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Continuity and Differentiability

JEE Main 2025 (Online) 23rd January Morning ShiftMCQMedium📈 Topic in 32% papers+4−1
Ques. 1JEE 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

Answer

(2) 10

Solution

Continuity: LHL = RHL =f(0)=4.

LHL, using axx=a: 2 [(k1+1)+(k2-1) ]=4 k1+k2=2

RHL, using (1+t) t: 2x (k1-k2)x2=4 k1-k2=4

k12+k22= (k1+k2)2+(k1-k2)22= 4+162=10

Answer: (2)

Keep practising — Continuity and Differentiability

Ques. 2JEE Main 2026, 8 April (Evening Shift)

Let f(x)= cases 13, & x 2\ b(1- x)( -2x)2, & x> 2 cases. If f is continuous at x= 2, then the value of 03b-6 |x2+2x-3 |dx is :

  1. 5
  2. 2
  3. 3
  4. 4
Ques. 3JEE Main 2026, 6 April (Evening Shift)

Let f(x)= casesx3+8; & x<0,\ x2-4; & x 0, cases and g(x)= cases(x-8)1/3; & x<0,\ (x+4)1/2; & x 0. cases Then the number of points, where the function g f is discontinuous, is .

Ques. 4JEE Main 2026, 4 April (Morning Shift)

If y= -1 ( 3 x-4 x4 x+3 x )+2 -1 ( x1+ 1-x2 ), then dydx at x= 32 is equal to :

  1. 3
  2. -1
  3. 1
  4. 2
Ques. 5JEE Main 2026, 4 April (Morning Shift)

Let f be a real polynomial of degree n such that f(x)=f'(x)f''(x), for all x R. If f(0)=0, then 36 (f'(2)+f''(2)+ 02 f(x) dx ) is equal to:

  1. 42
  2. 46
  3. 56
  4. 66
Ques. 6JEE Main 2026, 4 April (Morning Shift)

The number of points, at which the function f(x)= \6x, 2+3x2 \+|x-1| |x2- 14 |, x (- , ), is not differentiable, is .

Ques. 7JEE Main 2026, 4 April (Evening Shift)

Let f(x)= casesex-1, & x<0\ x2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)|. If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to .

Ques. 8JEE Main 2026, 2 April (Evening Shift)

The number of points in the interval [2, 4], at which the function f(x)= [x2-x- 12 ], where [ ] denotes the greatest integer function, is discontinuous, is .

Ques. 9JEE Main 2026, 28 January (Evening Shift)

Let f(x) = 0 ( x - x ( 2 ) (x - 1)1 + x ( 2 )(x - 1) ), x R. Consider the following two statements : (I) f(x) is discontinuous at x = 1. (II) f(x) is continuous at x = -1. Then,

  1. Neither (I) nor (II) is True
  2. Only (II) is True
  3. Only (I) is True
  4. Both (I) and (II) are True
Ques. 10JEE Main 2026, 24 January (Morning Shift)

Let , R be such that the function f(x) = cases 2 (x2 - 2 ) + 2 x, & x < 1 \ ( + 3)x + ( - ), & x 1 cases be differentiable at all x R. Then 34( + ) is equal to

  1. 48
  2. 84
  3. 36
  4. 24
Ques. 11JEE Main 2026, 24 January (Evening Shift)

Let [t] denote the greatest integer less than or equal to t. If the function f(x) = cases b2 ( 2 [ 2( x + x) x ] ), & x < 0 \ x - 12 2xx3, & x > 0 \ a, & x = 0 cases is continuous at x = 0, then a2 + b2 is equal to :

  1. 12
  2. 58
  3. 34
  4. 916

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