Question

The equation $\sin \left({\sin }^{-1}x\right)=2-x^2$ holds good for

• A. no negative value of $x$
• B. no real value of $x$
• C. one real value of $x$
• D. atleast one negative value of $x$

Answer 1

Answer: (A), (C)

one real value of $x$

$\sin \left({\sin }^{-1}x\right)=x\forall x\in [-1,1]$
$\therefore \sin \left({\sin }^{-1}x\right)=2-x^2$
$\Rightarrow x=2-x^2\text{and}x\in [-1,1]$
$\Rightarrow x^2+x-2=0$
$\Rightarrow x=-2,x=1$
$\therefore x=1$ is the solution as $-2\notin [-1,1]$