Q88 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then the value of $x$ is

  • A
    $\frac{\pi}{2}+\frac{1}{5}$
  • B
    $\frac{\pi}{2}-\frac{1}{5}$
  • C
    $-\frac{1}{5}$
  • Correct answer
    D
    $\frac{1}{5}$

Solution

Given that $\sin\left(\sin^{-1} \frac{1}{5} + \cos^{-1} x\right) = 1$, the argument must satisfy $\sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2}$.

Using the identity $\sin^{-1} y + \cos^{-1} y = \frac{\pi}{2}$ for $y \in [-1, 1]$, we identify $x = \frac{1}{5}$.

Since $\frac{1}{5}$ lies within the domain of $\cos^{-1} x$, this solution is valid.

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

Relevant Questions