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Continuity and Differentiability

JEE Main 2026 (Online) 23rd January Evening ShiftMCQEasy📈 Topic in 34% papers+4−1
Ques. 1JEE Main 2026, 23 January (Evening Shift)

If f(x) = cases a|x| + x2 - 2( |x|)( |x|)x & , x 0 \ b & , x = 0 cases is continuous at x = 0, then a + b is equal to :

  1. 1
  2. 2
  3. 0
  4. 4

Answer

(2) 2

Solution

Using 2 |x| |x| = 2|x|:

Right-hand limit (x > 0, |x| = x):

x 0+ ax + x2 - 2xx = a + 0 - 2 = a - 2

Left-hand limit (x < 0, |x| = -x, 2|x| = - 2x):

x 0- -ax + x2 + 2xx = -a + 0 + 2 = 2 - a

For continuity at x = 0, RHL = LHL = f(0), so

a - 2 = 2 - a a = 2

and b = a - 2 = 0.

a + b = 2 + 0 = 2

Answer: (2)

Keep practising — Continuity and Differentiability

Ques. 2JEE Main 2026, 22 January (Evening Shift)

Let [ ] denote the greatest integer function, and let f(x) = \ 2x, x2 \. Let S = \x (-2, 2) : the function g(x) = |x| [x2 ] is discontinuous at x\. Then x S f(x) equals

  1. 2 - 2
  2. 2 6 - 3 2
  3. 1 - 2
  4. 6 - 2 2
Ques. 3JEE Main 2025, 7 April (Morning Shift)

The number of points of discontinuity of the function f(x) = [ x22 ] - [ x], x [0, 4], where [ ] denotes the greatest integer function, is .

Ques. 4JEE Main 2025, 7 April (Evening Shift)

If the function f(x) = ( x) - ( x) x - x is continuous at x = 0, then f(0) is equal to .

Ques. 5JEE Main 2025, 4 April (Morning Shift)

Let m and n be the number of points at which the function f(x) = \x, x3, x5, , x21 \, x R, is not differentiable and not continuous, respectively. Then m + n is equal to .

Ques. 6JEE 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
Ques. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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