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 $x$
$\Rightarrow 2y{y}^{\prime }-2\tan x=y^{\prime }$
At $(0,0)y^{\prime }=0$
At $(0,1)y^{\prime }=0$
Differentiating wrt $x$
$2y{y}^{\prime \prime }+2{\left(y^{\prime }\right)}^2-2{\sec }^{2}x=y^{\prime \prime }$
At $(0,0)y^{\prime \prime }=-2$
At $(0,1)y^{\prime \prime }=2$
$\therefore \left|y^{\prime \prime }(0)\right|=2$