Question

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

• A. $\left|y^{\prime \prime}(0)\right|=2$
• B. $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=3$
• C. $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=1$
• D. $y ^{\prime \prime}(0)=0$

Answer 1

Answer: (A)

$y^{2}+\ln \left(\cos ^{2} x\right)=y$
$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
for $x=0$
$y =0$ or 1
Differentiating wrt $x$
$\Rightarrow \,\,\,\, 2 y y^{\prime}-2 \tan x=y^{\prime}$
At $(0,0) y ^{\prime}=0$
At $(0,1) y ^{\prime}=0$
Differentiating wrt $x$
$2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2}-2 \sec ^{2} x=y^{\prime \prime}$
At $(0,0) \,\,\,\, y^{\prime \prime}=-2$
At $(0,1) \,\,\,\, y ^{\prime \prime}=2$
$\therefore \,\,\,\, \left|y^{\prime \prime}(0)\right|=2$