Question

The number of solution of equation ${\sin }^{-1}x+n{\sin }^{-1}(1-x)=\frac{m\pi }{2},$where $n>0,m\le 0,$is

• A. $3$
• B. $1$
• C. $2$
• D. none of these

Answer 1

The correct option is D none of these

${\sin }^{-1}\left(x\right)+n{\sin }^{-1}(1-x)=\frac{m\pi }{2}\Rightarrow \pi m=2({\sin }^{-1}\left(x\right)+n{\sin }^{-1}(1-x))$
Solution exists only when
$m=1,x=1$ or $n=m,x=0$
But $n>0,m\le 0$