If the range of the function , is , then is equal to : [2025]

Question

If the range of the function $f (x) = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$ , is $(- \infty, α] \cup [β, \infty)$ , then $α^{2} + β^{2}$ is equal to : [2025]

Smart Achievers · Accepted Answer

(2)

$y = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$

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

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

Case I : If y = 0

$\Rightarrow$ x = 5

Case II : if y $\neq$ 0

For real solutions, $D \geq 0$

$\Rightarrow {(- 3 y + 1)}^{2} - 4 (y) (2 y - 5) \geq 0$

$\Rightarrow 9 y^{2} + 1 - 6 y - 8 y^{2} + 20 y \geq 0$

$\Rightarrow y^{2} + 14 y + 1 \geq 0$

$\Rightarrow {(y + 7)}^{2} - 48 \geq 0$

$\Rightarrow | y + 7 | \geq 4 \sqrt{3} \Rightarrow y + 7 \geq 4 \sqrt{3} or y + 7 \leq - 4 \sqrt{3}$

$\Rightarrow y \geq 4 \sqrt{3} - 7 or y \leq - 4 \sqrt{3} - 7$

From Case I and Case II, we have

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

$∴ α = - 4 \sqrt{3} - 7 and β = 4 \sqrt{3} - 7$

$\Rightarrow α^{2} + β^{2} = {(- 4 \sqrt{3} - 7)}^{2} + {(4 \sqrt{3} - 7)}^{2} = 2 (48 + 49) = 194$ .

Solution

Q.
If the range of the function $f (x) = \frac{5 - x}{x^{2} - 3 x + 2}, x \neq 1, 2$ , is $(- \infty, α] \cup [β, \infty)$ , then $α^{2} + β^{2}$ is equal to : [2025]

1 190

2 194

3 188

4 192

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