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Question

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


A

x+y=1

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B

x+y+1=0

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C

2xy+1=0

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D

x+2y+2=0

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Solution

The correct option is A

x+y=1


Explanation for the correct answer:

Step 1: Find the first order derivative of the equation of the given curve

Given, equation of curve y=(1+x)y+sin-1(sin2x)...i

On differentiating equation (i) w.r.t. x, we get

dydx=y(1+x)y1+ln(1+x)(1+x)ydydx+2sinxcosx(1sin4x)

Step 2:Find the slope of the normal at the required point

Now at x=0 [given]

y=1 [from eq (i)]

The first order derivative at a given point gives the slope of the tangent at that point

dydx0,1=1(1+0)11+0+1(1sin40)×2sin0cos0

=1+0=1

Tangent and normal are perpendicular to each other

Therefore, slope of normal is =-1dydx

=-1

Step 3: Find the equation of the normal using slope-point form

So, the equation of normal to the curve passing through 0,1 is

y1=-1x0y1=-xx+y=1

Hence, the correct answer is option (A).


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