JEE PYQ: Application Of Derivatives Question 76
Question 76 - 2019 (11 Jan Shift 1)
The maximum value of the function $f(x) = 3x^3 - 18x^2 + 27x - 40$ on the set $S = {x \in \mathbb{R} : x^2 + 30 \le 11x}$ is:
(1) $-122$
(2) $-222$
(3) $122$
(4) $222$
Show Answer
Answer: (3)
Solution
$S: x^2 - 11x + 30 \le 0 \Rightarrow x \in [5, 6]$. $f’(x) = 9(x-1)(x-3) > 0$ on $[5,6]$, so $f$ is increasing. Max at $x = 6$: $f(6) = 648 - 648 + 162 - 40 = 122$.
BETA
