The maximum value of the function f(x)=3x3−18x2+27x−40 on the set S=x∈R:x2+30≤11x is

Solution:

Here, $x^{2} - 11 x + 30 \leq 0$

$\begin{aligned} \Rightarrow x^{2} - 5 x - 6 x + 30 \leq 0 \Rightarrow (x - 5) (x - 6) \leq 0 \\ \Rightarrow 5 \leq x \leq 6 ∴ S = {x \in R, 5 \leq x \leq 6} \end{aligned}$

Now, $f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40$

$∴ f^{'} (x) = 9 x^{2} - 36 x + 27 = 9 (x - 1) (x - 3)$ ,

which is positive in [5, 6]. So $S o, f (x) i s i n c r e a \sin g i n [5, 6]$

Hence, maximum value of $f (x) = f (6) = 122$

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