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Netherlands Contests
Dutch Mathematical Olympiad
2001 Dutch Mathematical Olympiad
2
f(x + y) = f(x) + f(y) + xy, f(4) = 10, f(2001)=?
f(x + y) = f(x) + f(y) + xy, f(4) = 10, f(2001)=?
Source: Dutch NMO 2001 p2
September 21, 2019
Functional Equations
algebra
Problem Statement
The function f has the following properties :
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
+
x
y
f(x + y) = f(x) + f(y) + xy
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
+
x
y
for all real
x
x
x
and
y
y
y
f
(
4
)
=
10
f(4) = 10
f
(
4
)
=
10
Calculate
f
(
2001
)
f(2001)
f
(
2001
)
.
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