If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then :
Correct Option: 1
$y^{2}+\ln \left(\cos ^{2} x\right)=y \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
for $x=0$ $y=0$ or 1
Differentiating wrt $\mathrm{x}$
$\Rightarrow 2 y y^{\prime}-2 \tan x=y^{\prime}$
At $(0,0) \mathrm{y}^{\prime}=0$
At $(0,1) \mathrm{y}^{\prime}=0$
Differentiating wrt $x$
$2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2}-2 \sec ^{2} x=y^{\prime \prime}$
At $(0,0) \quad y^{\prime \prime}=-2$
At $(0,1) \quad y^{\prime \prime}=2$
$\therefore \quad\left|y^{\prime \prime}(0)\right|=2$
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