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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2016 · Mathematics · Continuity and Differentiability

Consider the following in respect of the function f(x)=\ array l...

Consider the following in respect of the function f(x)=\left\{2+x,x≥02−x,x<0\begin{array}{l} 2+x, \quad x \geq 0 \\ 2-x, \quad x < 0 \end{array}\right. 1. \lim _{x \rightarrow 1} \mathrm{f}(x) does not exist. 2. f(x)f(x) is differentiable at x=03.x=0 3.f(x)iscontinuousatis continuous atx=0Whichoftheabovestatementsis/arecorrect?Which of the above statements is/are correct?

Options

  1. A1 only
  2. B3 only
  3. C2 and 3 only
  4. D1 and 3 only

Correct answer

B. 3 only

Step-by-step solution

 For  x≥1lim⁡x→1f(x)=lim⁡x→12+x=2+1=3For x<1lim⁡x→1−f(x)=lim⁡x→1−2+x=2+1=3\begin{aligned} & \text{\text{ For } } x \geq 1 \\ & \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} 2+x=2+1=3 \\ & \text{For } x < 1 \\ & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 2+x=2+1=3\end{aligned} So, lim⁡x→1f(x)\lim _{x \rightarrow 1} f(x) exist. At x=0x=0 RHL: lim⁡h→0+f(0+h)=lim⁡h→02+h=2\lim _{h \rightarrow 0^{+}} f(0+h)=\lim _{h \rightarrow 0} 2+h=2 LHL lim⁡h→0−f(0−h)=lim⁡h→02−h=2f(0)=2+0=2\lim _{h \rightarrow 0^{-}} f(0-h)=\lim _{h \rightarrow 0} 2-h=2f(0)=2+0=2. So,\mathrm{So},\mathrm{RHL}=\mathrm{LHL}=f(0)⇒f(x)=f(0)\Rightarrow f(x) is continuous at x=0x=0 Differentiability at x=0x=0 LHD: lim⁡h→0−f(0−h)−f(0)−h=lim⁡h→0−2+h−2−h=−hh=−1\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0^{-}} \frac{2+h-2}{-h}=\frac{-h}{h}=-1 RHD: lim⁡h→0+f(0+h)−f(0)h=lim⁡h→0+2+h−2h=1\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2+h-2}{h}=1 Since LHD ≠ RHD So, f(x)f(x) is not differentiable at x=0x=0.
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