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Iran Team Selection Test
2012 Iran Team Selection Test
2
Functional equation-We need pco!!!
Functional equation-We need pco!!!
Source: Iran TST 2012-First exam-2nd day-P5
April 24, 2012
function
induction
ceiling function
binomial coefficients
algebra proposed
algebra
Problem Statement
The function
f
:
R
≥
0
⟶
R
≥
0
f:\mathbb R^{\ge 0} \longrightarrow \mathbb R^{\ge 0}
f
:
R
≥
0
⟶
R
≥
0
satisfies the following properties for all
a
,
b
∈
R
≥
0
a,b\in \mathbb R^{\ge 0}
a
,
b
∈
R
≥
0
:a)
f
(
a
)
=
0
⇔
a
=
0
f(a)=0 \Leftrightarrow a=0
f
(
a
)
=
0
⇔
a
=
0
b)
f
(
a
b
)
=
f
(
a
)
f
(
b
)
f(ab)=f(a)f(b)
f
(
ab
)
=
f
(
a
)
f
(
b
)
c)
f
(
a
+
b
)
≤
2
max
{
f
(
a
)
,
f
(
b
)
}
f(a+b)\le 2 \max \{f(a),f(b)\}
f
(
a
+
b
)
≤
2
max
{
f
(
a
)
,
f
(
b
)}
.Prove that for all
a
,
b
∈
R
≥
0
a,b\in \mathbb R^{\ge 0}
a
,
b
∈
R
≥
0
we have
f
(
a
+
b
)
≤
f
(
a
)
+
f
(
b
)
f(a+b)\le f(a)+f(b)
f
(
a
+
b
)
≤
f
(
a
)
+
f
(
b
)
.Proposed by Masoud Shafaei
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