Discussion
If \(Pe^{x} = Qe^{-x}\) for all real values of \(x\), which one of the following statements is true?
Answer: A
The true statement is \(P = Q = 0\).
Multiply both sides by \(e^{x}\): the equation becomes \(Pe^{2x} = Q\) for every real \(x\). The right side is a constant, while \(e^{2x}\) takes every positive value as \(x\) varies. A constant can equal \(Pe^{2x}\) for all \(x\) only if \(P = 0\), and then \(Q = 0\) too.
\(P = Q = 1\) would require \(e^{x} = e^{-x}\), true only at \(x = 0\). \(P = 1, Q = -1\) gives \(e^{x} = -e^{-x}\), impossible since both sides have opposite signs. \(P/Q = 0\) needs \(P = 0\) with \(Q \neq 0\), and then the equation would say \(Qe^{-x} = 0\), which fails for every \(x\).
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