Question

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

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

Answer 1

The correct option is A x + y = 1

Given curve is
$y=(1+x{)}^y+si{n}^{-1}(si{n}^{2}x)$

$y=e^{log(1+x{)}^y}+si{n}^{-1}\left(si{n}^{2}x\right)=e^{ylog(1+x)}+si{n}^{-1}\left(si{n}^{2}x\right)$

On differentiating w.r.t.x, we get

$\frac{dy}{dx}=e^{log(1+x{)}^y}[\frac{y}{1+x}+log(1+x\left)\frac{dy}{dx}\right]+\frac{2sinxcosx}{\sqrt{1-si{n}^{4}x}}$

$\frac{dy}{dx}=(1+x{)}^y[\frac{y}{1+x}+log(1+x\left)\frac{dy}{dx}\right]+\frac{2sinxcosx}{\sqrt{1-si{n}^{4}x}}$

$⟹{\left(\frac{dy}{dx}\right)}_{at(0,1)}=1[\because atx=0,y=1]$
Slope of normal at (x = 0) $=-1$

$\therefore$ Equation of normal at $x=0$ and $y=1$ is
$y-1=-1(x-0)$

$⟹y-1=-x⟹x+y=1$