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JEE Main 2019 · 11 Jan · Shift I

JEE Main 2019 Mathematics question on Application of Derivatives

The maximum value of the function f(x)=3x3−18x2+27x−40f(x) = 3x^3 - 18x^2 + 27x - 40 on the set S={x∈R:x2+30≤11x}S = \{x \in R : x^2 + 30 \leq 11x\} is
  1. A.
    122
  2. B.
    -122
  3. C.
    -222
  4. D.
    222
Show answer and solution

Correct option: A

Solution

The set S is defined by x² + 30 ≤ 11x, i.e., x² - 11x + 30 ≤ 0, giving (x-5)(x-6) ≤ 0, so S = [5, 6].

Now f(x) = 3x³ - 18x² + 27x - 40 = 3x(x²- 6x + 9) - 40 = 3x(x-3)² - 40.

f'(x) = 9x² - 36x + 27 = 9(x²- 4x + 3) = 9(x-1)(x-3), which is positive for x > 3, so f is strictly increasing on [5, 6].

Maximum occurs at x = 6: f(6) = 3(216) - 18(36) + 27(6) - 40 = 648 - 648 + 162 - 40 = 122.

[JEE Main 2019 · 11 Jan · Shift I]

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