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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, thenA. `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)` |
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Answer» Correct Answer - 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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