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Find the equation of the normal to the curve `y=(1+y)^y+sin^(-1)(sin^2x)a tx=0.`A. `x+y=2`B. `x+y=1`C. `x-y=1`D. none of these |
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Answer» Correct Answer - B 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` |
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