Let \( f ( x ) = \frac { x } { ( 2 x - 1 ) } \) and \( g ( x ) = \frac { x } { ( x - 1 ) } \). Then, the domain of the function \( h ( x ) = f ( g ( x ) ) + g ( f ( x ) ) \) is all real numbers except
Answer: A
Key idea: For a composite \(f(g(x))\), \(x\) must lie in the domain of \(g\) and \(g(x)\) must lie in the domain of \(f\).
\(f\) is undefined at \(x = \tfrac{1}{2}\); \(g\) is undefined at \(x = 1\).
\(f(g(x))\): need \(x \ne 1\) and \(g(x) \ne \tfrac12\). Solving \(\dfrac{x}{x-1} = \dfrac12\) gives \(2x = x - 1\), i.e. \(x = -1\). So exclude \(1\) and \(-1\).
\(g(f(x))\): need \(x \ne \tfrac12\) and \(f(x) \ne 1\). Solving \(\dfrac{x}{2x-1} = 1\) gives \(x = 2x - 1\), i.e. \(x = 1\). So exclude \(\tfrac12\) and \(1\).
\(h\) needs both parts defined, so the excluded values are \(-1, \tfrac12, 1\).
Hence, the answer is (A) \(-1, \tfrac12\), and 1.