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Date	May 2022	Marks available	6	Reference code	22M.2.AHL.TZ2.10
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	10	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 $A, h_{A} cm$ , at time $t$ weeks can be modelled by the function $h_{A} (t) = \sin (2 t + 6) + 9 t + 27$ , where $0 \leq t \leq 9$ .

The scientist found that the height of Plant $B, h_{B} cm$ , at time $t$ weeks can be modelled by the function $h_{B} (t) = 8 t + 32$ , where $0 \leq t \leq 9$ .

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 $h_{A} (t) = h_{B} (t)$ .

[3]

For $t > 6$ , prove that Plant $A$ was always taller than Plant $B$ .

[3]

For $0 \leq t \leq 9$ , 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]

Markscheme

$32$ (cm) A1

[1 mark]

a.i.

$h_{A} (0) = \sin (6) + 27$ (M1)

$= 26.7205 \dots$

$= 26.7$ (cm) A1

[2 marks]

a.ii.

attempts to solve $h_{A} (t) = h_{B} (t)$ for $t$ (M1)

$t = 4.0074 \dots, 4.7034 \dots, 5.88332 \dots$

$t = 4.01, 4.70, 5.88$ (weeks) A2

[3 marks]

$h_{A} (t) - h_{B} (t) = \sin (2 t + 6) + t - 5$ A1

EITHER

for $t > 6, t - 5 > 1$ A1

and as $\sin (2 t + 6) \geq - 1 \Rightarrow h_{A} (t) - h_{B} (t) > 0$ R1

the minimum value of $\sin (2 t + 6) = - 1$ R1

so for $t > 6, h_{A} (t) - h_{B} (t) = t - 6 > 0$ A1

THEN

hence for $t > 6$ , Plant $A$ was always taller than Plant $B$ AG

[3 marks]

recognises that $h_{A}' (t)$ and $h_{B}' (t)$ are required (M1)

attempts to solve $h_{A}' (t) = h_{B}' (t)$ for $t$ (M1)

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

Note: Award full marks for $t = \frac{4 π}{3} - 3, \frac{5 π}{3} - 3, (\frac{7 π}{3} - 3, \frac{8 π}{3} - 3 \frac{10 π}{3} - 3, \frac{11 π}{3} - 3)$ .

Award subsequent marks for correct use of these exact values.

$1.18879 \dots < t < 2.23598 \dots$ OR $4.33038 \dots < t < 5.37758 \dots$ OR $7.47197 \dots < t < 8.51917 \dots$ (A1)

attempts to calculate the total amount of time (M1)

$3 (2.2359 \dots - 1.1887 \dots) (= 3 ((\frac{5 π}{3} - 3) - (\frac{4 π}{3} - 3)))$

$= 3.14 (= π)$ (weeks) A1

[6 marks]

Examiners report

Part (a) In general, very well done, most students scored full marks. Some though had an incorrect answer for part(a)(ii) because they had their GDC in degrees.

Part (b) Well attempted. Some accuracy errors and not all candidates listed all three values.

Part (c) Most students tried a graphical approach (but this would only get them one out of three marks) and only some provided a convincing algebraic justification. Many candidates tried to explain in words without a convincing mathematical justification or used numerical calculations with specific time values. Some arrived at the correct simplified equation for the difference in heights but could not do much with it. Then only a few provided a correct mathematical proof.

Part (d) In general, well attempted by many candidates. The common error was giving the answer as 3.15 due to the pre-mature rounding. Some candidates only provided the values of time when the rates are equal, some intervals rather than the total time.

a.i.

[N/A]

a.ii.

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.1—Introduction of differential calculus

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