Advanced book on Mathematics Olympiad

Real Analysis 519

This implies thatβ=f′(c)(c−α)+f(c), which shows thatM(α, β)lies on the tangent

to the graph offat(c, f (c)), and we are done.

420.Consider the functionF:[a, b]→R,

F(x)=f′(x)e−λf (x),λ∈R.

Becausefis twice differentiable,Fis differentiable. We haveF(a)=F(b), which by

Rolle’s theorem implies that there existsc∈(a, b)withF′(c)=0. But

F′(x)=e−λf (x)(f′′(x)−λ(f′(x))^2 ),

sof′′(c)−λ(f′(c))^2 =0. We are done.

(D. Andrica)

421.First solution: Let us assume that such numbers do exist. Ifx=yit follows that

x( 2 x+ 2 −x)= 2 x, which impliesx=y=0. This is impossible becausexandyare

assumed to be positive.

Hencexshould be different fromy. Letx 1 >x 2 >x 3 >0 be such thaty=x 1 −x 2

andx=x 2 −x 3. The relation from the statement can be written as

2 x^1 −x^2 − 1

1 − 2 x^3 −x^2

=

x 1 −x 2

x 2 −x 3

,

or

2 x^1 − 2 x^2

x 1 −x 2

=

2 x^2 − 2 x^3

x 2 −x 3

.

Applying the mean value theorem to the exponential, we deduce the existence of the
numbersθ 1 ∈(x 2 ,x 1 )andθ 2 ∈(x 3 ,x 2 )such that

2 x^1 − 2 x^2

x 1 −x 2

= 2 θ^1 ln 2,

2 x^2 − 2 x^3

x 2 −x 3

= 2 θ^2 ln 2.

But this implies 2θ^1 ln 2= 2 θ^2 ln 2, orθ 1 =θ 2 , which is impossible since the two numbers

lie in disjoint intervals. This contradiction proves the claim.

Second solution: DefineF(z)=( 2 z− 1 )/z. Note that by L’Hôpital’s rule, defining

F( 0 )=log 2 extendsFcontinuously toz=0. Rearrange the equality to give

F(−x)=

2 −x− 1

−x

=

2 y− 1

y

=F(y).