Question:
The maximum value of the function
$f(x)=3 x^{3}-18 x^{2}+27 x-40$ on the set
$S=\left\{x \in R: x^{2}+30 \leq 11 x\right\}$ is :
Correct Option: 1
Solution:
$\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R}, \mathrm{x}^{2}+30-11 \mathrm{x} \leq 0\right\}$
$=\{x \in R, 5 \leq x \leq 6\}$
Now $f(x)=3 x^{3}-18 x^{2}+27 x-40$
$\Rightarrow f^{\prime}(x)=9(x-1)(x-3)$,
which is positive in $[5,6]$
$\Rightarrow f(x)$ increasing in $[5,6]$
Hence maximum value $=f(6)=122$
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