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Friday 12 August 2011

Math GRE - #11

For all real numbers $x$ and $y$, which of the answers below is the expression \[\frac{x+y+|x-y|}{2}\] equal to?

  • The maximum of $x$ and $y$
  • The minimum of $x$ and $y$
  • $|x+y|$
  • The average of $|x|$ and $|y|$
  • The average of $|x+y|$ and $x-y$

Solution :

The expression is equal to the maximum of $x$ and $y$.
Suppose $x>y$. Then $|x-y|=x-y$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y+x-y}{2}=x.\]
Now suppose $x<y$. Then  $|x-y|=-(x-y)$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y-(x-y)}{2}=y.\]
\[\therefore\quad\frac{x+y+|x-y|}{2}=\max\{x,y\}.\]

Can you find a representation for $\min\{x,y\}$?

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