# Class 11 Maths MCQ – Second Order Derivative

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/e2 if there is a given function sinx(sinx), 0 < x < π?
a) 1/e2
b) (1/2)e
c) Data not sufficient
d) Varies as the value of x changes

Explanation: Let, f(x) = sinx(sinx)
So, f(x) = esinx 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)[cos2x(1 + log sinx)2 – sinx log sinx + (cos2x/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/e2

2. The function y = f(x) is represented parametrically by x = t5 – 5t3 – 20t + 7 and y = 4t3 – 3t2 – 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

Explanation: We have dx/dt = Φ’(t)
Φ’(t) = 5(t2 – 4)(t2 + 1) ≠ 0 if -2 < t < 2
And dy/dt = 12t2 – 6t – 18
Also, dy/dt = 0
=> t = -1 or t = 3/2
Now, d2y/dt2 = 24t – 8
=> d2(-1)/dt2 = -30 and, d2(3/2)/dt2 = 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 x5 – a0x4 + 3ax3 + bx2 + cx + d = 0 if it is given that 2a02 < 15 and a0, a, b, c, d are real?
a) Can’t be real
b) Equal
c) Real
d) Depends on the value of x

Explanation: Let f(x) = x5 – a0x4 + 3ax3 + bx2 + cx + d so that,
Now, f’(x) = 5x5 – 4a0x3 + 9ax2 + 2bx + c
And, f”(x) = 20x3 – 12a0x2 + 18ax + 2b
And, f”(x) = 60x2 – 24a0x + 18a
= 6(10x2 – 4a0x + 3a)
Now, discriminant of 10x2 – 4a0x + 3a = 16a02 – 4
And it is given 8(2a02 – 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 3x + 3f(x) = minimum of Φ(t), where Φ(t) = minimum of (2t3 – 15t2 + 36t – 25, 2 + |sint|; 2 ≤ t ≤ 4} ?
a) (-∞, 1)
b) (-∞, loge3)
c) (0, loge2)
d) (-∞, loge2)

Explanation: Let, g(t) = 2t3 – 15t2 + 36t -25
g’(t) = 6t2 – 30t + 36 = 0
=> 6(t2 – 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, 3x + 3f(x) = 2
=>3f(x) = 2 – 3x
Thus, 3f(x) > 0
=> 3x > 0 and 3x < 2
Therefore, x € (-∞, loge2)

5. If, y = cos(2sin-1x), then what is the value of (1 – x2)y” – xy’ + 4y?
a) –1
b) 0
c) 1
d) Depends on the value of x

Explanation: We have, y = cos(2sin-1x)
Differentiating both the sides,
y’ = d/dxcos(2sin-1x)
= -sin(2sin-1x) * d/dx(2sin-1x)
or, y’ = -sin(2sin-1x)*2*1/√(1 – x2)
or, y’*√(1 – x2) = -2sin(2sin-1x)
Squaring both sides,
or, (y’)2 *(1 – x2) = 4sin2(2sin-1x)
or, (y’)2 *(1 – x2) = 4(1 – y2)
Differentiating again with respect to x, we get,
(1 – x2)*2y’y” + (y’)2 (-2x) = 4 * (-2yy’)
(1 – x2)y” – xy’ = – 4y
=> (1 – x2)y” – xy’ + 4y = 0

6. What will be the form of the equation after eliminating a and b from the equation y = a sin-1x + b cos-1x?
a) (1 – x2)y” + xy’ = 0
b) (1 – x2)y” – xy’ = 0
c) (1 + x2)y” – xy’ = 0
d) (1 + x2)y” + xy’ = 0

Explanation: We have, y = a sin-1x + b cos-1x
Differentiating both sides with respect to x we get,
dy/dx = 12∫ e2xcos3x dx – 5∫ e2x sin3x dx
y’ = a* 1/√(1 – x2) + b * (-1/√(1 – x2))
or, y’√(1 – x2) = a – b
now, squaring both sides,
or, y’2(1 – x2) = (a – b)2
Differentiating again with respect to x we get,
(1 – x2) * 2y’y” + y’2 . (-2x) = 0
As, y’ ≠ 0,
(1 – x2)y” – xy’ = 0

