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:
Show Hint
- 190
- 192
- 188
- 194
The Correct Option is D
Approach Solution - 1
To find the values of \( \alpha \) and \( \beta \) for the function \( f(x) = \frac{5 - x}{x^2 - 3x + 2} \), where \( x \neq 1, 2 \), we will first analyze the function to find its range.
The denominator \( x^2 - 3x + 2 \) can be factored as:
\(x^2 - 3x + 2 = (x - 1)(x - 2)\)
Since the denominator is zero at \( x = 1 \) and \( x = 2 \), these are vertical asymptotes of the function, and the function is not defined at these points.
We need to find values of \( y = f(x) \) such that:
\(\frac{5 - x}{(x - 1)(x - 2)} = y\)
Simplifying for \( x \), multiply both sides by the denominator:
\(5 - x = y(x^2 - 3x + 2)\)
Rearrange this equation to form a quadratic in terms of \( x \):
\(yx^2 - (3y + 1)x + (2y + 5) = 0\)
For the range of \( y \), the discriminant of this quadratic must be greater than or equal to zero:
\((3y + 1)^2 - 4y(2y + 5) \geq 0\)
Simplifying the discriminant,
\((3y + 1)^2 = 9y^2 + 6y + 1\) \(4y(2y + 5) = 8y^2 + 20y\)
Thus,
\(9y^2 + 6y + 1 - (8y^2 + 20y) \geq 0\) \(y^2 - 14y + 1 \geq 0\)
Solving for \( y \), the roots of the equation \( y^2 - 14y + 1 = 0 \) using the quadratic formula are:
\(y = \frac{14 \pm \sqrt{196 - 4}}{2}\) \(y = \frac{14 \pm \sqrt{192}}{2}\) \(y = \frac{14 \pm 8\sqrt{3}}{2}\) \(y = 7 \pm 4\sqrt{3}\)
This gives the critical points for the range: \( \alpha = 7 - 4\sqrt{3} \) and \( \beta = 7 + 4\sqrt{3} \).
The range is \( (-\infty, \alpha] \cup [\beta, \infty) \). Now, we calculate \( \alpha^2 + \beta^2 \):
First, observe:
\(\alpha + \beta = 14, \quad \alpha\beta = (7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 49 - (48) = 1\)
Using the identity:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) \(\alpha^2 + \beta^2 = 14^2 - 2 \times 1\) \(\alpha^2 + \beta^2 = 196 - 2 = 194\)
Therefore, the value of \( \alpha^2 + \beta^2 \) is 194.
Approach Solution -2
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 \(4\).
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