ISI MSQE prev year question

[XIPP]

SEE isi 2007 msqe paper, q26 of math section.

Suppose the real valued continuous function f defined on the set of non-negative real
numbers satisfies the condition f(x) = xf(x), then f(2) equals
a:1,
b:2,
c:3,
d:f(1)


And I can see by elimination that f(2)=2*f(2) condition is not satisfied for f(2)=1,2,3 and so f(1) should be answer. But, how is f(1) really the answer? that is how f(1) can be derived if I don't follow the method of elimination?

[Amit Goyal]

Note that f satisfies:
f(x) = xf(x) for all x
Above can be rewritten as:
(1-x)f(x) = 0 for all x
Thus when x ≠ 1, we must have f(x) = 0.
And since f is continuous, f(1) must also be 0.
So, we get f(2) = 0 = f(1).

[XIPP]

Thank you so much Amit Sir.