For real values of \( x \), the range of the function \( f ( x ) = \frac { 2 x - 3 } { 2 x ^ { 2 } + 4 x - 6 } \) is
Answer: B
Key idea: Set \( f(x) = y \) and require the resulting quadratic in \( x \) to have real roots.
1. \( y(2x^2 + 4x - 6) = 2x - 3 \Rightarrow 2y x^2 + (4y - 2)x + (3 - 6y) = 0 \).
2. Discriminant \( \ge 0 \): \( (4y - 2)^2 - 8y(3 - 6y) \ge 0 \Rightarrow 64y^2 - 40y + 4 \ge 0 \Rightarrow 16y^2 - 10y + 1 \ge 0 \).
3. \( (8y - 1)(2y - 1) \ge 0 \Rightarrow y \le \frac{1}{8} \) or \( y \ge \frac{1}{2} \).
4. Checks: \( y = 0 \) occurs at \( x = \frac{3}{2} \); \( y = \frac{1}{8} \) at \( x = 3 \); \( y = \frac{1}{2} \) at \( x = 0 \). The excluded points \( x = 1, -3 \) never satisfy the quadratic (substituting them gives 1 = 0 and 9 = 0), so nothing is lost.
So the range is \( \left(-\infty, \frac{1}{8}\right] \cup \left[\frac{1}{2}, \infty\right) \).
Shortcut: \( f(0) = \frac{-3}{-6} = \frac{1}{2} \) already rules out (A) and (D).
Hence, the answer is (B).