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Application of Derivatives

JEE Main 2026 (Online) 22nd January Morning ShiftMCQMedium📈 Topic in 13% papers+4−1
Ques. 1JEE Main 2026, 22 January (Morning Shift)

Let f(x) = x2025 - x2000, x [0, 1] and the minimum value of the function f(x) in the interval [0, 1] be (80)80(n)-81. Then n is equal to

  1. -40
  2. -41
  3. -80
  4. -81

Answer

(4) -81

Solution

f'(x) = 2025x2024 - 2000x1999 = x1999(2025x25 - 2000)

f'(x) = 0 x25 = 20002025 = 8081

f'(x) < 0 before and > 0 after this point, so it gives the minimum (f(0) = f(1) = 0).

f(x) = x2000(x25 - 1) = ( 8081 )80 ( 8081 - 1 )

= - 80808180 181 = - 80808181

= (80)80(-81)-81 ( (-81)81 = -8181)

n = -81

Answer: (4)

Keep practising — Application of Derivatives

Ques. 2JEE Main 2026, 21 January (Morning Shift)

Let f : R R be a twice differentiable function such that the quadratic equation f(x)m2 - 2f'(x)m + f''(x) = 0 in m, has two equal roots for every x R. If f(0) = 1, f'(0) = 2, and ( , ) is the largest interval in which the function f( e x - x) is increasing, then + is equal to .

Ques. 3JEE Main 2026, 21 January (Evening Shift)

Let f : R R be a twice differentiable function such that f''(x) > 0 for all x R and f'(a - 1) = 0, where a is a real number. Let g(x) = f( 2 x - 2 x + a), 0 < x < 2. Consider the following two statements: (I) g is increasing in (0, 4 ) (II) g is decreasing in ( 4, 2 ) Then,

  1. Both (I) and (II) are True
  2. Neither (I) nor (II) is True
  3. Only (I) is True
  4. Only (II) is True
Ques. 4JEE Main 2025, 8 April (Evening Shift)

Let the function f(x) = x3 + 3x + 3, x 0 be strictly increasing in (- , 1) ( 2, ) and strictly decreasing in ( 3, 4) ( 4, 5). Then i=15 i2 is equal to

  1. 48
  2. 40
  3. 36
  4. 28
Ques. 5JEE 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
Ques. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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