Find the domain of $$f(x) =\lfloor\cos x\rfloor\sin\left(\frac{\pi}{\lfloor x-1\rfloor}\right)$$
where $\lfloor\cdot\rfloor$ is greatest integer function.
Working : Since $\lfloor\cos x\rfloor$ is defined for all real. Only part is the denominator in $\sin\left(\frac{\pi}{\lfloor x-1\rfloor}\right)$ which is $\lfloor x-1\rfloor$ so this part should not be equal to $0$ so that the function should be defined. Also the domain of $\sin x$ is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ Therefore: $-\frac{\pi}{2} \leq \frac{\pi}{\lfloor x-1\rfloor} \leq \frac{\pi}{2}$ Please guide further to solve this..
Best Answer
The domain of sin is $\mathbb{R}$. So,$[x-1]\neq0$ as you yourself noted. To solve this we note that if $[x-1]=0$, $x\in(0,1]$
So, $\mathbb {R}\setminus(0,1]$ is the domain.