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CAT 2025Slot 3QAQues & Sol

AlgebraMedium

Question

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

Options

Solution

1. Concept Used

  • Topic: Algebra — Quadratic Equations, Discriminant, and Range of a Function

  • Formula:

For a quadratic equation (ax^2+bx+c=0) to have real roots,

$$ b^2-4ac\geq0 $$

2. Calculation

Let

$$ y=\frac{2x-3}{2x^2+4x-6} $$

Then,

$$ 2yx^2+4yx-6y=2x-3 $$

$$ 2yx^2+(4y-2)x-6y+3=0 $$

This is a quadratic in (x). Since (x) is real, its discriminant must be greater than or equal to zero.

$$ (4y-2)^2+4\times2y\times(6y-3)\geq0 $$

$$ 16y^2+4-16y-24y+48y^2\geq0 $$

$$ 64y^2-40y+4\geq0 $$

$$ 16y^2-10y+1\geq0 $$

The roots of the quadratic are:

$$ y=\frac{10\pm6}{32} $$

Therefore,

$$ y=\frac12\quad\text{or}\quad y=\frac18 $$

Since the coefficient of (y^2) is positive, the quadratic is less than zero in the interval

$$ \left(\frac18,\frac12\right) $$

and greater than or equal to zero otherwise.

Therefore, the possible values of (y), which represent the range of (f(x)), are:

$$ \left(-\infty,\frac18\right]\cup\left[\frac12,\infty\right) $$

3. Solution

Answer = Option C ✅

The range of (f(x)) is

$$ \left(-\infty,\frac18\right]\cup\left[\frac12,\infty\right) $$