Find the equation of the normal to the curve `y=(1+x)^y+sin^(-1)(s in^2x)` at `x=0` .

Options

A. `x+y=2`
B. `x+y=1`
C. `x-y=1`
D. none of these

Answer

B

Explanation

We have,
`y=(1+x)^(y)+sin^(-1) (sin^(2)x) " " `…(i)
When x = 0, we have y = 1
Differentiating (i) w.r.t. x, we get
`(dy)/(dx)=(1+x)^(y){(dy)/(dx)log(1+x)+(y)/(1+x)}+(sin2x)/(sqrt(1-sin^(4)x))`
`rArr ((dy)/(dx))_((0","1)) =1`
So, the equation of the normal at (0, 1) is
`y-1= -1 (x-0)rArr x+y=1`