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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2021 · Mathematics · Continuity and Differentiability

Consider the following statements in respect of f(x)=|x|-1 : 1....

Consider the following statements in respect of f(x)=∣x∣−1f(x)=|x|-1 : 1. f(x)f(x) is continuous at x=1x=1. 2. f(x)f(x) is differentiable at x=0x=0. Which of the above statements is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

Here, f(x)=|x|-1\therefore f(x)\left\{−x−1,x<0x−1,x≥0\begin{array}{cc} -x-1, & x < 0 \\ x-1, & x \geq 0 \end{array}\right. For continuity at x=1x=1. LHL=lim⁡x→1−f(x)=lim⁡x→1−(x−1)=1−1=0RHL=lim⁡x→1+f(x)=lim⁡x→1+(x−1)=1−1=0\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x-1)=1-1=0 \\ & \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-1)=1-1=0 \end{aligned}\therefore f(x)iscontinuousatis continuous atx=1.Fordifferentiabilityat. For differentiability atx=0..f′(x)={−1,x<01,x≥0LHD=lim⁡x→0−f′(x)=lim⁡x→0−(−1)=−1RHD=lim⁡x→0+f′(x)=lim⁡x→0+(1)=1\begin{aligned} & f^{\prime}(x)=\left\{\begin{array}{cc} -1, & x < 0 \\ 1, & x \geq 0 \end{array}\right. \\ & \mathrm{LHD}=\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1 \\ & \mathrm{RHD}=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}}(1)=1 \end{aligned}∴f(x)\therefore f(x) is not differentiable at x=0x=0
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