June 2022 Paper 3 Q5

OCR ACurrent spec14 marksDifferentiation Numerical Methods

5 In this question you must show detailed reasoning.

Curve y = (2x − 3)/(4x² + 1) with a tangent at point P, which lies below the x-axis for small positive x; the curve has a minimum just right of the y-axis and approaches the x-axis at both ends

The diagram shows the curve with equation \(y = \dfrac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.

(a) Show that the \(x\)-coordinate of \(P\) satisfies the equation \[4x^3 + 3x - 3 = 0.\] [5]

(b) Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]

(c) Show that the iteration \[x_{n+1} = \frac{3 - 4x_n^3}{3}\] cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]

(d) Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]

Mark scheme (a)

Scheme	Marks	AO
DR \(y = (2x - 3)(4x^2 + 1)^{-1}\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(4x^2 + 1)^{-1} + (2x - 3)(-1)(4x^2 + 1)^{-2}(8x)\)	M1*	1.1
\(y = \dfrac{2x - 3}{4x^2 + 1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(4x^2 + 1)(2) - (2x - 3)(8x)}{(4x^2 + 1)^2}\)	A1	1.1
\(\dfrac{2(1 + 12x - 4x^2)}{(4x^2 + 1)^2} = 2\)	M1dep*	3.1a
\(1 + 12x - 4x^2 = (4x^2 + 1)^2 \Rightarrow 16x^4 + 12x^2 - 12x = 0\)	M1	1.1
\(x(4x^3 + 3x - 3) = 0 \Rightarrow 4x^3 + 3x - 3 = 0\) as \(x \ne 0\)	A1	2.3
	[5]

Notes

M1*: Attempt use of quotient rule or equivalent (e.g. product rule). Condone one incorrect term only (of the five terms) but must be subtraction in the numerator (but allow subtraction the wrong way round); condone absence of brackets; no denominator (if using quotient rule) is M0
By the five terms we mean the four in the numerator and the fifth is the term in the denominator

A1: cao must include brackets as necessary
Any correct equivalent form

M1dep*: Sets their derivative (in any form) equal to 2 (M0 if equating to normal gradient)
May equate at any stage (even after incorrect manipulation of their derivative)

M1: Multiply both sides by \((4x^2 + 1)^2\) and simplify (so combining like terms) to obtain a quartic equation (must be expanded with at least three terms – condone lack of = 0 if all terms on the same side) – allow sign errors/minor slips but the expansion of \((4x^2 + 1)^2\) must be three terms of the form \(16x^4 + ax^2 + 1\) where \(a = \pm 4, \pm 8\)
Dependent on both previous M marks

A1: AG with explicit rejection of \(x = 0\) – as a minimum must indicate that \(x\) cannot equal 0
Just cancelling \(x\) is A0

Mark scheme (b)

Scheme	Marks	AO
DR Consider both \(\mathrm{f}(0.5)\) and \(\mathrm{f}(1)\) Where \(\mathrm{f}(x) = \pm(4x^3 + 3x - 3)\)	M1	1.1
\(\mathrm{f}(0.5) = -1 \lt 0\) and \(\mathrm{f}(1) = 4 \gt 0\) (or \(\mathrm{f}(0.5) = 1 \gt 0\) and \(\mathrm{f}(1) = -4 \lt 0\)) Change of sign indicates that the \(x\)-coordinate lies between 0.5 and 1	A1	2.4
	[2]

Notes

M1: Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen
Just stating that \(\mathrm{f}(0.5) \lt 0\) and \(\mathrm{f}(1) \gt 0\) is M0

A1: Correct values together with explanation (change of sign) and correct conclusion (as a minimum ‘root’ oe)

Alternative

Scheme	Marks
Considers both \(\mathrm{g}(0.5)\) and \(\mathrm{g}(1)\) where \(\mathrm{g}(x) = \dfrac{(4x^2 + 1)(2) - (2x - 3)(8x)}{(4x^2 + 1)^2}\)	M1
\(\mathrm{g}(0.5) = 3 \gt 2\) and \(\mathrm{g}(1) = 0.72 \lt 2\) Values either side of 2 indicates that the \(x\)-coordinate lies between 0.5 and 1	A1

M1: Must be using the correct derivative. Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen
Just stating that \(\mathrm{g}(0.5) \gt 2\) and \(\mathrm{g}(1) \lt 2\) is M0

A1: Correct values together with explanation (values either side of 2) and correct conclusion (as a minimum ‘root’ oe)

Mark scheme (c)

Scheme	Marks	AO
DR Let \(\mathrm{h}(x) = \dfrac{3 - 4x^3}{3} \Rightarrow \mathrm{h}'(x) = -4x^2\)	B1*	2.1
As the root \(\alpha\) lies in the interval \((0.5, 1) \Rightarrow \mathrm{h}'(\alpha) \lt -1\) so iterative formula cannot converge to the \(x\)-coordinate of \(P\)	B1dep*	2.2a
	[2]

Notes

B1*: Calculates correct derivative of rhs of given iterative formula

B1dep*: Correct explanation that any value in the given interval gives a gradient which is less than \(-1\)
No marks for just showing that the iteration doesn’t converge using different starting values

Mark scheme (d)

Scheme	Marks	AO
DR \(\mathrm{f}(x_n) = 4x_n^3 + 3x_n - 3 \Rightarrow \mathrm{f}'(x_n) = 12x_n^2 + 3\)	B1	1.1
\(x_{n+1} = x_n - \left\{\dfrac{4x_n^3 + 3x_n - 3}{12x_n^2 + 3}\right\}\)	M1	2.1
\(x_0 = 0.5\), \(x_1 = \frac{2}{3}\) or 0.666666…, \(x_2 = \frac{29}{45}\) or 0.644444…, \((x_3 = 0.64395510\ldots)\)	A1	1.1
\(x\) coordinate of \(P\) is 0.64395	A1	2.2a
\(y\) coordinate of \(P\) is \(-0.64395\)	B1	1.1
	[5]

Notes

B1: Correct derivative (possibly seen in N-R formula)
Condone \(x\) for \(x_n\) oe

M1: Correct N-R formula seen with correct \(\mathrm{f}(x_n)\) and their \(\mathrm{f}'(x_n)\) substituted
Condone \(x\) for \(x_n\) oe

A1: First two iterations correctly stated to at least 5 decimal places (or exact) (truncated or rounded)
The correct first two iterations can imply B1 M1

A1: Independent of previous A mark (but must have scored B1 M1) – must be stated to exactly 5 decimal places
This A mark does not imply the previous A mark

B1: Independent of all previous marks – must be stated to exactly 5 decimal places
The correct answers with no evidence of N-R (e.g. no iterations stated and no N-R formula) then B0M0A0A0B1 max.