Let's analyze the problem:

The population is given by the function $P(t) = 5.4^t$.

Part (i)

$P'(t)$ is the derivative of $P(t)$, which represents the instantaneous rate of change of the population at time $t$.

We calculate $P'(t)$ using the chain rule for exponentials:

$P'(t) = \frac{d}{dt} \left( 5.4^t \right) = 5.4^t \ln(5.4)$

So, $P'(t) = 5.4^t \ln(5.4)$.

Part (ii)

This statement claims that $P'(1)$ is less than the average rate of change of $P(t)$ between $t = 0$ and $t = 1$.

The average rate of change between $t = 0$ and $t = 1$ is calculated as:

$\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = P(1) - P(0)$

Since $P(1) = 5.4^1 = 5.4$ and $P(0) = 5.4^0 = 1$:

$\text{Average rate of change} = 5.4 - 1 = 4.4$

Now, calculate $P'(1)$:

$P'(1) = 5.4^1 \ln(5.4) = 5.4 \ln(5.4)$

The natural logarithm $\ln(5.4)$ is approximately 1.6864, so:

$P'(1) \approx 5.4 \times 1.6864 \approx 9.111$

Since $9.111 > 4.4$, the statement in (ii) is false.

Part (iii)

The average rate of change between $t = 0$ and $t = 1$ is 4.4, not 5.4, so statement (iii) is also false.

Conclusion

• (i) is true.
• (ii) is false.
• (iii) is false.

Thus, the correct answer is (A) (i) only.

Do you have any questions or would you like more details? Here are 5 related questions to expand on this:

1. How do you compute the derivative of other exponential functions like $a^t$?
2. What is the significance of the average rate of change compared to the instantaneous rate of change?
3. Can we apply similar reasoning to functions with different bases, like $2^t$ or $e^t$?
4. How would the problem change if the base of the exponential was smaller, like 3 or 2?
5. What is the interpretation of the logarithm $\ln(5.4)$ in the context of exponential growth?

Tip: For exponential growth, the derivative always involves the natural logarithm of the base, which scales the rate of growth.