Consider a function `f : R -> R; f(x^2 +yf(z)) = xf(x) + zf(y), AA x,y,z in R` If `f(x) = 0,AA x in R` is not considered a part of solution set, then

Options

A. `g(x)` is not continuous `AA xepsilonR`
B. `g(x)` is differentiable except at two points `AA x epsilonR`
C. `g(x)` is differentiable at `x=sinalpha(alpha!+npi, n epsilonI)`
D. `g(x)` is non differentiable at integers

Answer

C

Explanation

`f(x^(2)+yf(z))=x.f(x)+zf(y)`
`x=y=z=0`
`f(0)=0`
`x=0`
`f(y.f(z))=zf(y)`
`y=z=t`
`f(t.f(f))=t.f(t)`
Using this and susbtitution, we get `f(x)=x` and `f(x)=0`