Question

If $P e^{x = Q e^{-x}$ for all real values of $x$, which one of the following statements is true?}

• A. $P = Q = 0$
• B. $P = Q = 1$
• C. $P = 1; \; Q = -1$
• D. $\dfrac{P}{Q} = 0$

Hint:

In equations that must hold for \emph{all real $x$}, check if variable-dependent terms (like $e^{2x}$) can be made constant. If not, the only solution is forcing coefficients to zero.

Answer 1

The Correct Option is A

Step 1: Write the given condition.
We are told: \[ P e^{x} = Q e^{-x}, \quad \text{for all real values of } x. \] Step 2: Rearrange the equation.
Divide both sides by $e^{-x}$ (which is never zero): \[ P e^{x} \cdot e^{x} = Q \quad \Rightarrow \quad P e^{2x} = Q. \] Step 3: Analyze the dependence on $x$.
- The left-hand side is $P e^{2x}$, which depends on $x$. - The right-hand side is $Q$, a constant (independent of $x$). - For this equality to hold for \emph{all real $x$}, the exponential term must not introduce variation. The only way this is possible is if $P = 0$. Step 4: Substitute $P = 0$.
If $P = 0$, then: \[ 0 \cdot e^{2x} = Q \quad \Rightarrow \quad Q = 0. \] Step 5: Verify with other options.
- (A) $P = Q = 0$ → Satisfies condition for all $x$.
- (B) $P = Q = 1$ → Gives $e^{x} = e^{-x}$, true only if $x = 0$, not for all real $x$.
- (C) $P = 1, Q = -1$ → Gives $e^{x} = - e^{-x}$, impossible since LHS and RHS have opposite signs.
- (D) $\dfrac{P}{Q} = 0$ → Would imply $P=0, Q \neq 0$, but then equation becomes $0 = Q e^{-x}$, which forces $Q=0$. Contradiction. Step 6: Conclusion.
The only consistent solution is: \[ \boxed{P = Q = 0} \]