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(A) Given $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. We need to find $x$ such that $f(g(x)) = g(f(x))$. First,calculate $f(g(x))$: $f(g(x)) = f(x + 1

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-$1$

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(A) Given $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. We need to find $x$ such that $f(g(x)) = g(f(x))$. First,calculate $f(g(x))$: $f(g(x)) = f(x + 1) = (x + 1)^{2} + 2(x + 1) - 3 = (x^{2} + 2x + 1) + 2x + 2 - 3 = x^{2} + 4x$. Next,calculate $g(f(x))$: $g(f(x)) = g(x^{2} + 2x - 3) = (x^{2} + 2x - 3) + 1 = x^{2} + 2x - 2$. Equating both expressions: $x^{2} + 4x = x^{2} + 2x - 2$. Subtracting $x^{2}$ from both sides: $4x = 2x - 2$. $4x - 2x = -2$. $2x = -2$. $x = -1$.

Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. Then,the value of $x$ for which $f(g(x)) = g(f(x))$ is

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