Q68 / 150 · MHT CET 2025 (23 Apr Shift 2) · Mathematics

If $\left(\cos ^{-1} x\right)^2-\left(\sin ^{-1} x\right)^2>0$, then

  • A
    $x < \frac{1}{2}$
  • B
    $-1 < x < \sqrt{2}$
  • C
    $0 \leqslant x < \frac{1}{\sqrt{2}}$
  • Correct answer
    D
    $-1 \leqslant x < \frac{1}{\sqrt{2}}$

Solution

The inequality $\left(\cos^{-1} x\right)^2 - \left(\sin^{-1} x\right)^2 > 0$ factors as $(\cos^{-1} x - \sin^{-1} x)(\cos^{-1} x + \sin^{-1} x) > 0$.

Using the identity $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$, the inequality simplifies to $(\cos^{-1} x - \sin^{-1} x)\left(\frac{\pi}{2}\right) > 0$.

Since $\frac{\pi}{2} > 0$, this reduces to $\cos^{-1} x > \sin^{-1} x$.

Substituting $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$ gives $\sin^{-1} x < \frac{\pi}{4}$.

Taking sine of all terms, and noting $\sin(t)$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, yields $-1 \leq x < \frac{1}{\sqrt{2}}$.

This corresponds to option D.

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

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