Let \( \displaystyle f : R \to \left[\frac{1}{3}, 3\right] \) be defined by
\( \displaystyle f(x) = \frac{x^{2} - x + 1}{x^{2} + x + 1} \)
where R is the set of real numbers. Then f is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → onto but not one-one
The denominator \(x^{2} + x + 1\) is always positive, so f is defined for every real x.
Not one-one:
\( \displaystyle f(2) = \frac{4 - 2 + 1}{4 + 2 + 1} = \frac{3}{7} \)
\( \displaystyle f\left(\frac{1}{2}\right) = \frac{\frac{1}{4} - \frac{1}{2} + 1}{\frac{1}{4} + \frac{1}{2} + 1} = \frac{3}{7} \)
Onto: a value y is taken when \(f(x) = y\) has a real solution x:
\(x^{2} - x + 1 = y(x^{2} + x + 1)\)
\((y - 1)x^{2} + (y + 1)x + (y - 1) = 0\)
For \(y = 1\) this gives \(2x = 0\), so \(x = 0\) works. For \(y \ne 1\), a real root needs a non-negative discriminant:
\((y + 1)^{2} - 4(y - 1)^{2} \ge 0\)
\((y + 1 - 2y + 2)(y + 1 + 2y - 2) \ge 0\)
\((3 - y)(3y - 1) \ge 0\)
\( \displaystyle \frac{1}{3} \le y \le 3 \)
So the range of f is exactly \([1/3, 3]\), the given codomain, and f is onto but not one-one.