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

NDA 2016 · Mathematics · Continuity and Differentiability

Paragraph: Let f(x)=\ array ll -2, & -3 ≤ x ≤ 0 x-2, & 0 Question:...

Paragraph: Let \mathrm{f}(\mathrm{x})=\left\{−2,−3≤x≤0x−2,0<x≤3\begin{array}{ll}-2, & -3 \leq \mathrm{x} \leq 0 \\ \mathrm{x}-2, & 0 < \mathrm{x} \leq 3\end{array}\right. and g(x)=f(∣x∣)+∣f(x)∣\mathrm{g}(\mathrm{x})=\mathrm{f}(|\mathrm{x}|)+|\mathrm{f}(\mathrm{x})|
Question: Which of the following statements is/are correct? 1. g(x)g(x) is differentiable at x=0x=0. 2. g(x)\mathrm{g}(\mathrm{x}) is differentiable at x=2\mathrm{x}=2. Select the correct answer using the code given below:

Options

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

Correct answer

D. Neither 1 nor 2

Step-by-step solution

f(x)={−2,−3≤x≤0x−2,0<x≤3 and g(x)=f(∣x∣)+∣f(x)∣ At x=0 For  LHD  :g(x)=−2+∣−2∣=−2+2=0⇒g(x)=0 LHD =lim⁡x→0−g(x)−g(0)x−0=lim⁡h→0g(−h)−g(0)−h=lim⁡h→00−0−h=lim⁡h→00 LHD =0\begin{aligned} & f(x)=\left\{\begin{array}{cc} -2, & -3 \leq x \leq 0 \\ x-2, & 0 < x \leq 3 \end{array} \right. \text{ and } \\ & g(x)=f(|x|)+|f(x)| \\ & \begin{array}{l} \text{ At } x=0 \end{array} \\ & \text{\text{ For }\text{ LHD } }: g(x)=-2+|-2|=-2+2=0 \Rightarrow g(x)=0 \\ & \text{ LHD }=\lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0}=\lim _{h \rightarrow 0} \frac{g(-h)-g(0)}{-h} \\ & \quad=\lim _{h \rightarrow 0} \frac{0-0}{-h}=\lim _{h \rightarrow 0} 0 \\ & \text{ LHD }=0 \end{aligned} For RHD : g(x)=∣x∣−2+∣x−2∣g(x)=x−2−(x−2)x>0g(x)=|x|-2+|x-2|g(x)=x-2-(x-2) \quad x > 0 (and just greater than zero) g(x)=x−2−x+2=0g(x)=x-2-x+2=0 Now g\mathrm{g}(x\mathrm{x})isnotcontinuousatis not continuous atx=0,hence, henceg(x)isnotdifferentiableatis not differentiable atx=30AtAtx=2ForLHD:For LHD :g(x)=∣x∣−2+∣x−2∣=x−2−(x−2)=x−2−x+2=0\begin{aligned} g(x) & =|x|-2+|x-2|=x-2-(x-2) \\ & =x-2-x+2=0 \end{aligned}∴LHD=lim⁡x→2−g(x)−g(2)x−2=lim⁡x→20x−2=0\therefore \mathrm{LHD}=\lim _{x \rightarrow 2^{-}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{0}{x-2}=0 For RHD : \begin{aligned} & g(x)=|x|-2+|x-2|=x-2+x-2=2 x-4 \\ & \begin{aligned} \Rightarrow \mathrm{RHD} & =\lim _{x \rightarrow 2^{+}} \frac{g(x)-g(2)}{x-2}=\lim _{x \rightarrow 2} \frac{2 x-4-2(2)+4}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{2(x-2)}{x-2}=2 \end{aligned} \end{aligned}\Rightarrow \mathrm{LHD} \neq \mathrm{RHD} Thus g(x)g(x) is not differentiable at x=2x=2.$
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