Given y 2 + log e (cos 2 x) = y .....(1) Put x = 0, we get y 2 + log e (1) = y $\Rightarrow$ y 2 = 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'' – 2sec 2 x = 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