Question

The number of roots of the equation ${\sin }^{-1}(1-x)+2{\sin }^{-1}x=\frac{\pi }{2}$, is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

Let ${\sin }^{-1}(1-x)=A\sin A=1-x$

and ${\sin }^{-1}x=B,\sin B=x$

According to question,

$A-2B=\frac{x}{2}$

$⟹A-\frac{x}{2}=2B$

$⟹\cos (A-\frac{x}{2})=\cos 2B$

$⟹\sin A=1-2{\sin }^{2}B$ $[{\cos }^{2}B=1-2{\sin }^{2}B]$

$1-x=1-2x^2$

$2x^2-x=0$

$x(2x-1)=0$

$x=0$ or $2x-1=0x=\frac{1}{2}$