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Application of Derivatives · Question 16

JEE Main 2025 (Online) 4th April Evening ShiftMCQEasy📈 Topic in 32% papers+4−1
1

Let a > 0. If the function f(x) = 6x3 - 45ax2 + 108a2x + 1 attains its local maximum and minimum values at the points x1 and x2 respectively such that x1 x2 = 54, then a + x1 + x2 is equal to :

  1. 15
  2. 13
  3. 24
  4. 18

Answer

(4) 18

Solution

f'(x) = 18x2 - 90ax + 108a2 = 18(x2 - 5ax + 6a2) = 18(x - 2a)(x - 3a)

f'(x) = 0 x = 2a, 3a

f''(x) = 36x - 90a: f''(2a) = -18a < 0 (maximum), f''(3a) = 18a > 0 (minimum)

x1 = 2a, x2 = 3a

x1 x2 = 6a2 = 54 a2 = 9 a = 3 ( a > 0)

x1 = 6, x2 = 9

a + x1 + x2 = 3 + 6 + 9 = 18

Answer: (4)

JEE Main 2025 (Online) 4th April Evening Shift

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2

Let f : R R be a function defined by f(x) = ||x + 2| - 2|x| |. If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is

  1. 3
  2. 4
  3. 2
  4. 5
JEE Main 2025 (Online) 3rd April Evening Shift
3

If the function f(x) = 2x3 - 9ax2 + 12a2x + 1, where a > 0, attains its local maximum and local minimum values at p and q, respectively, such that p2 = q, then f(3) is equal to :

  1. 55
  2. 37
  3. 10
  4. 23
JEE Main 2025 (Online) 2nd April Morning Shift
4

The sum of all local minimum values of the function f(x) = cases 1 - 2x, & x < -1 \ 13(7 + 2|x|), & -1 x 2 \ 1118(x - 4)(x - 5), & x > 2 cases is

  1. 16772
  2. 15772
  3. 17172
  4. 13172
JEE Main 2025 (Online) 28th January Morning Shift

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