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Continuity and Differentiability · Question 12

JEE Main 2026 (Online) 23rd January Evening ShiftMCQEasy📈 Topic in 35% papers+4−1
1

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

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

Continuity at x = 0 RHL = LHL = f(0)

a - 2 = 2 - a a = 2

b = a - 2 = 0

a + b = 2 + 0 = 2

Answer: (2)

JEE Main 2026 (Online) 23rd January Evening Shift

Keep practising — Continuity and Differentiability

2

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
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3

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

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4

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

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5

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 .

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6

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
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7

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 .

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