Question

For real values of $x$, the range of the function $f(x) = \dfrac{2x - 3}{2x^2 + 4x - 6}$ is

• A. $(-\infty, \dfrac{1}{8}] \cup [1, \infty)$
• B. $(-\infty, \dfrac{1}{8}] \cup [\dfrac{1}{2}, \infty)$
• C. $(-\infty, \dfrac{1}{4}] \cup [\dfrac{1}{2}, \infty)$
• D. $(-\infty, \dfrac{1}{4}] \cup [1, \infty)$

Hint:

To find the range of a rational function:
Set $y = f(x)$ and rearrange into a quadratic in $x$.
Apply the discriminant condition $\Delta \ge 0$.
Solve the resulting inequality in $y$.
Check that boundary values are actually achieved by plugging back. This method works reliably for all rational functions with $x$ appearing up to degree 2 in the denominator.

Answer 1

The Correct Option is A

Approach: To find the range of a rational function, set \(y=f(x)\), cross-multiply to get a quadratic in \(x\), and demand its discriminant be \(\ge 0\) — those are exactly the \(y\)-values the function can actually hit.

Step 1: Set \(y=f(x)\) and clear the denominator.
\(y = \dfrac{2x-3}{2x^2+4x-6}\Rightarrow y(2x^2+4x-6) = 2x-3.\)
Rearrange into a quadratic in \(x\):
\[2y\,x^2 + (4y-2)x + (3-6y) = 0.\]

Step 2: Demand real \(x\) (discriminant \(\ge 0\)).
For \(y\ne 0\) this is a genuine quadratic; real \(x\) needs \(\Delta \ge 0\):
\(\Delta = (4y-2)^2 - 4(2y)(3-6y).\)
\((4y-2)^2 = 16y^2-16y+4;\quad 4(2y)(3-6y)=24y-48y^2.\)
\(\Delta = 16y^2-16y+4 - (24y-48y^2) = 64y^2 - 40y + 4.\)

Step 3: Solve \(\Delta \ge 0\).
\(64y^2-40y+4 = 4(16y^2-10y+1) = 4(2y-1)(8y-1).\)
Roots: \(y=\tfrac12\) and \(y=\tfrac18\). The parabola opens upward, so \(\Delta\ge 0\) outside the roots:
\[y \le \tfrac18 \quad\text{or}\quad y \ge \tfrac12.\]

Step 4: Check the boundary case \(y=0\).
At \(y=0\) the equation is linear: \(-2x+3=0\Rightarrow x=\tfrac32\) (real). Since \(0 \le \tfrac18\), it already lies in the first branch — consistent.

Conclusion: The range is
\[\boxed{\left(-\infty,\tfrac18\right] \cup \left[\tfrac12,\infty\right)}\]
which is the option \(\left(-\infty,\tfrac18\right] \cup \left[\tfrac12,\infty\right)\).

Answer 2

We are given: \[ f(x) = \frac{2x - 3}{2x^2 + 4x - 6}. \] Let the range be all real \(y\) such that the equation \[ y = \frac{2x - 3}{2x^2 + 4x - 6} \] has real solutions for \(x\).
Step 1: Rearranging. \[ y(2x^2 + 4x - 6) = 2x - 3. \] Expand: \[ 2yx^2 + 4yx - 6y = 2x - 3. \] Bring all terms to one side: \[ 2yx^2 + (4y - 2)x + (-6y + 3) = 0. \] For real \(x\), the discriminant must be \(\ge 0\): \[ (4y - 2)^2 - 4(2y)(-6y + 3) \ge 0. \]
Step 2: Simplify the discriminant. Compute: \[ (4y - 2)^2 = 16y^2 - 16y + 4. \] Next: \[ -4(2y)(-6y + 3) = -8y(-6y + 3) = 48y^2 - 24y. \] Thus: \[ \Delta = (16y^2 - 16y + 4) + (48y^2 - 24y) = 64y^2 - 40y + 4. \] We require: \[ 64y^2 - 40y + 4 \ge 0. \]
Step 3: Solve the quadratic inequality. Divide by 4: \[ 16y^2 - 10y + 1 \ge 0. \] The discriminant is: \[ (-10)^2 - 4 \cdot 16 \cdot 1 = 100 - 64 = 36. \] Roots: \[ y = \frac{10 \pm 6}{32}. \] Thus: \[ y_1 = \frac{16}{32} = \frac12, \qquad y_2 = \frac{4}{32} = \frac18. \] Since this quadratic opens upward (coefficient of \(y^2\) is positive), the inequality \[ 16y^2 - 10y + 1 \ge 0 \] holds for: \[ y \le \frac18 \quad \text{or} \quad y \ge \frac12. \]
Step 4: Check which endpoint values are actually attained. Evaluate whether \(y = \tfrac12\) belongs to the range. Solve: \[ \frac{2x - 3}{2x^2 + 4x - 6} = \frac12. \] Cross-multiply: \[ 2(2x - 3) = 2x^2 + 4x - 6. \] \[ 4x - 6 = 2x^2 + 4x - 6. \] Subtract LHS: \[ 0 = 2x^2. \] Thus \(x = 0\). Check denominator at \(x=0\): \[ 2(0)^2 + 4(0) - 6 = -6 \ne 0. \] Hence \(y = \frac12\) is included. Similarly for \(y = \frac18\): \[ \frac{2x - 3}{2x^2 + 4x - 6} = \frac18. \] Cross-multiply: \[ 8(2x - 3) = 2x^2 + 4x - 6. \] \[ 16x - 24 = 2x^2 + 4x - 6. \] Bring all to one side: \[ 2x^2 - 12x + 18 = 0. \] Divide by 2: \[ x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0. \] So \(x = 3\), but check the denominator at \(x = 3\): \[ 2(9) + 4(3) - 6 = 18 + 12 - 6 = 24 \ne 0, \] Hence \(y = \frac18\) is also included.
Final Range: \[ \boxed{(-\infty, \frac18] \cup [\frac12, \infty)}. \] Which matches option **(A)**.