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Function f(x) = tan (3x) will be discontinuous at x = 
1. (2n + 1) π
2. (n + 1) π/2
3. (2n + 1) π/6
4. (2n + 1) π/4
5. None of these

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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.

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