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CA Foundation · Quantitative Aptitude · Sets, Relations and Functions, Limits and Continuity

Let f(x) = x + 2 for x < 0 and f(x) = x² − 1 for x ≥ 0 on the real numbers. Which statement is correct?

f is onto R but not one-one. The first branch covers all values below 2 and the second covers all values from −1 upward, so every real is reached. However f(−2) = 0 and f(1) = 0, so the function repeats a value and is not injective.

  1. Af is one-one and onto R
  2. Bf is one-one but not onto R
  3. Cf is onto R but not one-oneCorrect
  4. Df is neither one-one nor onto R

Explanation

For x < 0, f takes values below 2 (all values in (−∞, 2)). For x ≥ 0, f takes values in [−1, ∞). Together they cover all reals, so f is onto. But f(−2) = 0 and f(1) = 0, so two inputs give the same output and f is not one-one. Hence onto but not one-one.

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