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: Instead of discriminants, find the turning values of $y$ by calculus-free reasoning on extrema: the boundary $y$-values of a rational function occur where $f'(x)=0$, i.e. where the line $y=$const is tangent to the curve. Equivalently, locate the $y$'s where the quadratic $2y\,x^2+(4y-2)x+(3-6y)=0$ has a repeated root.

Step 1: Tangency condition. A horizontal line $y=k$ touches the graph (gives the edge of the range) exactly when $2k\,x^2+(4k-2)x+(3-6k)=0$ has a double root, i.e. discriminant $=0$:
$(4k-2)^2 - 8k(3-6k) = 0.$

Step 2: Simplify. $16k^2-16k+4-24k+48k^2 = 64k^2-40k+4 = 0$, so $16k^2-10k+1=0 \Rightarrow (8k-1)(2k-1)=0.$
The two critical $y$-values are $k=\tfrac18$ and $k=\tfrac12.$

Step 3: Decide which side is included. Test a value between them, say $y=\tfrac14$: the discriminant $64(\tfrac1{16})-40(\tfrac14)+4 = 4-10+4 = -2 < 0$, so $y=\tfrac14$ is NOT attained. Hence the gap $\left(\tfrac18,\tfrac12\right)$ is excluded and the function reaches everything outside it. Test $y=1$: $64-40+4=28>0$, attainable — confirming the upper branch starts at $\tfrac12$, not $1$.

Step 4: Range $= \left(-\infty,\tfrac18\right] \cup \left[\tfrac12,\infty\right).$

\[ \left(-\infty,\tfrac18\right] \cup \left[\tfrac12,\infty\right) \]