MathDB
Problems
Contests
National and Regional Contests
Turkey Contests
National Olympiad First Round
1998 National Olympiad First Round
32
32
Part of
1998 National Olympiad First Round
Problems
(1)
f(x) + f(y) = f(x)f(y) + 1 - 1/xy
Source: 0
4/23/2009
For every x,y\in \Re ^{\plus{}}, the function f: \Re ^{\plus{}} \to \Re satisfies the condition f\left(x\right)\plus{}f\left(y\right)\equal{}f\left(x\right)f\left(y\right)\plus{}1\minus{}\frac{1}{xy}. If
f
(
2
)
<
1
f\left(2\right)<1
f
(
2
)
<
1
, then
f
(
3
)
f\left(3\right)
f
(
3
)
will be
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
/
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
4
/
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
More information needed
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
There is no
f
satisfying the condition above.
<span class='latex-bold'>(A)</span>\ 2/3 \\ <span class='latex-bold'>(B)</span>\ 4/3 \\ <span class='latex-bold'>(C)</span>\ 1 \\ <span class='latex-bold'>(D)</span>\ \text{More information needed} \\ <span class='latex-bold'>(E)</span>\ \text{There is no } f \text{ satisfying the condition above.}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2/3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
4/3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
More information needed
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
There is no
f
satisfying the condition above.
function