Range of Rational Functions Using the Discriminant

Introduction

In our last lesson, we introduced the discriminant method as a powerful tool for finding the range of rational functions. Today, we will focus exclusively on mastering this technique, which is a frequently tested concept in the JEE Advanced. Your goal is to become proficient in applying this method to various forms of rational expressions, including the critical special cases that often distinguish top performers.

This session is structured as a series of guided problems. We will start with standard applications and then move to a crucial subtlety: what happens when the expression you're analyzing temporarily stops being a quadratic.

The Discriminant Method: A Quick Recap

As we saw previously, to find the range of a rational function $f (x)$ for all real $x$ where it is defined, the core strategy is:

Let $y = f (x)$ .
Rearrange the equation to form a quadratic in $x$ , typically of the form $A (y) x^{2} + B (y) x + C (y) = 0$ .
For $x$ to be a real number, the discriminant of this quadratic, $D = B (y)^{2} - 4 A (y) C (y)$ , must be non-negative ( $D \geq 0$ ).
Solving this inequality for $y$ gives the set of all possible output values, which constitutes the range of the function.

Let's apply this to a couple of standard problem types.

Example 1: Linear Numerator, Quadratic Denominator

This is a common variant. Let's find the range of $f (x) = \frac{x + 2}{2 x ^{2} + 3 x + 6}$ . We set $y = f (x)$ , cross-multiply, and collect terms to form a quadratic in $x$ . Then we'll apply the $D \geq 0$ condition.

The BYJU'S article on Quadratic Equations provides a clear, step-by-step solution.

Quadratic Equation Definition - BYJU'S

This resource demonstrates the standard application of the discriminant method. Focus on how the equation is rearranged and how the condition D >= 0 is applied to find the range of 'y'.

In the article, locate "Example 4" under the main heading "How to Find the Range of a Quadratic Equation?". Study the full solution. Notice that solving the discriminant inequality is the final step.

The final step in that example required solving the quadratic inequality $39 y^{2} -10 y -1 \leq 0$ . As you know, this is done by finding the roots and using the wavy curve method. The image below shows the process for a similar inequality.

This image illustrates the wavy curve method for solving a quadratic inequality $2y^2 - 5y + 2 \le 0$. After factoring, the roots $1/2$ and $2$ are marked on a number line, and the sign of the expression is tested in each interval to find the solution set $y \in [1/2, 2]$.

Example 2: Quadratic Numerator, Quadratic Denominator

Now let's consider the case where both numerator and denominator are quadratic, such as $f (x) = \frac{x ^{2} + 14 x + 9}{x ^{2} + 2 x + 3}$ . The process remains the same.

Quadratic Equation Definition - BYJU'S

This example reinforces the same method for a slightly more complex rational function.

In the "Practice Questions for Solving Quadratic Equations" section, find "Question 2". Examine the provided solution.

The Critical Pitfall: The Vanishing Leading Coefficient

The discriminant method rests on the assumption that we are dealing with a quadratic equation in $x$ . But what if, for a certain value of $y$ , the coefficient of the $x^{2}$ term becomes zero? For that specific $y$ , the equation is no longer quadratic, and the $D \geq 0$ condition is meaningless. This is a classic trap in JEE Advanced problems.

When the coefficient of $x^{2}$ , $A (y)$ , becomes zero, you must investigate that value of $y$ separately.

Let $y_{0}$ be the value such that $A (y_{0}) = 0$ .
Substitute $y_{0}$ back into the rearranged equation $y Q (x) = P (x)$ .
The equation will now be linear in $x$ .
- If this linear equation has a valid solution for $x$ , then $y_{0}$ is part of the range.
- If it results in a contradiction (like $- 4 = 0$ ), then $y_{0}$ is not part of the range.

The following video provides an excellent explanation of this scenario. The instructor, Arvind Kalia, works through an example where this exact situation occurs.

Range of Function | One Shot | Refocus-Revise-Reward | Arvind Kalia Sir | Vedantu

This video by Vedantu JEE clearly explains the logic behind the discriminant method and, more importantly, how to handle the special case of the vanishing x² coefficient.

