Choice A is the correct answer. Use daily compounding.

1. Values: $P = 1000$, $r = 0.05$, $n = 365$, $t = 1$.
2. Formula: $A = 1000(1 + \frac{0.05}{365})^{365}$.
3. Calculate: $(1 + \frac{0.05}{365})^{365} \approx 1.05127$, so $A \approx 1051.27$.

💡 Strategic Tip: Daily compounding approaches continuous compounding ($e^r$).

Choice B is incorrect because this is annual compounding. Choice C is incorrect because this calculation is incorrect. Choice D is incorrect because this assumes 10% growth.