Math, asked by chanikyachowdary1, 1 year ago

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Answered by prajwal1697
0

f(x)f( \frac{1}{x} ) = f(x) + f( \frac{1}{x} ) \\  =  > f(x){  \large(}f( \frac{1}{x}) - 1{  \large)}= f( \frac{1}{x}) \\  =  >  f(x) =  \frac{f( \frac{1}{x}) }{f( \frac{1}{x}) - 1}  \:  \:  \:  \:  \:  \: (1) \:  \\  =  > replace \: x \: by \:  \frac{1}{x}  \: we \: get \\  =  > f( \frac{1}{x}) =  \frac{f(x)}{f(x) - 1}    \:  \:  \:  \:  \:  \:  \: (2)\\  =  > multiply \: 1 \: and \: 2 \: now \\  =  > f(x)f( \frac{1}{x} )  =  \frac{f(x)}{(f(x) - 1)} \frac{f( \frac{1}{x}) }{(f( \frac{1}{x}) - 1)}  \\  =  >  (f(x) - 1)(f( \frac{1}{x})) - 1) = 1 \\  =  > take \: f(x) - 1 = g(x) \\  =  > we \: get \:  \\  =  > g(x)g( \frac{1}{x} ) = 1 \\  =  > this \: is \: a \: defined \: funtion \: which \: gives \:  \\ g(x) =  {x}^{n}  \\ try \: with \: some \: example \: to \: cheak \: its \: correctness

hence we got

g(x)=  {x}^{n} \\ which\: is \:equal \:to\: \\f(x)-1= {x}^{n}\\ f(x)= {x}^{n}+1\\given  \: that \: f(-2)=33\\=>{x}^{n}+1=33\\=>{-2}^{n}=32\\ therefore \:n=5\\now \:we \:got \:\\f(x)={x}^{5}+1 \\f(1)={1}^{5}+1\\f(1)=2

alternative :

f(x)f( \frac{1}{x} ) = f(x) + f( \frac{1}{x} ) \\substitute\: x=1\\ we \:get \:f(1)f( \frac{1}{1} ) = f(1) + f( \frac{1}{1} ) \\{f(1)}^{2}=2f(1)\\f(1)=2

hope it helps you

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