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If p, q are positive integers, f is a function defined for positive numbers and attains only positive values such that `f(xf(y))=x^p y^q`, then prove that `p^2=q`. |
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Answer» Given `f(xf(y))=x^(p)y^(q)` `or x=({f(xf(y))}^(1//p))/(y^(q//p)) " (1)" ` Let `xf(y)=1 or x=(1)/(f(y))`. Then from (1), `f(y)=(y^(q//p))/({f(1)}^(1//p))` ` or f(1)=(1)/({f(1)}^(1//p))` ` :. f(1)=1` ` :. f(y)=y^(q//p)" (2)" ` Now, `f(xy^(q//p))=x^(p)y^(q). " Put "y^(q//P)=z.` Then `f(xz)=(xz)^(p)` `or f(x)=x^(p) " (3)" ` From (2) and (3), `x^(P)=x^(q//p) or p^(2)=q.` |
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