Solve the Following Questions

Question:

If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then :

  1. $\left|y^{\prime \prime}(0)\right|=2$

  2. $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=3$

  3. $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=1$

  4. $y^{\prime \prime}(0)=0$


Correct Option: 1

Solution:

$y^{2}+\ln \left(\cos ^{2} x\right)=y \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

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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