JEE Main 2020 — Differentiation Question with Solution

Given y² + log_e (cos²x) = y .....(1)

Put x = 0, we get

y² + log_e (1) = y

$$\Rightarrow$$ y² = y

$$\Rightarrow$$ y = 0, 1

Differentiating (1) we get

2yy' + $$\frac{1}{\cos x}\left(-\sin x\right)$$ = y'

$$\Rightarrow$$ 2yy' - 2tanx = y' ....(2)

From (2) when x = 0, y = 0 then y'(0) = 0

From (2) when x = 0, y = 1 then

2y' = y'

$$\Rightarrow$$ y'(0) = 0

Again differentiating (2) we get

2(y')² + 2yy'' – 2sec²x = y''

from (2) when x = 0, y = 0, y’(0) = 0 then

y”(0) = -2

Also from (2) when x = 0, y = 1, y’(0) = 0 then

y”(0) = 2

$$\therefore$$ |y''(0)| = 2