June 2023 Paper 1 Q11

OCR MEICurrent spec10 marksLogs & Exponentials Modelling

11 The height \(h\) cm of a sunflower plant \(t\) days after planting the seed is modelled by \(h = a + b\ln t\) for \(t \geqslant 9\), where \(a\) and \(b\) are constants. The sunflower is 10 cm tall 10 days after planting and 200 cm tall 85 days after planting.

(a)

(i) Show that the value of \(b\) which best models these values is 88.8 correct to 3 significant figures. [2]

(ii) Find the corresponding value of \(a\). [1]

(b)

(i) Explain why the model is not suitable for small positive values of \(t\). [1]

(ii) Explain why the model is not suitable for very large positive values of \(t\). [1]

(c) Show that the model indicates that the sunflower grows to 1 m in height in less than half the time it takes to grow to 2 m. [2]

(d) Find the value of \(t\) for which the rate of growth is 3 cm per day. [3]

Mark scheme (a)

Scheme	Marks	AO
(i) Substitute \(t = 10,\ h = 10\) and \(t = 85,\ h = 200\) \(10 = a + b\ln 10\) \(200 = a + b\ln 85\)	M1	3.3
Solve simultaneous equations to give \(b = \left[\dfrac{190}{\ln\frac{85}{10}} = \dfrac{190}{2.14}\right] = 88.8\)	E1	2.1
	[2]
(ii) \(a = -194\)	B1	3.3
	[1]

Notes

(i) M1: Forms two equations and attempt to solve simultaneously (BC)
Allow if \(10 = a + 2.303b\) and \(200 = a + 4.443b\) used

E1: AG must be 3 s.f.
If by simultaneous equations solved BC, 88.78…..or better must also be seen

(ii) B1: Accept awrt \(-194\) or \(-195\)

Mark scheme (b)

Scheme	Marks	AO
(i) For small values of \(t\) the model for \(h\) predicts a negative height [which is not possible]	E1	3.5b
	[1]
(ii) The model predicts that the sunflower would continue to increase in height for ever, which is not possible	E1	3.5b
	[1]

Notes

(i) E1: argument based on negativity

(ii) E1: argument based on contrast between ever increasing height predicted by the model and reality

Mark scheme (c)

Scheme	Marks	AO
height at \(\frac{85}{2}\) days \(a + 88.8\ln\frac{85}{2}\) cm	M1	3.4
[using given answers above] 139 cm which is more than 1 m	E1	2.2a
	[2]

Notes

M1: Also allow for 84.5 days used for 85

E1: Established using a value of \(h\) between 137.9 and 139 needed.

Alternative method

Scheme	Marks
time to reach 1 m: \(100 = a + 88.8\ln t\)	M1
[using given answers above] 27.4 days which is less than \(\frac{85}{2}\)	E1

M1: Equate to 100 and solve for \(t\). Condone \(h = 1\) used

E1: Established using a value of \(t\) between 27 and 28 needed.

Mark scheme (d)

Scheme	Marks	AO
rate of growth \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{b}{t}\)	M1	3.1b
rate 3 cm per day when \(\dfrac{b}{t} = 3\)	M1	3.4
\(t = \dfrac{b}{3} = 29.6\)	A1	1.1b
	[3]

Notes

M1: Attempt to differentiate to give expression of the form \(\frac{k}{t}\)

M1: equates their derivative to 3

A1: allow 29 or 30 days