Mathematics - Functions Question with Solution | TestHub

(i) $f(x)=\tan \frac{\pi }{2}[x]f(x+1)=\tan \frac{\pi }{2}[x+1]=-\cot \frac{\pi }{2}[x]$

$$\begin{array}{rlr}& f(x+2)=\tan \frac{\pi }{2}[x+2]=\tan \frac{\pi }{2}(2+[x])& \end{array}$$

$$=\tan \left(\pi +\frac{\pi }{2}[x]\right)=\tan \frac{\pi }{2}[x]=f(x)$$

$\therefore 2$ is period of $\mathrm{f}(\mathrm{x})$

1 is not a period of $\mathrm{f}(\mathrm{x})$

$\therefore 2$ is the fundamental period

(ii) $f(x)=\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\frac{2\sin 2x\cos x}{2\cos 2x\cdot \cos x}=\tan 2x$ where $x\ne (2n+1)\frac{\pi }{4},(2n+1)\frac{\pi }{2}$ domain is $\mathrm{f}\left(\mathrm{x}+\frac{\pi }{2}\right)=\tan 2\left(\mathrm{x}+\frac{\pi }{2}\right)=\tan (2\mathrm{x}+\pi )=\tan 2\mathrm{x}$ it seems that $\frac{\pi }{2}$ is a period but it is not because $f(0)$ is defined where as $f\left(\frac{\pi }{2}\right)$ is not defined.

$\therefore$ Period is $\pi$