Question

Each of the four inequalities given below defines a region in the xy-plane. Let $P$ be the property that, for any two points $(x_1,y_1)$ and $(x_2,y_2)$ in the region, the point $\left(\frac{1}{2}(x_1+x_2),\frac{1}{2}(y_1+y_2)\right)$ is also in the region. Then which of them does not satisfy property $P$?

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

Answer 1

Answer: (C)

$x^2-y^2\le 1$

Option A, $x^2+2y^2\le 1$
It can be written as
$\frac{x^2}{1}+\frac{y^2}{1/2}\le 1$
which represents inside region of an ellipse
We know for 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 lies in the region.
Now, mid-point is $(\frac{1}{4},0)$ which also lies in the region .
Now, option B, $max\{|x|,|y|\}\le 1$
Now, option C, $x^2-y^2\le 1$
which represents inside region of hyperbola.
A hyperbola has two parts . If we take 2 points , one in one part and other in other part, the mid-point need not to be inside the hyperbola.
Let's take $(-\frac{1}{2},0)$ and $(\frac{1}{4},0)$ as two points.
Their mid-point is $(-\frac{1}{8},0)$ is not in the region of hyperbola.
Hence, option C does not satisfy property P.
Option D , $y^2-x^2\le 0$
which represents inside region of pair of straight lines.
Mid -point of any two point in the region will lie in the region only.
Hence, satisfies property P.