Q227 / 270 · NDA 2016 (Phase 2) · Mathematics

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

  • Correct answer
    A
    1 and 2 only
  • B
    2 and 3 only
  • C
    1 and 3 only
  • D
    1,2 and 3

Solution

For continuity at $x=2$. RHL $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} 37-x=37-2=3 \sigma$ LHL $\begin{aligned} & \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 3 x^2+12 x-1 \\ & =3(2)^2+12 \times 2-1=12+24-1=35 \\ & f(2)=3 \times 4+12 \times 2-1=12+24-1=35 \end{aligned}$ $\mathrm{So}, \mathrm{RHL}=\mathrm{LHL}$ $\Rightarrow f(x)$ is continuous at $x=2$. For differentiability at $x=2$. $\begin{aligned} & \text { LHD }=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2} \\ & =\lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-(12+24-1)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=\lim _{h \rightarrow 0} 24-3 h=24 \\ & \text { RHD }=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2} \\ & =\lim _{h \rightarrow 0} \frac{37-2-h-35}{h}=-1 \end{aligned}$ $\mathrm{LHD}_{\neq} \mathrm{RHD}$ $\Rightarrow f(x)$ is not differentiable at $x=2$. To check 2. For $x=2$. $\begin{aligned} f(x) & =3 x^2+12 x-1 \\ & =3\left(x^2+4 x-\frac{1}{3}\right)=3\left((x+2)^2-\frac{13}{3}\right) \end{aligned}$ On putting $x=2$ $f(x)=3\left(16-\frac{13}{3}\right)=35$ On putting $x=1$ $f(1)=3\left((1+2)^2-\frac{13}{3}\right)=3\left(9-\frac{13}{3}\right)=14$ So $f(x)$ attains greatest value at $x=2$.

Hence, the correct answer is option: 'A'.

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