1. [Maximum mark: 13]
Scott purchases food for his dog in large bags and feeds the dog the same amount of dog food each day. The amount of dog food left in the bag at the end of each day can be modelled by an arithmetic sequence.
On a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the third day, there were 115.5 cups of dog food remaining in the bag, and at the end of the eighth day, there were 108 cups of dog food remaining in the bag.
(a) Find the number of cups of dog food:
(i) fed to the dog per day;
(ii) remaining in the bag at the end of the first day. [4]
(b) Calculate the number of days that Scott can feed his dog with one bag of food. [2]
In 2021, Scott spent $625 on dog food. Scott expects that the amount he spends on dog food will increase at an annual rate of 6.4%.
(c) Determine the amount that Scott expects to spend on dog food in 2025. Round your answer to the nearest dollar. [3]
(d) (i) Calculate the value of \( \sum_{n=1}^{10} (625 \times 1.064^{(n-1)}) \).
(ii) Describe what the value in part (d)(i) represents in this context. [3]
(e) Comment on the appropriateness of modelling this scenario with a geometric sequence. [1]
▶️Answer/Explanation
(a)(i) Using the arithmetic sequence formula \( u_n = u_1 + (n-1)d \):
For day 3: \( u_1 + 2d = 115.5 \)
For day 8: \( u_1 + 7d = 108 \)
Subtracting equations: \( 5d = -7.5 \Rightarrow d = -1.5 \).
Thus, the dog is fed 1.5 cups/day.
(a)(ii) Substitute d = − 1.5 d=−1.5 into u 1 + 2 d = 115.5 u 1 +2d=115.5: u 1 = 115.5 + 3 = 118.5 u 1 =115.5+3=118.5. Remaining at the end of day 1: 118.5 cups. (b) Solve 118.5 − 1.5 ( n − 1 ) ≤ 0 118.5−1.5(n−1)≤0: 1.5 ( n − 1 ) ≥ 118.5 ⇒ n − 1 ≥ 79 ⇒ n = 80 1.5(n−1)≥118.5⇒n−1≥79⇒n=80. The bag lasts 80 days. (c) Using geometric growth: 625 × 1.06 4 4 ≈ 801 625×1.064 4 ≈801. Expected spending in 2025: $801. (d)(i) Sum of geometric series: S 10 = 625 × 1.06 4 10 − 1 1.064 − 1 ≈ 8390 S 10 =625× 1.064−1 1.064 10 −1 ≈8390. Value: $8390. (d)(ii) The total amount Scott expects to spend on dog food from 2021 to 2030. (e) The geometric model assumes a constant annual growth rate, which may not account for economic fluctuations or changes in the dog’s consumption habits. However, it is appropriate if the inflation rate remains stable.
2. [Maximum mark: 15]
A cafe makes \( x \) litres of coffee each morning. The cafe’s profit each morning, \( C \), measured in dollars, is modelled by:
\[ C = \frac{x}{10} \left( k^2 – \frac{3}{100}x^2 \right) \]
where \( k \) is a positive constant.
(a) Find an expression for \( \frac{dC}{dx} \) in terms of \( k \) and \( x \). [3]
(b) Hence find the maximum value of \( C \) in terms of \( k \). Give your answer in the form \( pk^3 \), where \( p \) is a constant. [4]
The cafe’s manager knows that the cafe makes a profit of $426 when 20 litres of coffee are made in a morning.
(c) (i) Find the value of \( k \).
(ii) Use the model to find how much coffee the cafe should make each morning to maximize its profit. [3]
(d) Sketch the graph of \( C \) against \( x \), labelling the maximum point and the \( x \)-intercepts with their coordinates. [3]
(e) Determine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning. [2]
▶️Answer/Explanation
(a) Differentiate \( C \):
\( \frac{dC}{dx} = \frac{k^2}{10} – \frac{9x^2}{1000} \).
