Animated Solution for Mathematics - Functions: Let $f(x)=x^2,x\in R$. For any $A\subseteq R$, define $g(A)=\{x\in R,f(x)\in A\}$. If $S=[0,4]$, then which one of the following statements is not true ?

A Journey into Pre-images

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are dissecting the very anatomy of functions.

Many students view functions as simple machines: you put an $x$ in, you get an $f(x)$ out. But in JEE Advanced, we must look deeper. We must understand the relationship between the domain and the range, and specifically, the fascinating, often counter-intuitive world of pre-images.

Decoding the Operator

Let us look at our function $f(x)=x^2$. It is the classic parabola, symmetric about the y-axis. But the star of our show today is the operator $g(A)=\{x\in \mathbb{R}:f(x)\in A\}$.

Do not let the notation intimidate you. This is simply the definition of a pre-image. When we say $g(A)$, we are asking: "If I have a target set $A$ on the y-axis, which values on the x-axis are responsible for landing inside that target?"

It is a backward-looking operation. While $f$ maps domain to range, $g$ maps range back to domain. Understanding this directionality is the key to unlocking the entire problem.

The Geometry of $g(S)$

We are given the set $S=[0,4]$. This is our target zone on the y-axis. To find $g(S)$, we need to solve the inequality $f(x)\in [0,4]$.

Mathematically, this translates to:

Imagine the parabola $y=x^2$. We are looking for all $x$ such that the height of the curve is between $0$ and $4$. If you visualize the graph, you will see that the curve hits the height of $4$ at $x=2$ and $x=-2$.

Because the parabola is symmetric, every value between $-2$ and $2$ will result in a $y$-value between $0$ and $4$. Thus, $g(S)=[-2,2]$.

Notice how the set expanded! We started with a set of length $4$ on the y-axis, and our pre-image is a set of length $4$ on the x-axis. This symmetry is the heartbeat of the parabola.

The Forward vs. Backward Dance

Now, let us test the options. We need to compare $f(g(S))$ and $f(S)$.

First, consider $f(g(S))$. We take our pre-image $g(S)=[-2,2]$ and pass it through the function $f(x)=x^2$. Since $x$ ranges from $-2$ to $2$, the square $x^2$ ranges from $0$ to $4$.

So, $f(g(S))=[0,4]$. This is exactly $S$. Therefore, the statement $f(g(S))=S$ is true.

Next, let us look at $f(S)$. Here, $S$ is the input. We are taking the interval $[0,4]$ on the x-axis and squaring it.

Since the function is strictly increasing on this interval, the minimum is $0^2=0$ and the maximum is $4^2=16$. Thus, $f(S)=[0,16]$.

Comparing these, we see that $f(g(S))=[0,4]$ and $f(S)=[0,16]$. They are clearly not equal. This confirms that the statement $f(g(S)) eq f(S)$ is true.

The Final Verdict

Finally, we evaluate $g(f(S))$. We know $f(S)=[0,16]$. Now we find the pre-image of this new set by solving:

Taking the square root, we get $-4\le x\le 4$. So, $g(f(S))=[-4,4]$.

Let us check the remaining options: 1. Is $g(f(S)) eq S$? Yes, because $[-4, 4] eq [0, 4]$. This statement is true. 2. Is $g(f(S))=g(S)$? We have $g(f(S))=[-4,4]$ and $g(S)=[-2,2]$.

These are definitely not equal. Therefore, the statement $g(f(S))=g(S)$ is False.

We have navigated the landscape of sets and functions. We saw how the pre-image operator $g$ acts as a "widening" lens, capturing both the positive and negative roots of the parabola.

We saw how the forward image $f$ can expand a set significantly. By systematically calculating each set, we stripped away the complexity and found the imposter statement.

Remember, in JEE Advanced, the math is rarely just about calculation; it is about visualizing the transformation of sets. Keep practicing this visualization, and you will master the art of functions!