Question

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

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

Answer 1

The correct option is D

$\left|y\text{'}\text{'}\left(0\right)\right|=2$

Explanation for the correct option.

Step 1. Find the value of $y\left(0\right)$.

In the equation $y^2+{\log }_e\left({\cos }^{2}x\right)=y$ substitute $x=0$ and find the value of $y$.

$$\begin{gathered} y^2+{\log }_e\left({\cos }^{2}0\right)=y \\ \Rightarrow y^2+{\log }_e\left(1\right)=y \\ \Rightarrow y^2-y=0 \\ \Rightarrow y(y-1)=0 \end{gathered}$$

So either $y=0$ or $y=1$.

So at $x=0$ the solution is given by the ordered pair $\left(0,0\right)$ and $\left(0,1\right)$.

Step 2. Find the value of $\left|y\text{'}\left(0\right)\right|$.

Differentiate the equation $y^2+{\log }_e\left({\cos }^{2}x\right)=y$ with respect to $x$.

$$\begin{gathered} 2yy\text{'}+\frac{1}{{\cos }^{2}x}\left(2\cos x\right)(-\sin x)=y\text{'} \\ \Rightarrow 2yy\text{'}-2\tan x=y\text{'}...\left(1\right) \end{gathered}$$

Now for $\left(x,y\right)=\left(0,0\right)$ equation $1$ is written as:

$$\begin{gathered} 2\times 0\times y\text{'}-2\tan \left(0\right)=y\text{'} \\ \Rightarrow 0-0=y\text{'} \\ \Rightarrow 0=y\text{'} \end{gathered}$$

Again for $\left(x,y\right)=\left(0,1\right)$ equation $1$ is written as:

$$\begin{gathered} 2\times 1\times y\text{'}-2\tan \left(0\right)=y\text{'} \\ \Rightarrow 2y\text{'}-0=y\text{'} \\ \Rightarrow 2y\text{'}-y\text{'}=0 \\ \Rightarrow y\text{'}=0 \end{gathered}$$

So for both the solution at $x=0$, $\left|y\text{'}\left(0\right)\right|=0$.

Step 3. Find the value of $\left|y\text{'}\text{'}\left(0\right)\right|$.

Differentiate equation $1$, $2yy\text{'}-2\tan x=y\text{'}$ with respect to $x$.

$2{(y\text{'})}^2+2yy\text{'}\text{'}-2{\sec }^{2}x=y\text{'}\text{'}...\left(2\right)$

For $\left(x,y\right)=\left(0,0\right)$ substitute $0$ for $x$, $0$ for $y$ and $0$ for $y\text{'}$ in the equation $2$.

$$\begin{gathered} 2{\left(0\right)}^2+2\times 0\times y\text{'}\text{'}-2{\sec }^{2}0=y\text{'}\text{'} \\ \Rightarrow 0+0-2=y\text{'}\text{'} \\ \Rightarrow -2=y\text{'}\text{'} \end{gathered}$$

Again, for $\left(x,y\right)=\left(0,1\right)$ substitute $0$ for $x$, $1$ for $y$ and $0$ for $y\text{'}$ in the equation $2$.

$$\begin{gathered} 2{\left(0\right)}^2+2\times 1\times y\text{'}\text{'}-2{\sec }^{2}0=y\text{'}\text{'} \\ \Rightarrow 2y\text{'}\text{'}-2=y\text{'}\text{'} \\ \Rightarrow 2y\text{'}\text{'}-y\text{'}\text{'}=2 \\ \Rightarrow y\text{'}\text{'}=2 \end{gathered}$$

So, for $x=0$, the value of $y\text{'}\text{'}\left(0\right)$ is either $-2$ or $2$. So $\left|y\text{'}\text{'}\left(0\right)\right|=2$.

Hence, the correct option is D.