CAT 2025 Slot 3 · Quant · Question 15

Set y=f\left( x \right) and clear the denominator: y\left( 2{x}^{2}+4x-6 \right)=2x-3.

Rearranging into a quadratic in x: 2y{x}^{2}+\left( 4y-2 \right)x+\left( 3-6y \right)=0.

A value y is in the range exactly when this has a real solution x, so its discriminant must be non-negative.

The discriminant is {\left( 4y-2 \right)}^{2}-4\left( 2y \right)\left( 3-6y \right)=16{y}^{2}-16y+4-24y+48{y}^{2}.

That simplifies to 64{y}^{2}-40y+4, and dividing by 4, the condition is 16{y}^{2}-10y+1\ge 0.

Factorising, \left( 8y-1 \right)\left( 2y-1 \right)\ge 0.

The parabola opens upwards, so this holds outside the roots: y\le \frac{1}{8} or y\ge \frac{1}{2}.

Both endpoints are actually reached — y=\frac{1}{2} at x=0 and y=\frac{1}{8} at x=3 — so they are included.

The range is \left( -\infty ,\frac{1}{8}\right]\cup \left[ \frac{1}{2},\infty \right).

Shortcut. For any ratio of quadratics, cross-multiply, treat it as a quadratic in x, and impose discriminant \ge 0. The range always comes out as the solution of a quadratic inequality in y, with no need to analyse the graph.