + C

(ii) ∫\(\frac{\sin x}{\cos (x-\alpha)}\) dx
Solution:

(iii) ∫\(\frac{\tan x+\tan \alpha}{\tan x-\tan \alpha}\) dx
Solution:

 

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

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

Question 1.
Find the extreme points of the following functions. Specify if the extremum is a maximum or minimum.
Find the extreme values.
(i) y = x² + 2x + 3
Solution:

(ii) y = 5x² – 2x⁵
Solution:

∴ The function attains minimum at x = 0.
and maximum at x = 1 .
The minimum value = 0
The maximum value = 5 – 2 = 3

(iii) y = \(\frac{3 x}{x^2+1}\)
Solution:

(iv) y = x²\(\sqrt{1-x^2}\)
Solution:

(v) y = 2x³ – 15x² – 36x + 18
Solution:

The function attains maximum at x = -1 and minimum at x = 6.
Maximum value
= -2 – 15 + 36 + 18 = 37
Minimum value
= 2 × 216 – 15 × 36 – 36 × 6 + 13
= 432 – 540 – 216 + 18
= 450 – 756
= -306

(vi) y = 60/(x⁴ – x² + 25)
Solution:

(vii) y = (x – 1)³
Solution:

So the point x = 1 is neither a point of maximum nor a point of minimum. It is an inflexion point.

(viii) y = (x – 2)³ (x + 3)⁴
Solution:

(ix) y = x + \(\frac{1}{x}\)
Solution:

∴ Maximum at x = -1
and minimum at x = 1
Maximum value = -1 – 1 = -2
Minimum value = 2.

(x) y = 4 cos 2x – 3 sin 2x, x ∈ (-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Solution:

(xi) y = sin x cos x, x ∈ (\(\frac{\pi}{8}\), \(\frac{\pi}{2}\))
Solution:

But there is no minimum in the domain.

(xii) y = cos x (1 + sin x), x ∈ [0, 2π]
Solution:

(xiii) y = sin^p x cos^q x; p, q > 0, x ∈ [0, \(\frac{\pi}{2}\))
Solution:

(xiv) y = x e^-x ; x ∈ (-2, 2)
Solution:

Question 2.
Show that the following functions do not possess maximum or minimum.
(i) x³
Solution:
y = x³

Thus the function does not possess a maximum or a minimum. (Proved)

(ii) x⁵
Solution:
Processed as in (i)

(iii) 3x³ – 12x² + 16x – 15
Solution:
Let y = 3x³ – 12x2 + 16x – 5

So the function does not possess a maximum or a minimum. (Proved)

(iv) 4 – 3x + 3x² – x³
Solution:
Processed as in (iii).

(v) ln |x|, x ≠ 0
Solution:
y = ln |x|, x ≠ 0

Which have no solution. Hence the function has neither a maximum nor a minimum.
(Proved)

Question 3.
Use the function f(x) = \(x^{\frac{1}{x}}\), x > 0 to show that e^x > π^c.
Solution:
f(x) = \(x^{\frac{1}{x}}\), x > 0

Question 4.
Prove the inequality x² \(e^{-x^2}\) ≤ e^-1, x ∈ R.
Solution:

Question 5.
If f(x) = a ln x + bx² + x has extreme values at x = -1 and x = 2 then find a and b.
Solution:
f(x) = a ln x + bx² + x
f'(x) = \(\frac{a}{x}\) + 2b x + 1
Given that f(x) attains extreme values at x = -1 and x = 2
Then f(-1) = 0 and f(2) = 0
⇒ -a – 2b + 1 = 0 … (1)
\(\frac{a}{2}\) + 4b + 1 = 0 … (2)
From (1) we get a = -2b + 1
Putting it in (2)
we get \(\frac{1-2 b}{2}\) + 4b + 1 = 0
⇒ 1 – 2b + 8b + 2 = 0
⇒ 3 + 6b = 0
⇒ b = \(-\frac{3}{6}\) = \(-\frac{1}{2}\)
Again a = 1 – 2b = 1 + 1 = 2
∴ a = 2, b = \(-\frac{1}{2}\)

Question 6.
Show that \(\frac{x}{1+x \tan x}\), x ∈ (0, \(\frac{\pi}{2}\)) is maximum when x = cos x.
Solution:

Thus the function attains maximum when x = cos x. (Proved)

