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 $S={x \in R: x^{2}+30 \leq 11 x }$ 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 $S={x \in R: x^{2}+30 \leq 11 x }$
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
$$ f(6)=3(6)^{3}-18(6)^{2}+27(6)-40=122 $$
