CAT 2016 Quantitative Ability

(x + 1) x f(x + 1) + x x f(x) + (x — 1) x f(x — 1)= 0 ...(i)
In the above equation, replacing x by x - 1 , we get
x × f(x) + ( x - 1) × f ( x - 1) + (x -2 ) × f(x -2 ) = 0 ...(ii)
From equations (i) and (ii), we get
$\left(\mathit{x}+1\right)\times \mathit{f}\left(\mathit{x}+1\right)=\left(\mathit{x}-2\right)\times \mathit{f}\left(\mathit{x}-2\right)$ ...(iii)
Replacing x by x + 2 in equation (iii), we get
$\mathit{x}\times \mathit{f}\left(\mathit{x}\right)=\left(\mathit{x}+3\right)\times \mathit{f}\left(\mathit{x}+3\right)=\left(\mathit{x}+6\right)\times \mathit{f}\left(\mathit{x}+6\right)$ and so
on...
Hence, $\mathit{f}\left(1\right)=4\mathit{f}\left(4\right)=7\mathit{f}\left(7\right)...,2\mathit{f}\left(2\right)=5\mathit{f}\left(5\right)=8\mathit{f}\left(8\right)...$
and $3\mathit{f}\left(3\right)=6\mathit{f}\left(6\right)=9\mathit{f}\left(9\right)...$
Also, $3\mathit{f}\left(3\right)=6\mathit{f}\left(6\right)\Rightarrow \mathit{f}\left(3\right)=2\mathit{f}\left(6\right)\Rightarrow \mathit{f}\left(3\right)=180\times 2=360$
By putting the values of $\mathit{f}\left(1\right)$ and $\mathit{f}\left(3\right)$ in $\mathit{f}\left(1\right)+2\mathit{f}\left(2\right)+3\mathit{f}\left(3\right)=0$, we get f(2) = -560
Also, $2\mathit{f}\left(2\right)=14\mathit{f}\left(14\right)\Rightarrow \mathit{f}\left(14\right)=\frac{\mathit{f}\left(2\right)}{7}=-80$