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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
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Verified Answer
The correct answer is:
$122$
$\begin{aligned} & f(x)=3 x^3-18 x^2+27 x-40 \\ & \Rightarrow f^{\prime}(x)=9 x^2-36 x+27 \\ & =9(x-1)(x-3)\end{aligned}$
Sign scheme for $f^{\prime}(x)$
Now, $x^2+30 \leq 11 x$
$\begin{aligned} & \Rightarrow(x-5)(x-6) \leq 0 \\ & \Rightarrow x \in[5,6]\end{aligned}$
For the interval $[5,6] f(x)$ is increasing
Hence $f_{\max }=f(6)$
$\forall x \in[5,6]$ i.e. on the set S
$=3 \times 6^3-18 \times 6^2+27 \times 6-40=122$
Sign scheme for $f^{\prime}(x)$
Now, $x^2+30 \leq 11 x$
$\begin{aligned} & \Rightarrow(x-5)(x-6) \leq 0 \\ & \Rightarrow x \in[5,6]\end{aligned}$
For the interval $[5,6] f(x)$ is increasing
Hence $f_{\max }=f(6)$
$\forall x \in[5,6]$ i.e. on the set S
$=3 \times 6^3-18 \times 6^2+27 \times 6-40=122$
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