The maximum value of the function $f(x)=3 x^3-18 x^2+27 x-40$ on the set $\mathrm{S}=\left\{x \in \mathbb{R}…

The maximum value of the function $f(x)=3 x^3-18 x^2+27 x-40$ on the set $\mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\}$ is
  1. -122
  2. -222
  3. 222
  4. 122

Solution

$\begin{aligned} & \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\} \\ & x^2+30 \leq 11 x \\ & \Rightarrow x^2-11 x+30 \leq 0 \\ & \Rightarrow(x-5)(x-6) \leq 0 \\ & \Rightarrow x \in[5,6] \end{aligned}$
Now, $\mathrm{f}(x)=3 x^3-18 x^2+27 x-40$ $\begin{aligned} \mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \\ \mathrm{f}^{\prime}(x) & =9\left(x^2-4 x+3\right) \\ & =9\left[\left(x^2-4 x+4\right)-1\right] \\ & =9(x-2)^2-9 \end{aligned}$ $\therefore \quad \mathrm{f}^{\prime}(x)\gt0 \forall x \in[5,6]$ $\therefore \quad \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$ $\therefore \quad$ Maximum value of $\mathrm{f}(x)$ when $x \in[5,6]$ is $f(6)=122$

Asked in: MHT CET 2024 (02 May Shift 2)

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