Animated Solution for Mathematics - Functions: Let the function defined in column I have domain $(-\frac{\pi }{2},\frac{\pi }{2})$ and range $(-\infty ,\infty )$.

Understanding Mappings

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are stepping into a playground of functions.

Imagine you are standing on a coordinate plane, looking at two specific paths: a straight line, $f(x)=1+2x$, and the sweeping curve of $f(x)=\tan x$. Our mission is to understand how these functions behave within a restricted domain: $(-\frac{\pi }{2},\frac{\pi }{2})$.

Before we dive into the algebra, let's ground ourselves in the definitions. A function is a machine that takes an input and gives an output.

We are interested in the nature of this machine. Is it 'one-one' (injective)? This means every input has a unique output—no two different $x$ values share the same $y$.

Is it 'onto' (surjective)? This means every single value in our target co-domain, $(-\infty ,\infty )$, is reached by the function. If we miss even a single point, the function is not onto.

Analyzing $f(x)=1+2x$

Let's start with the first machine: $f(x)=1+2x$. This is a linear function.

To test if it is one-one, we look at its rate of change. We calculate the derivative:

Since $2>0$, the function is strictly increasing. It is always climbing, never looking back. Because it is strictly increasing, it can never hit the same $y$-value twice. Therefore, it is definitively one-one.

Now, for the 'onto' test. We need to see if the range of this function covers the entire co-domain $(-\infty ,\infty )$.

We evaluate the function at the boundaries of our domain $(-\frac{\pi }{2},\frac{\pi }{2})$. At the lower bound, we get:

At the upper bound, we get:

So, the range of our function is the interval $(1-\pi ,1+\pi )$. Compare this to the co-domain $(-\infty ,\infty )$.

The range is just a small segment of the real number line! It fails to cover the infinite expanse of the co-domain. Thus, $f(x)=1+2x$ is one-one, but not onto.

Analyzing $f(x)=\tan x$

Now, let's turn our attention to the more elegant curve: $f(x)=\tan x$. This function is a classic in the JEE syllabus.

To check if it is one-one, we again look at the derivative:

Within our domain $(-\frac{\pi }{2},\frac{\pi }{2})$, the value of ${\sec }^{2}x$ is always positive. Because the derivative is always positive, the function is strictly increasing. Just like our line, it never turns back, so it is one-one.

But what about the 'onto' property? This is where the magic happens.

As $x$ approaches $-\frac{\pi }{2}$ from the right, $\tan x$ dives down to $-\infty$. As $x$ approaches $\frac{\pi }{2}$ from the left, $\tan x$ shoots up to $+\infty$.

Because the function is continuous and covers every value between these two extremes, its range is $(-\infty ,\infty )$. This perfectly matches our co-domain! Because the range equals the co-domain, the function is onto.

We have successfully dissected both functions. The linear function $f(x)=1+2x$ is one-one but not onto, matching option (q). The trigonometric function $f(x)=\tan x$ is both one-one and onto, matching option (r).

Take a moment to appreciate this. We used the power of calculus—the derivative—to determine the behavior of these functions without even needing to draw them.

We used the concept of limits to understand their reach. This is the essence of JEE Advanced mathematics: taking complex, abstract definitions and using them as tools to reveal the hidden structure of the problem.

You have mastered the playground today. Keep this analytical mindset, and no function will ever be too intimidating for you.