2023 H2 Maths Paper 1 Q10 — Worked Solution

Part (a)

5 marks

Andrew’s energy intake is $C$ Calories per day. Since he expends $30$ Calories per day for every kg of his mass $M$ , his energy expenditure is $30M$ Calories per day. Since the rate of change of mass is proportional to (intake $\text{-}$ expenditure),

\frac{dM}{dt} = k(C - 30M)

where $k$ is the constant of proportionality. (shown)

Part (b)

1 mark

C = 3300

Explanation: for Andrew’s mass to remain constant at $110$ kg, $\frac{dM}{dt} = 0$ , so from part (a), $k\left( C - 30(110) \right) = 0$ . Since $k \neq 0$ , $C = 30(110) = 3300$ .

Part (c)

Approach: substitute the new intake into the differential equation from (a), separate variables, and use the initial mass at the start of the plan to fix the constant of integration.

Andrew’s new energy intake is $80\%$ of $3300$ , i.e. $C'= 0.8(3300) = 2640$ . Substituting into the differential equation from (a) with $C = 2640$ ,

\frac{dM}{dt} = k(2640 - 30M)

= 30k\left( \frac{2640}{30} - M \right)

= 30k(88 - M)

Separating variables,

\int\frac{1}{88 - M}\, dM = \int 30k\, dt

\Rightarrow - \ln|88 - M| = 30kt + c_{1}

\Rightarrow \ln|88 - M| = - 30kt - c_{1}

\Rightarrow 88 - M = Ae^{- 30kt}

where $A = e^{- c_{1}}$ . Since Andrew’s mass is $110$ kg at the start of the plan ( $t = 0$ ),

110 = 88 - Ae^{0}

\Rightarrow A = - 22

So $88 - M = - 22e^{- 30kt}$ , i.e.

M = 88 + 22e^{- 30kt}

To check your answer: at

t = 0

,

M = 88 + 22 = 110

, matching Andrew’s initial mass on the plan. ✓

as required. (shown)

Part (d)

4 marks

Approach: use the data point $t = 75$ , $M = 100$ to find $k$ , then solve for the (non-integer) time at which $M = 96$ , and round up since the mass must fall strictly below $96$ kg.

At $t = 75$ , $M = 100$ :

100 = 88 + 22e^{- 30k(75)}

\Rightarrow 12 = 22e^{- 2250k}

\Rightarrow e^{- 2250k} = \frac{6}{11}

\Rightarrow k = \frac{\ln(11/6)}{2250}

Setting $M = 96$ ,

96 = 88 + 22e^{- 30kt}

\Rightarrow e^{- 30kt} = \frac{4}{11}

\Rightarrow t = \frac{\ln(11/4)}{30k}

Substituting $k = \frac{\ln(11/6)}{2250}$ , so $30k = \frac{\ln(11/6)}{75}$ ,

t = \frac{\ln(11/4)}{\ln(11/6)/75}

= \frac{75\ln(11/4)}{\ln(11/6)}

\approx 125.2

This is the total time (from the start of the plan) for the mass to reach $96$ kg. The additional days needed after day $75$ :

t - 75 \approx 50.2

To check your answer: by GC,

M(125.17) = 88 + 22e^{- 30k(125.17)} \approx 96.0

, matching the value

t \approx 125.2

used above. ✓

Since $M$ decreases continuously and must fall strictly below (not just reach) $96$ kg, the smallest whole number of additional days is $51$ .

$\therefore$ Andrew needs $51$ additional days on the plan.

Part (e)(i)

2 marks

Since $M = 88 + 22e^{- 30kt}$ with $k > 0$ , the term $22e^{- 30kt}$ is strictly positive for every finite $t$ , however large, so $M > 88$ for all finite $t$ — the graph approaches the asymptote $M = 88$ but never reaches or crosses it.

$\therefore$ since $80 < 88$ and $M$ never falls to or below $88$ , Andrew cannot achieve a mass of $80$ kg using this plan.

Part (e)(ii)

1 mark

0 < C < 2400

Explanation: for a general intake $C$ , the differential equation $\frac{dM}{dt} = k(C - 30M)$ has the same structure as in part (c), with equilibrium value $\frac{C}{30}$ in place of $88$ . Starting from $M = 110$ , which is above the equilibrium whenever $C < 3300$ , $M$ decreases towards $\frac{C}{30}$ but never reaches it — exactly as in (e)(i). So Andrew’s mass reaches (and passes through) $80$ kg at some finite time only if the equilibrium itself lies strictly below $80$ :

\frac{C}{30} < 80 \Rightarrow C < 2400

Since $C$ is an energy intake it is also positive, giving $0 < C < 2400$ Calories per day.

2023 H2 Maths Paper 1, Question 10