If the range of the function $f(x) = \frac{5-x}{x^2-3x + 2}$ , $x \neq 1, 2$, is $(-\infty, \alpha] \cup [\beta, \infty)$, then $\alpha^2 + \beta^2$ is equal to :
We are given the function $f(x) = \frac{5-x}{x^2-3x + 2}$ where $x \neq 1, 2$. The range is given in the form $(-\infty, \alpha] \cup [\beta, \infty)$. We need to find the value of $\alpha^2 + \beta^2$.
Finding the Range of the Function
Let $y = f(x)$. We set up the equation:
$y = \frac{5-x}{x^2-3x + 2}$
Rearranging the equation to form a quadratic in terms of $x$:
$y(x^2 - 3x + 2) = 5 - x$
$yx^2 - 3yx + 2y = 5 - x$
$yx^2 + (1 - 3y)x + (2y - 5) = 0$
For $x$ to be a real number, the discriminant ($\Delta$) of this quadratic equation must be non-negative ($\Delta \ge 0$).
The discriminant is calculated as $\Delta = b^2 - 4ac$, where $a=y$, $b=(1-3y)$, and $c=(2y-5)$.
$\Delta = (1 - 3y)^2 - 4(y)(2y - 5)$
$\Delta = (1 - 6y + 9y^2) - (8y^2 - 20y)$
$\Delta = 1 - 6y + 9y^2 - 8y^2 + 20y$
$\Delta = y^2 + 14y + 1$
Since $\Delta \ge 0$, we have the inequality:
$y^2 + 14y + 1 \ge 0$
To solve this inequality, we first find the roots of the quadratic equation $y^2 + 14y + 1 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{-14 \pm \sqrt{14^2 - 4(1)(1)}}{2(1)}$
$y = \frac{-14 \pm \sqrt{196 - 4}}{2}$
$y = \frac{-14 \pm \sqrt{192}}{2}$
Simplifying $\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$:
$y = \frac{-14 \pm 8\sqrt{3}}{2}$
$y = -7 \pm 4\sqrt{3}$
The roots are $y_1 = -7 - 4\sqrt{3}$ and $y_2 = -7 + 4\sqrt{3}$. Since the quadratic $y^2 + 14y + 1$ opens upwards, the inequality $y^2 + 14y + 1 \ge 0$ holds when $y$ is less than or equal to the smaller root, or greater than or equal to the larger root.
Thus, the range of the function is:
$(-\infty, -7 - 4\sqrt{3}] \cup [-7 + 4\sqrt{3}, \infty)$
Identifying $\alpha$ and $\beta$
Comparing the obtained range with the given form $(-\infty, \alpha] \cup [\beta, \infty)$, we identify:
- $\alpha = -7 - 4\sqrt{3}$
- $\beta = -7 + 4\sqrt{3}$
Calculating $\alpha^2 + \beta^2$
Now, we calculate the squares of $\alpha$ and $\beta$:
$\alpha^2 = (-7 - 4\sqrt{3})^2 = (-(7 + 4\sqrt{3}))^2 = (7 + 4\sqrt{3})^2$
$\alpha^2 = 7^2 + 2(7)(4\sqrt{3}) + (4\sqrt{3})^2$
$\alpha^2 = 49 + 56\sqrt{3} + (16 \times 3)$
$\alpha^2 = 49 + 56\sqrt{3} + 48 = 97 + 56\sqrt{3}$
And for $\beta$:
$\beta^2 = (-7 + 4\sqrt{3})^2 = (-7)^2 + 2(-7)(4\sqrt{3}) + (4\sqrt{3})^2$
$\beta^2 = 49 - 56\sqrt{3} + (16 \times 3)$
$\beta^2 = 49 - 56\sqrt{3} + 48 = 97 - 56\sqrt{3}$
Finally, we compute $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = (97 + 56\sqrt{3}) + (97 - 56\sqrt{3})$
$\alpha^2 + \beta^2 = 97 + 97 = 194$
The value of $\alpha^2 + \beta^2$ is 194.
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