If the range of the function f(x) = frac{5-x} | vedclass

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(D) Let $y = \frac{5-x}{x^2-3x+2}$.$y(x^2-3x+2) = 5-x$$yx^2 - 3xy + 2y = 5-x$$yx^2 + (1-3y)x + (2y-5) = 0$.If $y=0$,then $x=5$,which is a valid value.

Vedclass · Accepted Answer

$194$

Vedclass · Answer

(D) Let $y = \frac{5-x}{x^2-3x+2}$.$y(x^2-3x+2) = 5-x$$yx^2 - 3xy + 2y = 5-x$$yx^2 + (1-3y)x + (2y-5) = 0$.If $y=0$,then $x=5$,which is a valid value.If $y 
eq 0$,for $x$ to be real,the discriminant $D \geq 0$.$D = (1-3y)^2 - 4(y)(2y-5) \geq 0$$1 + 9y^2 - 6y - 8y^2 + 20y \geq 0$$y^2 + 14y + 1 \geq 0$.Solving $y^2 + 14y + 1 = 0$ using the quadratic formula:$y = \frac{-14 \pm \sqrt{196-4}}{2} = \frac{-14 \pm \sqrt{192}}{2} = \frac{-14 \pm 8\sqrt{3}}{2} = -7 \pm 4\sqrt{3}$.Thus,$y \in (-\infty, -7-4\sqrt{3}] \cup [-7+4\sqrt{3}, \infty)$.Here,$\alpha = -7-4\sqrt{3}$ and $\beta = -7+4\sqrt{3}$.$\alpha^2 + \beta^2 = (-7-4\sqrt{3})^2 + (-7+4\sqrt{3})^2$$= (49 + 48 + 56\sqrt{3}) + (49 + 48 - 56\sqrt{3})$$= 97 + 97 = 194$.

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 :

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