C3 January 2012 Q7
7. The function f is defined by
\[\mathrm{f} : x \mapsto \frac{3(x + 1)}{2x^2 + 7x - 4} - \frac{1}{x + 4}, \qquad x \in \mathbb{R},\ x \gt \frac{1}{2}\]
\[\mathrm{g}(x) = \ln(x + 1)\]
| Scheme | Marks |
|---|---|
| \(2x^2 + 7x - 4 = (2x - 1)(x + 4)\) | B1 |
| \(\dfrac{3(x + 1)}{(2x - 1)(x + 4)} - \dfrac{1}{(x + 4)} = \dfrac{3(x + 1) - (2x - 1)}{(2x - 1)(x + 4)}\) | M1 |
| \(= \dfrac{x + 4}{(2x - 1)(x + 4)}\) | M1 |
| \(= \dfrac{1}{2x - 1}\) | A1* |
| (4) |
Notes
M1 Combines the two fractions to form a single fraction with a common denominator. Cubic denominators are fine for this mark. Allow slips on the numerator but one must have been adapted. Allow ‘invisible’ brackets. Accept two separate fractions with the same denominator. Amongst many possible options are
Correct \(\dfrac{3(x + 1) - (2x - 1)}{(2x - 1)(x + 4)}\), Invisible bracket \(\dfrac{3x + 1 - 2x - 1}{(2x - 1)(x + 4)}\),
Cubic and separate \(\dfrac{3(x + 1)(x + 4)}{(2x^2 + 7x - 4)(x + 4)} - \dfrac{2x^2 + 7x - 4}{(2x^2 + 7x - 4)(x + 4)}\)
M1 Simplifies the (now) single fraction to one with a linear numerator divided by a quadratic factorised denominator. Any cubic denominator must have been fully factorised (check first and last terms) and cancelled with terms on a fully factorised numerator (check first and last terms).
A1* Cso. This is a given solution and it must be fully correct. All bracketing/algebra must have been correct.
You can however accept \(\dfrac{x + 4}{(2x - 1)(x + 4)}\) going to \(\dfrac{1}{2x - 1}\) without the need for ‘seeing’ the cancelling
For example \(\dfrac{3(x + 1) - 2x - 1}{(2x - 1)(x + 4)} = \dfrac{x + 4}{(2x - 1)(x + 4)} = \dfrac{1}{2x - 1}\) scores B1,M1,M1,A0. Incorrect line leading to solution.
Whereas \(\dfrac{3(x + 1) - (2x - 1)}{(2x - 1)(x + 4)} = \dfrac{x + 4}{(2x - 1)(x + 4)} = \dfrac{1}{2x - 1}\) scores B1,M1,M1,A1
| Scheme | Marks |
|---|---|
| \(y = \dfrac{1}{2x - 1} \Rightarrow y(2x - 1) = 1 \Rightarrow 2xy - y = 1\) | |
| \(2xy = 1 + y \Rightarrow x = \dfrac{1 + y}{2y}\) | M1M1 |
| \(y \text{ OR } f^{-1}(x) = \dfrac{1 + x}{2x}\) | A1 |
| (3) |
Notes
M1 This is awarded for an attempt to make x or a swapped y the subject of the formula. The minimum criteria is that they start by multiplying by (2x-1) and finish with x= or swapped y=. Allow ‘invisible’ brackets.
M1 For applying the order of operations correctly. Allow maximum of one ‘slip’. Examples of this are
\(y = \frac{1}{2x - 1} \to y(2x - 1) = 1 \to 2x - 1 = \frac{1}{y} \to x = \dfrac{\frac{1}{y} \pm 1}{2}\) (allow slip on sign)
\(y = \frac{1}{2x - 1} \to y(2x - 1) = 1 \to 2xy - y = 1 \to 2xy = 1 \pm y \to x = \frac{1 \pm y}{2y}\) (allow slip on sign)
\(y = \frac{1}{2x - 1} \to 2x - 1 = \frac{1}{y} \to 2x = \frac{1}{y} + 1 \to x = \frac{1}{2y} + 1\) (allow slip on \(\div 2\))
A1 Must be written in terms of x but can be \(y = \dfrac{1 + x}{2x}\) or equivalent inc \(y = \dfrac{\frac{1}{x} + 1}{2}\), \(y = \dfrac{x^{-1} + 1}{2}\), \(y = \dfrac{1}{2x} + \dfrac{1}{2}\)
| Scheme | Marks |
|---|---|
| x>0 | B1 |
| (1) |
Notes
B1 Accept x>0, (0,\(\infty\)), domain is all values more than 0. Do not accept x\(\geqslant\) 0 , y>0, [0, \(\infty\)], \(f^{-1}(x) \gt 0\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2\ln(x + 1) - 1} = \dfrac{1}{7}\) | M1 |
| \(\ln(x + 1) = 4\) | A1 |
| \(x = e^4 - 1\) | M1A1 |
| (4) | |
| (12 marks) |
Notes
M1 Attempt to write down fg(x) and set it equal to 1/7.
The order must be correct but accept incorrect or lack of bracketing. Eg \(\dfrac{1}{2\ln x + 1 - 1} = \dfrac{1}{7}\)
A1 Achieving correctly the line \(\ln(x + 1) = 4\). Accept also \(\ln(x + 1)^2 = 8\)
M1 Moving from \(\ln(x \pm A) = c \quad A \neq 0 \text{ to } x =\) The ln work must be correct
Alternatively moving from \(\ln(x + 1)^2 = c\) to \(x = \cdots\)
Full solutions to calculate \(x\) leading from \(gf(x) = \frac{1}{7}\), that is \(\ln\left(\frac{1}{2x - 1} + 1\right) = \frac{1}{7}\) can score this mark.
A1 Correct answer only \(= e^4 - 1\). Accept \(e^4 - e^0\)