Question

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

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

Answer 1

The correct option is A $-\frac{1}{2}\le x\le \frac{1}{2}$

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