JEE Main 2023 · 11 April · morning · Single correct · +4 / −1

Let f(x)=[x²-x]+|-x+[x]|, where x ∈ ℝ and [t] denotes the greatest integer less than or equal to t. Then, f is :

  1. (a) continuous at x=0, but not continuous at x=1
  2. (b) continuous at x=0 and x=1
  3. (c) continuous at x=1, but not continuous at x=0
  4. (d) not continuous at x=0 and x=1
Answer and solution

Answer (c) continuous at x=1, but not continuous at x=0

Solution

Maths sir · 4 steps

Rewrite the second term as the fractional part, then check both sides of 0 and of 1 separately.

  1. Tidy up. Since x - [x] ≥ 0, we get |-x+[x]| = x - [x] = {x}, the fractional part. So f(x) = [x² - x] + {x}, and f(0) = f(1) = [0] + 0 = 0.
  2. Sign of the product. x² - x = x(x-1) is a small positive number just left of 0 and just right of 1, so its greatest integer is 0 there. It is a small negative number just right of 0 and just left of 1, so its greatest integer is -1 there.
    Approach[x² - x]{x}Limit of f
    x → 0⁻0→ 11
    x → 0⁺-1→ 0-1
    x → 1⁻-1→ 10
    x → 1⁺0→ 00
  3. Compare. At x = 0 the left limit is 1 and the right limit is -1, so f breaks there. At x = 1 the two jumps cancel: both limits are 0, equal to f(1).
  4. So. f is continuous at x = 1 but not at x = 0, which is option (c).

The idea to keep. Near an integer, check each piece from each side. A jump in the greatest integer part can exactly cancel the opposite jump in the fractional part, so the sum may stay continuous.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Continuity & Differentiability

Asked about 90 times in JEE Main since 2020. This question is from 2023.