Question 7.
Determine the absolute maximum and absolute minimum of the following function on [-1, 1]
f(x) = \(\left\{\begin{array}{l}
(x+1)^2, x \leq 0 \\
(x-1)^2, x>0
\end{array}\right.\)
Solution:
f(x) = \(\left\{\begin{array}{l}
(x+1)^2, x \leq 0 \\
(x-1)^2, x>0
\end{array}\right.\)
f(x) = \(\left\{\begin{array}{l}
2(x+1), x \leq 0 \\
2(x-1), x>0
\end{array}\right.\)
For extremum f'(x) = 0
⇒ x + 1 = 0 if x < 0
or, x – 1 = 0 if x > 0
⇒ x = -1 or x = 1
Now f(-1) = 0, f(1) = 0
Again f is not differentiable at x = 0 and f(0) = 1
The function attains absolute minimum at x = ±1. Absolute minimum value = 0.
The function attains absolute maximum at x = 0. Absolute Maximum value = 1.

Question 8.
Find extreme values of
f(x) = \(\left\{\begin{array}{l}
\frac{x}{1-x^2},-1<x<0 \\
x^3-x, 0 \leq x<2
\end{array}\right.\) on (-1, 2)
Solution:
Given that

Question 9.
Find two numbers x and y whose sum is 15 such that xy² is maximum.
Solution:
Let z = xy²
Given that x + y = 15
Then y = 15 – x
Thus z = x (15 – x)²
\(\frac{d z}{d x}\) = (15 – x)² + x² (15 – x) . (-1)
= (15 – x) (15 – x – 2x)
= (15 – x) (15 – 3x)
= 3(15 – x) (5 – x)
\(\frac{d^2 z}{d x^2}\) = -3(5 – x) – 3(15 – x)
= -3 (20 – 2x)
= -6 (10 – x)
For extreme points
\(\frac{d z}{d x}\) = 0
⇒ 3(15 – x) (5 – x) = 0
⇒ x = 15 or 5.
\(\left.\frac{d^2 z}{d x^2}\right]_{x=15}\) = 30 > 0
\(\left.\frac{d^2 z}{d x^2}\right]_{x=5}\) = 30 < 0
∴ z = xy² is maximum when x = 5.
In this case y = 10
∴The two numbers are 5, 10.

Question 10.
If the sum of two positive numbers in constant then show that their product is maximum when they are equal.
Solution:
Let x and y be two positive numbers such that x + y = c (constant)
Then y = c – x
Let p = xy = x(c – x) = cx – x²

∴ The two numbers are equal. (Proved)

Question 11.
Determine a rectangle of area 25 sq. units which has minimum perimeter.
Solution:
Let x and y be the length and breadth of a rectangle of area 25 sq. units.

∴ The perimeter is minimum when the rectangle is a square of side 5 units.

Question 12.
Find the altitude of a right circular cylinder of maximum volume inscribed in a sphere of radius r.
Solution:
Let ABCD be a right circular cylinder inscribed in a sphere with centre at O and radius ‘r’. Let x be the altitude of the cylinder.

Question 13.
Show that the radius of the right circular cylinder of greatest curved surface that can be inscribed in a given cone is half the radius of the base of the cone.

Solution:
Let ABC be a cone, Let r be the radius of the base and h be the height of the cone. Let PQRS be a cylinder inscribed in the cone whose height is x.
In the diagram OT = x.
Then CT = h – x
Now Δ CTR and Δ RQB are similar.

∴ Radius of the right circular cylinder of greatest curved surface is half the base of the cone. (Proved)

Question 14.
Show that the semi vertical angle of a cone of given slant height is tan^-1√2 when its volume is maximum.
Solution:

Question 15.
A cylindrical open water tank with a circular base is to be made out of 30 sq. metres of metal sheet. Find the dimensions so that it can hold maximum water. (Neglect thickness of sheet)
Solution:

Let ABCD be an open cylindrical tank with circular base.
Let r be the radius of the base and x be the height of the tank.
Given that the tank is made out of 30 square metres of metal sheet.
The surface area of the tank
= 27πrx + πr² [∵ The tank is open.
So 2πrx + πr² = 30
⇒ 2πrx = 30 – πr²

Question 16.
A cylindrical vessel of capacity 500 cubic metres open at the top is to be constructed. Find the dimensions of the vessel if the material used is minimum given that the thickness of the material used is 2 cm.
Solution:

Question 17.
Find the coordinates of the point on the curve x²y – x + y = 0 where the slope of the tangent is maximum.
Solution:
Let (x, y) be the point on the curve
x²y – x + y = 0.

