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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Functions

If the range of the function f(x)=(5-x)/(x^2-3 x+2) , x ≠ 1,2 , is...

If the range of the function f(x)=5−xx2−3x+2f(x)=\frac{5-x}{x^2-3 x+2}, x≠1,2x \neq 1,2, is (−∞,α]∪[β,∞)(-\infty, \alpha] \cup[\beta, \infty), then α2+β2\alpha^2+\beta^2 is equal to :

Options

  1. A190
  2. B192
  3. C188
  4. D194

Correct answer

D. 194

Step-by-step solution

y=5−xx2−3x+2yx2−3xy+2y+x−5=0yz2+(−3y+1)x+(2y−5)=0\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=0y=0 (Accepted)⇒x=5\Rightarrow x=5Case II : If y≠0y \neq 0D≥0(−3y+1)2−4(y)(2y−5)≥09y2+1−6y−8y2+20y≥0y2+14y+1≥0(y+7)2−48≥0∣y+7∣≥43⇒y+7≥43 or y+7≤−43⇒y≥43−7 or y≤−43−7\begin{aligned}& \mathrm{D} \geq 0 \\ & (-3 y+1)^2-4(y)(2 y-5) \geq 0 \\ & 9 y^2+1-6 y-8 y^2+20 y \geq 0 \\ & \mathrm{y}^2+14 y+1 \geq 0 \\ & (y+7)^2-48 \geq 0 \\ & |y+7| \geq 4 \sqrt{3} \\ & \Rightarrow y+7 \geq 4 \sqrt{3} \text{ or } \mathrm{y}+7 \leq-4 \sqrt{3} \\ & \Rightarrow \mathrm{y} \geq 4 \sqrt{3}-7 \text{ or } \mathrm{y} \leq-4 \sqrt{3}-7\end{aligned}From Case I and Case II$\mathrm{y} \in(-\infty,-4 \sqrt{3}-7] \cup[4 \sqrt{3}-7, \infty)$So $\alpha=-4 \sqrt{3}-7 \beta=4 \sqrt{3}-7\begin{aligned}\Rightarrow a^2+b^2 & =(-4 \sqrt{3}-7)^2+(4 \sqrt{3}-7)^2 \\ & =2(48+49) \\ & =194\end{aligned}$$$
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