Question

If $x=a$ is a solution of the equation $\sin ^{-1} \frac{x}{3}+\sin ^{-1} \frac{2 x}{3}=\sin ^{-1} x$, then the roots of the equation $x^{2}-a x-1=0$ are

• A. $\pm 1$
• B. $\frac{1}{2}, 1$
• C. $\pm \frac{1}{2}$
• D. $-\frac{1}{2}, 1$

Answer 1

Answer: (A)

Given,
$\sin ^{-1} \frac{x}{3}+\sin ^{-1} \frac{2 x}{3}=\sin ^{-1} x$
$\Rightarrow \sin ^{-1}\left(\frac{x}{3} \sqrt{1-\frac{4 x^{2}}{9}}+\frac{2 x}{3} \sqrt{1-\frac{x^{2}}{9}}\right)=\sin ^{-1} x$
$\Rightarrow x\left[\frac{1}{9} \sqrt{9-4 x^{2}}+\frac{2}{9} \sqrt{9-x^{2}}-1\right]=0$
$\Rightarrow 54=18 \sqrt{9-x^{2}} $
$\Rightarrow 3=\sqrt{9-x^{2}}$
$\Rightarrow 9=9-x^{2} $
$ \Rightarrow x^{2}=0$
$\Rightarrow x=0 $
$\Rightarrow a=0$
$\therefore $ Equation becomes $x^{2}-1=0$
$ \Rightarrow x=\pm 1$
Hence, roots are $\pm 1$.