A student is asked to memorise a long list of digits, and then write down the list some time later - Leaving Cert Mathematics - Question 10 - 2022

We need to solve for $t$ when $P(t) = 0.55$:

$0.55 = 0.82 - 0.12 \ln(t + 1)$

Rearranging gives:

$0.12 \ln(t + 1) = 0.82 - 0.55$ $0.12 \ln(t + 1) = 0.27$ $\ln(t + 1) = \frac{0.27}{0.12} = 2.25$

Exponentiating both sides:

$t + 1 = e^{2.25}$ $t + 1 \approx 9.4877$ $t \approx 9.4877 - 1 \approx 8.4877$

So, it would take approximately 8.49 hours.

To find $P'(t)$, we differentiate the function:

$P'(t) = -0.12 \cdot \frac{1}{t + 1}$

Now, calculate $P'(1)$:

$P'(1) = -0.12 \cdot \frac{1}{1 + 1} = -0.12 \cdot \frac{1}{2} = -0.06.$

Hence, $P'(1) = -0.06$.

Since $P''(t)$ is always negative (as proven in part (d)), it indicates that the proportion of digits recalled correctly decreases as $t$ increases. This suggests that the student's ability to recall digits diminishes over time.

To show that the graph has no points of inflection, we can find the second derivative:

$P''(t) = 0.12 \cdot \frac{1}{(t + 1)^2}$

Since $P''(t)$ is always positive for $0 \leq t \leq 12$, this means the graph of $y = P(t)$ is concave up throughout this interval. Therefore, there are no points of inflection.

From the equation, $A = B (c + 1)$ Taking logarithms: $\log_{10} A = \log_{10} B + \log_{10} (c + 1)$ Therefore, c = \frac{\log_{10} A - \log_{10} B}{\log_{10} (t + 1)}.$

Using the provided values: $\log_{10} A = 0.1652 \quad (80\%)$ $\log_{10} B = 0.6722 \quad (47\%)$ Substituting these into the equation for c:

$c = \frac{0.1652 - 0.6722}{\log_{10} (24 + 1)}$

Calculating: $\log_{10} 25 \approx 1.3979$ So, $c = \frac{-0.507}{1.3979} \approx -0.363$

Hence, $c \approx -0.363$.