Let \(f : R \to A\) be defined by
\( \displaystyle f(x) = \frac{2x^{2}}{x^{2} + 1} \)
where R is the set of real numbers. If f is onto, then A is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [0, 2)
f is onto A exactly when A is the range of f. Rewriting f:
\( \displaystyle f(x) = \frac{2(x^{2} + 1) - 2}{x^{2} + 1} = 2 - \frac{2}{x^{2} + 1} \)
Since \(x^{2} + 1 \ge 1\), the fraction satisfies
\( \displaystyle 0 \lt \frac{2}{x^{2} + 1} \le 2 \)
So
\(0 \le f(x) \lt 2\)
The value 0 is taken at \(x = 0\), and 2 is never reached. Every y with \(0 \le y \lt 2\) is taken, at \(x = \sqrt{\displaystyle\frac{y}{2 - y}}\).
So the range is [0, 2), and \(A = [0, 2)\).