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Application of Derivatives · Question 10

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

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)

JEE Main 2026 (Online) 22nd January Morning Shift

Keep practising — Application of Derivatives

2

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 .

JEE Main 2026 (Online) 21st January Morning Shift
3

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
JEE Main 2026 (Online) 21st January Evening Shift
4

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
JEE Main 2025 (Online) 8th April Evening Shift
5

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
JEE Main 2025 (Online) 7th April Morning Shift
6

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
JEE Main 2025 (Online) 7th April Evening Shift
7

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
8

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
9

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
10

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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