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

(A) ( - ∞ , ( 1 )/( 8 ) ] [ 1 , ∞ )

(C) ( - ∞ , ( 1 )/( 4 ) ] [ ( 1 )/( 2 ) , ∞ )

(D) ( - ∞ , ( 1 )/( 4 ) ] [ 1 , ∞ )

CAT 2025 Slot 3 QA Q15 Functions & Graphs

Question 15 of the Quantitative Ability (QA) section, CAT 2025 Slot 3, with CAT's official answer and a worked solution.

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

A
\( \left( - \infty , \frac { 1 } { 8 } \right] \cup [ 1 , \infty ) \)
B
\( \left( - \infty , \frac { 1 } { 8 } \right] \cup \left[ \frac { 1 } { 2 } , \infty \right) \)
C
\( \left( - \infty , \frac { 1 } { 4 } \right] \cup \left[ \frac { 1 } { 2 } , \infty \right) \)
D
\( \left( - \infty , \frac { 1 } { 4 } \right] \cup [ 1 , \infty ) \)

Answer: (B) ( - ∞ , ( 1 )/( 8 ) ] [ ( 1 )/( 2 ) , ∞ ) Functions & Graphs MCQ

Set \(y = \frac{2x - 3}{2x^2 + 4x - 6}\): \(2yx^2 + (4y - 2)x + (3 - 6y) = 0\).
Real \(x\) needs discriminant \(\ge 0\): \((4y - 2)^2 - 8y(3 - 6y) = 64y^2 - 40y + 4 \ge 0\).
\(16y^2 - 10y + 1 = (8y - 1)(2y - 1) \ge 0 \Rightarrow y \le \frac{1}{8}\) or \(y \ge \frac{1}{2}\).
End values are attained at allowed \(x\) (\(y = \frac{1}{8}\) at \(x = 3\), \(y = \frac{1}{2}\) at \(x = 0\)), so the range is \((-\infty, \frac{1}{8}] \cup [\frac{1}{2}, \infty)\).

Shortcut. \(f(0) = \frac{1}{2}\) is in the range: this rules out C and D; \(f(3) = \frac{3}{24} = \frac{1}{8}\) picks B.