Question

The function f (x) = tan x is discontinuous on the set

 

विकल्प

• {n π : n ∈ Z}

• {2n π : n ∈ Z}

• \[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\]

• \[\left\{ \frac{n\pi}{2}: n \in Z \right\}\]

Answer 1

\[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\]

When

\[\tan\left( 2n + 1 \right)\frac{\pi}{2} = \tan\left( n\pi + \frac{\pi}{2} \right) = - \cot\left( n\pi \right)\]

 , it is not defined at the integral points.

 

\[\left[ n \in Z \right]\]

Hence, 

\[f\left( x \right)\]  is discontinuous on the set 

\[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\].