Advanced book on Mathematics Olympiad

518 Real Analysis

416.Replacingfby−fif necessary, we may assumef(b)>f(c), hencef(a)>f(c)
as well. Letξbe an absolute minimum offon[a, b], which exists because the function
is continuous. Thenξ∈(a, b)and thereforef′(ξ )=0.

417.Consider the functionf:[ 2 ,∞)→R,f(x)=xcosπx. By the mean value theorem
there existsu∈[x, x+ 1 ]such thatf′(u)=f(x+ 1 )−f(x). The inequality from the
statement will follow from the fact thatf′(u) >1. Sincef′(u)=cosπu+πusinπu,we
have to prove that

cos

π

u

+

π

u

sin

π

u

> 1 ,

for allu∈[ 2 ,∞). Note thatf′′(u)=−π
2
u^3 cos

π

u<0, foru∈[^2 ,∞),sof

′is strictly

decreasing. This implies thatf′(u) >limv→∞f′(v)=1 for allu, as desired. The
conclusion follows.
(Romanian college admission exam, 1987)

418.Letαbe the slope of the line through the collinear points(ai,f(ai)),i= 0 , 1 ,...,n,
on the graph off. Then

f(ai)−f(ai− 1 )

ai−ai− 1

=α, i= 1 , 2 ,...,n.

From the mean value theorem it follows that there exist pointsci∈(ai− 1 ,ai)such that
f′(ci)=α,i= 1 , 2 ,...,n. Consider the functionF:[a 0 ,an]→R,F(x)=f′(x)−α.
It is continuous,(n− 1 )-times differentiable, and hasnzeros in[a 0 ,an]. Applying
successively Rolle’s theorem, we conclude thatF(n−^1 )=f(n)has a zero in[a, b], and
the problem is solved.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by G. Sire ̧tchi)

419.The functionsφ,ψ :[a, b]→R,φ(x)= f(x)x−α andψ(x)= x−^1 α satisfy the
conditions of Cauchy’s theorem. Hence there existsc∈(a, b)such that

φ(b)−φ(a)

ψ(b)−ψ(a)

=

φ′(c)

ψ′(c)

.

Replacingφandψwith their formulas gives

(a−α)f (b)−(b−α)f (a)

a−b

=f(c)−(c−α)f′(c).

On the other hand, sinceMlies on the line determined by(a, f (a)),(b, f (b)), the
coordinates ofMare related by

β=

(a−α)f (b)−(b−α)f (a)

a−b

.