CUET Preparation Today
If $f:\left[\frac{π}{2},\frac{3π}{2}\right]→[-1,1]$ is defined by $f(x)=\sin x$, then $f^{-1}(x)$ is given by |
$\sin^{-1}x$ $π+\sin^{-1}x$ $π-\sin^{-1}x$ none of these |
$π-\sin^{-1}x$ |
Obviously f is one-one and onto, thus $f^{-1}$ exists. Since $-\frac{π}{2}≤\sin^{-1}x≤\frac{π}{2},\frac{π}{2}≤π-\sin^{-1}x≤\frac{3π}{2}$ |
