Question

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

• A. $x+y=0$
• B. $x-y=1$
• C. $x+y=1$
• D. None of these

Answer 1

The Correct Option is C

Explanation:
Given: Equation of the curve$y=(1+x{)}^y+{\sin }^{?1}?({\sin }^2?x)$.....(i)
We have to find the equation of the normal to the curve
at $x=0$Consider, $y=(1+x{)}^y+{\sin }^{?1}?({\sin }^2?x)$At
$x=0$ we get $y=1$
Therefore, the point of intersection at which the normal is drawn is $\mathrm{P}(0,1)$
Differentiating (i) w.r.t. x, we have[ Using product rule, chain rule, standard derivatives-9 and standard derivatives- $16]$$\frac{dy}{dx}=(1+x{)}^Y[\log ?(1+x)?\frac{dy}{dx}+\frac{y}{1+x}]+{\frac{1}{\sqrt{1?\sin }}}^{4}x?2\sin ?x?\cos ?x$$?(1+x{)}^y[\log ?(1+x)?\frac{dy}{dx}+\frac{y}{1+x}]?\frac{dy}{dx}+\frac{2\sin ?x?\cos ?x}{\sqrt{1?{\sin }^4?x}}$$[\text{since,}1?{\sin }^4?x=1?{({\sin }^2?x)}^2$$=(1?{\sin }^2?x)(1+{\sin }^2?x)$[ Using trigonometric identities ]$={\cos }^2?x?(1+{\sin }^2?x)]$