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Q.

The equation of the normal to the curve  y=(1+x)y+sin-1sin2x at x=0 is

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a

x+y=1

b

x+y+1=0

c

2 x-y+1=0

d

x+2 y+2=0

answer is A.

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Detailed Solution

Given y=1+xy+sin−1sin2xWhen x=0,y=1To get the slope of the tangent, differentiate both sides dydx=1+xydydxlog1+x+y1+x+sin2x1−sin4xdydx|0,1=1The slope of the normal is −1Equation of the normal is x+y−1=0
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