Question

The maximum value of the function $f\left(x\right)=2x^3-18x^2+48x-11$ over the set $S=\{x\in R:x^2+42\le 13x\}$ is

• A. $115$
• B. $21$
• C. $29$
• D. $129$

Answer 1

The correct option is D $129$

$x^2+42\le 13x\Rightarrow x^2-13x+42\le 0\Rightarrow (x-6)(x-7)\le 0\Rightarrow x\in [6,7]$

Now, $f\left(x\right)=2x^3-18x^2+48x-11$
$f^{\prime }\left(x\right)=6x^2-36x+48\Rightarrow f^{\prime }\left(x\right)=6(x^2-6x+8)\Rightarrow f^{\prime }\left(x\right)=6(x-2)(x-4)\therefore f^{\prime }\left(x\right)>0(\because x\in [6,7\left]\right)$

So, $f$ is strictly increasing in $[6,7]$.
So, $f\left(7\right)$ is the maximum value.

Hence, the maximum is
$f\left(7\right)=2\times 7^3-18\times 7^2+48\times 7-11=7(98-126+48)-11=7\times 20-11=129$