Question

If \( \alpha, \beta \in \mathbb{R} \) and \( \alpha \leq \frac{x^2 - 3x + 4}{x^2 - 4x + 5} \leq \beta \) for all \( x \in \mathbb{R} \), then \( (2\alpha - 3)^2 + (2\beta - 3)^2 = \)

• A. \( 10 \)
• B. \( 8 \)
• C. \( 6 \)
• D. \( 4 \)

Hint:

For rational functions where the numerator and denominator are close in value, the range is often centered around 1. Note that \( \frac{3 \pm \sqrt{2}}{2} \) is approximately \( 1.5 \pm 0.7 \).

Answer 1

The Correct Option is D

Concept: The range of a rational function \( y = \frac{ax^2+bx+c}{px^2+qx+r} \) can be found by expressing the equation as a quadratic in \( x \) and setting the discriminant \( D \geq 0 \) for real \( x \).
• Rewrite as: \( (y-a)x^2 + (qy-b)x + (ry-c) = 0 \).
• Condition for real \( x \): \( B^2 - 4AC \geq 0 \).

Step 1: Setting up the quadratic in \( x \).
Let \( y = \frac{x^2 - 3x + 4}{x^2 - 4x + 5} \). Multiplying across: \[ y(x^2 - 4x + 5) = x^2 - 3x + 4 \implies (y-1)x^2 + (3-4y)x + (5y-4) = 0 \]

Step 2: Finding the range using the discriminant.
For \( x \) to be real, \( D = (3-4y)^2 - 4(y-1)(5y-4) \geq 0 \): \[ 9 + 16y^2 - 24y - 4(5y^2 - 9y + 4) \geq 0 \] \[ -4y^2 + 12y - 7 \geq 0 \implies 4y^2 - 12y + 7 \leq 0 \] Roots of \( 4y^2 - 12y + 7 = 0 \) are \( y = \frac{12 \pm \sqrt{144 - 112}}{8} = \frac{3 \pm \sqrt{2}}{2} \). Thus, \( \alpha = \frac{3-\sqrt{2}}{2} \) and \( \beta = \frac{3+\sqrt{2}}{2} \).

Step 3: Calculating the final expression.
From \( \alpha, \beta \), we have \( 2\alpha-3 = -\sqrt{2} \) and \( 2\beta-3 = \sqrt{2} \). \[ (2\alpha - 3)^2 + (2\beta - 3)^2 = (-\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4 \]