Question

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

• A. $x+y=1$
• B. $x-y=1$
• C. $x+y=-1$
• D. $x-y=-1$

Answer 1

Answer: (A)

Given curve is
$y=(1+x{)}^y+{\sin }^{-1}({\sin }^{2}x)$
On differentiating w.r.t x, we get
$\frac{dy}{dx}={\left(1+x\right)}^y\left[\frac{y}{1+x}+\log \left(1+x\right)\frac{dy}{dx}\right]$
$+\frac{2\sin x\cos x}{\sqrt{1-{\sin }^{4}x}}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{at\left(0,1\right)}=1$
$\left[\because atx=0,y=1\right]$
Slope of normal at $(x=0)=-1$
$\therefore$ Equation of normal at $x=0$ and $y=1$ is
$y-1=-1(x-0)$
$\Rightarrow y-1=-x$
$\Rightarrow x+y=1$