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Consider a function f : R -> R; f(x^2 +y...

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

A

`f(alpha) lt alpha^(4)AA alpha epsilon(0,1)`

B

`f(alpha) lt alpha^(2)AA alpha epsilon(0,1)`

C

`f(alpha)=alpha^(3)` for some `alpha epsilon R^(+)`

D

`lim_(alphato 0^(+)) (f(alpha))/(alpha) lt lim_(alpha to 0^(+)) (sinalpha)/(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
C
`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`
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