October 2020 Paper 1 Q4

EdexcelCurrent spec5 marksFunctions (including |mod|)

4. The function f is defined by\[\mathrm{f}(x) = \frac{3x-7}{x-2} \qquad x \in \mathbb{R},\ x \neq 2\]

(a) Find \(\mathrm{f}^{-1}(7)\) (2)

(b) Show that \(\mathrm{ff}(x) = \dfrac{ax+b}{x-3}\) where \(a\) and \(b\) are integers to be found. (3)

Mark scheme (a)

Scheme	Marks	AO
Either attempts \(\dfrac{3x-7}{x-2} = 7 \Rightarrow x = \ldots\) Or attempts \(\mathrm{f}^{-1}(x)\) and substitutes in \(x = 7\)	M1	3.1a
\(\dfrac{7}{4}\) oe	A1	1.1b
	(2)

Notes

M1: For either attempting to solve \(\dfrac{3x-7}{x-2} = 7\). Look for an attempt to multiply by the \((x-2)\) leading to a value for \(x\).
Or score for substituting in \(x = 7\) in \(\mathrm{f}^{-1}(x)\). FYI \(\mathrm{f}^{-1}(x) = \dfrac{2x-7}{x-3}\)
The method for finding \(\mathrm{f}^{-1}(x)\) should be sound, but you can condone slips.

A1: \(\dfrac{7}{4}\)

Mark scheme (b)

Scheme	Marks	AO
Attempts \(\mathrm{ff}(x) = \dfrac{3 \times \left(\dfrac{3x-7}{x-2}\right) - 7}{\left(\dfrac{3x-7}{x-2}\right) - 2} = \dfrac{3 \times (3x-7) - 7(x-2)}{3x - 7 - 2(x-2)}\)	M1, dM1	1.1b 1.1b
\(= \dfrac{2x-7}{x-3}\)	A1	2.1
	(3)
	(5 marks)

Notes

M1: For an attempt at fully substituting \(\dfrac{3x-7}{x-2}\) into \(\mathrm{f}(x)\). Condone slips but the expression must have a correct form. E.g. \(\dfrac{3 \times \left(\frac{* - *}{* - *}\right) - a}{\left(\frac{* - *}{* - *}\right) - b}\) where \(a\) and \(b\) are positive constants.

dM1: Attempts to multiply all terms on the numerator and denominator by \((x-2)\) to create a fraction \(\dfrac{P(x)}{Q(x)}\) where both \(P(x)\) and \(Q(x)\) are linear expressions. Condone \(\dfrac{P(x)}{Q(x)} \times \dfrac{x-2}{x-2}\)

A1: Reaches \(\dfrac{2x-7}{x-3}\) via careful and accurate work. Implied by \(a = 2, b = -7\) following correct work.

Methods involving \(\dfrac{3x-7}{x-2} \equiv a + \dfrac{b}{x-2}\) may be seen. The scheme can be applied in a similar way
FYI \(\dfrac{3x-7}{x-2} \equiv 3 - \dfrac{1}{x-2}\)