This set of Class 11 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Second Order Derivative”.

1. Whichis greater (1/2)^{e} or 1/e^{2} if there is a given function sinx^{(sinx)}, 0 < x < π?

a) 1/e^{2}

b) (1/2)^{e}

c) Data not sufficient

d) Varies as the value of x changes

View Answer

Explanation: Let, f(x) = sinx

^{(sinx)}

So, f(x) = e

^{sinx log sinx}

So, on differentiating it we get,

f’(x) = sinx

^{(sinx)}[cosx log sinx + cos x]

So, f’(x) = 0 when, x = sin

^{-1}(1/e) or x = π/2

Also, f”(x) = sinx

^{(sinx)}[cos

^{2}x(1 + log sinx)

^{2}– sinx log sinx + (cos

^{2}x/sinx) – sinx]

So that, f”(sin

^{-1}(1/e)) = (e – 1/e)e

^{-1/e}

And, f”(π/2) = -1

Hence, f(x) has local as well as global minimum at x = sin

^{-1}(1/e) also have local and global maximum at x = π/2

Global minimum value of f(x) is (1/e)

^{1/e}

It therefore follows that ,

sin π/6

^{(sinπ/6)}> (1/e)

^{1/e}

=>(1/2)

^{1/2}> (1/e)

^{1/e}

=>(1/2)

^{e}> 1/e

^{2}

2. The function y = f(x) is represented parametrically by x = t^{5} – 5t^{3} – 20t + 7 and y = 4t^{3} – 3t^{2} – 18t + 3 (-2 < t < 2). At which point does y = f(x) has a minimum value?

a) t = -1

b) t = 0

c) t = 1/2

d) t = 3/2

View Answer

Explanation: We have dx/dt = Φ’(t)

Φ’(t) = 5(t

^{2}– 4)(t

^{2}+ 1) ≠ 0 if -2 < t < 2

And dy/dt = 12t

^{2}– 6t – 18

Also, dy/dt = 0

=> t = -1 or t = 3/2

Now, d

^{2}y/dt

^{2}= 24t – 8

=> d

^{2}(-1)/dt

^{2}= -30 and, d

^{2}(3/2)/dt

^{2}= 30

Consequently, y = f(x) has maximum at t = -1 minimum at t = 3/2

3. Which of the following is correct for the nature of the roots x^{5} – a_{0}x^{4} + 3ax^{3} + bx^{2} + cx + d = 0 if it is given that 2a_{0}^{2} < 15 and a_{0}, a, b, c, d are real?

a) Can’t be real

b) Equal

c) Real

d) Depends on the value of x

View Answer

Explanation: Let f(x) = x

^{5}– a

_{0}x

^{4}+ 3ax

^{3}+ bx

^{2}+ cx + d so that,

Now, f’(x) = 5x

^{5}– 4a

_{0}x

^{3}+ 9ax

^{2}+ 2bx + c

And, f”(x) = 20x

^{3}– 12a

_{0}x

^{2}+ 18ax + 2b

And, f”(x) = 60x

^{2}– 24a

_{0}x + 18a

= 6(10x

^{2}– 4a

_{0}x + 3a)

Now, discriminant of 10x

^{2}– 4a

_{0}x + 3a = 16a

_{0}

^{2}– 4

And it is given 8(2a

_{0}

^{2}– 15a) < 0

Hence, root of f”’(x) can’t be real

=>all the roots can’t be real.

4. What will be the domain of the function, if 3^{x} + 3^{f(x)} = minimum of Φ(t), where Φ(t) = minimum of (2t^{3} – 15t^{2} + 36t – 25, 2 + |sint|; 2 ≤ t ≤ 4} ?

a) (-∞, 1)

b) (-∞, log_{e}3)

c) (0, log_{e}2)

d) (-∞, log_{e}2)

View Answer

Explanation: Let, g(t) = 2t

^{3}– 15t

^{2}+ 36t -25

g’(t) = 6t

^{2}– 30t + 36 = 0

=> 6(t

^{2}– 5t + 6) = 0

=> t = 2, 3

For, 2 ≤ t ≤ 4, at t = 3, g”(t) > 0

So, g(t) is minimum.

g(t)

