smc

First, note that the following inequality $$|x|+|y|\le \sqrt{2(x^2+y^2)}\le 2\text{mbox}Max(|x|,|y|)$$ represents the graphs $|x|+|y|=a$, $\sqrt{2(x^2+y^2)}=a$, and $2\text{mbox}Max(|x|,|y|)=a$. We don't actually have to worry about the inequality, we just want to find the picture that graphs those $3$ equations for some constant $a$. WLOG, we can set $a=1$. To make our life easier, we can also focus entirely on the first quadrant, since all $4$ graphs have different shapes in the first quadrant. Therefore, $x\ge 0$ and $y\ge 0$. First, we have $|x|+|y|=1$. Since $x\ge 0$ and $y\ge 0$, this becomes $x+y=1$. After a bit of rearranging, we get $$y=-x+1$$. This eliminates $III$, which does not have a diagonal line in the first quadrant. Next, we have $\sqrt{2(x^2+y^2)}=1$. Squaring both sides and dividing by $2$, we get $$x^2+y^2={\frac{\sqrt{2}}{2}}^2$$ Note that this describes a circle with the center $(0,0)$ and radius $\frac{\sqrt{2}}{2}$. We can find that $(\frac{1}{2},\frac{1}{2})$ is on the circle, which is also on the line $y=-x+1$. Therefore, the circle intersects the line at that point. We can graph this as follows: This eliminates $I$, which does not include the intersection point listed above. Finally, we have $2\text{mbox}Max(|x|,|y|)=1$. Rearranging and substituting $x$ for $|x|$ and likewise for $y$, we get $$\text{mbox}Max(x,y)=\frac{1}{2}$$ Note that this equation means that either $x\le y=\frac{1}{2}$ or $y\le x=\frac{1}{2}$. Therefore, from $x=0$ to $x=\frac{1}{2}$, the graph looks like the line $y=\frac{1}{2}$. Then, from $y=0$ to $y=\frac{1}{2}$, the graph looks like the line $x=\frac{1}{2}$. Graphing this, we get: Thus, we can eliminate $IV$. However, at this point, we don't know if the answer is $II$ or none of the graphs, since this entire time we were only graphing in the first quadrant to simplify things. However, note that no matter what sign $x$ or $y$ is, the equations all surround the variables with absolute value signs or exponents and make the values positive. Therefore, the graphs are symmetric about the $x$ and $y$ axes. As a result, the final graph looks like this: Therefore, we can choose $\boxed{II}$ as our answer.