Question

Which of the following is the solution set of equation ${\sin }^{-1}x={\cos }^{-1}x+{\sin }^{-1}(3x-1)$?

• A. $[0,\frac{1}{3}]$
• B. $[\frac{1}{3},\frac{2}{3}]$
• C. $[0,\frac{2}{3}]$
• D. none of these

Answer 1

Answer: (D)

none of these

For the existence of the given, we must have
$-1\le x\le 1and-1\le 3x-1\le 1\Rightarrow 0\le x\le \frac{2}{3}$
Now,
${\sin }^{-1}x={\cos }^{-1}x+{\sin }^{-1}(3x-1)$
$\Rightarrow {\sin }^{-1}x-{\cos }^{-1}x={\sin }^{-1}(3x-1)$
$\Rightarrow 2{\sin }^{-1}x-\frac{\pi }{2}={\sin }^{-1}(3x-1)$
$\Rightarrow \sin (2{\sin }^{-1}x-\frac{\pi }{2})x=\sin \left({\sin }^{-1}(3x-1)\right)$
$\Rightarrow -\cos \left(2{\sin }^{-1}x\right)=3x-1$
$\Rightarrow -\{1-2{\sin }^2{\sin }^{-1}x\}=3x-1$
$\Rightarrow -(1-{2x}^2)=3x-1$
$\Rightarrow 2x^2-3x=0\Rightarrow x=0,\frac{3}{2}$
$\Rightarrow x=0[\because 0\le x\le 2/3]$