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If f: R->R is an odd function such that ...

If `f: R->R` is an odd function such that `f(1+x)=1+f(x)`and `x^2f(1/x)=f(x),x!=0` then find `f(x)dot`

Text Solution

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`x^(2)f((1)/(x))=f(x), x ne 0 `
Replace x by `x+1`. Then,
`(x+1)^(2)f((1)/(1+x))=f(x+1)`
`f((1)/(1+x))=(1+f(x))/((1+x)^(2)) " (1)" `
`f((1)/(1+x))=f(1-(x)/(1+x))`
`=1+f(-(x)/(1+x))`
`=1-f((x)/(1+x)) " " ( :' f(x) " is odd")`
`=1-((x)/(1+x))^(2)f((1+x)/(x))`
`=1-((x)/(1+x))^(2)f(1+(1)/(x))`
`=1-((x)/(1+x))^(2)(1+(f(x))/(x^(2))) " (2) " `
From (1) and (2),
`(1+f(x))/((1+x)^(2))=1-((x)/(1+x))^(2)(1+(f(x))/(x^(2)))`
or `1+f(x)=(1+x)^(2)-x^(2)-f(x)`
or `f(x)=x`
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