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

NDA 2016 · Mathematics · Continuity and Differentiability

Paragraph: f(x)=\ array lr 3 x^2+12 x-1 & -1 ≤ x ≤ 2 37-x, & 2...

Paragraph: f(\mathrm{x})=\left\{3x2+12x−1−1≤x≤237−x,2<x≤3\begin{array}{lr} 3 x^2+12 x-1 & -1 \leq x \leq 2 \\ 37-x, & 2 < x \leq 3 \end{array}\right.
Question: Which of the following statements are correct? 1. f(x)f(x) is continuous at x=2x=2. 2. f(x)f(x) attains greatest value at x=2x=2. 3. f(x)\mathrm{f}(\mathrm{x}) is differentiable at x=2\mathrm{x}=2. Select the correct answer using the code given below:

Options

  1. A1 and 2 only
  2. B2 and 3 only
  3. C1 and 3 only
  4. D1,2 and 3

Correct answer

A. 1 and 2 only

Step-by-step solution

For continuity at x=2x=2. RHL lim⁡x→2+f(x)=lim⁡x→2+37−x=37−2=3σ\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} 37-x=37-2=3 \sigma LHL lim⁡x→2−f(x)=lim⁡x→2−3x2+12x−1=3(2)2+12×2−1=12+24−1=35f(2)=3×4+12×2−1=12+24−1=35\begin{aligned} & \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 3 x^2+12 x-1 \\ & =3(2)^2+12 \times 2-1=12+24-1=35 \\ & f(2)=3 \times 4+12 \times 2-1=12+24-1=35 \end{aligned}\mathrm{So}, RHL=LHL\mathrm{RHL}=\mathrm{LHL}\Rightarrow f(x)iscontinuousatis continuous atx=2.Fordifferentiabilityat. For differentiability atx=2.. LHD =lim⁡x→2−f(x)−f(2)x−2=lim⁡h→0f(2−h)−f(2)2−h−2=lim⁡h→03(2−h)2+12(2−h)−(12+24−1)−h=lim⁡h→03h2−24h−h=lim⁡h→024−3h=24 RHD =lim⁡x→2+f(x)−f(2)x−2=lim⁡h→0f(2+h)−f(2)2+h−2=lim⁡h→037−2−h−35h=−1\begin{aligned} & \text{ LHD }=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2} \\ & =\lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-(12+24-1)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=\lim _{h \rightarrow 0} 24-3 h=24 \\ & \text{ RHD }=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2} \\ & =\lim _{h \rightarrow 0} \frac{37-2-h-35}{h}=-1 \end{aligned}LHD≠RHD⇒f(x)\mathrm{LHD}_{\neq} \mathrm{RHD}\Rightarrow f(x) is not differentiable at x=2x=2. To check 2. For x=2x=2. f(x)=3x2+12x−1=3(x2+4x−13)=3((x+2)2−133)\begin{aligned} f(x) & =3 x^2+12 x-1 \\ & =3\left(x^2+4 x-\frac{1}{3}\right)=3\left((x+2)^2-\frac{13}{3}\right) \end{aligned} On putting x=2f(x)=3(16−133)=35x=2f(x)=3\left(16-\frac{13}{3}\right)=35 On putting x=1f(1)=3((1+2)2−133)=3(9−133)=14x=1f(1)=3\left((1+2)^2-\frac{13}{3}\right)=3\left(9-\frac{13}{3}\right)=14 So f(x)f(x) attains greatest value at x=2x=2.
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