June 2024 Paper 1 Q6

OCR ACurrent spec8 marksLogs & Exponentials Polynomials

6 In this question you must show detailed reasoning.

The cubic polynomial \(\mathrm{f}(x)\) is defined by \(\mathrm{f}(x) = 4x^3 - 25x^2 - 58x + 16\).

(a) Show that \(x = \tfrac{1}{4}\) is a root of the equation \(\mathrm{f}(x) = 0\). [1]

(b) Hence express \(\mathrm{f}(x)\) as the product of a linear factor and a quadratic factor, with all terms in the factors having integer coefficients. [3]

(c) Solve the equation \(4\mathrm{e}^{3y} - 25\mathrm{e}^{2y} - 58\mathrm{e}^{y} + 16 = 0\), giving each root in the form \(y = k\ln 2\) where \(k\) is a constant. [4]

Mark scheme (a)

Scheme	Marks	AO
DR \(\mathrm{f}\left(\tfrac{1}{4}\right) = 4\left(\tfrac{1}{4}\right)^3 - 25\left(\tfrac{1}{4}\right)^2 - 58\left(\tfrac{1}{4}\right) + 16\) \(= \tfrac{1}{16} - \tfrac{25}{16} - \tfrac{29}{2} + 16 = 0\)	B1	2.1
	[1]

Notes

B1: Show \(\mathrm{f}\left(\tfrac{1}{4}\right) = 0\), some detail required.
B0 for just f(0.25) = 0, but condone seeing either just the substitution or just the evaluated terms
Could use division by \((4x - 1)\) or \(\left(x - \tfrac{1}{4}\right)\) but must identify remainder of 0

Mark scheme (b)

Scheme	Marks	AO
DR \((4x - 1)(x^2 - 6x - 16)\)	B1	2.2a
Obtain complete division by \((4x - 1)\) or \((1 - 4x)\)	M1	1.1
Obtain correct product	A1	1.1
	[3]

Notes

B1: Identify factor of \((4x - 1)\)
Allow factor of \(\left(x - \tfrac{1}{4}\right)\)

M1: Must be a complete method ie attempt all 3 terms to obtain \(x^2\) and one other correct term (allow one slip in method)
Could be implied by \(A = 1\) and one other correct if using coefficient matching
Condone division by \(\left(x - \tfrac{1}{4}\right)\), to obtain \(4x^2\) and either \(-24x\) or \(-64\)

A1: Integer coefficients now required
Must be written as a product, so cannot be implied by eg correct quotient appearing following division by \((4x - 1)\) but the two factors never combined
Could be \((1 - 4x)(-x^2 + 6x + 16)\)

If division was used in part (a) then quotient must appear in part (b), but evidence for B1M1 could be in (a)
If \((x - 8)(x + 2)\) seen before the quadratic factor then both roots must be justified (eg factor theorem), otherwise M0 (but could still get B1)

Mark scheme (c)

Scheme	Marks	AO
DR \((4\mathrm{e}^y - 1)(\mathrm{e}^y - 8)(\mathrm{e}^y + 2)\) \(\mathrm{e}^y = \tfrac{1}{4},\ \mathrm{e}^y = 8,\ \mathrm{e}^y = -2\) \(y = \ln\tfrac{1}{4},\ y = \ln 8\)	M1	3.1a
\(y = -2\ln 2,\ y = 3\ln 2\)	A1	1.1
Obtain both correct solutions in required form	A1	1.1
\(\mathrm{e}^y = -2\) has no solutions as \(\mathrm{e}^y \gt 0\) for all \(y\)	B1	2.3
	[4]

Notes

M1: Attempt to find \(y\) from at least one positive root for \(\mathrm{e}^y\)
Attempt to link \(\mathrm{e}^y\) to the root(s) of the cubic in \(x\), and then solve \(\mathrm{e}^y = k\) to obtain \(y = \ln k\), where \(k\) is one of their positive roots

A1: Obtain at least one correct solution in the required form
\(y = -2\ln 2\) comes from the given root, but \(y = 3\ln 2\) must come from the correct solution of the correct quadratic

A1: Must come from the correct solution of the correct quadratic
Allow BOD if \(\ln(-2)\) also seen

B1: Reject \(\mathrm{e}^y = -2\) with a reason
Must have some reason, eg ‘\(\mathrm{e}^y\) is always positive’, ‘\(\mathrm{e}^y\) cannot be negative’, ‘cannot take log of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’
B0 for ‘math error’, ‘does not work’, ‘not possible’, N/A etc
\(\mathrm{e}^y = -2\) must come from the correct solution of the correct quadratic