Question

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

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

Answer 1

Answer: (B)

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