If $\mathrm{A}=\left\{\mathrm{x} / 9 \mathrm{x} \geq \mathrm{x}^2+20\right\}$ and $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{R}$ is defined by $f(x)=2 x^3-15 x^2+36 x-48$, then the maximum value of $f(x)$ is

• A

  -20
• Correct answer
  B

  7
• C

  20
• D

  -16

Solution

$$ \begin{aligned} & \text { Given } \mathrm{A}=\left\{x: 9 \mathrm{x} \geq \mathrm{x}^2+20\right\} \\ & \mathrm{A}=\{x: x \in[4,5]\}=[4,5] \\ & \text { Given } f(x)=2 x^3-15 x^2+36 x-48 \\ & f^{\prime}(x)=6\left(x^2-5 x+6\right)=6(x-3)(x-2) \end{aligned} $$ For maxima or minima $f^{\prime}(x)=0$
$\begin{aligned} & \Rightarrow x=3,2 \\ & \text { but } A=[4,5]\end{aligned}$ Hence $f(x)$ is increasing function in $\mathrm{A}=[4,5]$ Therefore maximum value of $f(x)$ will be obtained at $x=5$ $\therefore f(5)=7$

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