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Application of Derivatives

JEE Main 2026 (Online) 2nd April Morning ShiftMCQHard📈 Topic in 32% papers+4−1
Ques. 1JEE Main 2026, 2 April (Morning Shift)

The number of critical points of the function f(x)= cases | xx |, & x 0\ 1, & x=0 cases in the interval (-2 , 2 ) is equal to :

  1. 1
  2. 3
  3. 5
  4. 7

Answer

(3) 5

Solution

Critical points are points where f'(x)=0 or f'(x) does not exist. f is even, so study x 0.

At x=0: near 0, f(x)= xx (positive, continuous since f(0)=1), differentiable with f'(0)=0 → critical point.

On (0, ): f(x)= xx, f'(x)= x x- xx2.

Let g(x)=x x- x: g(0)=0, g'(x)=-x x<0 ⇒ g<0 ⇒ f' 0 in (0, ).

At x= : f( )=0 is a minimum of | | with a sharp corner (left derivative - 1 , right derivative + 1 ) → f' does not exist → critical point.

On ( ,2 ): f(x)=- xx, f'(x)=0 x=x, which has exactly one root (x 4.49) in ( ,2 ) → critical point.

By symmetry the same happens at x=- and x -4.49.

[figure]

Critical points: 0, , 4.49 ⇒ total =5

Answer: (3)

Keep practising — Application of Derivatives

Ques. 2JEE Main 2026, 2 April (Evening Shift)

Let f(x) be a polynomial of degree 5, and have extrema at x=1 and x=-1. If x 0 ( f(x)x3 )=-5, then f(2)-f(-2) is equal to:

  1. 0
  2. 50
  3. 92
  4. 112
Ques. 3JEE Main 2026, 28 January (Evening Shift)

Let f be a differentiable function satisfying f(x) = 1 - 2x + 0x e(x-t) f(t) dt, x R and let g(x) = 0x (f(t) + 2)15(t - 4)6(t + 12)17 dt, x R. If p and q are respectively the points of local minima and local maxima of g, then the value of |p + q| is equal to .

Ques. 4JEE Main 2026, 24 January (Morning Shift)

Let (2 , ) be the largest interval in which the function f(t) = |t+1|t2, t < 0, is strictly decreasing. Then the local maximum value of the function g(x) = 2 e(x-2) + x2 + 4x - , x > 2, is

Ques. 5JEE Main 2026, 24 January (Evening Shift)

Consider the following three statements for the function f : (0, ) R defined by f(x) = | e x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, ). Then

  1. Only (I) is TRUE.
  2. Only (I) and (III) are TRUE.
  3. Only (II) and (III) are TRUE.
  4. All (I), (II) and (III) are TRUE.
Ques. 6JEE 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
Ques. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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