Question

If the range of the function $ f(x) = \frac{5 - x}{x^2 - 3x + 2}, \quad x \neq 1, 2 $ is $ (-\infty, \alpha] \cup [\beta, \infty) $, then $ \alpha^2 + \beta^2 $ is equal to:

• A. 190
• B. 192
• C. 188
• D. 194

Hint:

When finding the range of rational functions with vertical asymptotes, analyze the behavior near the points of discontinuity, and use derivatives to locate critical points.

Answer 1

The Correct Option is D

Step 1: Factorize the Denominator
First, factorize the denominator: \[ x^2 - 3x + 2 = (x - 1)(x - 2) \] Thus, the function can be rewritten as: \[ f(x) = \frac{5 - x}{(x - 1)(x - 2)} \] 
Step 2: Find Critical Points
To find the extrema of \( f(x) \), compute its derivative using the quotient rule: \[ f'(x) = \frac{(-1)(x^2 - 3x + 2) - (5 - x)(2x - 3)}{(x^2 - 3x + 2)^2} \] Simplify the numerator: \[\begin{align} \text{Numerator} &= -x^2 + 3x - 2 - (10x - 15 - 2x^2 + 3x) &= -x^2 + 3x - 2 - 10x + 15 + 2x^2 - 3x &=x^2 - 10x + 13 \end{align}\] Thus: \[ f'(x) = \frac{x^2 - 10x + 13}{(x^2 - 3x + 2)^2} \] Set \( f'(x) = 0 \) to find critical points: \[ x^2 - 10x + 13 = 0 \] Solve the quadratic equation: \[ x = \frac{10 \pm \sqrt{100 - 52}}{2} = \frac{10 \pm \sqrt{48}}{2} = 5 \pm 2\sqrt{3} \] 
Step 3: Determine the Range
Let \( y = f(x) \). Rewrite the equation: \[ y = \frac{5 - x}{(x - 1)(x - 2)} \] Express as a quadratic in \( x \): \[ y(x^2 - 3x + 2) = 5 - x \] \[ yx^2 + (-3y + 1)x + (2y - 5) = 0 \] For real \( x \), the discriminant must be non-negative: \[ D = (-3y + 1)^2 - 4 \cdot y \cdot (2y - 5) \geq 0 \] \[ D = 9y^2 - 6y + 1 - 8y^2 + 20y \geq 0 \] \[ D = y^2 + 14y + 1 \geq 0 \] Find the roots of \( D \): \[ y = \frac{-14 \pm \sqrt{196 - 4}}{2} = \frac{-14 \pm \sqrt{192}}{2} = -7 \pm 4\sqrt{3} \] Since the coefficient of \( y^2 \) is positive, \( D \geq 0 \) when: \[ y \leq -7 - 4\sqrt{3} \quad \text{or} \quad y \geq -7 + 4\sqrt{3} \] Thus, the range of \( f(x) \) is: \[ (-\infty, -7 - 4\sqrt{3}] \cup [-7 + 4\sqrt{3}, \infty) \] Comparing with the given range \( (-\infty, \alpha] \cup [\beta, \infty) \), we identify: \[ \alpha = -7 - 4\sqrt{3} \] \[ \beta = -7 + 4\sqrt{3} \] 
Step 4: Compute \( \alpha^2 + \beta^2 \)
Calculate \( \alpha^2 \) and \( \beta^2 \): 
\[\begin{align} \alpha^2 &= (-7 - 4\sqrt{3})^2 = 49 + 56\sqrt{3} + 48 = 97 + 56\sqrt{3} \beta^2 &= (-7 + 4\sqrt{3})^2 = 49 - 56\sqrt{3} + 48 = 97- 56\sqrt{3} \end{align}\] Add them together: \[ \alpha^2 + \beta^2 = (97 + 56\sqrt{3}) + (97 - 56\sqrt{3}) = 194 \] 
Conclusion
The correct answer is \(4\).