JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If the range of the function \(f(x)=\frac{5-x}{x^2-3 x+2}\), \(x \neq 1,2\), is \((-\infty, \alpha] \cup[\beta, \infty)\), then \(\alpha^2+\beta^2\) is equal to:
- A 190
- B 192
- C 188
- D 194
Answer & Solution
Correct Answer
(D) 194
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\frac{5-x}{x^2-3 x+2} \\ & y x^2-3 x y+2 y+x-5=0 \\ & y z^2+(-3 y+1) x+(2 y-5)=0 \end{aligned}\) Case I : If \(y=0\) (Accepted) \(\Rightarrow x=5\) Case II : If \(y \neq 0\)…
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