Given equation

The relation is $y^2 + \ln(\cos 2x)=y.$

Rearranging, $y^2-y+\ln(\cos 2x)=0.$

We need to find $y'(0)$ and $y''(0)$.

Find $y(0)$

At $x=0$, $\ln(\cos 0)=\ln 1=0.$ So, $y^2-y=0$ $y(y-1)=0.$ Hence at $x=0$, possible values are $y(0)=0 \quad \text{or} \quad y(0)=1.$

Differentiate implicitly

From $y^2-y+\ln(\cos 2x)=0,$ differentiate w.r.t. $x$: $2yy'-y'+\frac{d}{dx}\ln(\cos 2x)=0.$

Now, $\frac{d}{dx}\ln(\cos 2x)=\frac{-2\sin 2x}{\cos 2x}=-2\tan 2x.$

Therefore, $(2y-1)y'-2\tan 2x=0$ $(2y-1)y'=2\tan 2x.$

Find $y'(0)$

At $x=0$, $\tan 0=0$, so $(2y(0)-1)y'(0)=0.$

Since $y(0)=0$ or $1$, in either case $2y(0)-1=\pm 1 \neq 0$. Hence, $y'(0)=0.$

Differentiate again

From $(2y-1)y'=2\tan 2x,$ differentiate again: $\frac{d}{dx}[(2y-1)y']=\frac{d}{dx}[2\tan 2x].$

Using product rule, $2(y')^2+(2y-1)y''=4\sec^2 2x.$

Find $y''(0)$

At $x=0$, $y'(0)=0, \quad \sec^2 0=1.$ So, $2(0)^2+(2y(0)-1)y''(0)=4$ $(2y(0)-1)y''(0)=4.$

Now consider the two possible branches:

• If $y(0)=1$, then $y''(0)=4.$
• If $y(0)=0$, then $-y''(0)=4 \implies y''(0)=-4.$

In both cases, $|y''(0)|=4.$

Also, $|y'(0)|+|y''(0)|=0+4=4.$

Check options

• A: $|y''(0)|=2$ → false
• B: $|y'(0)|+|y''(0)|=3$ → false
• C: $y''(0)=0$ → false
• D: $|y'(0)|+|y''(0)|=1$ → false

Conclusion

None of the given options is correct. The correct result is $|y''(0)|=4.$