Question

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

• A. $x+y=1$
• B. $x+y+1=0$
• C. $2x–y+1=0$
• D. $x+2y+2=0$

Answer 1

The correct option is A

$x+y=1$

Explanation for the correct answer:

Step 1: Find the first order derivative of the equation of the given curve

Given, equation of curve $y={(1+x)}^y+{\sin }^{-1}({\sin }^{2}x)...\left(\mathrm{i}\right)$

On differentiating equation $\left(\mathrm{i}\right)$ w.r.t. $x$, we get

$\frac{dy}{dx}=y{(1+x)}^{y–1}+\ln (1+x){(1+x)}^y\frac{dy}{dx}+\frac{2\sin x\cos x}{\sqrt{(1–{\sin }^{4}x)}}$

Step 2:Find the slope of the normal at the required point

Now at $x=0$ [given]

$y=1$ [from eq $\left(\mathrm{i}\right)$]

The first order derivative at a given point gives the slope of the tangent at that point

$\therefore$ ${\left(\frac{dy}{dx}\right)}_{\left(0,1\right)}=1{(1+0)}^{1–1}+0+\frac{1}{\sqrt{(1–{\sin }^4{0}^{\circ })}}\times 2\sin 0^{\circ }\cos 0^{\circ }$

$$\begin{gathered} =1+0 \\ =1 \end{gathered}$$

Tangent and normal are perpendicular to each other

Therefore, slope of normal is $=\frac{-1}{{\displaystyle \frac{dy}{dx}}}$

$=-1$

Step 3: Find the equation of the normal using slope-point form

So, the equation of normal to the curve 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, the correct answer is option (A).