A hospital patient is given a pill containing 50 units of antibiotic at 12pm. By 1pm, the number of units of antibiotic in the patient's body has dropped by 12%. By 2pm, a further 12% of the antibiotic remaining in the blood at 1pm is lost. Given that the rate of decrease is such that the amount of antibiotic drops by 12% every one hour, find the amount of antibiotic remaining in the blood at 6pm. A doctor considers prescribing a course of treatment which involves the patient taking the pill every 6 hours over a long period of time. It is given that $x_{n+1}=50+(0.88)^6 x_n,$ where $x_n$ denotes the number of units of antibiotic in the body after the $n^{\text{th}}$ pill is taken. As $n \to \infty$, $x_n \to L$. Determine the value of $L$. Show that $x_{n+1}>x_n$ if $x_n<L$; $x_{n+1}<x_n\text{ if }x_n>L.$

1. Determine the time elapsed and the constant decay multiplier.

The time elapsed from 12pm to 6pm is $6$ hours.
The amount of antibiotic drops by $12\%$ every hour, so the multiplier is $1 - 0.12 = 0.88$.

2. Formulate and evaluate the amount remaining after 6 hours.

$\text{Amount remaining} = 50 \times (0.88)^6$
$= 23.220... \approx 23.2 \text{ units}$

1. Apply the limit as $n \to \infty$ to the recurrence relation

As $n \to \infty$, both $x_n \to L$ and $x_{n+1} \to L$.
Substituting this into the given recurrence relation:
$L = 50 + (0.88)^6 L$

2. Solve the linear equation for $L$

$L - (0.88)^6 L = 50$
$L(1 - 0.88^6) = 50$
$L = \frac{50}{1 - 0.88^6}$
$L \approx 93.3539$
$L \approx 93.4 \text{ (to 3 s.f.)}$

1. Form an expression for the difference $x_{n+1} - x_n$.

$x_{n+1} - x_n = 50 + (0.88)^6 x_n - x_n$
$x_{n+1} - x_n = 50 - x_n(1 - (0.88)^6)$

2. Substitute the exact relation for the limit $L$ into the difference expression.

From part (ii)(a), as $n \to \infty$, $x_n \to L$, so the limit satisfies:
$L = 50 + (0.88)^6 L \implies 50 = L(1 - (0.88)^6)$
Substitute this equivalent of 50 into the expression for $x_{n+1} - x_n$:
$x_{n+1} - x_n = L(1 - (0.88)^6) - x_n(1 - (0.88)^6)$
$x_{n+1} - x_n = (L - x_n)(1 - (0.88)^6)$

3. Analyze the sign of the difference to deduce the inequalities.

Since $(0.88)^6 < 1$, the constant term $(1 - (0.88)^6) > 0$.

• If $x_n < L$, then $L - x_n > 0$.
  $(L - x_n)(1 - (0.88)^6) > 0 \implies x_{n+1} - x_n > 0 \implies x_{n+1} > x_n$

• If $x_n > L$, then $L - x_n < 0$.
  $(L - x_n)(1 - (0.88)^6) < 0 \implies x_{n+1} - x_n < 0 \implies x_{n+1} < x_n$

(Both cases are shown)