_{min}= g(3) = 2*27 – 15*9 + 36*3 – 25 = 2

Also, 2 + |sint| ≥ 2

Hence, minimum Φ(t) = 2

Therefore, 3

^{x}+ 3

^{f(x)}= 2

=>3

^{f(x)}= 2 – 3

^{x}

Thus, 3

^{f(x)}> 0

=> 3

^{x}> 0 and 3

^{x}< 2

Therefore, x € (-∞, log

_{e}2)

5. If, y = cos(2sin^{-1}x), then what is the value of (1 – x^{2})y” – xy’ + 4y?

a) –1

b) 0

c) 1

d) Depends on the value of x

View Answer

Explanation: We have, y = cos(2sin

^{-1}x)

Differentiating both the sides,

y’ = d/dxcos(2sin

^{-1}x)

= -sin(2sin

^{-1}x) * d/dx(2sin

^{-1}x)

or, y’ = -sin(2sin

^{-1}x)*2*1/√(1 – x

^{2})

or, y’*√(1 – x

^{2}) = -2sin(2sin

^{-1}x)

Squaring both sides,

or, (y’)

^{2}*(1 – x

^{2}) = 4sin

^{2}(2sin

^{-1}x)

or, (y’)

^{2}*(1 – x

^{2}) = 4(1 – y

^{2})

Differentiating again with respect to x, we get,

(1 – x

^{2})*2y’y” + (y’)

^{2}(-2x) = 4 * (-2yy’)

(1 – x

^{2})y” – xy’ = – 4y

=> (1 – x

^{2})y” – xy’ + 4y = 0

6. What will be the form of the equation after eliminating a and b from the equation y = a sin^{-1}x + b cos^{-1}x?

a) (1 – x^{2})y” + xy’ = 0

b) (1 – x^{2})y” – xy’ = 0

c) (1 + x^{2})y” – xy’ = 0

d) (1 + x^{2})y” + xy’ = 0

View Answer

Explanation: We have, y = a sin

^{-1}x + b cos

^{-1}x

Differentiating both sides with respect to x we get,

dy/dx = 12∫ e

^{2x}cos3x dx – 5∫ e

^{2x}sin3x dx

y’ = a* 1/√(1 – x

^{2}) + b * (-1/√(1 – x

^{2}))

or, y’√(1 – x

^{2}) = a – b

now, squaring both sides,

or, y’

^{2}(1 – x

^{2}) = (a – b)

^{2}

Differentiating again with respect to x we get,

(1 – x

^{2}) * 2y’y” + y’

^{2}. (-2x) = 0

As, y’ ≠ 0,

(1 – x

^{2})y” – xy’ = 0

7. What is the value of 4a cos3θ(d^{2}y/dx^{2}) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?

a) [1 + (d^{2}y/dx^{2})^{2}]^{3/2}

b) [1 – (d^{2}y/dx^{2})^{2}]^{3/2}

c) [1 + (d^{2}y/dx^{2})^{2}]^{1/2}

d) [1 – (d^{2}y/dx^{2})^{2}]^{1/2}

View Answer

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2a cos

^{2}θ*sin2θ and y = 2a sin

^{2}θ*cos2θ

Now differentiating x and y with respect to θ we get,

dx/dθ = 2a[cos

^{2}θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ)

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin

^{2}θ (-2sin2θ)]

= 4a sinθ (cosθ cos2θ – sinθ sin2θ )

= 4a sinθ cos(θ + 2θ)

= 4a sinθ cos3θ

Thus, dy/dx = (dy/dθ)/(dx/dθ)

= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)

= tanθ

So, (d

^{2}y/dx

^{2}) = d/dx(tanθ)

= sec

^{2}θ*dθ/dx

= sec

^{2}θ*(1/(dx/dθ))

= sec

^{2}θ*1/(4a cosθ cos3θ)

Or, 4a cos3θ (d

^{2}y/dx

^{2}) = sec

^{3}θ

= (sec

^{2}θ)

^{3/2}

= (1 + tan2θ)

^{3/2}

As, dy/dx = tanθ

So, 4a cos3θ(d

^{2}y/dx

^{2}) = [1 + (d

^{2}y/dx

^{2})

^{2}]

