Question

The range of x for which the formula $3 \, \sin^{-1} x = \sin^{-1} \, [x(3 - 4x^2)]$ hold is

• A. $ - \frac{2}{3} \leq x \leq \frac{2}{3} $
• B. $ - \frac{1}{2} \leq x \leq \frac{1}{2} $
• C. $ - \frac{1}{3} \leq x \leq \frac{1}{3} $
• D. $ - \frac{1}{4} \leq x \leq \frac{2}{3} $

Answer 1

Answer: (B)

Given,
$ 3 \,\sin ^{-1} \,x =\sin ^{-1}\left[X\left(3-4 x^{2}\right)\right] \,...(i)$
Let $\sin ^{-1} \,x=\theta$
$\therefore \, X =\sin \,\theta \,...(ii)$
Now, from Eq. (i)
$3 \theta=\sin ^{-1}\left[\sin \theta\left(3-4 \sin ^{2} \theta\right)\right]$
Here, $-\frac{\pi}{2} \leq 3 \theta \leq \frac{\pi}{2}$
$\left[\because\right.$ Domain $\sin ^{-1} \,x$ is in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \, -\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$
$ \therefore \, -\frac{1}{2} \leq \sin \theta \leq \frac{1}{2}$
$\because \, X=\sin \,\theta$[from Eq. (ii)]
$\therefore \, -\frac{1}{2} \leq x \leq \frac{1}{2}$