Question

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

• A. $x+y-1=0$
• B. $x+y-2=0$
• C. $x+y-3=0$
• D. $x+y-4=0$

Answer 1

The correct option is A $x+y-1=0$

Clearly at $x=0,y=1$

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

Differentiating w.r.t $x$

$\frac{dy}{dx}=(1+x{)}^y[\frac{y}{1+x}+\log (1+x).\frac{dy}{dx}]+\frac{2\sin x.\cos x}{\sqrt{1+{\sin }^{4}x}}$

Putting $x=0,y=1$, we get, $\frac{dy}{dx}=1=m$ (say)

Thus slope of the normal is $=-\frac{1}{m}=-1$

Hence require equation is given by,

$(y-1)=-1(x-0)$

$\Rightarrow x+y=1$