Question

If ${\sin }^{-1}x+{\cos }^{-1}(1-x)={\sin }^{-1}(-x)$ , then $x$ satisfies the equation

• A. $2x^2-x+2=0$
• B. $2x^2-3x=0$
• C. $2x^2+x-1=0$
• D. none of these

Answer 1

Answer: (B)

$si{n}^{-1}x+co{s}^{-1}\left(1-x\right)=si{n}^{-1}\left(-x\right)$
$\Rightarrow co{s}^{-1}\left(1-x\right)=-si{n}^{-1}\left(x\right)+si{n}^{-1}\left(-x\right)$
$\Rightarrow co{s}^{-1}\left(1-x\right)=-2si{n}^{-1}x$
$\Rightarrow 1-x=cos\left(-2si{n}^{-1}x\right)$
$=cos\left(2si{n}^{-1}x\right)=1-2si{n}^2\left(si{n}^{-1}x\right)=1-2x^2$
$\Rightarrow$ Either $x=0$ or $x=1/2$
But $x=1/2$ does not satisfy the given equation
$\therefore x=0$