Mathematics Questions and Answers – Second Order Derivative

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This set of Mathematics 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) = 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
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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.
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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)
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, 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
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∫ 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
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
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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 = et sint and y = et 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 = 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)
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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)

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter