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Continuity and Differentiability

JEE Main 2025 (Online) 7th April Evening ShiftNumericalMedium📈 Topic in 32% papers+4
Ques. 1JEE 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 .

Answer

2

Solution

For continuity, f(0) = x 0 f(x).

Expansions near x = 0:

x = x + x33 + , x = x - x36 +

( x) = x + 3 x3 + = x + x33 + x33 + = x + 2x33 +

( x) = x - 3 x6 + = x - x36 - x36 + = x - x33 +

numerator = x3 +

Denominator: x - x = x33 + x36 + = x32 +

f(0) = x 0 x3 x32 = 2

Answer: 2

Keep practising — Continuity and Differentiability

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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

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