If the range of the function f(x) = 5-x/x²−3x+2, x ≠ 1,2, is (−∞, α] ∪ [β, ∞), then α² + β² is equal to:

 Question

If the range of the function

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

is $(-\infty, \alpha] \cup [\beta, \infty)$, then find the value of $\alpha^{2}+\beta^{2}$.

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Question Image

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Short Solution

1. Let $y = \dfrac{5 - x}{x^{2} - 3x + 2}$'
   Multiply through to get:

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

2. Rearrange to a quadratic in $x$:

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

   or

   $$y{x}^2+(-3y+1)x+(2y-5)=0$$

3. For $x$ to be real, discriminant $\Delta$ must be non-negative:

   $$\mathrm{\Delta }=(-3y+1{)}^2-4y(2y-5)\ge 0$$

4. Simplify $\Delta$:

   $$\mathrm{\Delta }=9y^2-6y+1-8y^2+20y=y^2+14y+1$$

5. Solve $\Delta =0$:

   $$y = \frac{-14\pm \sqrt{196-4}}{2} = \frac{-14\pm \sqrt{192}}{2} = \frac{-14\pm 8\sqrt3}{2}$$ $$y = -7 \pm 4\sqrt3$$
   

6. Because the quadratic in $x$ is valid for $\Delta\ge0$, the range of $y$ is:

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

   So:

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

7. Compute $\alpha^{2}+\beta^{2}$:

   $${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$$

   • $\alpha+\beta = -14$
     

   • $\alpha\beta = (-7)^{2}-(4\sqrt3)^{2} =49-48=1$
     

   Therefore:

   $${\alpha }^2+{\beta }^2=196-2(1)=196-2=194$$

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Image Solution

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Conclusion

For the function

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

• The range is:

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

• Here:

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

So:

$$\boxed{{\displaystyle {\alpha }^2+{\beta }^2=\mathrm{194}}}$$

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Video Solution