Correct Answer - Option 3 : (2n + 1) π/6

__Concept: __

- f(c) must be defined. You can’t have a hole in the function (denominator is zero i.e. infinity)
- cos x = 0, where x = odd multiple of π/2 = (2n + 1) π/2

__Calculation:__

Here, \(\tan \left( {3{\rm{x}}} \right) = \frac{{\sin \left( {3{\rm{x}}} \right)}}{{\cos \left( {3{\rm{x}}} \right)}}\)

Check where denominator become zero

We know cos x = 0, where x = (2n + 1) π/2

So for cos 3x = 0 where, 3x = (2n + 1) π/2

x = (2n + 1) π/6

∴ Given function is discontinuous at x = (2n + 1) π/6

Hence, option (3) is correct.

__Hint:__

- When dealing with a
**rational expression** in which both the numerator and denominator are continuous.
- The only points in which the rational expression will be discontinuous where denominator becomes zero.