Question:
Themaximum value of the function $f(x)=3 x^{3}-18 x^{2}+27 x-40$ on
the set $\mathrm{S}=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$ is :
Correct Option: 3,
Solution:
Consider the function,
$f(x)=3 x(x-3)^{2}-40$
Now $S=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$
So $x^{2}-11 x+30 \leq 0$
$\Rightarrow \quad x \in[5,6]$
$\therefore f(x)$ will have maximum value
for $x=6$
The maximum value of function is,
$f(6)=3 \times 6 \times 3 \times 3-40=122$