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Application of Derivatives · Question 4

JEE Main 2026 (Online) 4th April Evening ShiftMCQMedium📈 Topic in 13% papers+4−1
1

0 x (16 ( x2 ) 3 ( x2 ) ) is equal to :

  1. 3 32
  2. 3 3
  3. 4 3
  4. 6 3

Answer

(2) 3 3

Solution

Let t= x2; 0 x 0 t 2

g(t)=16 t 3t

g'(t)=16 ( 4t-3 2t 2t )=16 2t ( 2t-3 2t )

g'(t)=0 in (0, 2 ) 2t= 13 t= 6

g ( 6 )=16 12 ( 32 )3=16 12 3 38=3 3

End values: g(0)=0, g ( 2 )=0

maximum value =3 3 (at x= 3)

Answer: (2)

JEE Main 2026 (Online) 4th April Evening Shift

Keep practising — Application of Derivatives

2

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
JEE Main 2026 (Online) 2nd April Morning Shift
3

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
JEE Main 2026 (Online) 2nd April Evening Shift
4

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 .

JEE Main 2026 (Online) 28th January Evening Shift
5

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

JEE Main 2026 (Online) 24th January Morning Shift
6

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.
JEE Main 2026 (Online) 24th January Evening Shift
7

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
JEE Main 2026 (Online) 22nd January Morning Shift
8

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
9

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
10

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
11

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

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