October 2021 Paper 1 Q7

OCR ACurrent spec9 marksDifferentiation Numerical Methods

7 The curve \(y = (x^2 - 2)\ln x\) has one stationary point which is close to \(x = 1\).

(a) Show that the \(x\)-coordinate of this stationary point satisfies the equation \(2x^2\ln x + x^2 - 2 = 0\). [2]

(b) Show that the Newton-Raphson iterative formula for finding the root of the equation in part (a) can be written in the form \(x_{n+1} = \dfrac{2x_n^2\ln x_n + 3x_n^2 + 2}{4x_n(\ln x_n + 1)}\). [4]

(d) Find the coordinates of this stationary point, giving each coordinate correct to 3 decimal places. [2]

Mark scheme (a)

Scheme	Marks	AO
\(2x\ln x + \dfrac{x^2 - 2}{x}\)	M1	3.1a
\(2x\ln x + \dfrac{x^2 - 2}{x} = 0\) \(2x^2\ln x + x^2 - 2 = 0\) A.G.	A1	1.1
	[2]

Notes

M1: Attempt differentiation using product rule
May expand first to give \(2x\ln x + \dfrac{x^2}{x} - \dfrac{2}{x}\)
(allow middle term as just \(x\))

A1: Equate to 0 and obtain given answer
Must be equated to 0 before clearing the fractions
Must be equation ie … = 0

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}'(x) = 4x\ln x + 2x^2 . \frac{1}{x} + 2x\)	B1	1.1
\(x_{n+1} = x_n - \dfrac{2x_n^2\ln x_n + x_n^2 - 2}{4x_n\ln x_n + 2x_n^2 . \frac{1}{x_n} + 2x_n}\)	M1	1.1
\(x_{n+1} = \dfrac{x_n(4x_n\ln x_n + 4x_n) - (2x_n^2\ln x_n + x_n^2 - 2)}{4x_n\ln x_n + 4x_n}\)	M1	1.1
\(x_{n+1} = \dfrac{4x_n^2\ln x_n + 4x_n^2 - 2x_n^2\ln x_n - x_n^2 + 2}{4x_n\ln x_n + 4x_n}\)
\(x_{n+1} = \dfrac{2x_n^2\ln x_n + 3x_n^2 + 2}{4x_n(\ln x_n + 1)}\) A.G.	A1	2.1
	[4]

Notes

B1: Correct derivative seen
Allow simplified middle term of \(2x\)

M1: Use correct Newton-Raphson formula, with numerator correct and their derivative in the denominator
Allow fractional term without subscripts
SC Condone use of N-R on \((x^2 - 2)\ln x\)

M1: Attempt rearrangement into single fraction with brackets expanded
Allow without subscripts
N-R not necessarily correct, but must be recognisable attempt
SC Rearrange their N-R on \((x^2 - 2)\ln x\)

A1: Obtain given answer, with no errors seen
Subscripts needed on RHS at least one step before AG
LHS needs \(x_{n+1}\) seen

Scheme	Marks	AO
\(x_2 = 1.25\), \(x_3 = 1.2075\)	B1	1.1
	[1]

Mark scheme (d)

Scheme	Marks	AO
\((1.206,\ -0.102)\)	B1 B1	2.2a 2.2a
	[2]

Notes

B1: Correct \(x\)-coordinate
B1: Correct \(y\)-coordinate

Must be 3dp or better
Could be given as single coordinate or \(x = 1.206\), \(y = -0.102\)
Allow BOD if 1.206 given but not identified as \(x\)-value