Let a 0, a neq 1. Then,the set S of all pos | vedclass

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(A) Given the equation: $(1+a^2)(1+b^2) = 4ab$Expanding the left side: $1 + b^2 + a^2 + a^2b^2 = 4ab$Rearranging the terms: $a^2 - 2ab + b^2 + a^2b^2

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an empty set

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(A) Given the equation: $(1+a^2)(1+b^2) = 4ab$Expanding the left side: $1 + b^2 + a^2 + a^2b^2 = 4ab$Rearranging the terms: $a^2 - 2ab + b^2 + a^2b^2 - 2ab + 1 = 0$This can be written as: $(a-b)^2 + (ab-1)^2 = 0$Since $a$ and $b$ are real numbers,the sum of squares is zero if and only if each term is zero:$(a-b)^2 = 0 \implies a = b$$(ab-1)^2 = 0 \implies ab = 1$Substituting $b=a$ into $ab=1$,we get $a^2 = 1$,which implies $a = 1$ or $a = -1$.However,the problem states $a > 0$ and $a 
eq 1$. Thus,there is no value of $b$ that satisfies the equation under the given constraints.Therefore,the set $S$ is an empty set.

Let $a > 0, a \neq 1$. Then,the set $S$ of all positive real numbers $b$ satisfying $(1+a^2)(1+b^2) = 4ab$ is

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