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Application of Derivatives

JEE Main 2025 (Online) 7th April Morning ShiftMCQMedium📈 Topic in 12% papers+4−1
Ques. 1JEE Main 2025, 7 April (Morning Shift)

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)

Keep practising — Application of Derivatives

Ques. 2JEE Main 2025, 7 April (Evening Shift)

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
Ques. 3JEE Main 2025, 4 April (Evening Shift)

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
Ques. 4JEE Main 2025, 3 April (Evening Shift)

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
Ques. 5JEE Main 2025, 2 April (Morning Shift)

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
Ques. 6JEE Main 2025, 29 January (Evening Shift)

Let f(x)= 0x t (t2-9t+20 )dt, 1 x 5. If the range of f is [ , ], then 4( + ) equals :

  1. 253
  2. 157
  3. 154
  4. 125
Ques. 7JEE Main 2025, 28 January (Morning Shift)

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
Ques. 8JEE Main 2025, 24 January (Morning Shift)

Consider the region R= \(x,y):x y 9- 113x2, x 0 \. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is:

  1. 821123
  2. 567121
  3. 730119
  4. 625111
Ques. 9JEE Main 2025, 24 January (Evening Shift)

Let (2,3) be the largest open interval in which the function f(x)=2 e(x-2)-x2+ax+1 is strictly increasing and (b,c) be the largest open interval, in which the function g(x)=(x-1)3(x+2-a)2 is strictly decreasing. Then 100(a+b-c) is equal to :

  1. 360
  2. 420
  3. 160
  4. 280
Ques. 10JEE Main 2025, 23 January (Morning Shift)

If the set of all values of a, for which the equation 5x3-15x-a=0 has three distinct real roots, is the interval ( , ), then -2 is equal to .

Ques. 11JEE Main 2025, 23 January (Evening Shift)

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 cm3/min and the thickness of the ice-cream layer decreases at the rate of 14 cm/min. The surface area (in cm2) of the chocolate ball (without the ice-cream layer) is :

  1. 128
  2. 196
  3. 225
  4. 256

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