June 2024 Paper 2 Q4

OCR ACurrent spec10 marksIntegration Numerical Methods

4 The diagram shows part of the graph of \(y = x\mathrm{e}^{1-3x}\).

Graph of y = x e to the power 1 minus 3x for x from 0 to 2: rises from the origin to a maximum of about 0.33 near x = 0.33, then decreases towards the x-axis; grid lines at 0.5 intervals

(a) Use the sign change method to determine, correct to 2 decimal places, the root of the equation \(x\mathrm{e}^{1-3x} - 0.2 = 0\), that lies between \(x = 0.5\) and \(x = 1\). [3]

(b) Determine the exact \(x\)-coordinate of the maximum point of the curve \(y = x\mathrm{e}^{1-3x}\). [3]

(c) In this question you must show detailed reasoning.
Determine the exact area of the region enclosed by the curve \(y = x\mathrm{e}^{1-3x}\), the \(x\)-axis and the line \(x = 1\). [4]

Mark scheme (a)

Scheme	Marks	AO
\(x\mathrm{e}^{1-3x} - 0.2\) evaluated for any two values of \(x\) that give results with opposite signs.	M1	3.1a
\(x = 0.79\) (2dp)	B1	1.1
(\(x = 0.79\) to 2 dp) because the change of sign occurs between 0.785 and 0.795.	A1	2.2a
	[3]

Notes

M1: Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc. but signs (and values if given) must be correct to 1sf – see table.

B1: cao (Allow this mark even with no/insufficient working).

A1: By showing two values between 0.785-0.795 with opposite signs
Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc, but if given must be correct to 1sf
e.g. \(x = 0.79 \Rightarrow y = 0.0007\) and \(x = 0.795 \Rightarrow y = -0.0009\)

\(x\)	\(x\mathrm{e}^{1-3x} - 0.2\)
0.1	0.00138
0.5	0.10327
0.7	0.03301
0.75	0.01488
0.78	0.00424
0.785	0.00249
0.786	0.00214
0.787	0.00179
0.788	0.00144
0.789	0.00109
0.79	0.00074
0.791	0.00040
0.792	0.00005
0.793	−0.00030
0.794	−0.00065
0.795	−0.00099
0.796	−0.00134
0.797	−0.00169
0.798	−0.00203
0.799	−0.00238
0.8	−0.00272
1	−0.06466

(Note that exact root is 0.792…)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{1-3x} + x(-3)\mathrm{e}^{1-3x}\ (= 0)\)	M1	3.1a
\(\mathrm{e}^{1-3x}\) is never 0, hence can divide by it \(\left(\mathrm{e}^{1-3x}(1 - 3x) = 0\right)\) \((\rightarrow\ 1 - 3x = 0)\)	B1	2.1
\(x = \dfrac{1}{3}\)	A1	1.1
	[3]

Notes

M1: For an attempt at differentiating using the product rule, with at least one term correct (their derivative must have two terms).

B1: May be implied by not giving a corresponding solution (provided another solution for \(x\) is reached). (NB this mark can be gained following M0 provided their derivative has \(\mathrm{e}^{1-3x}\) as a factor)

A1: Not decimal, www (i.e. must come from a correct derivative)

Mark scheme (c)

Scheme	Marks	AO
DR \(\displaystyle\int_0^1 x\mathrm{e}^{1-3x}\,\mathrm{d}x = \left[x\frac{\mathrm{e}^{1-3x}}{-3}\right]_0^1 - \int_0^1 \frac{\mathrm{e}^{1-3x}}{-3}\,\mathrm{d}x\) oe	M1	3.1a
A1 for both terms correct	A1	2.1
\(\displaystyle = \left[x\frac{\mathrm{e}^{1-3x}}{-3}\right]_0^1 - \left[\frac{\mathrm{e}^{1-3x}}{9}\right]_0^1\)	A1	2.1
\(= \dfrac{1}{9}\mathrm{e} - \dfrac{4}{9}\mathrm{e}^{-2}\) or \(\dfrac{\mathrm{e}^3 - 4}{9\mathrm{e}^2}\) oe	A1	1.1
	[4]

Notes

M1: M1 for attempting integration by parts with at least one term correct, must see limits (condone swapped limits, may appear later). Must have two terms.

A1: A1 for both terms correct after 2^nd integral performed

A1: Need not be simplified, isw any incorrect attempts to simplify