Question:
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 :
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 :
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Solve the inequality to get a closed interval, then check where f is monotonic on it.
Updated On: Nov 5, 2026
- \(-222\)
- \(-122\)
- \(122\)
- \(222\)
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The Correct Option is C
Solution and Explanation
\(x^2 - 11x + 30 \le 0 \Rightarrow (x-5)(x-6)\le0 \Rightarrow S = [5, 6]\).
\(f'(x) = 9x^2 - 36x + 27 = 9(x-1)(x-3)\), which is positive for \(x\gt3\). So f is increasing on \([5,6]\).
Maximum at \(x = 6\): \(f(6) = 3(216) - 18(36) + 27(6) - 40 = 648 - 648 + 162 - 40 = 122\).
Answer: (3) \(122\)
\(f'(x) = 9x^2 - 36x + 27 = 9(x-1)(x-3)\), which is positive for \(x\gt3\). So f is increasing on \([5,6]\).
Maximum at \(x = 6\): \(f(6) = 3(216) - 18(36) + 27(6) - 40 = 648 - 648 + 162 - 40 = 122\).
Answer: (3) \(122\)
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