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Question: Consider a function $f(x) = \sin^{-1}\frac{2x}{1+x^2} + \cos^{-1}\frac{1-x^2}{1+x^2} + \tan^{-1}\fra...

Consider a function f(x)=sin⁡−12x1+x2+cos⁡−11−x21+x2+tan⁡−12x1−x2−atan⁡−1x(a∈R)f(x) = \sin^{-1}\frac{2x}{1+x^2} + \cos^{-1}\frac{1-x^2}{1+x^2} + \tan^{-1}\frac{2x}{1-x^2} - a\tan^{-1}x (a \in R), the possible value of aa for which f′(x)=0f'(x) = 0 over some interval of xx, is

A

6

B

-6

C

2

D

-2

Answer

6, 2, -2

Explanation

Solution

To find the value of aa for which f′(x)=0f'(x) = 0 over some interval of xx, we need to simplify the function f(x)f(x) using standard inverse trigonometric identities. The identities depend on the range of xx. Let x=tan⁡θx = \tan\theta, where θ∈(−π/2,π/2)\theta \in (-\pi/2, \pi/2).

The terms in f(x)f(x) are:

  1. sin⁡−12x1+x2\sin^{-1}\frac{2x}{1+x^2}
  2. cos⁡−11−x21+x2\cos^{-1}\frac{1-x^2}{1+x^2}
  3. tan⁡−12x1−x2\tan^{-1}\frac{2x}{1-x^2}

Let's analyze each term for different intervals of xx:

Identity 1: sin⁡−1(2x1+x2)\sin^{-1}\left(\frac{2x}{1+x^2}\right) Substituting x=tan⁡θx=\tan\theta, we get sin⁡−1(sin⁡(2θ))\sin^{-1}(\sin(2\theta)).

  • If ∣x∣≤1|x| \le 1 (i.e., −π/4≤θ≤π/4-\pi/4 \le \theta \le \pi/4), then 2θ∈[−π/2,π/2]2\theta \in [-\pi/2, \pi/2]. So, sin⁡−1(sin⁡(2θ))=2θ=2tan⁡−1x\sin^{-1}(\sin(2\theta)) = 2\theta = 2\tan^{-1}x.
  • If x>1x > 1 (i.e., π/4<θ<π/2\pi/4 < \theta < \pi/2), then 2θ∈(π/2,π)2\theta \in (\pi/2, \pi). So, sin⁡−1(sin⁡(2θ))=π−2θ=π−2tan⁡−1x\sin^{-1}(\sin(2\theta)) = \pi - 2\theta = \pi - 2\tan^{-1}x.
  • If x<−1x < -1 (i.e., −π/2<θ<−π/4-\pi/2 < \theta < -\pi/4), then 2θ∈(−π,−π/2)2\theta \in (-\pi, -\pi/2). So, sin⁡−1(sin⁡(2θ))=−π−2θ=−π−2tan⁡−1x\sin^{-1}(\sin(2\theta)) = -\pi - 2\theta = -\pi - 2\tan^{-1}x.

Identity 2: cos⁡−1(1−x21+x2)\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) Substituting x=tan⁡θx=\tan\theta, we get cos⁡−1(cos⁡(2θ))\cos^{-1}(\cos(2\theta)).

  • If x≥0x \ge 0 (i.e., 0≤θ<π/20 \le \theta < \pi/2), then 2θ∈[0,π)2\theta \in [0, \pi). So, cos⁡−1(cos⁡(2θ))=2θ=2tan⁡−1x\cos^{-1}(\cos(2\theta)) = 2\theta = 2\tan^{-1}x.
  • If x<0x < 0 (i.e., −π/2<θ<0-\pi/2 < \theta < 0), then 2θ∈(−π,0)2\theta \in (-\pi, 0). So, cos⁡−1(cos⁡(2θ))=−2θ=−2tan⁡−1x\cos^{-1}(\cos(2\theta)) = -2\theta = -2\tan^{-1}x (since cos⁡(−y)=cos⁡y\cos(-y) = \cos y and −2θ∈(0,π)-2\theta \in (0, \pi)).

Identity 3: tan⁡−1(2x1−x2)\tan^{-1}\left(\frac{2x}{1-x^2}\right) Substituting x=tan⁡θx=\tan\theta, we get tan⁡−1(tan⁡(2θ))\tan^{-1}(\tan(2\theta)).

