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Let `f: R->R` where, `f(x)=(x^2+4x+7)/(x^2+x+1)` Is `f(x)` one-one? |
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Answer» `f(x) = (x^(2)+4x+7)/(x^(2)+x+1)=1+(3(x+2))/(x^(2)+x+1)` Let `f(x_(1))=f(x_(2))` `implies 1+(3(x_(1)+2))/(x_(1)^(2)+x_(1)+1)=1+(3(x_(2)+2))/(x_(2)^(2)+x_(2)+1)` `implies x_(1)x_(2)^(2)+x_(1)x_(2)+x_(1)+2x_(2)^(2)+2x_(2)+2` `=x_(1)^(2)x_(2)+x_(1)x_(2)+x_(2)+2x_(1)^(2)+2x_(1)+2` `implies (x_(1)-x_(2))(2x_(1)+2x_(2)+x_(1)x_(2)+1)=0` Let us consider `2x_(1)+2x_(2)+x_(1)x_(2)+1=0` `implies x_(2)=(1+2x_(1))/(2+x_(1))` This relation is satisfied by infinite number of pairs `(x_(1), x_(2))," where " x_(1) ne x_(2), e.g., (0,-1//2),(1,-1) ` etc. Hence f(x) is many-one. |
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