Second Order Derivative - Class 11 Maths MCQ

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
View Answer

Answer: b
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)[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
View Answer

Answer: d
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
View Answer

Answer: a
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 3^x + 3^f(x) = minimum of Φ(t), where Φ(t) = minimum of (2t³ – 15t² + 36t – 25, 2 + |sint|; 2 ≤ t ≤ 4} ?
a) (-∞, 1)
b) (-∞, log_e3)
c) (0, log_e2)
d) (-∞, log_e2)
View Answer

Answer: d
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, 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_e2)

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
View Answer

Answer: b
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

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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
View Answer

Answer: b
Explanation: We have, y = a sin^-1x + b cos^-1x
Differentiating both sides with respect to x we get,
dy/dx = 12∫ e^2xcos3x dx – 5∫ e^2x 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
View Answer

Answer: a
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
View Answer

Answer: b
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 = 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

Answer: d
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)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
View Answer

Answer: c
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)
View Answer

Answer: d
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)

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