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Let `f(x) =(alphax)/(x+1)` Then the value of `alpha` for which `f(f(x) = x` is |
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Answer» Correct Answer - -1 `f(x)=(alpha x)/(x+1), x ne -1` Now `f(f(x))=x` `implies(alpha((alpha x)/(x+1)))/((alpha x)/(x+1)+1)=x` `implies(alpha^(2)x)/((alpha +1)x+1)=x` `implies (alpha+1)x^(2)+(1-alpha^(2))x=0 " ...(1)" ` Now this is true for all real x. `implies alpha +1=0 " and " 1-alpha^(2)=0` `implies alpha = -1 ` (common value) |
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