Which of the following functions is onto?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(f: R \rightarrow R,\ f(x) = 2x + 1\)
A function f: R → R is onto when every real number y is the image of some real x, that is, when the range equals the codomain R.
For f(x) = 2x + 1, take any real y and solve f(x) = y:
\(2x + 1 = y\)
\(\displaystyle x = \frac{y - 1}{2}\)
This x is a real number for every real y, and \(f\left(\frac{y - 1}{2}\right) = y\). So f is onto.
Why the other options are incorrect
Since \(x^2 \ge 0\) for every real x, the three quadratic functions miss some real numbers.
Option (1) \(f: R \rightarrow R,\ f(x) = x^2\) — Incorrect. Its range is [0, ∞); a negative number such as −1 is never taken.
Option (2) \(f: R \rightarrow R,\ f(x) = x^2 + 1\) — Incorrect. Its range is [1, ∞); for example 0 is never taken.
Option (3) \(f: R \rightarrow R,\ f(x) = x^2 - 1\) — Incorrect. Its range is [−1, ∞); for example −2 is never taken.