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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f (x) = \ cases e^ [x] + |x| - 1 [x] + |x| & : x ≠ 0 - 1 & : x =...

If ff\left(x\right) = \left\{{e[x]+∣x∣−1[x]+∣x∣:x≠0−1:x=0\begin{cases}\dfrac{e^{\left[x\right] + \left|x\right|} - 1}{ \left[x\right] + \left|x\right|} & : x \ne 0 \\ - 1 & : x = 0\end{cases}\right) (where [.]\left[.\right]denotes the greatest integer function), then

Options

  1. Aff\left(x\right)iscontinuousatis continuous atx = 0
  2. Blim⁡x→0+f(x)=−1\lim_{x \to {0}^{+}}f\left(x\right) = - 1
  3. Clim⁡x→0−f(x)=1\lim_{x \to {0}^{-}}f\left(x\right) = 1
  4. Dlim⁡x→0+f(x)=1\lim_{x \to {0}^{+}}f\left(x\right) = 1

Correct answer

D. lim⁡x→0+f(x)=1\lim_{x \to {0}^{+}}f\left(x\right) = 1

Step-by-step solution

f\left(x\right) = \left\{\begin{cases}\dfrac{e^{\left[x\right] + \left|x\right|} - 1}{ \left[x\right] + \left|x\right|} & x \ne 0 \\ - 1 & x = 0\end{cases}\right)<br />\lim_{x \to {0}^{-}}f\left(x\right) = \lim_{x \to {0}^{-}}\ddfrace[x]+∣x∣−1dfrac{e^{\left[x\right]+ \left|x\right|} - 1}{\left[x\right]+ \left|x\right|} = \dfrac{e^{- 1} - 1}{- 1} = \dfrac{e - 1}{e}<br/>lim⁡x→0+f(x)=lim⁡x→0+\d<br />\lim_{x \to {0}^{+}}f\left(x\right) = \lim_{x \to {0}^{+}}\d dfrac{e^{\left[x\right]+∣x∣+ \left|x\right|} - 1}{\left[x\right]+∣x∣+ \left|x\right|}=lim⁡x→0+ex−1x=1= \lim_{x \to {0}^{+}}\dfrac{e^{x} - 1}{x} =1
∵LHL≠RHL∵ LHL \ne RHL at x=0⇒x = 0 ⇒f \left(x\right)isdiscontinuousatis discontinuous atx = 0
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