## BdMO National 2013: Higher Secondary 2

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Higher Secondary 2

Let \$g\$ be a function from the set of ordered pairs of real numbers to the same set such that \$g(x, y)=-g(y, x)\$ for all real numbers \$x\$ and \$y\$. Find a real number \$r\$ such that \$g(x, x)=r\$ for all real numbers \$x\$.

ahmedulkavi
Posts: 14
Joined: Tue Feb 01, 2011 11:20 am

### Re: BdMO National 2013: Higher Secondary 2

How can I solve this problem?

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### Re: BdMO National 2013: Higher Secondary 2

The statement essentially means that for any pair of real numbers \$(x,y)\$, we have \$g(x,y)=-g(y,x)\$. Note that, we can let \$x,y\$ be \$\textbf{any}\$ pair of real numbers. Substituting \$(x,x)\$ implies that,
\[g(x,x)=-g(x,x)\]
\[2g(x,x)=0\]
\[g(x,x)=0\]
Hence the problem is solved.

N.B. If you still have confusion in understanding, note that \$(x+y)^2=x^2+2xy+y^2\$. Plugging in \$(x,x)\$ implies \$(x+x)^2=x^2+2xx+x^2\$ or \$(2x)^2=x^2+2x^2+x^2\$ or \$4x^2=4x^2\$, which is true.
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