$\large\frac{dy}{\sqrt {1-y^2}}=\frac{dx}{1+x^2}$
It satisfied $(0, \large\frac{1}{\sqrt 2})$
$\sin ^{-1}y =\tan^{-1} x +c$
$\sin ^{-1}y =\tan ^{-1} x +\large\frac{\pi}{4}$
$\large\frac{-\pi}{2}$$ \leq \tan^{-1}x +\large\frac{\pi}{4} \leq$$ \large\frac{\pi}{2}$
$\large\frac{-3 \pi}{4}$$ \leq \tan^{-1}x \leq$$ \large\frac{\pi}{2}$
Variation of $\tan ^{-1}x$ is $\bigg(\large\frac{-\pi}{2},\frac{\pi}{2}\bigg)$
So variation of x is
$-\infty < x \leq 1$
Hence b is the correct answer.