Animated Solution for Mathematics - Functions: If the range of the function $f(x)=\frac{5-x}{x^2-3x+2},x\ne 1,2$, is $(-\infty ,\alpha ]\cup [\beta ,\infty )$, then ${\alpha }^2+{\beta }^2$ is equal to :

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel the mystery of a rational function.

You might look at $f(x)=\frac{5-x}{x^2-3x+2}$ and feel a bit intimidated by the denominator. But remember, in the world of JEE, every complex expression is just a puzzle waiting for the right key.

Our goal is to find the range—the set of all possible values that $y$ can take as $x$ varies across all real numbers (excluding the points where the function is undefined).

To find the range, we need to bridge the gap between $x$ and $y$. We start by setting:

Now, let's perform the magic of cross-multiplication. We get $y(x^2-3x+2)=5-x$.

Expanding this, we arrive at $y{x}^2-3yx+2y=5-x$. Bringing everything to one side, we form a beautiful quadratic equation in $x$:

This is the heart of our solution. We have transformed a function into an equation where $x$ is the variable and $y$ acts as a parameter.

Here is the crucial realization: for any value of $y$ to be in the range, there must exist at least one real value of $x$ that satisfies this equation.

And how do we guarantee real roots for a quadratic equation? We look to the discriminant, $D=B^2-4AC$. For real roots, we must have $D\ge 0$.

Let's identify our coefficients: $A=y$, $B=-(3y-1)$, and $C=(2y-5)$. Substituting these into our condition, we get:

Expanding this, we find $(9y^2-6y+1)-(8y^2-20y)\ge 0$, which simplifies elegantly to:

Now, we solve the inequality $y^2+14y+1\ge 0$. The roots of $y^2+14y+1=0$ are given by the quadratic formula:

Since the inequality is $\ge 0$, the solution is $y\in (-\infty ,-7-4\sqrt{3}]\cup [-7+4\sqrt{3},\infty )$.

Comparing this to the given range $(-\infty ,\alpha ]\cup [\beta ,\infty )$, we identify $\alpha =-7-4\sqrt{3}$ and $\beta =-7+4\sqrt{3}$.

Finally, we calculate ${\alpha }^2+{\beta }^2$. Using the identity $(a-b{)}^2+(a+b{)}^2=2(a^2+b^2)$, we get:

We have conquered the problem! The final answer is 194. Keep this elegance in your toolkit, and you will face any JEE problem with confidence.