Question:
If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then :
Correct Option: , 3
Solution:
$y^{2}+2 \log _{e}(\cos x)=y$........(1)
$\Rightarrow 2 y y^{\prime}-2 \tan x=y^{\prime}$.....(2)
From(1), $y(0)=0$ or 1
$\therefore y^{\prime}(0)=0$
Again differentiating (2) we get,
$2\left(y^{\prime}\right)^{2}+2 y y^{\prime \prime}-2 \sec ^{2} x=y^{\prime \prime}$
Put $x=0, y(0)=0,1$ and $y^{\prime}(0)=0$,
we get, $\left|y^{\prime \prime}(0)\right|=2$
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