Q100 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics
If $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ and $y=1$ is
- A$-2$
- Correct answerB$-1$
- C$1$
- D$2$
Solution
$$
y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}
$$
Taking log on both sides, we get
$$
\log y=\frac{1}{2}\left[\log \left(1-\sin ^{-1} x\right)-\log \left(1+\sin ^{-1} x\right)\right]
$$
Differentiating w. r. t. $x$, we get
$$
\begin{aligned}
& \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2}\left[\frac{1}{1-\sin ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}}-\frac{1}{1+\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}\right] \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{2 \sqrt{1-x^2}}\left(\frac{1}{1-\sin ^{-1} x}+\frac{1}{1+\sin ^{-1} x}\right) \\
& \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(0,1)}=\frac{1}{2(1)}\left(\frac{1}{1-0}+\frac{1}{1+0}\right)=-1
\end{aligned}
$$
Hence, the correct answer is option: 'B'.