Question

Find the equation of normal to the curvey=(1+x)y+sin−1⁡(sin2⁡x) at x=0

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

Answer 1

Answer: (C)

x+y=1

Answer 2

Explanation:
Given: Equation of the curve y=(1+x)y+sin−1⁡(sin2⁡x).....(i)
We have to find the equation of the normal to the curve at x=0
Consider, y=(1+x)y+sin−1⁡(sin2⁡x)At x=0 we get y=1
Therefore, the point of intersection at which the normal is drawn is P(0,1)
Differentiating (i) w.r.t. x, we have
[ Using product rule, chain rule, standard derivatives-9 and standard derivatives- 16]dydx=(1+x)Y[log⁡(1+x)⋅dydx+y1+x]+11−sin4x⋅2sin⁡x⋅cos⁡x
⇒(1+x)y[log⁡(1+x)⋅dydx+y1+x]−dydx+2sin⁡x⋅cos⁡x1−sin4⁡x[
since, 1−sin4⁡x=1−(sin2⁡x)2=(1−sin2⁡x)(1+sin2⁡x)[ Using trigonometric identities ]=cos2⁡x⋅(1+sin2⁡x)]