Animated Solution for Mathematics - Sets and Relations: Consider the following relations in the set of real numbers $R$. $R=\{(x,y):x\in R,y\in R,x^2+y^2\le 25\}$ $R^{\prime }=\{(x,y):x\in R,y\in R,y\ge \frac{4}{9}{x}^2\}$ Find the domain and range of $R\cap R^{\prime }$. Is the relation $R\cap R^{\prime }$ a function?

Welcome, future engineer. Today, we are going to peel back the layers of a classic coordinate geometry problem. Many students look at a problem involving relations and immediately start scribbling algebraic equations. But I want you to pause.

Before we touch a single variable, let us build the mental image. Imagine you are standing on a coordinate plane. You have a circle, $x^2+y^2\le 25$, which is a beautiful, symmetric disk centered at the origin with a radius of $5$.

Now, overlay that with a parabola, $y\ge \frac{4}{9}{x}^2$. This parabola is a cup opening upwards, starting at the origin. The problem asks for the intersection, $R\cap R^{\prime }$. This is the 'sweet spot'—the region that is simultaneously inside the circle and above the parabola.

To find the boundaries of this region, we need to know exactly where these two shapes meet. We take the boundary equations: $x^2+y^2=25$ and $y=\frac{4}{9}{x}^2$.

We want to find the intersection points. The easiest path is substitution. From the parabola, we know that $x^2=\frac{9}{4}y$.

Let us take this expression and drop it into the circle's equation. Suddenly, the circle equation transforms:

If we multiply the entire equation by $4$ to clear the fraction, we get:

This is a standard quadratic equation. Factoring it, we find $(4y+25)(y-4)=0$. This gives us two potential values for $y$: $y=4$ and $y=-\frac{25}{4}$.

But here is where your intuition as a physicist must kick in. Look at the parabola $y=\frac{4}{9}{x}^2$. Since $x^2$ is always non-negative, $y$ must be non-negative. Therefore, we must reject the negative root. We are left with $y=4$.

Now that we have the height of the intersection, let us find the width. We substitute $y=4$ back into our parabola equation: $x^2=\frac{9}{4}(4)$, which simplifies to $x^2=9$.

Taking the square root, we get $x=\pm 3$. This tells us that the intersection points are $(-3,4)$ and $(3,4)$. This is the horizontal limit of our region.

The domain is the set of all $x$ values for which the region exists, which is the closed interval $[-3,3]$. The problem statement gives this as $16x^4+81x^2-2025\le 0$, which is just a fancy way of saying $x^2\le 9$.

Now, for the range. The range is the vertical spread. The lowest point of our region is the vertex of the parabola at $(0,0)$, so $y_{min}=0$. The highest point is the top of the circle at $(0,5)$, so $y_{max}=5$. Thus, our range is $[0,5]$.

Is it a Function?

Finally, we address the conceptual question: is this relation a function? Recall the definition of a function: every input $x$ must map to exactly one output $y$.

If you take a vertical line and sweep it across our region, what happens? At any $x$ between $-3$ and $3$, the vertical line hits the parabola at one point and the circle at another.

Because a single $x$ value corresponds to multiple $y$ values, the relation fails the Vertical Line Test. It is not a function.

You have successfully navigated the intersection of geometry and algebra. Remember, in JEE Advanced, the math is just the language; the visualization is the story. Keep practicing, keep visualizing, and the answers will reveal themselves.