maths

[manvi ahuja]

The number of continuous functions ,f,satisfying..
x.f(y)+y.f(x)=(x+y).f(x).f(y),where x & y are  any real numbers is..
..
A.1
B.2
C.3
D.none of these

[anon_econ]

Is the answer B?

[manvi ahuja]

Yes bt how did u solve it...???nd vasudha pls try my query on probability as well...i posted it few days back...

[anon_econ]

The equality holds for x=y as well. So u must have 2xf(x)=2x[f(x)]^2. Solve this to get f(x)[f(x)-1]=0, when x is not =0.
This means that when x is not equal to 0, f(x) is either 0 or 1. Since f is a continuous function, f(0) also has to have the same value as f(x) when x is not =0.
Thus u get 2 continuous functions: f(x)=0 and f(x)=1.

I don't know how to do that probability question.

[manvi ahuja]

Thanx a lot yar..