Question

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

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

Answer 1

Answer: (A)

At $x=0,y=1$.
Hence, the point at which normal is drawn is $P(0,1)$.
Differentiating the given equation w.r.t. $x$, we have
$(1+x{)}^y\{\log (1+x)\frac{dy}{dx}+\frac{y}{1+x}\}$
$-\frac{dy}{dx}+\frac{1}{\sqrt{1-{\sin }^{4}x}}2\sin x\cos x=0$
or ${\left(\frac{dy}{dx}\right)}_{(0,1)}=\frac{(1+0{)}^1\times \frac{1}{1+0}-\frac{2\sin 0}{\sqrt{1-{\sin }^{2}0}}}{1-(1+0{)}^1\log 1}=1$
$\therefore$ Slope of the normal $=-1$
Therefore, equation of the normal having slope $-1$ at point $P(0,1)$ is given by
$y-1=-(x-0)$ or $x+y=1$