Question

Each of the four inequalities given below defines a region in the $XY$ plane. One of these $4$ regions does not have the following property. For any $2$ points $\left(x_{1,}{y}_1\right)$ and $\left(x_{2,}{y}_2\right)$ 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

• A. $x^2+2y^2\le 1$
• B. $Max\{\left|x\right|,\left|y\right|\}\le 1$
• C. $x^2-y^2\le 1$
• D. $y^2-x\le 0$

Answer 1

The correct option is D

$y^2-x\le 0$

Explanation for the correct option:

Consider Option $\left(D\right)$, $y^2–x\le 0$

which represents the inside region of a pair of straight lines.

The midpoint of any two-point in the region will lie in the region only.

It satisfies property P.

Explanation for the incorrect options:

1) Consider Option$\left(A\right)$, $x^2+2y^2\le 1$

It can be written as $\frac{x^2}{1}+\frac{y^2}{{\displaystyle \frac{1}{2}}}\le 1$ represents an inside region of an ellipse.

In an ellipse, the mid-point of any two points in the region is also in the region.

Let’s take two points $(0,0)$ and $(\frac{1}{2},0)$ which lie in the region.

The mid-point is $(\frac{1}{4},0)$ also lies in the region.

2)Consider Choice $\left(C\right)$,$x^2–y^2\le 1$

It represents the inside region of the hyperbola.

Take $(\frac{-1}{2},0)$ and $(\frac{1}{4},0)$ as two points.

The mid-point is $(\frac{-1}{8},0)$ is not in the region of the hyperbola.

Option $\left(C\right)$ does not satisfy property P.

Hence, correct option is $\left(D\right)$