October 2021 Paper 3 Q8

OCR ACurrent spec11 marksDifferentiation Integration

Curve M through the origin O rising to a maximum and then decreasing towards the x-axis; straight line L through O meets M again at P; the region R between the curve and the line, from O to P, is shaded

The diagram shows the curve \(M\) with equation \(y = x\mathrm{e}^{-2x}\).

(a) Show that \(M\) has a point of inflection at the point \(P\) where \(x = 1\). [5]

The line \(L\) passes through the origin \(O\) and the point \(P\). The shaded region \(R\) is enclosed by the curve \(M\) and the line \(L\).

(b) Show that the area of \(R\) is given by \[\tfrac{1}{4}\left(a + b\mathrm{e}^{-2}\right),\] where \(a\) and \(b\) are integers to be determined. [6]

Mark scheme (a)

Scheme	Marks	AO
\(y' = \mathrm{e}^{-2x}(1 - 2x)\)	M1* A1	2.1 1.1
\(y'' = \mathrm{e}^{-2x}(-4 + 4x)\)	A1ft	1.1
\(y'' = 0\) at \(x = 1\) and \(y''(0.5) = -2\mathrm{e}^{-1} \lt 0\), \(y''(1.5) = 2\mathrm{e}^{-3} \gt 0\) (so change of sign indicates a point of inflection at \(x = 1\))	M1dep* A1	3.1a 2.2a
	[5]

Notes

M1*: Differentiates \(y\) with respect to \(x\) – answer of the form \(\pm\mathrm{e}^{-2x} \pm \lambda x\mathrm{e}^{-2x}\)
\(\lambda \neq 0\)

A1ft: Follow through their first derivative

M1dep*: Solves \(y'' = 0\) (or attempts to verify \(y'' = 0\) by substituting \(x = 1\)) or considers sign change either side of \(y''\)

A1: Conclusion not required for this mark

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\int x\mathrm{e}^{-2x}\,\mathrm{d}x = -\frac{1}{2}x\mathrm{e}^{-2x} + \frac{1}{2}\int \mathrm{e}^{-2x}\,\mathrm{d}x\)	M1*	2.1
\(\displaystyle\int x\mathrm{e}^{-2x}\,\mathrm{d}x = -\frac{1}{2}x\mathrm{e}^{-2x} - \frac{1}{4}\mathrm{e}^{-2x}\)	A1	1.1
\(\displaystyle\int_0^1 x\mathrm{e}^{-2x}\,\mathrm{d}x = \left[-\frac{1}{2}x\mathrm{e}^{-2x} - \frac{1}{4}\mathrm{e}^{-2x}\right]_0^1\) \(= \left(-\dfrac{1}{2}\mathrm{e}^{-2} - \dfrac{1}{4}\mathrm{e}^{-2}\right) - \left(0 - \dfrac{1}{4}\right)\)	M1dep*	1.1
\(\dfrac{1}{4} - \dfrac{3}{4}\mathrm{e}^{-2}\)	A1	1.1
Area of triangle below \(OP = \dfrac{1}{2}\mathrm{e}^{-2}\)	B1	1.1
\(= \frac{1}{4}\left(1 - 5\mathrm{e}^{-2}\right)\)	A1	2.2a
	[6]

Notes

M1*: Integration by parts – of the form \(\pm\alpha x\mathrm{e}^{-2x} \pm \beta\int \mathrm{e}^{-2x}\,\mathrm{d}x\)
Where \(\alpha, \beta = 2, \frac{1}{2}\)

M1dep*: Use of correct limits in their fully integrated expression – need not be simplified (or equivalent)

A1: Allow unsimplified

B1: Or by correctly evaluating \(\int_0^1 \mathrm{e}^{-2}x\,\mathrm{d}x\)
Allow unsimplified

A1: \(a = 1,\ b = -5\) (must be in this form)