Question

A function $f\left(x\right)$ is defined for all real $x.$ If $f(a+b)=f\left(ab\right)\forall a,b$ and $f(-\frac{1}{2})=-\frac{1}{2}$, then $f\left(2003\right)$ is equal to

• A. -2003
• B. 2003
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $-\frac{1}{2}$

Let $f\left(0\right)=K,$ and $a=0$
We get $f\left(b\right)=f\left(0\right)=K$ and again if $b=0$ then $f\left(a\right)=K$
$\Rightarrow f\left(a\right)=f\left(b\right)=K\forall a,b\Rightarrow f\left(x\right)$ is a constant function.
Hence, $f\left(2003\right)=-\frac{1}{2}$