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Consider the function f(x) satisfying the identityf(x) +f((x-1)/(x))=1+x AA x in R -{0,1}, and g(x)=2f(x)-x+1. The domain of y=sqrt(g(x)) is |
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Answer» `(-oo,(1-sqrt(5))/(2)] CUP [1,(1+sqrt(5))/(2)]` In (1), replace x by `(x-1)/(x)`. Then `f((x-1)/(x))+f(((x-1)/(x)-1)/((x-1)/(x)))=1+(x-1)/(x)` `ORF((x-1)/(x))+f((1)/(1-x))=1+(x-1)/(x) "(2)" ` Now, from `(1) -(2)`, we have `f(x)-f((1)/(1-x))=x-(x-1)/(x) "(3)" ` In (3), replace x by `(1)/(x-1)`. Then `f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-((1)/(1-x)-1)/((1)/(1-x))` ` orf((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-x "(4)" ` Now, from `(1)+(3)+(4)`, we have `2f(x)=1+x+x-(x-1)/(x)+(1)/(1-x)-x` ` orf(x)=(x^(3)-x^(2)-1)/(2x(x-1))` `g(x)=(x^(3)-x^(2)-1)/(x(x-1))-x+1=(x^(2)-x-1)/(x(x-1))` Now, for `y=sqrt(g(x)),` we must have `(x^(2)-x-1)/(x(x-1)) GE 0` `or ((x-(1-sqrt(5))/(2))(x-(1+sqrt(5))/(2)))/(x(x-1)) ge 0` `orx in (-oo,(1-sqrt(5))/(2)] cup (0,1)cup [(1+sqrt(5))/(2),oo)` |
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