JEE Challenger
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MathMedMarks to AllAdv 2017 • Paper 2

Evaluate Derivative of Variable Upper and Lower Limit Integral

Suppose g(x)=∫sin⁡xsin⁡(2x)sin⁡−1(t) dtg(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt. Which of the following equations hold true?

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

g′(π2)=−2πg'\left(\frac{\pi}{2}\right) = -2\pi

B

g′(−π2)=2πg'\left(-\frac{\pi}{2}\right) = 2\pi

C

g′(π2)=2πg'\left(\frac{\pi}{2}\right) = 2\pi

D

g′(−π2)=−2πg'\left(-\frac{\pi}{2}\right) = -2\pi

Topics & Concepts

IntegralsDefinite IntegrationDifferentiation

Step-by-Step Solution

To evaluate the derivative of the given integral function g(x)g(x), we use the Leibniz Rule for differentiation under the integral sign:

Given: g(x)=∫sin⁡xsin⁡(2x)sin⁡−1(t) dtg(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt

Differentiating g(x)g(x) with respect to xx: g′(x)=sin⁡−1(sin⁡(2x))⋅ddx(sin⁡(2x))−sin⁡−1(sin⁡x)⋅ddx(sin⁡x)g'(x) = \sin^{-1}(\sin(2x)) \cdot \frac{d}{dx}(\sin(2x)) - \sin^{-1}(\sin x) \cdot \frac{d}{dx}(\sin x)

g′(x)=2cos⁡(2x)sin⁡−1(sin⁡(2x))−cos⁡(x)sin⁡−1(sin⁡x)g'(x) = 2\cos(2x) \sin^{-1}(\sin(2x)) - \cos(x) \sin^{-1}(\sin x)

Step 1: Evaluate g′(π2)g'\left(\frac{\pi}{2}\right)

Substitute x=π2x = \frac{\pi}{2} into g′(x)g'(x):

  1. Upper limit terms: sin⁡(2x)=sin⁡(π)=0  ⟹  sin⁡−1(0)=0\sin(2x) = \sin(\pi) = 0 \implies \sin^{-1}(0) = 0 cos⁡(2x)=cos⁡(π)=−1\cos(2x) = \cos(\pi) = -1

  2. Lower limit terms: sin⁡x=sin⁡(π2)=1  ⟹  sin⁡−1(1)=π2\sin x = \sin\left(\frac{\pi}{2}\right) = 1 \implies \sin^{-1}(1) = \frac{\pi}{2} cos⁡x=cos⁡(π2)=0\cos x = \cos\left(\frac{\pi}{2}\right) = 0

Substituting these values into the derivative formula: g′(π2)=2(−1)(0)−(0)(π2)=0g'\left(\frac{\pi}{2}\right) = 2(-1)(0) - (0)\left(\frac{\pi}{2}\right) = 0

Step 2: Evaluate g′(−π2)g'\left(-\frac{\pi}{2}\right)

Substitute x=−π2x = -\frac{\pi}{2} into g′(x)g'(x):

  1. Upper limit terms: sin⁡(2x)=sin⁡(−π)=0  ⟹  sin⁡−1(0)=0\sin(2x) = \sin(-\pi) = 0 \implies \sin^{-1}(0) = 0 cos⁡(2x)=cos⁡(−π)=−1\cos(2x) = \cos(-\pi) = -1

  2. Lower limit terms: sin⁡x=sin⁡(−π2)=−1  ⟹  sin⁡−1(−1)=−π2\sin x = \sin\left(-\frac{\pi}{2}\right) = -1 \implies \sin^{-1}(-1) = -\frac{\pi}{2} cos⁡x=cos⁡(−π2)=0\cos x = \cos\left(-\frac{\pi}{2}\right) = 0

Substituting these values into the derivative formula: g′(−π2)=2(−1)(0)−(0)(−π2)=0g'\left(-\frac{\pi}{2}\right) = 2(-1)(0) - (0)\left(-\frac{\pi}{2}\right) = 0

Conclusion

The correct mathematical values are: g′(π2)=0andg′(−π2)=0g'\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad g'\left(-\frac{\pi}{2}\right) = 0

Since none of the options [A], [B], [C], or [D] match the calculated values of 00, none of the given options are correct. Consequently, full marks are awarded to all candidates for this question.

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