3413/4?
how on earth do you get these crafty probs? Do you mind revealing the source?
341
Hari Shankar
·2011-03-08 01:18:30
you are right. i only referred to the lower bound. The max is of course 1.
@kunl - i didnt get the Harvard reference. Could you be more specific? Are you referring to the Harvard-MIT tournament? The AOPS site only has the 2008 tournament, and that did not feature this prob
1The maximum value of " f " is " 1 " , hence , it's evident that ,
0∫1 f ( x ) dx ≤ 0∫1 1 dx ≤ 1 ......... [ 1 ]
Since " 1 " is attainable by " f " , and " f ( f ( x ) ) = 1 " ;
f ( 1 ) = 1 ............. [ 2 ]
Let " m " be the smallest value of " f " . , so that ,
f ( m ) = 1 ............. [ 3 ]
Now ,
0∫1 f ( x ) dx
= 0∫m f ( x ) dx + m∫1 f ( x ) dx
= 0∫m f ( x ) dx + 1 - m...............( combining [ 2 ] and [ 3 ] )
> m ( m - 0 ) + 1 - m
> m 2 - m + 1
> ( m - 12 ) 2 + 34 > 34 ................... [ 4 ]
From [ 1 ] and [ 4 ] , we get the answer .
341
Hari Shankar
·2011-03-08 02:08:59
Solution:(a lil more complete)
Let R be the range of f, with R being a subset of [0,1]
If x ε R, then f(x)=1.
Since f is a continuous function defined over a closed interval, we have
0≤m≤f(x)≤M≤1, i.e. it attains both its max and min. Here obviously M=1.
Further, f attains all values between m and 1.
Hence for 0≤m≤x≤1, we have f(x)=1.
Therefore 0∫1f(x) dx = 0∫mf(x) dx+m∫1f(x) dx
= 0∫mf(x) dx + 1-m
Now we have by the well known inequality that
m(m-0)≤0∫mf(x) dx≤1(m-0)
So that m2-m+1≤0∫1f(x) dx≤1
We further have m2-m+1≥3/4 with equality when m=1/2
Now, here we note that since f is continuous, the left bound is never attained, though it can be brought as close as we please. Drawing a graph of this function can help see why this is true. A function that is almost asymptotically equal to 1 at x=m, will have the value of the integral close to m2-m+1.
1how 0≤m≤x≤1, f(x)=1 ???????
341
Hari Shankar
·2011-03-12 03:38:59
If R is the range of f, then f(f(x))=1 means that f(R)=1.
Now, if you set aside that f is continuous, then you will imagine that if R is some set of points in [0,1], then at all these points the value of f is 1.
The fact that f is continuous simplifies things by making this set of points a closed interval. This is because:
(a) a continuous function in a closed interval is bounded above and below and in fact attains those bounds. Call these two bounds as m and M. Here since 1 ε R, we simply have M=1
(b) Further, by Mean Value Theorem, the range will consist of every value between m and 1.
That means R =[m,1]
Then as a corollary we have that for all x \in [m,1], f(x)=1
1
kunl
·2011-03-22 23:04:46
@prophet sir
http://web.mit.edu/hmmt/www/datafiles/
