Question

The maximum value of function $f\left(x\right)=3x^3-18x^2+27x-40$ on the set $S=\{x\in R:x^2+30\le 11x\}$ is:

• A. $122$
• B. $-222$
• C. $-122$
• D. $222$

Answer 1

The correct option is A $122$

$S=\{x\in R,x^2+30-11x\le 0\}$
$=\{x\in R,5\le x\le 6\}$
Now $f\left(x\right)=3x^3-18x^2+27x-40$
$⟹f^{\prime }\left(x\right)=9(x-1)(x-3)$, which is positive in $[5,6]$
$⟹f\left(x\right)$ increasing in $[5,6]$
Hence maximum value $=f\left(6\right)=122$