Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined as \(f(x) = \frac{2x^2 - 3x + 2}{3x^2 + x + 3}\). Then f is:
The correct answer is neither one-one nor onto.
One-one? \(f(2) = \frac{8 - 6 + 2}{12 + 2 + 3} = \frac{4}{17}\) and \(f\left(\frac{1}{2}\right) = \frac{\frac{1}{2} - \frac{3}{2} + 2}{\frac{3}{4} + \frac{1}{2} + 3} = \frac{4}{17}\). Two different inputs give the same output, so f is not one-one.
Onto? Put y = f(x): \((3y - 2)x^2 + (y + 3)x + (3y - 2) = 0\). For real x, discriminant ≥ 0: \((y + 3)^2 - 4(3y - 2)^2 \ge 0 \Rightarrow 35y^2 - 54y + 7 \le 0 \Rightarrow \frac{1}{7} \le y \le \frac{7}{5}\). The range \(\left[\frac{1}{7}, \frac{7}{5}\right]\) is not all of ℝ, so f is not onto.
Short trick: Here f(x) = f(1/x), so f cannot be one-one; a bounded range means it cannot be onto ℝ.
Similar Questions
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