Q66 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

The trigonometric equation $\sin ^{-1} x=2 \sin ^{-1} a$, has a solution

  • A
    only when $\frac{1}{\sqrt{2}} \lt a \lt \frac{1}{2}$
  • B
    for all real values of ' $a$ '
  • Correct answer
    C
    only when $|a| \leqslant \frac{1}{\sqrt{2}}$
  • D
    only when $|a| \geq \frac{1}{\sqrt{2}}$

Solution

$\sin ^{-1} x=2 \sin ^{-1} a$ will have a solution if $2 \sin ^{-1} a \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \Rightarrow \sin ^{-1} a \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$ $\Rightarrow a \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \Rightarrow|a| \leq \frac{1}{\sqrt{2}}$

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

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