Each of the four inequalities given below defines a region in the xy plane. One of these four region

Q: Wait, why do we reject the hyperbola? It looks like a valid region.

The hyperbola has a 'dent' or a concave region. The midpoint of two points on the same branch can fall into the gap between the branches, which is outside the region. This 'gap' is exactly what makes it non-convex.

Q: Is a square always convex? How can I be sure?

Yes, a square is a polygon, and all polygons without dents are convex. The midpoint of any two points in a square is always inside the square because the boundary is straight, not curved inward.

Q: Do I need to check every single point in the region to prove it's non-convex?

Absolutely not! That would be impossible. You only need one counter-example. If you find just one pair of points whose midpoint lies outside, you have successfully proven the region is non-convex.

Question

Each of the four inequalities given below defines a region in the xy plane. One of these four regions does not have the following property: For any two points $(x_1, y_1)$ and $(x_2, y_2)$ in the region, the point $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ is also in the region. The inequality defining this region is

Options:

(A) $x^2 + 2y^2 \le 1$

(B) Max $\{|x|, |y|\} \le 1$

(C) $x^2 - y^2 \le 1$

(D) $y^2 - x \le 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The condition given—that for any two points $(x_1, y_1)$ and $(x_2, y_2)$ in a region, their midpoint

$$M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$$

must also be in the region—is the formal definition of a **convex set**.

Imagine you are standing inside a shape. If you can draw a straight line to any other point in that shape without ever stepping outside the boundary, that shape is convex. If you have to 'jump' over a dent or a hole, it is non-convex. We are hunting for the shape with a 'dent.'

### Evaluating the Candidates

First, we have $x^2 + 2y^2 \le 1$. This is the interior of an ellipse. Ellipses are perfectly smooth, rounded, and convex. No matter where you pick two points, the line connecting them stays safely inside.

Next, consider $\max\{|x|, |y|\} \le 1$. This inequality is equivalent to $|x| \le 1$ and $|y| \le 1$, which defines a square centered at the origin. A square is a polygon, and all polygons without inward dents are convex.

Finally, consider $y^2 - x \le 0$, which rearranges to $y^2 \le x$. This is the interior of a parabola opening to the right. Like the ellipse, it curves outward, making it a perfectly convex region.

### The Culprit: The Hyperbola

Now, we arrive at our target: $x^2 - y^2 \le 1$. This represents the region containing the origin, bounded by a rectangular hyperbola.

Notice the shape of the boundary—it curves inward toward the origin. This is our 'dent.' To prove it fails the midpoint property, we need a counter-example.

Let's pick two points on the right branch of the hyperbola: $P_1(1.2, 1)$ and $P_2(1.2, -1)$.

### Verifying the Failure

First, let's verify they are in the region. For $P_1$, we have:

$$1.2^2 - 1^2 = 1.44 - 1 = 0.44 \le 1$$

The same holds for $P_2$. Both points are inside the region. Now, let's find their midpoint $M$:

$$M = \left( \frac{1.2 + 1.2}{2}, \frac{1 + (-1)}{2} \right) = (1.2, 0)$$

### The Moment of Truth

Now, we test $M(1.2, 0)$ against the inequality $x^2 - y^2 \le 1$. Substituting the values, we get:

$$1.2^2 - 0^2 = 1.44$$

Since $1.44 > 1$, the midpoint $M$ lies outside the region. The line segment connecting our two points has crossed the boundary and entered the 'gap' between the hyperbola's branches.

This confirms that the region defined by **$x^2 - y^2 \le 1$** is **non-convex**. Keep this intuition of 'convexity' in your toolkit—it is a powerful way to visualize inequalities in the JEE Advanced exam!

Select Answer:

Visualized Solution

The Midpoint Property

Geometric Meaning: Convexity

Option A: $x^{2} + 2 y^{2} \leq 1$

Option B: $max {∣ x ∣, ∣ y ∣} \leq 1$

Option D: $y^{2} - x \leq 0$

Option C: $x^{2} - y^{2} \leq 1$

Selecting Test Points

The Midpoint Test

Midpoint is Outside!

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Evaluating the Candidates

The Culprit

Verifying the Failure

The Moment of Truth

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