Q13 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If fx=e2x+2x+1-12x+2x+1:x≠01:x=0, then (where, . represents the greatest integer function)

  • A
    limx→0+⁡⁡fx=1
  • B
    limx→0-⁡⁡fx=e-1
  • C
    fx is continuous at x=0
  • Correct answer
    D
    fx is discontinuous at x=0

Solution

limx→0+⁡fx⁡=limh→0+⁡f0+h
=limh→0+⁡e2h+2h+1-12h+2h+1
=limh→0+⁡e2h+1-12h+1=e-1
And, limx→0-⁡fx=limh→0+⁡f0-h
=limh→0+⁡e-2h-2h+1-1-2h-2h+1
=limh→0+⁡e-2h-1-2h=1
Since, limx→0+⁡fx⁡≠limx→0-⁡fx, therefore, fx is not continuous at x=0

Hence, the correct answer is option: 'D'.

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