Q99 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

If $f(x)=3 x+\frac{12}{x}$ is continuous on $\mathbb{R}-\{0\}$ and $M$ is its maximum value, then $\lim _{x \rightarrow M}(f x)=$

  • A
    37
  • Correct answer
    B
    -37
  • C
    2
  • D
    -2

Solution

$\because f(x)=3 x+\frac{12}{x} \Rightarrow f^{\prime}(x)=3-\frac{12}{x^2}$ For critical point: $f^{\prime}(x)=0$ $\begin{aligned} & \Rightarrow 3-\frac{12}{x^2}=0 \Rightarrow x^2=4 \Rightarrow x=-2,2 \\ & f^{\prime \prime}(x)=\frac{24}{x^3}\end{aligned}$ At $x=-2, f^{\prime \prime}(-2)=-3 < 0$ $\therefore f(x)$ has maxima at $x=-2$ and maximum value is $\begin{aligned} & f(-2)=-6-6=-12 \\ & \therefore M=-12\end{aligned}$ $\lim _{x \rightarrow M} f(x)=\lim _{x \rightarrow-12}\left(3 x+\frac{12}{x}\right)=-37$

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

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