The greatest value of the function $f(x)=\sin ^{-1} x^2$ in the interval $[-1 / \sqrt{2}, 1 / \sqrt{2}]$, is
Answer & explanation
Correct answer: option 4
We have,
$f(x)=\sin ^{-1} x^2 \Rightarrow f^{\prime}(x)=\frac{2 x}{\sqrt{1-x^4}}$
∴ $f^{\prime}(x)=0 \Rightarrow x=0$
Now,
$f\left(-\frac{1}{\sqrt{2}}\right)=\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}, f\left(\frac{1}{\sqrt{2}}\right)=\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}$
and, $f(0)=\sin ^{-1}(0)=0$.
Hence, the greatest value is $\frac{\pi}{6}$.