LARGETNI

3/4?

how on earth do you get these crafty probs? Do you mind revealing the source?

@prophet sir
Harvard

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

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.

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

@prophet sir
http://web.mit.edu/hmmt/www/datafiles/