(b) Set d C d x = 0 dx dC =0: k 2 10 = 9 x 2 1000 ⇒ x = 10 k 3 10 k 2 = 1000 9x 2 ⇒x= 3 10k . Substitute x x into C C: C max = 2 k 3 9 C max = 9 2k 3 . (c)(i) Substitute x = 20 x=20 and C = 426 C=426: 426 = 2 ( k 2 − 12 ) ⇒ k = 15 426=2(k 2 −12)⇒k=15. (c)(ii) Maximum at x = 10 × 15 3 = 50 x= 3 10×15 =50 litres. (d) Graph: Parabola opening downward with maximum at ( 50 , 750 ) (50,750), intercepts at ( 0 , 0 ) (0,0) and ( 86.6 , 0 ) (86.6,0). (e) Solve C = 0 C=0: x = 86.6 x=86.6 litres (maximum without loss).
3. [Maximum mark: 18]
The Voronoi diagram shows four supermarkets at points A(0, 0), B(6, 0), C(0, 6), and D(2, 2). The vertices X, Y, Z are shown. The equations of edges are given.
(a) Find the midpoint of [BD]. [2]
(b) Find the equation of (XZ). [4]
(c) Find the coordinates of X. [3]
(d) Determine the exact length of [YZ]. [2]
(e) Find the size of \( \angle XYZ \) in degrees. [4]
(f) Hence find the area of triangle XYZ. [2]
(g) State one criticism of the interpretation linking Voronoi cell area to supermarket popularity. [1]
▶️Answer/Explanation
(a) Midpoint of BD: \( \left( \frac{6+2}{2}, \frac{0+2}{2} \right) = (4, 1) \).
(b) Gradient of BD: 2 − 0 2 − 6 = − 1 2 2−6 2−0 =− 2 1 . Perpendicular bisector gradient: 2. Equation: y − 1 = 2 ( x − 4 ) ⇒ y = 2 x − 7 y−1=2(x−4)⇒y=2x−7. (c) Solve y = 2 − x y=2−x and y = 2 x − 7 y=2x−7: Intersection X ( 3 , − 1 ) X(3,−1). (d) Distance YZ: ( 7 − ( − 1 ) ) 2 + ( 7 − 3 ) 2 = 80 = 4 5 (7−(−1)) 2 +(7−3) 2 = 80 =4 5 . (e) Using cosine rule: cos θ = 80 + 32 − 80 2 × 80 × 32 ⇒ θ ≈ 71. 6 ∘ cosθ= 2× 80 × 32 80+32−80 ⇒θ≈71.6 ∘ . (f) Area: 1 2 × 80 × 32 × sin 71. 6 ∘ = 24 km 2 2 1 × 80 × 32 ×sin71.6 ∘ =24km 2 . (g) Criticism: Proximity isn’t the only factor influencing shopping choices (e.g., pricing, brand preference).
4. [Maximum mark: 15]
The Arrhenius equation \( k = Ae^{-\frac{c}{T}} \) links reaction rate \( k \) with temperature \( T \).
(a) Show that \( \frac{dk}{dT} \) is always positive. [3]
(b) Sketch the graph of \( k \) against \( T \). [3]
(c) Write down the gradient and \( y \)-intercept of \( \ln k \) vs \( \frac{1}{T} \). [4]
(d) Find the regression line equation for given data. [3]
(e) Estimate constants \( c \) and \( A \). [3]
▶️Answer/Explanation
(a) Differentiate \( k = Ae^{-\frac{c}{T}} \):
\( \frac{dk}{dT} = \frac{Ac}{T^2} e^{-\frac{c}{T}} > 0 \) (since \( A, c, T > 0 \)).
(b) Graph: Increasing curve approaching asymptote k = A k=A as T → ∞ T→∞, approaching 0 as T → 0 T→0. (c)(i) Gradient: − c −c. (c)(ii) y y-intercept: ln A lnA. (d) Regression line: ln k = − c ( 1 T ) + ln A lnk=−c( T 1 )+lnA. (e) From regression: c ≈ c≈ slope value, A ≈ e intercept A≈e intercept .
5. [Maximum mark: 12]
A geneticist uses a Markov chain model with matrix \( M = \begin{pmatrix} 0.94 & b \\ 0.06 & 0.98 \end{pmatrix} \).
