Question

If $f^{\prime }\left(x\right)=\frac{1-2{\sin }^{2}x}{f\left(x\right)}$, $f\left(x\right)>0,\forall x\in R$ and $f\left(0\right)=1$. Then $f\left(x\right)$ is a periodic function with the period

• A. $2\pi$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. not periodic

Answer 1

The correct option is B $\pi$

Let

$f^{\prime }\left(x\right)=\frac{1-2{\sin }^{2}x}{f\left(x\right)}$

or, $f\left(x\right)f^{\prime }\left(x\right)=\cos 2x$

Integrating both sides we get,

$\frac{\left(f\right(x){)}^2}{2}=\frac{\sin 2x}{2}+c$ [$c$ being integrating constant]

Using $f\left(0\right)=1$ we get $c=\frac{1}{2}$.

So, we have

$\left(f\right(x){)}^2=1+\sin 2x$.

As given $f\left(x\right)>0,\forall x\in R$ we have $f\left(x\right)=\sqrt{1+\sin 2x},\forall x\in R$.

It is clear that $f(\pi +x)=f\left(x\right)$, this means that $f\left(x\right)$ is a periodic function of period $\pi$.

So, option (B) is correct.