Question

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

• A. $\mathrm{x}+\mathrm{y}=1$
• B. $\mathrm{x}-\mathrm{y}=1$
• C. $\mathrm{x}+\mathrm{y}=-1$
• D. $\mathrm{x}-\mathrm{y}=-1$

Answer 1

The correct option is A

$\mathrm{x}+\mathrm{y}=1$

Explanation for correct option:

Finding the equation of normal:

Given equation of the curve,

$\mathrm{y}={\left(1+\mathrm{x}\right)}^{\mathrm{y}}+{\sin }^{-1}\left({\sin }^2\mathrm{x}\right)$ at $\mathrm{x}=0$

Substituting $\mathrm{u}={\left(1+\mathrm{x}\right)}^{\mathrm{y}}$

$\Rightarrow \ln \left(\mathrm{u}\right)=\mathrm{yln}\left(1+\mathrm{x}\right)$

On differentiating both sides, we get

$$\begin{gathered} \left(\frac{1}{\mathrm{u}}\right)\frac{d\mathrm{u}}{d\mathrm{x}}=\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)\ln \left(1+\mathrm{x}\right)+\frac{\mathrm{y}}{1+\mathrm{x}} \\ \Rightarrow \frac{d\mathrm{u}}{d\mathrm{x}}={\left(1+\mathrm{x}\right)}^{\mathrm{y}}\left[\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)\ln \left(1+\mathrm{x}\right)+\frac{\mathrm{y}}{1+\mathrm{x}}\right] \end{gathered}$$

Substituting $\mathrm{v}={\sin }^{-1}\left({\sin }^2\left(\mathrm{x}\right)\right)$ we get

$\Rightarrow \frac{d\mathrm{v}}{d\mathrm{x}}=\left[\frac{1}{\sqrt{1-{\sin }^4\left(\mathrm{x}\right)}}\right]\times 2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)$

On differentiating the equation of the curve,

$$\begin{gathered} \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{d\mathrm{u}}{d\mathrm{x}}+\frac{d\mathrm{v}}{d\mathrm{x}} \\ \Rightarrow \left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)={\left(1+\mathrm{x}\right)}^{\mathrm{y}}\left[\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)\ln \left(1+\mathrm{x}\right)+\frac{\mathrm{y}}{1+\mathrm{x}}\right]+\left[\frac{1}{\sqrt{1-{\sin }^4\left(\mathrm{x}\right)}}\right]\times 2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right) \\ \Rightarrow {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{0,1}={\left(1+0\right)}^1\left[\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)\ln \left(1+0\right)+\frac{1}{1+0}\right]+\left[\frac{1}{\sqrt{1-{\sin }^4\left(0\right)}}\right]\times 2\sin \left(0\right)\cos \left(0\right) \\ \Rightarrow {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{0,1}=1\left[\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)\ln \left(1\right)+\frac{1}{1}\right]+\left[\frac{1}{\sqrt{1-0}}\right]\times 0 \\ \Rightarrow {\left(\frac{d\mathrm{y}}{d\mathrm{x}}\right)}_{0,1}=1\left(1\right)+0=1 \end{gathered}$$

Since we get the slope of tangent as $\left(\frac{dy}{dx}\right)=1$ and we know that tangent and normal are both perpendicular to each other so, using perpendicular lines slope condition such that product of slope of tangent and normal will be $-1$ hence, the slope of normal will be $=-1$

So, the equation of the normal is given by

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

Hence, option (A) is the correct answer.