Application of Derivatives 4 Question 8
####9. 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 (2019 Main, 11 Jan I)
(a) 122
(b) -122
(c) -222
(d) 222
Show Answer
Answer:
Correct Answer: 9. (a)
Solution:
- We have,
$ \begin{aligned} f(x) & =3 x^{3}-18 x^{2}+27 x-40 \\ \Rightarrow \quad f^{\prime}(x) & =9 x^{2}-36 x+27 \\ = & 9\left(x^{2}-4 x+3\right)=9(x-1)(x-3) \end{aligned} $
Also, we have $S=\left{x \in R: x^{2}+30 \leq 11 x\right}$
Clearly, $\quad x^{2}+30 \leq 11 x$
$ \begin{array}{llrl} \Rightarrow & & x^{2}-11 x+30 & \leq 0 \\ \Rightarrow & & (x-5)(x-6) & \leq 0 \Rightarrow x \in[5,6] \\ \text { So, } & & S & =[5,6] \end{array} $
Note that $f(x)$ is increasing in $[5,6]$
$\left[\because f^{\prime}(x)>0\right.$ for $x \in[5,6]$
$\therefore f(6)$ is maximum, where
$ f(6)=3(6)^{3}-18(6)^{2}+27(6)-40=122 $
