If \(Pe^{x} = Qe^{-x}\) for all real values of \(x\), which one of the following statements is true?
Answer
(A)
Explanation
Multiply both sides by \(e^{x}\) to get \(Pe^{2x} = Q\), which must hold for every real \(x\).
At \(x = 0\) this gives \(P = Q\), and at \(x = \frac{1}{2}\ln 2\) it gives \(2P = Q\). Together these force \(P = 0\), and then \(Q = 0\). In short, \(e^{2x}\) keeps changing, so \(Pe^{2x}\) can stay equal to the constant \(Q\) only when \(P = 0\).
- B holds only at \(x = 0\).
- C would make the positive number \(e^{x}\) equal the negative number \(-e^{-x}\).
- D needs \(P = 0\) with \(Q \neq 0\), but \(P = 0\) forces \(Q = 0\) as well.
Answer: A