^{3/2}

8. Given, y = tan^{-1} √(x^{2} – 1) then what is the value of (2x^{2} – 1)y’ + x(x^{2} – 1)y”?

a) -1

b) 0

c) 1

d) 2

View Answer

Explanation: We have, y = tan

^{-1}√(x

^{2}– 1)

Differentiating both sides with respect to x, we get,

y’ = 1/(1 + x

^{2}– 1)* d/dx(x

^{2}– 1)

^{1/2}

1/x

^{2}* 1/2((x

^{2}– 1))

^{1/2}* 2x

Squaring both sides,

x

^{2}(x

^{2}– 1)(y’)

^{2}= 1

Differentiating again with respect to x we get,

x

^{2}(x

^{2}– 1)(y’)

^{2}+ (y’)

^{2}d/dx(x

^{4}– x

^{2}) = d/dx(1)

Or, x

^{2}(x

^{2}– 1)

^{2}y’y” + (y’)

^{2}(4x

^{3}– 2x) = 0

Or, x

^{2}(x

^{2}– 1)y” + (y’)(2x

^{3}– x) = 0

9. What is the value of (x + y)^{2}y” if x = e^{t} sint and y = e^{t} cost?

a) x/2(y’ + y)

b) x/2(y’ – y)

c) 2(xy’ + y)

d) 2(xy’ – y)

View Answer

Explanation: Since, x = e

^{t}sint and y = e

^{t}cost

Therefore, dx/dt = e

^{t}sint + e

^{t}cost

= y + x

And, dy/dt = e

^{t}cost – e

^{t}sint

= y – x

So, y’ = dy/dx = (dy/dt)/(dx/dt) = (y – x)/(y + x)

Thus, y” = [(x + y)(y’ – 1) – (y – x)(y’ + 1)]/(x + y)

^{2}

Or, (x + y)

^{2}y” = (x + y – y + x)y’ – x – y + x

= 2xy’ – 2y

= 2(xy’ – y)

10. If, y = (sin^{-1}x)^{2}, then what is the value of (1 – x^{2})y” – xy’ + 4?

a) 2

b) 4

c) 6

d) 8

View Answer

Explanation: We have, y = (sin

^{-1}x)

^{2}………..(1)

Differentiating with respect to x, we get,

y’ = 2(sin

^{-1}x)*1/(1 – x

^{2})

^{1/2}

or, y’(1 – x

^{2})

^{1/2}= 2(sin

^{-1}x)

Squaring both sides,

(1 – x

^{2})(y’)

^{2}= 4(sin

^{-1}x)

^{2}

From (1),

(1 – x

^{2})( y’)

^{2}= 4y

Differentiating with respect to x, we get

(1 – x

^{2})y” + (y’)

^{2}d/dx(1 – x

^{2}) = 4 y’

=>(1 – x

^{2})2y’y” + (y’)

^{2}(-2x) = 4y’

Or, (1 – x

^{2})y” – xy’ = 2

Or, (1 – x

^{2})y” – xy’ + 4 = 2 + 4 = 6

11. If, y = tan^{-1}(x/a), then what is the value of y’?

a) (2ax/(x^{2} – a^{2})^{2})

b) -(2ax/(x^{2} – a^{2})^{2})

c) (2ax/(x^{2} + a^{2})^{2})

d) -(2ax/(x^{2} + a^{2})^{2})

View Answer

Explanation: We have, y = tan

^{-1}(x/a)

Differentiating two times, with respect to x, we get,

y’ = d/dx(tan

^{-1}(x/a))

= 1/(1 + (x/a)

^{2})*d/dx(x/a)

= a

^{2}/(x

^{2}+ a

^{2})* 1/a

= a/(x

^{2}+ a

^{2})

y” = d/dx (a/(x

^{2}+ a

^{2}))

= a* d/dx(x

^{2}+ a

^{2})

^{-1}

= a(-1) (x

^{2}+ a

^{2})

^{-2}*d/dx(x

^{2}+ a

^{2})

Now solving we get,

= -a/(x

^{2}+ a

^{2})

^{2}*2x

= -(2ax/(x

^{2}+ a

^{2})

^{2})

**Sanfoundry Global Education & Learning Series – Mathematics – Class 11**.

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