JEE Main 2022 · 28 July · evening · Single correct · +4 / −1

Let S={x ∈[-6,3]-{-2,2}: (|x+3|-1)/(|x|-2) ≥ 0} and
T={x ∈ ℤ: x²-7|x|+9 ≤ 0} .
Then the number of elements in S ∩ T is :

  1. (a) 7
  2. (b) 5
  3. (c) 4
  4. (d) 3
Answer and solution

Answer (d) 3

Solution

Maths sir · 3 steps

Find T first: it is a small set of integers, so you only test a handful of values in S.

  1. Solve T. Put t = |x|, so t² - 7t + 9 ≤ 0. The roots are t = (7 ± √13)/2, about 1.70 and 5.30, so |x| must lie between them.

    |x| ∈ {2, 3, 4, 5} ⇒ T = {± 2, ± 3, ± 4, ± 5}

  2. Trim to S's domain. S allows only x ∈ [-6, 3] with ± 2 removed. That rules out 4 and 5, which are too large, and ± 2. Only -5, -4, -3, 3 are left to test.
    x|x+3| - 1|x| - 2Ratio ≥ 0?
    -513yes
    -402yes, ratio is 0
    -3-11no, ratio is -1
    351yes
  3. So. S ∩ T = {-5, -4, 3}, which has 3 elements. That is option (d).

The idea to keep. When one set is a short list of integers, solve that one first. Then test only those values in the other condition instead of solving the full inequality.

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

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Asked about 40 times in JEE Main since 2020. This question is from 2022.