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Problems
Contests
National and Regional Contests
Switzerland Contests
Switzerland Team Selection Test
1998 Switzerland Team Selection Test
10
10
Part of
1998 Switzerland Team Selection Test
Problems
(1)
f(x+3/42)+ f(x) = f(x+1/6)+f\left(x+1/7), periodic
Source: Switzerland - Swiss TST 1998 p10
2/19/2020
5. Let
f
:
R
→
R
f : R \to R
f
:
R
→
R
be a function that satisfies for all
x
∈
R
x \in R
x
∈
R
(i)
∣
f
(
x
)
∣
≤
1
| f(x)| \le 1
∣
f
(
x
)
∣
≤
1
, and (ii)
f
(
x
+
13
42
)
+
f
(
x
)
=
f
(
x
+
1
6
)
+
f
(
x
+
1
7
)
f\left(x+\frac{13}{42}\right)+ f(x) = f\left(x+\frac{1}{6}\right)+f\left(x+\frac{1}{7}\right)
f
(
x
+
42
13
)
+
f
(
x
)
=
f
(
x
+
6
1
)
+
f
(
x
+
7
1
)
Prove that
f
f
f
is a periodic function
periodic
function
functional