CAT Exam Model Paper 6 with solutions for free online practice

It is given that:f(x + y) = f(x) + f(y) ... (i)
Putting y = x in the above:
f(2x) = 2f(x)
Hence, f$\left(\frac{1}{2}\right)$ = 2f$\left(\frac{1}{3}\right)$ ...(ii)
Also, from equation (i):
$\mathrm{f}\left(1\right)=1\left(\frac{2}{3}+\frac{1}{3}\right)=\mathrm{f}\left(\frac{2}{3}\right)+\mathrm{f}\left(\frac{1}{3}\right)$
$=1\left(\frac{2}{3}\right)+\frac{1}{2}\mathrm{f}\left(\frac{2}{3}\right)$(from equation (ii)
⇒ $\mathrm{f}\left(1\right)=\frac{3}{2}\mathrm{f}\left(\frac{2}{3}\right)\mathrm{or}\mathrm{f}\left(\frac{2}{3}\right)=\frac{2}{3}\mathrm{f}\left(1\right)$
Alternate Solution:
Note Hint in such cases the given function f(x) must be a linear function with the constant term zero.
Lot f(x) = Kx, where K is a constant,
f(1) = K and $\mathrm{f}\left(\frac{2}{3}\right)=\frac{2\mathrm{k}}{3}=\frac{2}{3}\mathrm{f}\left(1\right)$