Q51 / 120 · VITEEE 2014 · Mathematics
The equation of normal to the curve $y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$ at $x=0$ is
- Correct answerA$x+y=1$
- B$x-y=1$
- C$x+y=-1$
- D$x-y=-1$
Solution
Given curve is
$y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$
On differentiating w.r.t $x$, we get
$\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=(1+\mathrm{x})^{\mathrm{y}}\left[\frac{\mathrm{y}}{1+\mathrm{x}}+\log (1+\mathrm{x}) \frac{\mathrm{dy}}{\mathrm{dx}}\right] \\
\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{at}(0,1)}=\frac{2 \sin \mathrm{x} \cos \mathrm{x}}{\sqrt{1-\sin ^{4} \mathrm{x}}}
\end{array}$
Slope of normal at $(x=0)=-1$
$\therefore \quad$ Equation of normal at $x=0$ and $y=1$ is $\begin{aligned} y-1 &=-1(x-0) \\ \Rightarrow y-1 &=-x \Rightarrow x+y=1 \end{aligned}$
Hence, the correct answer is option: 'A'.