7. What is the value of 4a cos3θ(d2y/dx2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?
a) [1 + (d2y/dx2)2]3/2
b) [1 – (d2y/dx2)2]3/2
c) [1 + (d2y/dx2)2]1/2
d) [1 – (d2y/dx2)2]1/2

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)
=> x = 2a cos2θ*sin2θ and y = 2a sin2θ*cos2θ
Now differentiating x and y with respect to θ we get,
dx/dθ = 2a[cos2θ*2cos2θ + sin2θ*2cos2θ cos2θ]
= 4a cosθ (cosθ cos2θ – sinθ sin2θ)
= 4a cosθ cos(θ + 2θ)
= 4a cosθ cos3θ
dy/dθ = 2a[cos2θ*2cosθ sinθ + sin2θ (-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, (d2y/dx2) = d/dx(tanθ)
= sec2θ*dθ/dx
= sec2θ*(1/(dx/dθ))
= sec2θ*1/(4a cosθ cos3θ)
Or, 4a cos3θ (d2y/dx2) = sec3θ
= (sec2θ)3/2
= (1 + tan2θ)3/2
As, dy/dx = tanθ
So, 4a cos3θ(d2y/dx2) = [1 + (d2y/dx2)2]3/2

8. Given, y = tan-1 √(x2 – 1) then what is the value of (2x2 – 1)y’ + x(x2 – 1)y”?
a) -1
b) 0
c) 1
d) 2

Explanation: We have, y = tan-1 √(x2 – 1)
Differentiating both sides with respect to x, we get,
y’ = 1/(1 + x2 – 1)* d/dx(x2 – 1)1/2
1/x2 * 1/2((x2 – 1))1/2 * 2x
Squaring both sides,
x2(x2 – 1)(y’)2 = 1
Differentiating again with respect to x we get,
x2(x2 – 1)(y’)2 + (y’)2d/dx(x4 – x2) = d/dx(1)
Or, x2(x2 – 1)2y’y” + (y’)2(4x3 – 2x) = 0
Or, x2(x2 – 1)y” + (y’)(2x3 – x) = 0

9. What is the value of (x + y)2y” if x = et sint and y = et cost?
a) x/2(y’ + y)
b) x/2(y’ – y)
c) 2(xy’ + y)
d) 2(xy’ – y)

Explanation: Since, x = et sint and y = et cost
Therefore, dx/dt = et sint + et cost
= y + x
And, dy/dt = et cost – et 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)2y” = (x + y – y + x)y’ – x – y + x
= 2xy’ – 2y
= 2(xy’ – y)

10. If, y = (sin-1x)2, then what is the value of (1 – x2)y” – xy’ + 4?
a) 2
b) 4
c) 6
d) 8

Explanation: We have, y = (sin-1x)2 ………..(1)
Differentiating with respect to x, we get,
y’ = 2(sin-1x)*1/(1 – x2)1/2
or, y’(1 – x2)1/2 = 2(sin-1x)
Squaring both sides,
(1 – x2)(y’)2 = 4(sin-1x)2
From (1),
(1 – x2)( y’)2 = 4y
Differentiating with respect to x, we get
(1 – x2)y” + (y’)2 d/dx(1 – x2) = 4 y’
=>(1 – x2)2y’y” + (y’)2 (-2x) = 4y’
Or, (1 – x2)y” – xy’ = 2
Or, (1 – x2)y” – xy’ + 4 = 2 + 4 = 6

11. If, y = tan-1(x/a), then what is the value of y’?
a) (2ax/(x2 – a2)2)
b) -(2ax/(x2 – a2)2)
c) (2ax/(x2 + a2)2)
d) -(2ax/(x2 + a2)2)

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)
= a2/(x2 + a2)* 1/a
= a/(x2 + a2)
y” = d/dx (a/(x2 + a2))
= a* d/dx(x2 + a2)-1
= a(-1) (x2 + a2)-2*d/dx(x2 + a2)
Now solving we get,
= -a/(x2 + a2)2*2x
= -(2ax/(x2 + a2)2)

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

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