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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 9 April · Evening Shift · Mathematics · Limits

If f(x) = [x] - [(x)/(4)], x ∈ R, where [x] denotes the greatest...

If f(x)=[x]−[x4],x∈R,f\left(x\right) = \left[x\right] - \left[\dfrac{x}{4}\right], x \in R, where \left[x\right]$ denotes the greatest integer function, then:

Options

  1. Alim⁡x→4+f(x)\lim_{x \to 4 +}f\left(x\right) exists but lim⁡x→4−f(x)\lim_{x \to 4 -}f\left(x\right) does not exist
  2. Bff is continuous at x=4x = 4
  3. Clim⁡x→4−f(x)\lim_{x \to 4 -}f\left(x\right) exists but lim⁡x→4+f(x)\lim_{x \to 4 +}f\left(x\right) does not exist
  4. DBoth lim⁡x→4−f(x)\lim_{x \to 4 -}f\left(x\right) and lim⁡x→4+f(x)\lim_{x \to 4 +}f\left(x\right) exist but are not equal

Correct answer

B. ff is continuous at x=4x = 4

Step-by-step solution

R.H.L.=lim⁡x→4+[x]−[x4]=4−1=3R . H . L . = \lim_{x \to {4}^{+}}\left[x\right] - \left[\dfrac{x}{4}\right] = 4 - 1 = 3

L.H.L.=lim⁡x→4−[x]−[x4]=3−0=3L . H . L . = \lim_{x \to {4}^{-}}\left[x\right] - \left[\dfrac{x}{4}\right] = 3 - 0 = 3

f(x)=4−1=3f\left(x\right) = 4 - 1 = 3

L.H.L.=f(4)=R.H.L.L . H . L . = f\left(4\right) = R . H . L .

Hence, given function is continuous at x=4.x = 4 .

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