The maximum value of the function $\mathrm{f}(x)=2 x^3-15 x^2+36 x-48$ on the set $\mathrm{A}=\left\{x…

The maximum value of the function $\mathrm{f}(x)=2 x^3-15 x^2+36 x-48$ on the set $\mathrm{A}=\left\{x |x^2+20 \leq 9 x\right\}$ is
  1. -16
  2. -7
  3. 16
  4. 7

Solution

$\begin{array}{ll} & \text { Let } \mathrm{f}(x)=2 x^3-15 x^2+36 x-48 \\ \therefore & \mathrm{f}^{\prime}(x)=6 x^2-30 x+36=0 \text { at } x=3,2 \\ \therefore & \mathrm{f}^{\prime \prime}(x)=12 x-30 \\ & \mathrm{~A}=\left\{x \mid x^2-9 x+20 \leq 0\right\}=[4,5] \\ \therefore & 2,3 \notin \mathrm{~A} \\ \therefore \quad & \text { At } x=4, \mathrm{f}(x)=-16 \text { and at } x=5, \mathrm{f}(x)=7 \\ \therefore \quad & \text { Maximum value of } \mathrm{f}(x) \text { is at } x=5 \\ \therefore \quad & \text { Maximum value of } \mathrm{f}(x) \text { is } 7 .\end{array}$

Asked in: MHT CET 2024 (11 May Shift 2)

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