Let \(f : R \to R\) be defined by
\( \displaystyle f(x) = \begin{cases} 2x + 1, & x \le 1 \\ x + k, & x \gt 1 \end{cases} \)
where R is the set of real numbers. The value of k for which f is both one-one and onto is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Both pieces are increasing. Their ranges are:
\(x \le 1\): \(2x + 1 \le 3\), giving \((-\infty, 3]\)
\(x \gt 1\): \(x + k \gt 1 + k\), giving \((1 + k, \infty)\)
For f to be one-one, the two ranges must not overlap, and for f to be onto, together they must cover all of R, with no gap. Both hold exactly when the second range starts where the first ends:
\(1 + k = 3\)
\(k = 2\)
Why the other options are incorrect
Option (1) 0 — Incorrect. The ranges \((-\infty, 3]\) and \((1, \infty)\) overlap: \(f(0.5) = 2 = f(2)\), so f is not one-one.
Option (2) 1 — Incorrect. The ranges \((-\infty, 3]\) and \((2, \infty)\) overlap: \(f(0.75) = 2.5 = f(1.5)\), so f is not one-one.
Option (4) 3 — Incorrect. The ranges \((-\infty, 3]\) and \((4, \infty)\) leave a gap: 3.5 is not an image, so f is not onto.