Q65 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text { is equal to }\end{aligned}$

  • A
    $4 \sin ^{-1} x$, when $x \in[-1,1]$
  • Correct answer
    B
    $\pi$, when $x \in\left[-1,-\frac{1}{\sqrt{2}}\right]$
  • C
    $-\pi$, when $x \in\left[\frac{-1}{2}, \frac{1}{2}\right]$
  • D
    $4 \sin ^{-1} x+2 \cos ^{-1}\left(4 x^3-3 x\right), x \in\left[\frac{1}{\sqrt{2}}, 1\right]$

Solution

$\begin{aligned} & \text { } 2 \sin ^{-1} x+\sin ^{-1}\left[2 x \sqrt{1-x^2}\right] \\ & +3 \cos ^{-1} x-\cos ^{-1}\left(4 x^3-3 x\right) \\ & \text { Put } x=\cos \theta \Rightarrow \theta=\cos ^{-1} x \\ & =2 \sin ^{-1}(\cos \theta)+\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^2 \theta}\right) \\ & \qquad \quad+3 \cos ^{-1}(\cos \theta)-\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right) \\ & =2 \sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\theta\right)+\sin ^{-1} \mid 2 \cos \theta \sin \theta\right] \\ & =2\left(\frac{\pi}{2}-\theta\right)+2 \theta+3 \theta-3 \theta \\ & =\pi-2 \theta+2 \theta+3 \theta-3 \theta \\ & =\pi \text {, when } x \in\left\{-1, \frac{-1}{\sqrt{2}}\right\}\end{aligned}$

Hence, the correct answer is option: 'B'.

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