to the formula.","highlights":[{"text":"initial value","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$V=17100e^{-0.2t}+2000$"},{"latex":"$$V=17100\\htmlData{anchor=exp_zero}{e^{-0.2(0)}}+2000$"},{"latex":"$$V=17100\\htmlData{anchor=one_factor}{(1)}+2000$"},{"latex":"$$V=\\htmlData{anchor=sum_terms}{17100+2000}=19100$"},{"latex":"$$V=\\htmlData{anchor=final_value}{\\pounds19100}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"exp_zero"},{"color":"orange","style":"text-color","anchor":"one_factor"},{"color":"green","style":"text-color","anchor":"sum_terms"},{"color":"green","style":"text-color","anchor":"final_value"}]},"order":1,"title":"Substitute zero","description":"At the initial time, substitute t = 0. Then use e^0 = 1 and show the addition explicitly. This earns B1 for substituting t = 0 and obtaining £19100."}]}],"generatedAt":"2026-07-20T16:47:26.965Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]},{"id":"6ca9259a-c4fc-4fc7-bc94-df3d62e2e9c8","label":"b","content":"Given the model predicts that the value of the car is decreasing at a rate of £1000 per year at the instant when $t = T$, (i) Show that $3420e^{-0.2T} = 1000$ and (ii) find the age of the car at this instant, giving your answer in years and months to the nearest month.","markscheme":"- (i) $\\frac{dV}{dt} = -0.2(17100)e^{-0.2t} = -3420e^{-0.2t}$ M1\n- Rate of decrease is 1000: $3420e^{-0.2T} = 1000$ A1\n- (ii) $e^{-0.2T} = \\frac{1000}{3420} = \\frac{50}{171}$ M1\n- $-0.2T = \\ln(\\frac{50}{171}) \\Rightarrow T = 6.15$ years M1\n- $0.15 \\times 12 = 1.8$ months $\\approx 2$ months M1\n- 6 years 2 months A1","marks":6,"order":1,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Differentiates $V$ with respect to $t$.","code":"M"},{"id":"p2","marks":1,"description":"Sets the rate of decrease to 1000 to show the given equation.","code":"A"},{"id":"p3","marks":1,"description":"Isolates the exponential term.","code":"M"},{"id":"p4","marks":1,"description":"Uses logs to solve for $T$.","code":"M"},{"id":"p5","marks":1,"description":"Converts decimal years to months.","code":"M"},{"id":"p6","marks":1,"description":"Final answer 6 years 2 months.","code":"A"}],"finalAnswers":["6 years 2 months"],"generatedAt":"2026-05-12T02:14:06.626Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"logarithmic_solution","label":"Use logarithms","steps":[{"type":"text","order":0,"title":"Differentiate the model","content":"The constant term $2000$ differentiates to $0$. For the exponential term, use $\frac{d}{dt}(e^{kt})=ke^{kt}$, so the rate of change of value is\n\n$$\n\\frac{dV}{dt}=-0.2(17100)e^{-0.2t}=-3420e^{-0.2t}.\n$$\n\nThis earns the first method mark. The negative sign shows that the value is decreasing.","highlights":[{"text":"decreasing at a rate of £1000 per year","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\left.\\frac{dV}{dt}\\right|_{t=T}=\\htmlData{anchor=rate_negative}{-3420e^{-0.2T}}$"},{"latex":"$$\\htmlData{anchor=rate_negative_eq}{-3420e^{-0.2T}}= -1000$"},{"latex":"$$\\htmlData{anchor=rate_positive}{3420e^{-0.2T}}=1000$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rate_negative"},{"color":"green","style":"text-color","anchor":"rate_positive"}],"connections":[{"to":"rate_positive","from":"rate_negative_eq","color":"green","label":"multiply by −1"}]},"order":1,"title":"Apply the rate","description":"At t = T, the derivative is negative because the value is decreasing. Set it equal to −1000, then multiply both sides by −1. Obtaining 3420e^(−0.2T) = 1000 earns the A1 mark: 1000 is the positive magnitude of the decrease, while the derivative itself is −1000."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=coefficient}{3420e^{-0.2T}}=1000$"},{"latex":"$$\\htmlData{anchor=isolated_exp}{e^{-0.2T}}=\\frac{1000}{3420}$"},{"latex":"$$\\htmlData{anchor=simpl

1.9 F: Exponentials and logarithms

Question 28

1.9 F: Exponentials and logarithms Questions

1.9 F: Exponentials and logarithms

Question 28

1.9 F: Exponentials and logarithms Questions