If the range of the function f(x) = frac5 - xx^2 - 3x + 2$f(x) = \frac{5 - x}{x^2 - 3x + 2}$, x neq 1, 2$x \neq 1, 2$, is (-infty , alpha ] cup [ beta , infty)$(-\infty , \alpha ] \cup [ \beta , \infty)$, then alpha^2 +beta^2$\alpha^2 +\beta^2$ is equal to :

### Related Formula For a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ to yield real roots, its discriminant must satisfy: D = B^2 - 4AC ge 0$$D = B^2 - 4AC \ge 0$$ ### Core Logic Set y = frac5 - xx^2 - 3x + 2$y = \frac{5 - x}{x^2 - 3x + 2}$: y(x^2 - 3x + 2) = 5 - x$$y(x^2 - 3x + 2) = 5 - x$$ yx^2 - 3xy + 2y + x - 5 = 0$$yx^2 - 3xy + 2y + x - 5 = 0$$ Rearranging into a standard quadratic equation in terms of x$x$: yx^2 + (1 - 3y)x + (2y - 5) = 0$$yx^2 + (1 - 3y)x + (2y - 5) = 0$$ ### Step 1: Discriminant Method Case I: If y = 0$y = 0$, the equation simplifies to x - 5 = 0 implies x = 5$x - 5 = 0 \implies x = 5$, which is a valid part of the domain. Thus, 0$0$ belongs to the range. Case II: If y neq 0$y \neq 0$, for x$x$ to be real, D ge 0$D \ge 0$: (1 - 3y)^2 - 4(y)(2y - 5) ge 0$$(1 - 3y)^2 - 4(y)(2y - 5) \ge 0$$ 9y^2 + 1 - 6y - 8y^2 + 20y ge 0$$9y^2 + 1 - 6y - 8y^2 + 20y \ge 0$$ y^2 + 14y + 1 ge 0$$y^2 + 14y + 1 \ge 0$$ ### Step 2: Solving the Inequality Completing the square for y^2 + 14y + 1 ge 0$y^2 + 14y + 1 \ge 0$: (y + 7)^2 - 48 ge 0 implies (y + 7)^2 ge (4sqrt3)^2$$(y + 7)^2 - 48 \ge 0 \implies (y + 7)^2 \ge (4\sqrt{3})^2$$ This gives: y le -7 - 4sqrt3 quad textor quad y ge -7 + 4sqrt3$$y \le -7 - 4\sqrt{3} \quad \text{or} \quad y \ge -7 + 4\sqrt{3}$$ Comparing with the interval (-infty , alpha ] cup [ beta , infty)$(-\infty , \alpha ] \cup [ \beta , \infty)$: alpha = -7 - 4sqrt3$$\alpha = -7 - 4\sqrt{3}$$ beta = -7 + 4sqrt3$$\beta = -7 + 4\sqrt{3}$$ ### Step 3: Finding alpha^2 + beta^2 Using algebraic identities: alpha^2 + beta^2 = (-7 - 4sqrt3)^2 + (-7 + 4sqrt3)^2$$\alpha^2 + \beta^2 = (-7 - 4\sqrt{3})^2 + (-7 + 4\sqrt{3})^2$$ = 2(7^2 + (4sqrt3)^2) = 2(49 + 48) = 2(97) = 194$$= 2(7^2 + (4\sqrt{3})^2) = 2(49 + 48) = 2(97) = 194$$ ### Pattern Recognition For rational expressions of the form fractextLineartextQuadratic$\frac{\text{Linear}}{\text{Quadratic}}$, converting to a quadratic in x$x$ and forcing D ge 0$D \ge 0$ establishes the range boundaries elegantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions