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Application of Derivatives · Question 15

JEE Main 2025 (Online) 7th April Evening ShiftMCQMedium📈 Topic in 32% papers+4−1
1

Let f : R R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If x 0 f(x)x2 = 5, then f(2) is equal to :

  1. 8
  2. 10
  3. 12
  4. 14

Answer

(2) 10

Solution

x 0 f(x)x2 = 5 is finite, f(x) has no constant or x term and the x2 coefficient is 5:

f(x) = 5x2 + ax3 + bx4

f'(x) = 10x + 3ax2 + 4bx3 = x(4bx2 + 3ax + 10)

Extreme values at x = 4, 5 4 and 5 are roots of 4bx2 + 3ax + 10 = 0

Product of roots: 104b = 20 b = 18

Sum of roots: - 3a4b = 9 3a = - 92 a = - 32

f(x) = 5x2 - 32x3 + 18x4

f(2) = 20 - 12 + 2 = 10

Answer: (2)

JEE Main 2025 (Online) 7th April Evening Shift

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2

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
JEE Main 2025 (Online) 4th April Evening Shift
3

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
4

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
5

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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