(a) (i) Write down the value of \( b \).
(ii) Interpret \( b \). [2]
(b) Find the eigenvalues of \( M \). [3]
(c) Find the eigenvectors of \( M \). [3]
(d) Calculate the probability of the gene being normal:
(i) after 5 divisions;
(ii) in the long term. [4]
▶️Answer/Explanation
(a)(i) \( b = 0.02 \) (columns sum to 1).
(a)(ii) \( b \) represents the mutation probability from another state to normal.
(b) Eigenvalues: Solve det ( M − λ I ) = 0 ⇒ λ 1 = 1 det(M−λI)=0⇒λ 1 =1, λ 2 = 0.92 λ 2 =0.92. (c) Eigenvectors: For λ = 1 λ=1, eigenvector ( 1 3 ) ( 1 3 ); for λ = 0.92 λ=0.92, eigenvector ( 1 − 1 ) ( 1 −1 ). (d)(i) Using X 5 = M 5 ( 1 0 ) ≈ 0.735 X 5 =M 5 ( 1 0 )≈0.735. (d)(ii) Steady-state probability: 0.06 0.06 + 0.02 = 0.75 0.06+0.02 0.06 =0.75.
6. [Maximum mark: 21]
A ball’s path is modelled by \( x = 5 + 8t \), \( y = 10t – 5t^2 \). An arrow is shot from (0, 2) at 60 m/s, 10° elevation.
(a) Find the ball’s initial speed and angle. [4]
(b) Find the ball’s maximum height. [3]
(c) Find where the ball lands. [3]
(d) Express \( y \) in terms of \( x \). [3]
(e) Determine intersection points of the arrow and ball paths. [4]
(f) Find the release time to hit the ball before its maximum height. [4]
▶️Answer/Explanation
(a)(i) Initial speed: \( \sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8 \, \text{m/s} \).
(a)(ii) Angle: \( \tan^{-1}\left( \frac{10}{8} \right) \approx 51.3^\circ \).
(b) Maximum height: 1 0 2 2 × 10 = 5 m 2×10 10 2 =5m. (c) Solve y = 0 y=0: t = 2 t=2. x = 5 + 16 = 21 m x=5+16=21m. (d) Eliminate t t: y = 10 ( x − 5 ) 8 − 5 ( x − 5 8 ) 2 y= 8 10(x−5) −5( 8 x−5 ) 2 . (e) Arrow path: y = 2 + 60 sin 1 0 ∘ t − 5 t 2 y=2+60sin10 ∘ t−5t 2 . Solve with ball’s y y-equation for intersections. (f) Solve for t t when arrow’s height equals ball’s height before t = 1 s t=1s.
7. [Maximum mark: 16]
A differential equation models mercury concentration: \( \frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 2x = 0 \).
(a) Show the system \( \frac{dx}{dt} = y \), \( \frac{dy}{dt} = -2x – 3y \). [2]
(b) Find eigenvalues. [3]
(c) Solve the differential equation. [5]
(d) Sketch \( x(t) \) with maximum point. [2]
(e) Calculate total unsafe fishing time. [3]
(f) Justify extending the fishing ban by 10%. [1]
▶️Answer/Explanation
(a) Substitute \( y = \frac{dx}{dt} \) into \( \frac{dy}{dt} = -2x – 3y \), confirming the second-order equation.
(b) Eigenvalues: Solve λ 2 + 3 λ + 2 = 0 ⇒ λ = − 1 , − 2 λ 2 +3λ+2=0⇒λ=−1,−2. (c) General solution: x ( t ) = A e − t + B e − 2 t x(t)=Ae −t +Be −2t . Apply initial conditions x ( 0 ) = 0 x(0)=0, x ′ ( 0 ) = 1 x ′ (0)=1: x ( t ) = e − t − e − 2 t x(t)=e −t −e −2t . (d) Graph peaks at t = ln 2 t=ln2, then decays to 0. (e) Solve e − t − e − 2 t > 0.1 e −t −e −2t >0.1: Total unsafe time ≈1.07 days. (f) To account for model uncertainties or delayed recovery.