QJEEM20Mad124: JEE Main 2020 Mathematics - Class 12 JEEM20Mad124 Solution: If y 2 + log e (cos 2 x) = y, , then :

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' + ${1 \over {\cos x}}\left( { - \sin x} \right)$ = 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