2007 Cyprus MO/Lyceum/Problem 12

Problem

The function $f : \mathbb{R} \rightarrow \mathbb{R}$ has the properties $f(0) = -1$ and $f(xy)+f(x)+f(y)=x+y+xy+k\ \ \ \forall x,y \in \Re$, where $k \in \Re$ is a constant. The value of $f(-1)$ is

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } -2\qquad \mathrm{(E) \ } 3$

Solution

First, to determine the value of $k$, let $x=y=0$.

$f(0\cdot0)+f(0)+f(0)=0+0+0\cdot0+k$, so $\displaystyle k = (-1)+(-1)+(-1) = - 3$.

Now, to determine the value of $\displaystyle f(-1)$, let $x=-1$ and $y=0$.

$\displaystyle f(-1\cdot0)+f(-1)+f(0)=-1+0+0\cdot0-3$

$\displaystyle (-1)+f(-1)+(-1)=-4$

$\displaystyle f(-1)=-2\Longrightarrow\mathrm{ D}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 11
Followed by
Problem 13
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