C3 June 2007 Q2

EdexcelOld spec10 marksAlgebraic Fractions Differentiation

2.\[\mathrm{f}(x) = \frac{2x + 3}{x + 2} - \frac{9 + 2x}{2x^2 + 3x - 2}, \qquad x > \frac{1}{2}.\]

(a) Show that \(\mathrm{f}(x) = \dfrac{4x - 6}{2x - 1}\). (7)

(b) Hence, or otherwise, find \(\mathrm{f}^{\prime}(x)\) in its simplest form. (3)

Mark scheme (a)

Scheme	Marks
\(2x^2 + 3x - 2 = (2x - 1)(x + 2)\) at any stage	B1
\(\mathrm{f}(x) = \dfrac{(2x + 3)(2x - 1) - (9 + 2x)}{(2x - 1)(x + 2)}\) f.t. on error in denominator factors (need not be single fraction)	M1, A1ft
Simplifying numerator to quadratic form \(\left[= \dfrac{4x^2 + 4x - 3 - 9 - 2x}{(2x - 1)(x + 2)}\right]\)	M1
Correct numerator \(= \dfrac{4x^2 + 2x - 12}{\left[(2x - 1)(x + 2)\right]}\)	A1
Factorising numerator, with a denominator \(= \dfrac{2(2x - 3)(x + 2)}{(2x - 1)(x + 2)}\) o.e.	M1
\(\left[= \dfrac{2(2x - 3)}{2x - 1}\right]\) \(= \dfrac{4x - 6}{2x - 1}\) (✱)	A1 cso
	(7)

Notes

1^st M1 in either version is for correct method

1^st A1 Allow \(\dfrac{2x + 3(2x - 1) - (9 + 2x)}{(2x - 1)(x + 2)}\) or \(\dfrac{(2x + 3)(2x - 1) - 9 + 2x}{(2x - 1)(x + 2)}\) or \(\dfrac{2x + 3(2x - 1) - 9 + 2x}{(2x - 1)(x + 2)}\) (fractions)

2^nd M1 in (main a) is for forming 3 term quadratic in numerator

3^rd M1 is for factorising their quadratic (usual rules); factor of 2 need not be extracted

(✱) A1 is given answer so is cso

Alternative

\(2x^2 + 3x - 2 = (2x - 1)(x + 2)\) at any stage	B1
\(\mathrm{f}(x) = \dfrac{(2x + 3)(2x^2 + 3x - 2) - (9 + 2x)(x + 2)}{(x + 2)(2x^2 + 3x - 2)}\)	M1A1 f.t.
\(= \dfrac{4x^3 + 10x^2 - 8x - 24}{(x + 2)(2x^2 + 3x - 2)}\)
\(= \dfrac{2(x + 2)(2x^2 + x - 6)}{(x + 2)(2x^2 + 3x - 2)}\) or \(\dfrac{2(2x - 3)(x^2 + 4x + 4)}{(x + 2)(2x^2 + 3x - 2)}\) o.e. Any one linear factor \(\times\) quadratic factor in numerator	M1, A1
\(= \dfrac{2(x + 2)(x + 2)(2x - 3)}{(x + 2)(2x^2 + 3x - 2)}\) o.e.	M1
\(= \dfrac{2(2x - 3)}{2x - 1}\) \(\dfrac{4x - 6}{2x - 1}\) (✱)	A1

(corrected from the printed mark scheme: the second form in the fourth line is printed with denominator \((x + 2)(2x^2 + 3x + 2)\); the denominator is \((x + 2)(2x^2 + 3x - 2)\) throughout)

Alt (a) 3^rd M1 is for factorising resulting quadratic
Notice that B1 likely to be scored very late but on ePen scored first

Mark scheme (b)

Scheme	Marks
Complete method for \(\mathrm{f}^{\prime}(x)\); e.g. \(\mathrm{f}^{\prime}(x) = \dfrac{(2x - 1)\times 4 - (4x - 6)\times 2}{(2x - 1)^2}\) o.e	M1 A1
\(= \dfrac{8}{(2x - 1)^2}\) or \(8(2x - 1)^{-2}\)	A1
Not treating \(\mathrm{f}^{-1}\) (for \(\mathrm{f}^{\prime}\)) as misread
	(3)
	(10 marks)

Notes

SC: For M allow \(\pm\) given expression or one error in product rule

Alt: Attempt at \(\mathrm{f}(x) = 2 - 4(2x - 1)^{-1}\) and diff. M1; \(k(2x - 1)^{-2}\) A1; A1 as above

Accept \(8(4x^2 - 4x + 1)^{-1}\). Differentiating original function – mark as scheme.

(corrected from the printed mark scheme: printed as “Accept \(8(4x^2 - 4x + 1)^{-2}\)”; since \((2x - 1)^2 = 4x^2 - 4x + 1\), the equivalent form is \(8(4x^2 - 4x + 1)^{-1}\))