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]$$\frac{dy}{dx}=(1+x{)}^Y[\log (1+x)\cdot \frac{dy}{dx}+\frac{y}{1+x}]+{\frac{1}{\sqrt{1-\sin }}}^{4}x\cdot 2\sin x\cdot \cos x$
$\Rightarrow (1+x{)}^y[\log (1+x)\cdot \frac{dy}{dx}+\frac{y}{1+x}]-\frac{dy}{dx}+\frac{2\sin x\cdot \cos x}{\sqrt{1-{\sin }^{4}x}}$$[$
since, 1−sin4⁡x=1−(sin2⁡x)2$=(1-{\sin }^{2}x)(1+{\sin }^{2}x)$[ Using trigonometric identities ]$={\cos }^{2}x\cdot (1+{\sin }^{2}x)]$