If Pe^x = Qe^−x for all real values of x, which one of the following statements is true?

The correct option is P = Q = 0.

To understand why this is correct, we are given the equation:

$P{e}^x=Q{e}^{-x}$

This equation is an identity, meaning it must hold true for all real values of $x$. We can determine the constants $P$ and $Q$ by substituting specific values for $x$.

First, let us substitute $x=0$ into the equation:

$P{e}^0=Q{e}^{-0}$

Since any non-zero number raised to the power of 0 is 1 ($e^0=1$), the equation simplifies to:

$P\left(1\right)=Q\left(1\right)$

which gives:

$P=Q$

Next, let us substitute $x=1$ into the original equation:

$P{e}^1=Q{e}^{-1}$

Since we have established that $P=Q$, we can replace $Q$ with $P$ in this equation:

$Pe=P{e}^{-1}$

Subtracting $P{e}^{-1}$ from both sides gives:

$Pe-P{e}^{-1}=0$

Factoring out $P$:

$P(e-e^{-1})=0$

Since the mathematical constant $e\approx 2.718$ and $e^{-1}\approx 0.368$, their difference is non-zero:

$e-e^{-1}\ne 0$

Therefore, for the product $P(e-e^{-1})$ to equal 0, we must have:

$P=0$

Using our relation $P=Q$, we also obtain:

$Q=0$

Thus, the only values for which the equality holds for all real $x$ are indeed $P=Q=0$.