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Continuity and Differentiability

JEE Main 2025 (Online) 3rd April Morning ShiftMCQMedium📈 Topic in 34% papers+4−1
Ques. 1JEE Main 2025, 3 April (Morning Shift)

Let f(x) = cases (1 + ax)1/x, & x < 0 \ 1 + b, & x = 0 \ (x+4)1/2 - 2(x+c)1/3 - 2, & x > 0 cases be continuous at x = 0. Then ea bc is equal to:

  1. 64
  2. 48
  3. 36
  4. 72

Answer

(2) 48

Solution

LHL: x 0- (1 + ax)1/x = ea

RHL: numerator 0 as x 0+, so for a finite non-zero limit the denominator must also 0

⇒ c1/3 = 2 ⇒ c = 8

RHL = x 0+ (x+4)1/2 - 2(x+8)1/3 - 2

= 12(4)-1/2 13(8)-2/3 (L'Hôpital's rule)

= 1/41/12 = 3

Continuity ⇒ ea = 1 + b = 3 ⇒ b = 2

∴ ea bc = 3 2 8 = 48

Answer: (2)

Keep practising — Continuity and Differentiability

Ques. 2JEE Main 2025, 29 January (Evening Shift)

Let the function f(x)= (x2-1 ) |x2-ax+2 |+ |x| be not differentiable at the two points x= =2 and x= . Then the distance of the point ( , ) from the line 12x+5y+10=0 is equal to :

  1. 5
  2. 2
  3. 4
  4. 3
Ques. 3JEE Main 2025, 28 January (Morning Shift)

Let f(x) = cases 3x, & x < 0 \ \1 + x + [x], x + 2[x]\, & 0 x 2 \ 5, & x > 2 cases where [.] denotes greatest integer function. If and are the number of points, where f is not continuous and is not differentiable, respectively, then + equals .

Ques. 4JEE Main 2025, 24 January (Evening Shift)

Let [x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+|x-2|, -2<x<3, is not continuous and not differentiable. Then m+n is equal to :

  1. 6
  2. 9
  3. 8
  4. 7
Ques. 5JEE Main 2025, 23 January (Morning Shift)

If the function f(x)= cases 2x \ (k1+1)x+ (k2-1)x \, & x<0\\[2mm] 4, & x=0\\[2mm] 2x e ( 2+k1x2+k2x ), & x>0 cases is continuous at x=0, then k12+k22 is equal to :

  1. 5
  2. 10
  3. 20
  4. 8
Ques. 6JEE Main 2024, 9 April (Morning Shift)

Let f(x) = ax3 + bx2 + cx + 41 be such that f(1) = 40, f'(1) = 2 and f''(1) = 4. Then a2 + b2 + c2 is equal to:

  1. 54
  2. 51
  3. 73
  4. 62
Ques. 7JEE Main 2024, 9 April (Morning Shift)

Let f : (0, ) R be a function given by f(x) = cases ( 87 ) 8x 7x, & 0 < x < 2 \ a - 8, & x = 2 \ (1 + | x|) ba| x|, & 2 < x < cases where a, b Z. If f is continuous at x = 2, then a2 + b2 is equal to .

Ques. 8JEE Main 2024, 9 April (Evening Shift)

If e y = 3 -1x, then (1 - x2)y'' - xy' at x = 12 is equal to

  1. 9e /2
  2. 9e /6
  3. 3e /2
  4. 3e /6
Ques. 9JEE Main 2024, 8 April (Evening Shift)

For a, b > 0, let f(x) = cases ((a + 1)x) + b xx, & x < 0 \ 3, & x = 0 \ ax + b2x2 - axb a x x, & x > 0 cases be a continuous function at x = 0. Then ba is equal to :

  1. 4
  2. 5
  3. 8
  4. 6
Ques. 10JEE Main 2024, 6 April (Morning Shift)

If f(x) = cases x3 ( 1x ), & x 0 \ 0, & x = 0 cases, then

  1. f''(0) = 0
  2. f''(0) = 1
  3. f'' ( 2 ) = 24 - 22
  4. f'' ( 2 ) = 12 - 22
Ques. 11JEE Main 2024, 6 April (Morning Shift)

Let f : (- , ) - \0\ R be a differentiable function such that f'(1) = a a2 f ( 1a ). Then a a(a + 1)2 -1 ( 1a ) + a2 - 2 e a is equal to

  1. 52 + 8
  2. 38 + 4
  3. 34 + 8
  4. 32 + 4

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