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Date May 2022 Marks available 2 Reference code 22M.2.SL.TZ2.8
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 8 Adapted from N/A

Question

A scientist conducted a nine-week experiment on two plants, A and B, of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant A was given fertilizer regularly, while Plant B was not.

The scientist found that the height of Plant A, hA cm, at time t weeks can be modelled by the function hA(t)=sin(2t+6)+9t+27, where 0t9.

The scientist found that the height of Plant B, hB cm, at time t weeks can be modelled by the function hB(t)=8t+32, where 0t9.

Use the scientist’s models to find the initial height of

Plant B.

[1]
a.i.

Plant A correct to three significant figures.

[2]
a.ii.

Find the values of t when hAt=hBt.

[3]
b.

For 0t9, find the total amount of time when the rate of growth of Plant B was greater than the rate of growth of Plant A.

[6]
c.

Markscheme

32 (cm)          A1

 

[1 mark]

a.i.

hA0=sin6+27          (M1)

=26.7205

=26.7 (cm)          A1

 

[2 marks]

a.ii.

attempts to solve hAt=hBt for t          (M1)

t=4.00746,4.70343,5.88332

t=4.01,4.70,5.88 (weeks)          A2

 

[3 marks]

b.

recognises that hA't and hB't are required          (M1)

attempts to solve hA't=hB't for t          (M1)

t=1.18879 and 2.23598  OR  4.33038 and 5.37758   OR  7.47197 and 8.51917          (A1)

 

Note: Award full marks for t=4π3-3, 5π3-3, 7π3-3, 8π3-3 10π3-3, 11π3-3.

Award subsequent marks for correct use of these exact values.

 

1.18879<t<2.23598  OR  4.33038<t<5.37758  OR

7.47197<t<8.51917          (A1)

attempts to calculate the total amount of time          (M1)

32.2359-1.1887  =35π3-3-4π3-3

=3.14 =π (weeks)          A1

 

[6 marks]

c.

Examiners report

Many students did not change their calculators back to radian mode. This meant they had no chance of correctly answering parts (c) and (d), since even if follow through was given, there were not enough intersections on the graphs.

Most managed part (a) and some attempted to equate the functions in part b) but few recognised that 'rate of growth' was the derivatives of the given functions, and of those who did, most were unable to find them.

Almost all the candidates who did solve part (c) gave the answer 3×1.05=3.15, when working with more significant figures would have given them 3.14. They lost the last mark.

a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology
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Topic 2—Functions

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