The slope is maximum when x = 0. when x = 0, y = 0
The point is (0, 0) at which the slope of the tangent is maximum.

Question 18.
Find the points on the curve y = x² + 1 which are nearest to the point (0, 2).
Solution:
Let A = (0, 2)
Let P (x, y) be any point on the curve
y = x² + 1. …(1)

Question 19.
Show that the minimum distance of a point on the curve \(\frac{a^2}{x^2}\) + \(\frac{b^2}{y^2}\) = 1
from the origin is a + b.
Solution:
Let P(x, y) be any point on the curve

Question 20.
Show that the vertical angle of a right circular cone of minimum curved surface that circumscribes a given sphere is 2 sin^-1(√2 – 1).
Solution:

Let r is the radius of the given sphere, R is the radius of the base, H is the height and l is the C lant height of the cone ABC. Let θ is the semi vertical angle.

Question 21.
Show that the semi-vertical angle of a right circular cone of minimum volume that circumscribes a given sphere is sin^-1(\(\frac{1}{3}\)).
Solution:

Let ABC be a right circular cone which circumscribes a sphere of radius ‘r’.
Let θ be the semi-vertical angle of the cone.
Let D be the centre of the sphere. By symmetry D must lie on AO, where O is the centre of the base.
In the diagram DO = r.
Again in the Δ ADE

Question 22.
Show that the shortest distance of the point (0, 8a) from the curve ax² = y³ is 2a√11.
Solution:
Given curve is 2x² = y³ … (1)
Let any point on it is P(x, y)
Distance of P from A (0, 8a)

Question 23.
Show that the triangle of greatest area that can be inscribed in a circle is equilateral.
Hints: It BC is any chord then for the Δ ABC to have maximum area, the point A must be on the perpendicular bisector of BC so as to have the largest height AD. Let mDBAD = a. Let BO = OA = r. Then area of Δ ABC = \(\frac{1}{2}\)|BC| |AD| = \(\frac{1}{2}\) . 2r sin 2α (r + r + cos 2α) = r² sin 2a . (1 + cos 2a). Then maximise Δ to obtain 2a = \(\frac{\pi}{3}\)
Solution:

Let ABC be a triangle inscribed in a circle with centre at O and radius ‘r’.
The triangle ABC has maximum area if the point A is on the perpendicular bisector of BC.
Let ∠BAD = α, Let BO = AO = r,
Then ∠BOD = 2α.
In Δ OBD, sin 2α = \(\frac{\mathrm{BD}}{\mathrm{OB}}\)
⇒ BD = OB sin 2α = r sin 2α
Again OD = r cos 2α.
Thus AD = r + r cos 2α
Let Z be the area of the triangle ABC.

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

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

Find \(\frac{d y}{d x}\)
Question 1.
xy² + x²y + 1 = 0
Solution:
xy² + x²y + 1 = 0

Question 2.
\(x^{\frac{1}{2}} y^{-\frac{1}{2}}\) + \(x^{\frac{3}{2}} y^{-\frac{3}{2}}\) = 0
Solution:

Question 3.
x² + 3y² = 5
Solution:
x² + 3y² = 5
⇒ \(\frac{d}{d x}\)(x²) + 3 \(\frac{d}{d x}\)(y²) = 0
⇒ 2x + 6y\(\frac{d y}{d x}\) = 0
⇒ \(\frac{d}{d x}\) = –\(\frac{x}{3 y}\)

Question 4.
y² cot x = x² cot y.
Solution:
y² cot x = x² cot y

Question 5.
y = tan xy
Solution:
y = tan xy

Question 6.
x = y In (xy).
Solution:
x = y In (xy).

Question 7.
e^xy + y sin x = 1
Solution:
e^xy + y sin x = 1

Question 8.
In \(\sqrt{x^2+y^2}\) = tan^-1 \(\frac{y}{x}\)
Solution:

Question 9.
y^x = x^{sin y}
Solution:
y^x = x^{sin y}

Question 10.
If sin (x + y) = y cos (x + y) then prove that \(\frac{d y}{d x}\) = –\(\frac{1+y^2}{y^2}\).
Solution:

Question 11.
If \(\sqrt{1-x^4}\) + \(\sqrt{1-y^4}\) = k(x² – y²) then show that \(\frac{d y}{d x}\) = \(\frac{x \sqrt{1-y^4}}{y \sqrt{1-x^4}}\).
Solution:

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.