Question

The equation of the normal to the curve $y={(1+x)}^y+{\sin }^{-1}\left({\sin }^2\left(x\right)\right)$ at $x=0$.

• A. $x+y=2$
• B. $x+y=1$
• C. $x–y=1$
• D. $x^2–y^2=2$

Answer 1

The correct option is B

$x+y=1$

Explanation of the correct option.

Step $1$: Find the intersection point.

Given: Curve $y={(1+x)}^y+{\sin }^{-1}\left({\sin }^2\left(x\right)\right)$

For intersecting point put $x=0$ in equation of curve,

$$\begin{gathered} y={(1+0)}^y \\ \Rightarrow \ln y=y\ln 1 \\ \Rightarrow \ln y=0 \\ \Rightarrow y=e^0 \\ \Rightarrow y=1 \end{gathered}$$ $$\begin{gathered} \left[{\mathrm{lna}}^x=x\mathrm{lna} \\ \ln 1=0\right] \end{gathered}$$

Hence the point of intersection is $\left(0,1\right)$

Step $2$: Find the slope.

$y={(1+x)}^y+{\sin }^{-1}\left({\sin }^2\left(x\right)\right)$

Differentiate with respect to $x$.

$\frac{dy}{dx}=y{(1+x)}^{y-1}+{(1+x)}^y\ln (1+x)\frac{dy}{dx}+\frac{1}{\sqrt{\left(1-{\sin }^4\left(x\right)\right)}}\left[2\sin \left(x\right)\cos \left(x\right)\right]$

slope at the intersection $\left(0,1\right)$ point

$\frac{dy}{dx}=1+0$

$\Rightarrow \frac{dy}{dx}=1$

Since the slope of perpendicular is $-\frac{1}{m}$.

Therefore the slope of the required equation is $-1$.

Step $3$: Find the equation of the line.

We know that equation of line passing through $\left(x_{1,}{y}_1\right)$ is given by,

$y-y_1=m(x-x_1)$

Therefore, the equation of line passing through $\left(0,1\right)$ is,

$$\begin{gathered} y-1=-1\left(x-0\right) \\ \Rightarrow y-1=-x \\ \Rightarrow x+y=1 \end{gathered}$$

Hence Option $\left(\mathrm{B}\right)$ is the correct answer.