October 2021 Paper 2 Q14

OCR MEICurrent spec13 marksDifferentiation Sketching & Transforming Graphs

14 The equation of a curve is

\(y = x^2(x-2)^3\).

(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), giving your answer in factorised form. [4]

(b) Determine the coordinates of the stationary points on the curve. [4]

In part (c) you may use the result \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4(x-2)(5x^2 - 8x + 2)\).

(d) Sketch the curve. [2]

Mark scheme (a)

Scheme	Marks	AO
\(2x(x-2)^3 + x^2 \times 3(x-2)^2\)	M1 A1	3.1a 1.1
\((x-2)\) identified as factor	M1	2.1
\(x(x-2)^2(5x-4)\) cao	A1	1.1
	[4]

Notes

M1: product rule & chain rule; allow one error

Alternative

Scheme	Marks
\(5x^4 - 24x^3 + 36x^2 - 16x\)	M1 A1
\((x-2)\) identified as factor	M1
\(x(x-2)^2(5x-4)\) cao	A1

M1: NB from \(x^5 - 6x^4 + 12x^3 - 8x^2\)
expand brackets and differentiate; allow one error

Mark scheme (b)

Scheme	Marks	AO
their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) soi	M1	2.1
\(x = 0,\ 2,\ \frac{4}{5}\)	A1	1.1
(0,0) and (2.0)	A1	1.1
\((0.8,\ -1.10592)\) or \(\left(0.8,\ -\dfrac{3456}{3125}\right)\)	A1	2.2a
	[4]

Notes

A1: accept – 1.10592 to 2 sf or better

Mark scheme (c)

Scheme	Marks	AO
2^nd derivative = – 16 at (0,0) so max	B1	1.1
2^nd derivative = 5.76 at (0.8, – 1.10592) so min	B1	1.1
eg gradient = 1 at \(x = 1\) and 33 at \(x = 3\) so inflection at \(x = 2\) eg \(y = -1\) at \(x = 1\) and \(y = 9\) at \(x = 3\) so inflection at \(x = 2\) NB 2^nd derivative test is indecisive at \(x = 2\)	B1	3.1a
	[3]

Notes

or for any of the three points: considers \(y\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) either side of correct stationary point accompanied by suitable commentary; must see numerical values