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

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

Solution

Consider the function, $f(x)=3 x(x-3)^{2}-40$ $\operatorname{Now} S=\left\{x \in \infty k: x^{2}+30 \leq 11 x\right\}$ So $x^{2}-11 x+30 \leq 0 \quad \Rightarrow \quad x \circ \in[5,6]$ $\therefore f(x)$ will have maximum value for $x=6$ The maximum value of function is, $f(6)=3 \times 6 \times 3 \times 3-40=122$

Asked in: JEE Main 2019 (11 Jan Shift 1)

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