  • If ∣x∣<1|x| < 1 (i.e., −π/4<θ<π/4-\pi/4 < \theta < \pi/4), then 2θ∈(−π/2,π/2)2\theta \in (-\pi/2, \pi/2). So, tan⁡−1(tan⁡(2θ))=2θ=2tan⁡−1x\tan^{-1}(\tan(2\theta)) = 2\theta = 2\tan^{-1}x.
  • If x>1x > 1 (i.e., π/4<θ<π/2\pi/4 < \theta < \pi/2), then 2θ∈(π/2,π)2\theta \in (\pi/2, \pi). So, tan⁡−1(tan⁡(2θ))=2θ−π=2tan⁡−1x−π\tan^{-1}(\tan(2\theta)) = 2\theta - \pi = 2\tan^{-1}x - \pi.
  • If x<−1x < -1 (i.e., −π/2<θ<−π/4-\pi/2 < \theta < -\pi/4), then 2θ∈(−π,−π/2)2\theta \in (-\pi, -\pi/2). So, tan⁡−1(tan⁡(2θ))=2θ+π=2tan⁡−1x+π\tan^{-1}(\tan(2\theta)) = 2\theta + \pi = 2\tan^{-1}x + \pi.

Now, let's substitute these simplified forms into f(x)f(x) for different intervals:

Case 1: For x∈(0,1)x \in (0, 1)

  • sin⁡−12x1+x2=2tan⁡−1x\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x
  • cos⁡−11−x21+x2=2tan⁡−1x\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x
  • tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x

f(x)=2tan⁡−1x+2tan⁡−1x+2tan⁡−1x−atan⁡−1x=(6−a)tan⁡−1xf(x) = 2\tan^{-1}x + 2\tan^{-1}x + 2\tan^{-1}x - a\tan^{-1}x = (6-a)\tan^{-1}x. For f′(x)=0f'(x) = 0 over this interval, the coefficient of tan⁡−1x\tan^{-1}x must be zero, i.e., 6−a=0  ⟹  a=66-a = 0 \implies a=6.

Case 2: For x∈(−1,0)x \in (-1, 0)

  • sin⁡−12x1+x2=2tan⁡−1x\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x
  • cos⁡−11−x21+x2=−2tan⁡−1x\cos^{-1}\frac{1-x^2}{1+x^2} = -2\tan^{-1}x
  • tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x

f(x)=2tan⁡−1x−2tan⁡−1x+2tan⁡−1x−atan⁡−1x=(2−a)tan⁡−1xf(x) = 2\tan^{-1}x - 2\tan^{-1}x + 2\tan^{-1}x - a\tan^{-1}x = (2-a)\tan^{-1}x. For f′(x)=0f'(x) = 0 over this interval, 2−a=0  ⟹  a=22-a = 0 \implies a=2.

Case 3: For x>1x > 1

  • sin⁡−12x1+x2=π−2tan⁡−1x\sin^{-1}\frac{2x}{1+x^2} = \pi - 2\tan^{-1}x
  • cos⁡−11−x21+x2=2tan⁡−1x\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x
  • tan⁡−12x1−x2=2tan⁡−1x−π\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x - \pi

f(x)=(π−2tan⁡−1x)+(2tan⁡−1x)+(2tan⁡−1x−π)−atan⁡−1xf(x) = (\pi - 2\tan^{-1}x) + (2\tan^{-1}x) + (2\tan^{-1}x - \pi) - a\tan^{-1}x f(x)=π−2tan⁡−1x+2tan⁡−1x+2tan⁡−1x−π−atan⁡−1x=(2−a)tan⁡−1xf(x) = \pi - 2\tan^{-1}x + 2\tan^{-1}x + 2\tan^{-1}x - \pi - a\tan^{-1}x = (2-a)\tan^{-1}x. For f′(x)=0f'(x) = 0 over this interval, 2−a=0  ⟹  a=22-a = 0 \implies a=2.

Case 4: For x<−1x < -1

  • sin⁡−12x1+x2=−π−2tan⁡−1x\sin^{-1}\frac{2x}{1+x^2} = -\pi - 2\tan^{-1}x
  • cos⁡−11−x21+x2=−2tan⁡−1x\cos^{-1}\frac{1-x^2}{1+x^2} = -2\tan^{-1}x
  • tan⁡−12x1−x2=2tan⁡−1x+π\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x + \pi

f(x)=(−π−2tan⁡−1x)+(−2tan⁡−1x)+(2tan⁡−1x+π)−atan⁡−1xf(x) = (-\pi - 2\tan^{-1}x) + (-2\tan^{-1}x) + (2\tan^{-1}x + \pi) - a\tan^{-1}x f(x)=−π−2tan⁡−1x−2tan⁡−1x+2tan⁡−1x+π−atan⁡−1x=(−2−a)tan⁡−1xf(x) = -\pi - 2\tan^{-1}x - 2\tan^{-1}x + 2\tan^{-1}x + \pi - a\tan^{-1}x = (-2-a)\tan^{-1}x. For f′(x)=0f'(x) = 0 over this interval, −2−a=0  ⟹  a=−2-2-a = 0 \implies a=-2.

The possible values of aa for which f′(x)=0f'(x)=0 over some interval are 6,2,−26, 2, -2.