Question

If $f(x+y)=f\left(x\right)+f\left(y\right)-xy-3$ and $f\left(1\right)=3$, then the number of solutions for $f\left(n\right)=3n$, where $n\in N$ is

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

Putting $x=n$ and $y=1$
$f(n+1)=f\left(n\right)+f\left(1\right)-n\times 1-3\Rightarrow f(n+1)=f\left(n\right)-n<f\left(n\right)\Rightarrow f\left(n\right)>f(n+1)\Rightarrow f\left(1\right)>f\left(n\right)$ for $n>1$
$\Rightarrow 3>3n$, which is not possible for $n>1$

So, there is no solution for $n>1$.
The only possible solution is $n=1$