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CA Foundation · Quantitative Aptitude · Linear Inequalities

If the inequality (3x − 1)/(x + 2) ≥ 2 is solved for real x, the solution set is:

The inequality reduces to (x − 5)/(x + 2) ≥ 0. This is true for x ≥ 5 or x < −2, with x = −2 excluded since the denominator would be zero.

  1. Ax ≥ 5
  2. Bx > 5 or x < −2
  3. Cx ≥ 5 or x < −2Correct
  4. D−2 < x ≤ 5

Explanation

Rewrite as (3x − 1 − 2x − 4)/(x + 2) = (x − 5)/(x + 2) ≥ 0. This holds when x ≥ 5 or x < −2; x = −2 is excluded as the denominator is zero, and x = 5 is included. Cross-multiplying without considering the sign of x + 2 gives x ≥ 5 only.

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