mathematics olympiad 4
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Solusi
Solusi OSN Matematika 2014 Part 4
f (x) =2
2 + 4x
maka:
f( a
2014
)=
2
2 + 4a
2014
f
(2014− a
2014
)=
2
2 + 42014−a2014
f( a
2014
)+ f
(2014− a
2014
)=
2
2 + 4a
2014
+2
2 + 42014−a2014
Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 1 / 1
Solusi
f( a
2014
)+ f
(2014− a
2014
)=
2
2 + 4a
2014
+2
2 + 42014−a2014
=2(2 + 4
2014−a2014
)+ 2
(2 + 4
a2014
)(2 + 4
a2014
)(2 + 4
2014−a2014
)
=4 + 2
(4
2014−a2014
)+ 4 + 2
(4
a2014
)4 + 2
(4
2014−a2014
)+ 2
(4
a2014
)+(4
2014−a2014
)(4
a2014
)
Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 2 / 1
Solusi
f( a
2014
)+ f
(2014− a
2014
)=
8 + 2(4
2014−a2014
)+ 2
(4
a2014
)4 + 2
(4
2014−a2014
)+ 2
(4
a2014
)+ 4(
2014−a2014
+ a2014)
=8 + 2
(4
2014−a2014
)+ 2
(4
a2014
)4 + 2
(4
2014−a2014
)+ 2
(4
a2014
)+ 41
=8 + 2
(4
2014−a2014
)+ 2
(4
a2014
)8 + 2
(4
2014−a2014
)+ 2
(4
a2014
)= 1
Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 3 / 1
Solusi
f( a
2014
)+ f
(2014− a
2014
)= 1
Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 4 / 1
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