### Archive

Posts Tagged ‘数学题’

## NOIP2010提高组初赛问题求解第三题推广

0—m—2m—…—m*[n/m]
1—m+1—2m+1—..—m*[n/m]+1
……
n%m—m+n%m—…—n

n%m+1—n%m+1+m—…—n-m+1

m-1—2m-1—…—n-n%m+1

k=([32/9]/2+1)*(32%9+1)+[(32/9]+1)/2]*(9-1-32%9)=2*6+2*3=18

Categories: OI Tags: , , , , ,

## A math problem concerned about monotonicity

Given that $f(x)$ is a monotonously increasing function and $g(x)=f(x)-f(1-x)$. Now if for certain real numbers $x_1$, $x_2$, $g(x_1)+g(x_2)>0$, compare $x_1+x_2$ and $1$.

Here we will prove that \$latex  x_1+x_2>1\$.

Obviously, when $x_1>=1-x_1$, $x_2>=1-x_2$ while the equalities don’t hold simultaneously, we can easily deduce the condition; at the same time we can get $x_1+x_2>1$.

While if $x_1<=1-x_1$ and $x_2<=1-x_2$, apparently we have

$f(x_1)<=f(1-x_1)$, $f(x_2)<=f(1-x_2)$, so $f(x_1)+f(x_2)<=f(1-x_1)+f(1-x_2)$, so $g(x_1)+g(x_2)<=0$.

Now consider if $x_1>=1-x_1$ and $x_2<=1-x_2$. Thus we have $x_1>=1/2$ and $x_2<=1/2$. So at the same time we have
$x_2<=x_1$ and $x_2<=1-x_2$

Now if $x_1+x_2<=1$, then $x_1<=1-x_2$, then $x_2<=1-x_1$.

So we have $x_2<=1-x_1<=x_1<=1-x_2$

So $f(x_2)<=f(1-x_1)<=f(x_1)<=f(1-x_2)$

So $f(x_2)+f(x_1)<=f(1-x_1)+f(1-x_2)$

Which indicates that the condition holds only if $x_1+x_2>1$.

Symmetrically we can prove the same when $x_1<=1-x_1$ and $x_2>=1-x_2$.