Animated Solution for Mathematics - Functions: Range of the function $f(x)=\frac{x^2+x+2}{x^2+x+1},x\in R$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, might seem like a standard, perhaps even tedious, exercise in finding the range of a rational function.

But I want you to pause. Before you reach for the discriminant method—that heavy-duty tool that often leads to messy calculations—I want you to look at the function $f(x)=\frac{x^2+x+2}{x^2+x+1}$ with the eyes of an artist.

What do you see? You see a numerator and a denominator that are almost identical. This is not a coincidence; it is an invitation.

In mathematics, as in life, sometimes the most complex problems yield to the simplest interventions. We have $f(x)=\frac{x^2+x+2}{x^2+x+1}$.

Let us perform a bit of algebraic surgery. We can rewrite the numerator $x^2+x+2$ as $(x^2+x+1)+1$.

Why would we do this? Because it allows us to split the fraction into two distinct parts:

Suddenly, the fog clears. The first term is simply $1$. Our function is now $f(x)=1+\frac{1}{x^2+x+1}$.

This is the beauty of simplification. We have transformed a daunting rational function into a simple transformation of a quadratic expression.

Now, let us focus our attention on the denominator, $g(x)=x^2+x+1$. This is a classic upward-opening parabola because the coefficient of $x^2$ is positive.

Since it opens upward, it possesses a minimum value at its vertex. To find this vertex, we use the formula $x=-\frac{b}{2a}$.

With $a=1$ and $b=1$, we find the minimum occurs at $x=-\frac{1}{2}$.

Let us calculate the value of $g(x)$ at this point:

So, the smallest value the denominator can ever take is $\frac{3}{4}$.

We are almost there. We know that $f(x)=1+\frac{1}{g(x)}$.

To find the maximum value of $f(x)$, we need to make the fraction $\frac{1}{g(x)}$ as large as possible. This happens when $g(x)$ is at its minimum.

Substituting $g_{min}=\frac{3}{4}$ into our expression, we get:

This is our ceiling. Now, what about the floor?

As $x$ grows very large in either the positive or negative direction, the denominator $x^2+x+1$ grows without bound. Consequently, the fraction $\frac{1}{x^2+x+1}$ shrinks toward zero.

Thus, $f(x)$ approaches $1+0=1$. Since the denominator is always positive, the fraction is always positive, meaning $f(x)$ is always strictly greater than $1$.

We have found our boundaries: the function is strictly greater than $1$ and reaches a maximum of $\frac{7}{3}$.

Therefore, the range is $(1,\frac{7}{3}]$. Take a moment to appreciate the elegance of this result. We didn't need to solve any complex inequalities; we simply observed, simplified, and analyzed. This is the JEE way.