JEE Main 2022 · 27 June · morning · Single correct · +4 / −1

Let a be an integer such that lim _(x → 7) (18 - [1 - x])/[x - 3a] exists, where [t] is greatest integer ≤ t. Then a is equal to :

  1. (a) -6
  2. (b) -2
  3. (c) 2
  4. (d) 6
Answer and solution

Answer (a) -6

Solution

Maths sir · 4 steps

The greatest integer function jumps at x = 7, so find both one-sided limits and make them equal.

  1. Numerator near 7. For x → 7⁺, 1 - x is slightly less than -6, so [1 - x] = -7 and the numerator is 18 + 7 = 25. For x → 7⁻, 1 - x is slightly more than -6, so [1 - x] = -6 and the numerator is 24.
  2. Denominator near 7. Since 3a is an integer, [x - 3a] = [x] - 3a. So it is 7 - 3a for x → 7⁺ and 6 - 3a for x → 7⁻.

    25/(7 - 3a) = 24/(6 - 3a)

  3. Solve. Cross-multiply: 150 - 75a = 168 - 72a, so -3a = 18 and a = -6. Check: both sides become 25/25 = 24/24 = 1.
  4. So. The limit exists only when a = -6, which is option (a).

The idea to keep. When a box function sits at an integer point, find the left-hand and right-hand limits separately. Pull integer shifts out: [x + n] = [x] + n for any integer n.

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

Study Limits

Asked about 110 times in JEE Main since 2020. This question is from 2022.