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### Step 1: Understanding the Problem
The problem describes a bacterial infection modeled by the equation \( N'(t) = N_0 \exp(t) \), where \( N(t) \) is the bacterial population at time \( t \) in hours. After 1 hour, an antibiotic is administered, which causes the bacterial population to decrease according to the rate equation \( \frac{dN}{dt} = -5N^2 \).
### Step 2: Analyzing the Population Growth
For \( t < 1 \):
- The bacterial population grows exponentially. The equation \( N(t) = N_0 \exp(t) \) indicates that the population increases rapidly as time progresses.
For \( t = 1 \):
- At this point, the population reaches its maximum before the antibiotic takes effect.
### Step 3: Effect of Antibiotic
For \( t > 1 \):
- The antibiotic takes 1 hour to reach the wound. Therefore, for \( t > 1 \), the population starts to decrease according to the equation \( \frac{dN}{dt} = -5N^2 \).
- This indicates that the rate of decrease is proportional to the square of the population, which means that as the population decreases, the rate of decrease also slows down.
### Step 4: Analyzing the Ratio \( \frac{N_0}{N} \)
- Initially, \( N \) is increasing. After 1 hour, \( N \) starts to decrease.
- Since \( N_0 \) is a constant (the initial population), the ratio \( \frac{N_0}{N} \) will initially be small (as \( N \) is large) and will increase as \( N \) decreases.
### Step 5: Conclusion
- After 1 hour, as the antibiotic takes effect, \( N \) decreases, causing \( \frac{N_0}{N} \) to increase.
- Therefore, the plot of \( \frac{N_0}{N} \) versus \( t \) will show an increasing trend after the 1-hour mark.
### Final Answer
Thus, the correct option is (D), which indicates that \( \frac{N_0}{N} \) increases after 1 hour.