Watch the section from this timestamp. Pay close attention to the logic explained around y=1. The instructor shows that if the coefficients of both x² and x become zero simultaneously, it can lead to a contradiction (-1=0 in his example), forcing you to exclude that y value from the final range.

Practice Problems

Now it's time to apply the complete method, including the critical check for the vanishing $x^{2}$ coefficient.

Problem 1: Find the range of the function $f (x) = \frac{x ^{2} + x + 1}{x ^{2} - x + 1}$ .

Hint: Set up the equation $y = f (x)$ and rearrange it into a quadratic in $x$ . The coefficient of $x^{2}$ will be $(y - 1)$ . First, find the range using $D \geq 0$ . Then, separately test the case where $y = 1$ .

Problem 2: Find the range of $f (x) = \frac{x ^{2}}{x ^{2} - 4}$ .

Hint: The domain of $x$ is all real numbers except $x = \pm 2$ . Follow the standard procedure. The coefficient of $x^{2}$ will again be $(y - 1)$ . Check what happens when you substitute $y = 1$ back into your rearranged equation. Does it lead to a valid solution for $x$ or a contradiction?

Solution to Problem 2

Let $y = \frac{x ^{2}}{x ^{2} - 4}$ .
Cross-multiplying gives $y (x^{2} - 4) = x^{2}$ .
Rearranging into a quadratic in $x$ : $(y - 1) x^{2} - 4 y = 0$ .

Case 1: $y \neq = 1$
This is a quadratic equation. For $x$ to be real, we require $x^{2} \geq 0$ .
From the equation, $x^{2} = \frac{4 y}{y - 1}$ .
So we must solve the inequality $\frac{4 y}{y - 1} \geq 0$ .
The critical points are $y = 0$ and $y = 1$ . Using the wavy curve method, the solution is $y \in (- \infty, 0] \cup (1, \infty)$ .

Case 2: $y = 1$
We must check this case separately. The coefficient of $x^{2}$ is zero.
Substituting $y = 1$ into the rearranged equation $(y - 1) x^{2} - 4 y = 0$ gives:
$(1 - 1) x^{2} - 4 (1) = 0$
$0 \cdot x^{2} - 4 = 0$
$- 4 = 0$
This is a contradiction. Therefore, $y = 1$ is not a possible value.

Combining both cases, the range is $(- \infty, 0] \cup (1, \infty)$ .

Problem 3: Find the range of $y = \frac{x - 1}{x ^{2} - 2 x + 1}$ .

Hint: Be very careful with the domain of this function. Simplify the expression first. What does that tell you about the possible values of $y$ ?

Solution to Problem 3

The function is $y = \frac{x - 1}{( x - 1 ) ^{2}}$ . The domain is all real numbers except $x = 1$ .
For $x \neq = 1$ , the function simplifies to $y = \frac{1}{x - 1}$ .
We are now finding the range of a simpler function, $g (x) = \frac{1}{x - 1}$ .
As $x$ takes all real values except 1, the term $x - 1$ takes all real values except 0.
Therefore, its reciprocal, $y$ , can take any real value except 0.
The range is $(- \infty, 0) \cup (0, \infty)$ .

Conclusion

You have now done a deep dive into using the discriminant to find the range of rational functions. This is a core skill for JEE Advanced.

Key Takeaways:

The fundamental method is to rearrange $y = f (x)$ into a quadratic in $x$ and set the discriminant $D \geq 0$ .
The solution to the inequality $D \geq 0$ gives the range of $y$ .
Crucial Check: Always identify the value(s) of $y$ that make the $x^{2}$ coefficient zero. Test these values separately by substituting them back into the rearranged equation to see if they yield a valid solution or a contradiction.

In our next lesson, we will tackle the final topic in this module: solving problems with parametric quadratic equations. This involves using discriminant analysis to find the values of a parameter for which the roots of a quadratic equation satisfy specific conditions, a natural extension of the analytical skills we've sharpened today.

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