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Show that the function `f: R rarr { x in R: -1 lt x lt 1 } ` defined by `f(x)=(x)/(1+|x|), x in R` is one- one and onto function . |
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Answer» Let x, `y in R` and , f(x)=f(y) `rArr x/(1+|x|)=y/(1+|y|)` If x is positve and y is negative then `x gt y ` `rArr x-y gt 0 and 2xy lt 0 ` `therefore x/(1+x)=y/(1-y)` `rArr y+ xy =x -xy` `rArr 2xy = x-y` which is impossible then `f(x)=f(y)rArr (x)/(1+x)=(y)/(1+y)` `rArr x+y` If x and y both are negative then `f(x)=f(y)rArr x/(1-x)=y/(1-y)` `rArr x-xy = y-xy` `rArr x=y` Therefore , f is one - one Let y `in ` R be such that `-1 lt y lt 1` If y is negative then `x=y/(1+y) in R` is such that `f(x)=f((y)/(1+y))=((y)/(1+y))/(1+|(y)/(1+y)|)=((y)/(1+y))/(1-y/(1+y))=y` If y is positive then `x=y/(1-y) in R` is such that `f(x)=f(y/(1-y))=((y)/(1-y))/(1+((y)/(1-y)))=((y)/(1-y))/(1-y/(1+y))=y` `therefore` f is onto Therefore f is one - one onto . |
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