Q68 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$, where $0 \leq \cos ^{-1} x \leq \pi$ and $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$, is

  • A
    $-\frac{\sqrt{6}}{5}$
  • B
    $\frac{\sqrt{6}}{5}$
  • Correct answer
    C
    $-\frac{2 \sqrt{6}}{5}$
  • D
    $\frac{2 \sqrt{6}}{5}$

Solution

$\begin{aligned} & \cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)=\cos \left(\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right) \\ & =\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right) \quad\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right] \\ & =-\sin \left(\cos ^{-1} x\right) \\ & =-\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) \\ & =-\sqrt{1-x^2} \\ & =-\sqrt{1-\left(\frac{1}{5}\right)^2} \quad\left[\because x=\frac{1}{5}\right] \\ & =-\frac{2 \sqrt{6}}{5}\end{aligned}$

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

Relevant Questions