GATE CS 2025 Set 2 Q3 — General Aptitude

Given the equation:

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

for all real values of $x$. Since this identity holds for all $x \in \mathbb{R}$, we can substitute specific values of $x$ to determine $P$ and $Q$:

1. Substitute $x = 0$:
   $$P e^{0} = Q e^{0} \implies P = Q$$
2. Substitute $x = 1$:
   $$P e^{1} = Q e^{-1} \implies P e = \frac{Q}{e}$$

Substituting $Q = P$ into the second equation gives:

$$P e = \frac{P}{e} \implies P \left(e - \frac{1}{e}\right) = 0$$

Since $e - \frac{1}{e} \neq 0$, we must have $P = 0$. Since $P = Q$, it follows that $Q = 0$. Thus, the only solution is $P = Q = 0$.