P6.1 Exponential functions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates’ performances in this exponential modelling question were varied with some able to obtain full or nearly full marks, whilst others made very little progress. Candidates who did Many candidates realised that in (b) the required limit to the height of the tree was 31m, although many omitted the units losing the mark. For some candidates this was the only mark scored in this question.

Common mistakes

• as t → ∞, e^(−kt) → 0 so h → 31 m (units required)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested the candidates’ skill in various areas, i.e. finding partial fractions, integration, rules of logarithms, manipulating equations to eliminate logarithms using exponentials, indices, and commentating on the expected behaviour of a variable as another tends to infinity. This was a challenging question which in a number of instances, candidates struggled with after part (a). Those however, who were well prepared, were able to score full marks often with relatively concise solutions. In part (d), candidates sometimes identified that L = 25, even when little else of credit was presented throughout, but usually those who had not attempted part (c) also made no attempt at (d). The reasons given were often not rigorous enough to gain credit, even when 25 had been established as the required value of L. When correct explanations were seen they usually stated that L = 25 because 2.5 e 0 t - → as t → or that L = 25 because d 0. d V t =

Common mistakes

• L = 25
• reason: as t → ∞, e^(−2.5t) → 0 (or dV/dt = 0 when V = 25)
• reason must be rigorous

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on exponential functions caused problems for the majority in part (b), as they failed to recognise that differentiation was required. However, generally parts (a) and (c) were answered well. In part (a), most candidates identified that they needed to substitute t = 0 into the given equation and successfully reached h = 0.6 metres, which was often seen as 60cm or 3 5 , both of which were allowed. A few failed to use e0 = 1 and as a result arrived at an incorrect answer.

Common mistakes

• t = 0 and e⁰ = 1 gives h = 0.6 m (60 cm)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on exponential functions caused problems for the majority in part (b), as they failed to recognise that differentiation was required. However, generally parts (a) and (c) were answered well. In part (c), most candidates identified that the limiting value was 2.3 m, although a number missed this part out, especially after a poor attempt at part (b).

Common mistakes

• as t → ∞, e^(−0.2t) → 0 so the limit is 2.3 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was reasonably well attempted. Logs continue to be an area of challenge for many candidates but despite this, a significant proportion were able to earn a good number of marks here. In part (a), many candidates were able to earn the mark for recognising that it was necessary to set 0 t = and correctly obtained 0.8 k = . The most common error here occurred when candidates assumed, in error, that 0e 0 = and so obtained 2.2 k =

Common mistakes

• t = 0 gives k + 1.4 = 2.2, k = 0.8
• e⁰ = 1, not 0 (giving k = 2.2)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on an exponential function saw full marks for most. Of the small number of errors that were seen in part (a), a common one was an incorrect intercept of (0, 4) from replacing 40 with 4 instead of 1. The intercept was occasionally seen as (1, 0) although this was condoned if it was positioned correctly on the sketch. A few graphs did not go into the second quadrant and others were not carefully drawn – some did not get close enough to the asymptote of y = 0, sometimes due to a poor choice of scale. Other graphs bent back on themselves in the first quadrant. Some students offered an incorrect or extra horizontal or even a vertical asymptote. Although almost all knew the correct shape, a few responses had a straight line graph.

Common mistakes

• intercept (0, 1) not (0, 4)
• curve in quadrants 1 and 2 only, asymptotic to the x-axis (no other asymptote)
• must not bend back or be U-shaped

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was generally answered well by most candidates. It was typically silly mistakes or not reading the question which resulted in full marks not being scored. In part (a), most candidates were successful in finding the area to be 35 km2. Some just opted for 80 and the units were also incorrect, at times, such as being given as trees.

Common mistakes

• t = 0 gives A = 80 − 45 = 35 km² (not 80)
• correct units needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was another which focussed on problem solving, with the majority of candidates only able to score around half of the marks available. Part (d), in particular proved to be too difficult for most candidates so it was rare to see full marks. In part (a), many candidates could score at least one mark here by calculating the value of 83, however, it was not uncommon for units to be omitted which meant that the mark was not awarded.

Common mistakes

• t = 0 gives 83 °C
• units needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was another which focussed on problem solving, with the majority of candidates only able to score around half of the marks available. Part (d), in particular proved to be too difficult for most candidates so it was rare to see full marks. Most candidates were successful in part (c), which may have been due to the graph having been provided. Many stated 18 C as being the minimum temperature of the model which was condoned for the mark, although a significant number stated that “15 C is the room temperature” which did not show an understanding of the model. The other most popular way to answer the question was to substitute 15 into the equation and concluded that you cannot log a negative. Just stating that “you can’t log a negative” without providing the working to 8 back this up was not sufficient as it was not enough evidence that the candidate had engaged with the model.

Common mistakes

• as t → ∞ θ → 18 from above (minimum 18 °C)
• "cannot log a negative" needs the working 15 − 18 = −3
• "15 °C is room temperature" scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was another which focussed on problem solving, with the majority of candidates only able to score around half of the marks available. Part (d), in particular proved to be too difficult for most candidates so it was rare to see full marks. Part (d) caused more issues than the rest of the question for candidates. Several of them assumed A was equal to 94 and therefore did not score highly in this part. Some candidates attempted to find the gradient between the given coordinates and therefore did not score any marks. However, once candidates formed the two simultaneous equations, they seemed to find it quite easy to use the elimination method to eliminate A, and then go on to correctly solve for A and B and therefore work out the equation of the asymptote.

Common mistakes

• form A + B = 94 and A + Be^(−1) = 50 and solve
• asymptote μ = A ≈ 24.4 (not A = 94)
• gradient between the points is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Parts (a) and (b) of question 9 were standard bookwork involving the product rule of differentiation using an exponential function. Well rehearsed candidates were very successful here scoring the majority of marks. The diagram should have helped in determining the maximum and minimum points of the function.

Common mistakes

• f′(x) = 0 when 2 + x − x² = 0 (e^(−2x) is never 0), x = −1 and 2
• exact points (−1, −4e²) and (2, 8e⁻⁴)
• extra point from x = 0 penalised

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general parts (a), (c) and (d) were well attempted, with many scoring the three available marks here and nothing else. A good understanding of the exponential graph was clear in part (c) as the vast majority scored the mark here for (£)2300. Occasionally 500 was seen here, perhaps misinterpreting the question. Responses to (d) were many and varied. Frequent correct answers included mentioning damage to the vehicle, variability in mileage, and the potential for the car to become an antique. For those who weren’t awarded the mark the most popular responses were that it only made predictions based on year, not months or days, claiming the car’s value may not follow a linear pattern of decrease and predicting that the car’s value will eventually become zero.

Common mistakes

• as t → ∞ the exponential term → 0, so A = 2300 (not 500)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between students of all abilities with part (a) and (b) more accessible to students compared to part (c) and (d). Part (b) was the most successfully answered part of the question. Students realised that the graph of |f(x)| should be above the x-axis. Where students did not gain full credit, this was usually due to poor drawing of loops (not decreasing heights, or rounding at the cusps). A few students chose to draw over the original figure which made it difficult in many cases to determine which pieces formed their graph. They would be well advised to sketch separate graphs in similar questions.

Common mistakes

• modulus graph: negative loops reflected above the axis
• cusps (not rounded) on the t-axis
• decreasing heights