Advanced book on Mathematics Olympiad

544 Real Analysis

We now recognize a perfect square and write this as

∫ 1

0

(

f(x)−

x

2

) 2

dx= 0.

The integral of the nonnegative continuous function(f (x)−x 2 )^2 is strictly positive, unless
the function is identically equal to zero. It follows that the only function satisfying the
condition from the statement isf(x)=x 2 ,x∈[ 0 , 1 ].
(Revista de Matematic ̆a din Timi ̧soara(Timi ̧soara Mathematics Gazette), proposed
by T. Andreescu)

474.Performing the substitutionx

(^1) k
=t, the given conditions become
∫ 1
0
(f (t))n−ktk−^1 dt=

1

n

,k= 1 , 2 ,...,n− 1.

Observe that this equality also holds fork=n. With this in mind we write

∫ 1

0

(f(t)−t)n−^1 dt=

∫ 1

0

n∑− 1

k= 0

(

n− 1

k

)

(− 1 )k(f (t))n−^1 −ktkdt

=

∫ 1

0

∑n

k= 1

(

n− 1

k− 1

)

(− 1 )k−^1 (f (t))n−ktk−^1 dt

=

∑n

k= 1

(− 1 )k−^1

(

n− 1

k− 1

)∫ 1

0

(f (t))n−ktk−^1 dt

=

∑n

k= 1

(− 1 )k−^1

(

n− 1

k− 1

)

1

n

=

1

n

( 1 − 1 )n−^1 = 0.

Becausen−1 is even,(f (t)−t)n−^1 ≥0. The integral of this function can be zero only if
f(t)−t=0 for allt∈[ 0 , 1 ]. Hence the only solution to the problem isf:[ 0 , 1 ]→R,
f(x)=x.
(Romanian Mathematical Olympiad, 2002, proposed by T. Andreescu)

475.Note that the linear functiong(x)= 6 x−2 satisfies the same conditions asf.
Therefore,
∫ 1

0

(f (x)−g(x))dx=

∫ 1

0

x(f (x)−g(x))dx= 0.

Considering the appropriate linear combination of the two integrals, we obtain

∫ 1

0

p(x)(f (x)−g(x))dx= 0.