Question

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

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

Answer 1

Answer: (A)

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