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Application of Derivatives · Question 14

JEE Main 2025 (Online) 7th April Morning ShiftMCQMedium📈 Topic in 13% papers+4−1
1

Let x = -1 and x = 2 be the critical points of the function f(x) = x3 + ax2 + b e |x| + 1, x 0. Let m and M respectively be the absolute minimum and the absolute maximum values of f in the interval [-2, - 12 ]. Then |M + m| is equal to (Take e 2 = 0.7):

  1. 21.1
  2. 19.8
  3. 22.1
  4. 20.9

Answer

(1) 21.1

Solution

f'(x) = 3x2 + 2ax + bx

f'(-1) = 0 3 - 2a - b = 0 ... (i)

f'(2) = 0 12 + 4a + b2 = 0 8a + b = -24 ... (ii)

Adding (i) and (ii): 6a + 3 = -24 a = - 92, b = 3 - 2a = 12

f(x) = x3 - 92x2 + 12 e |x| + 1

In [-2, - 12 ] the only critical point is x = -1.

f(-2) = -8 - 18 + 12 e 2 + 1 = -25 + 8.4 = -16.6

f(-1) = -1 - 92 + 0 + 1 = -4.5

f (- 12 ) = - 18 - 98 - 12 e 2 + 1 = -1.25 - 8.4 + 1 = -8.65

M = -4.5, m = -16.6

|M + m| = |-21.1| = 21.1

Answer: (1)

JEE Main 2025 (Online) 7th April Morning Shift

Keep practising — Application of Derivatives

2

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
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  3. 12
  4. 14
JEE Main 2025 (Online) 7th April Evening Shift
3

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  1. 15
  2. 13
  3. 24
  4. 18
JEE Main 2025 (Online) 4th April Evening Shift
4

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
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JEE Main 2025 (Online) 3rd April Evening Shift
5

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  1. 55
  2. 37
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JEE Main 2025 (Online) 2nd April Morning Shift
6

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
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  4. 13172
JEE Main 2025 (Online) 28th January Morning Shift

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