Question

The function f (x) = tan x is discontinuous on the set
(a) {n π : n ∈ Z}
(b) {2n π : n ∈ Z}
(c) $\left\{\left(2n+1\right)\frac{\mathrm{\pi }}{2}:n\in Z\right\}$
(d) $\left\{\frac{n\mathrm{\pi }}{2}:n\in Z\right\}$

Answer 1

$\left(c\right)\left\{\left(2n+1\right)\frac{\mathrm{\pi }}{2}:n\in Z\right\}$

When$\tan \left(2n+1\right)\frac{\mathrm{\pi }}{2}=\tan \left(n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}\right)=-\cot \left(n\mathrm{\pi }\right)$, it is not defined at the integral points. $\left[n\in Z\right]$

Hence, $f\left(x\right)$ is discontinuous on the set $\left\{\left(2n+1\right)\frac{\mathrm{\pi }}{2}:n\in Z\right\}$.