1. Given function

We need the range of

$$f(x)=\frac{5-x}{x^2-3x+2}, \qquad x\ne 1,2.$$

Factor the denominator:

$$x^2-3x+2=(x-1)(x-2).$$

So

$$y=\frac{5-x}{x^2-3x+2}.$$

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2. Find range by solving for $x$ in terms of $y$

Let

$$y=\frac{5-x}{x^2-3x+2}.$$

Then

$$y(x^2-3x+2)=5-x.$$

Rearrange:

$$yx^2-3yx+2y-5+x=0.$$

So as a quadratic in $x$,

$$yx^2+(1-3y)x+(2y-5)=0.$$

For real $x$, this quadratic must have real solution(s). Hence its discriminant must be non-negative.

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3. Discriminant condition

Discriminant:

$$\Delta=(1-3y)^2-4y(2y-5).$$

Compute:

$$(1-3y)^2=1-6y+9y^2,$$ $$4y(2y-5)=8y^2-20y.$$

Thus

$$\Delta=1-6y+9y^2-(8y^2-20y)=y^2+14y+1.$$

For real $x$,

$$y^2+14y+1\ge 0.$$

Solve the quadratic inequality:

$$y^2+14y+1=0.$$

Roots are

$$y=\frac{-14\pm\sqrt{196-4}}{2}=\frac{-14\pm\sqrt{192}}{2}=-7\pm 4\sqrt{3}.$$

Therefore,

$$y\le -7-4\sqrt{3} \quad \text{or} \quad y\ge -7+4\sqrt{3}.$$

So the range is

$$(-\infty,-7-4\sqrt{3}]\cup[-7+4\sqrt{3},\infty).$$

Hence

$$\alpha=-7-4\sqrt{3}, \qquad \beta=-7+4\sqrt{3}.$$

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4. Compute $\alpha^2+\beta^2$

Use

$$(a-b)^2+(a+b)^2=2(a^2+b^2).$$

Here $a=-7$, $b=4\sqrt{3}$:

$$\alpha^2+\beta^2 = (-7-4\sqrt{3})^2+(-7+4\sqrt{3})^2$$ $$=2\left( (-7)^2+(4\sqrt{3})^2 \right)$$ $$=2(49+48)=2\cdot 97=194.$$

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5. Option check

The value is

$$\alpha^2+\beta^2=194.$$

So the correct option is:

D: 194

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6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.