$\text{Each of the four inequalities given below defines a region in the xy plane. One of these four regions does not have }$

$\mathrm{the} \mathrm{following} \mathrm{property}.\text{ For any two points }\left(x_1,y_1\right)\text{ and }$$\left(x_2,y_2\right)\text{ in the region, the point }\left(\left(x_1+x_2\right)/2,\left(y_1+y_2\right)/2\right)\text{ is }$

$\text{ also in the region. The inequality defining this region is }$

$x^2+2y^2\le 1\text{ represents the interior region of a ellipse, where on taking any two points,}$

$\text{the midpoint of that segment will also lie inside that circle.}$

$\begin{array}{l}\mathrm{Max}\left\{\right|x|,|y\left|\right\}\le 1\\ \text{ or }\left|x\right|\le 1,\left|y\right|\le 1\\ \text{ or }-1\le x\le 1\text{ and }-1\le y\le 1\end{array}$

$\text{which represents the interior region of a square with its sides x=±1 and y =±1, in which for any two points, }$

$\mathrm{their} \mathrm{midpoint} \mathrm{also} \mathrm{lies} \mathrm{inside} \mathrm{the} \mathrm{region}.$

$x^2-y^2\le 1,\text{ represents the exterior region of a hyperbola in }$

$\text{ which we take two points }(4,3)\text{ and }(4,-3)\text{ . Then their mid- point }(4,0)\text{ does not lie in the same region (as shown in the figure)}$

$y^2\le x\text{ represents the interior region of a parabola in which for }$

$\text{ any two points, their midpoint also lies inside the region. }$