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Continuity and Differentiability · Question 11

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

Let f(x) = cases ax2 + 2ax + 34x2 + 4x - 3 & , x - 32, 12 \ b & , x = - 32, 12 cases be continuous at x = - 32. If f f(x) = 75, then x is equal to:

  1. 1.4
  2. 2
  3. 1
  4. 0

Answer

(3) 1

Solution

4x2 + 4x - 3 = (2x + 3)(2x - 1) vanishes at x = - 32, so for a finite limit the numerator must also vanish there:

a 94 - 3a + 3 = 0 a = 4

4x2 + 8x + 3 = (2x + 3)(2x + 1)

f(x) = 2x + 12x - 1, x - 32, 12

Continuity at x = - 32: b = -3 + 1-3 - 1 = 12

Let f(x) = t. f(t) = 75 b, so t - 32, 12:

2t + 12t - 1 = 75 10t + 5 = 14t - 7 t = 3

2x + 12x - 1 = 3 2x + 1 = 6x - 3 x = 1

Answer: (3)

JEE Main 2026 (Online) 23rd January Morning Shift

Keep practising — Continuity and Differentiability

2

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
JEE Main 2026 (Online) 23rd January Evening Shift
3

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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4

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5

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6

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7

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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8

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 .

JEE Main 2025 (Online) 28th January Morning Shift

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