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Application of Derivatives

JEE Main 2025 (Online) 3rd April Evening ShiftMCQMedium📈 Topic in 33% papers+4−1
Ques. 1JEE 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

Answer

(1) 3

Solution

Let g(x) = |x + 2| - 2|x|.

For x -2: g(x) = -(x + 2) + 2x = x - 2

For -2 x 0: g(x) = (x + 2) + 2x = 3x + 2

For x 0: g(x) = (x + 2) - 2x = 2 - x

So f(x) = |g(x)| behaves as follows:

x -2: f(x) = 2 - x, decreasing to f(-2) = 4

-2 x 0: f(x) = |3x + 2|, decreasing to 0 at x = - 23, then increasing to f(0) = 2

x 0: f(x) = |2 - x|, decreasing to 0 at x = 2, then increasing

At x = -2, f keeps decreasing on both sides, so it is not an extremum. Hence there are local minima at x = - 23 and x = 2, so m = 2, and a local maximum at x = 0, so n = 1.

∴ m + n = 3

Answer: (1)

Keep practising — Application of Derivatives

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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
Ques. 9JEE Main 2025, 22 January (Evening Shift)

Let f(x)= 0x2 t2-8t+15et dt, x . Then the numbers of local maximum and local minimum points of f, respectively, are :

  1. 3 and 2
  2. 2 and 2
  3. 2 and 3
  4. 1 and 3
Ques. 10JEE Main 2024, 9 April (Morning Shift)

Let the set of all positive values of , for which the point of local minimum of the function (1 + x ( 2 - x2 ) ) satisfies x2 + x + 2x2 + 5x + 6 < 0, be ( , ). Then 2 + 2 is equal to .

Ques. 11JEE Main 2024, 9 April (Evening Shift)

Let the set of all values of p, for which f(x) = (p2 - 6p + 8 ) ( 2 2x - 2 2x ) + 2(2 - p)x + 7 does not have any critical point, be the interval (a, b). Then 16ab is equal to .

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