Question

If period of function $f\left(x\right)=\cos \left(nx\right)\sin \frac{5x}{n}$ is $3\pi$, then number of integral values of $n$ must be

• A. 4
• B. 8
• C. 6
• D. 2

Answer 1

The correct option is B

8

Given : Period of $f\left(x\right)=\cos \left(nx\right)\sin \left(\frac{5x}{n}\right)$ is $3\pi$

$\because f\left(x\right)$ is periodic $\therefore f(x+T)=f\left(x\right)$

$\Rightarrow \cos \left(n\right(x+T\left)\right)\sin \left(\frac{5(x+T)}{n}\right)=\cos \left(nx\right)\sin \left(\frac{5x}{n}\right)$ (Where $T=3\pi$)

At $x=0\cos nT\sin \left(\frac{5T}{n}\right)=0$

$\Rightarrow \cos nT=0$ or $\sin \frac{5T}{n}=0$

$\Rightarrow nT=r\pi +\frac{\pi }{2};rϵI$ or $\frac{5T}{n}=p\pi ;pϵI$

Putting the value of $T$, we get,

$\Rightarrow (3n-r)=\frac{1}{2}$ which is not possible or $\frac{5\times 3\pi }{n}=p\pi \Rightarrow n=\frac{15}{p}\Rightarrow p=\pm 1,\pm 3,\pm 5,\pm 15$

So, number of integral values of $n$ is 8.