JEE Main20202 Sep 2020Evening ShiftMathematicsApplication of DerivativesActual

The equation of the normal to the curve y = 1 + x 2 y + cos 2 sin - 1 x , at x = 0 is

Options

A  y + 4 x = 2
By = 4 x + 2
Cx   +   4 y   =   8
D2 y   +   x   =   4

Correct answer

C. x   +   4 y   =   8

Step-by-step solution

y = 1 + x 2 y + cos 2 sin - 1 x Now at x = 0 ,   y = 2 Let sin - 1 x = t   ⇒ sin t = x ⇒ cos 2 sin - 1 x = cos 2 t = 1 - sin 2 t = 1 - x 2 y = 1 + x 2 y + 1 - x 2 y = e 2 y ln 1 + x + 1 - x 2 d y d x = e 2 y ln ( 1 + x ) 2 y 1 + x + ln ( 1 + x ) · 2 d y d x - 2 x Now at x = 0 , d y d x = 2 × 2 1 + 0 + 0 d y d x = 4 y - 2 = - 1 4 x - 0 4 y - 8 = - x x + 4 y = 8

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