October 2021 Paper 1 Q9
9. \[\mathrm{f}(x) = \frac{50x^2 + 38x + 9}{(5x + 2)^2(1 - 2x)} \qquad x \neq -\frac{2}{5} \quad x \neq \frac{1}{2}\]
Given that \(\mathrm{f}(x)\) can be expressed in the form
\[\frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 - 2x}\]where \(A\), \(B\) and \(C\) are constants
| Scheme | Marks | AO |
|---|---|---|
| (i) \(50x^2 + 38x + 9 \equiv A(5x + 2)(1 - 2x) + B(1 - 2x) + C(5x + 2)^2\) \(\Rightarrow B = \ldots\) or \(C = \ldots\) | M1 | 1.1b |
| \(B = 1\) and \(C = 2\) | A1 | 1.1b |
| (ii) E.g. \(x = 0\) \(\ x = 0 \Rightarrow 9 = 2A + B + 4C\) \(\Rightarrow 9 = 2A + 1 + 8 \Rightarrow A = \ldots\) | M1 | 2.1 |
| \(A = 0\,*\) | A1* | 1.1b |
| (4) |
Notes
(a)(i) M1: Uses a correct identity and makes progress using an appropriate strategy (e.g. sub \(x = \dfrac{1}{2}\)) to find a value for \(B\) or \(C\). May be implied by one correct value (cover up rule).
A1: Both values correct
(a)(ii) M1: Uses an appropriate method to establish an equation connecting \(A\) with \(B\) and/or \(C\) and uses their values of \(B\) and/or \(C\) to find a suitable equation in \(A\).
Amongst many different methods are:
Compare terms in \(x^2 \Rightarrow 50 = -10A + 25C\) which would be implied by \(50 = -10A + 25 \times \text{``}2\text{''}\)
Compare constant terms or substitute \(x = 0 \Rightarrow 9 = 2A + B + 4C\) implied by \(9 = 2A + 1 + 4 \times 2\)
A1*: Fully correct proof with no errors.
Note: The second part is a proof so it is important that a suitable proof/show that is seen.
Candidates who write down 3 equations followed by three answers (with no working) will score M1 A1 M0 A0
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{1}{(5x + 2)^2} = (5x + 2)^{-2} = 2^{-2}\left(1 + \dfrac{5}{2}x\right)^{-2}\) or \((5x + 2)^{-2} = 2^{-2} + \ldots\) | M1 | 3.1a |
| \(\left(1 + \dfrac{5}{2}x\right)^{-2} = 1 - 2\left(\dfrac{5}{2}x\right) + \dfrac{-2(-2 - 1)}{2!}\left(\dfrac{5}{2}x\right)^2 + \ldots\) | M1 | 1.1b |
| \(2^{-2}\left(1 + \dfrac{5}{2}x\right)^{-2} = \dfrac{1}{4} - \dfrac{5}{4}x + \dfrac{75}{16}x^2 + \ldots\) | A1 | 1.1b |
| \(\dfrac{1}{(1 - 2x)} = (1 - 2x)^{-1} = 1 + 2x + \dfrac{-1(-1 - 1)}{2!}(2x)^2 + \ldots\) | M1 | 1.1b |
| \(\dfrac{1}{(5x + 2)^2} + \dfrac{2}{1 - 2x} = \dfrac{1}{4} - \dfrac{5}{4}x + \dfrac{75}{16}x^2 + \ldots + 2 + 4x + 8x^2 + \ldots\) | dM1 | 2.1 |
| \(= \dfrac{9}{4} + \dfrac{11}{4}x + \dfrac{203}{16}x^2 + \ldots\) | A1 | 1.1b |
| (ii) \(|x| \lt \dfrac{2}{5}\) | B1 | 2.2a |
| (7) | ||
| (11 marks) |
Notes
(b)(i) M1: Applies the key steps of writing \(\dfrac{1}{(5x + 2)^2}\) as \((5x + 2)^{-2}\) and takes out a factor of \(2^{-2}\) to form an expression of the form \((5x + 2)^{-2} = 2^{-2}(1 + {*}x)^{-2}\) where \(*\) is not 1 or 5
Alternatively uses direct expansion to obtain \(2^{-2} + \ldots\)
M1: Correct attempt at the binomial expansion of \((1 + {*}x)^{-2}\) up to the term in \(x^2\)
Look for \(1 + (-2){*}x + \dfrac{(-2)(-3)}{2}{*}x^2\) where \(*\) is not 5 or 1.
Condone sign slips and lack of \(*^2\) on term 3. ....
Alt Look for correct structure for 2nd and 3rd terms by direct expansion. See below
A1: For a fully correct expansion of \((2 + 5x)^{-2}\) which may be unsimplified. This may have been combined with their ‘B’
A direct expansion would look like \((2 + 5x)^{-2} = 2^{-2} + (-2)2^{-3} \times 5x + \dfrac{(-2)(-3)}{2}2^{-4} \times (5x)^2\)
M1: Correct attempt at the binomial expansion of \((1 - 2x)^{-1}\)
Look for \(1 + (-1){*}x + \dfrac{(-1)(-2)}{2}{*}x^2\) where \(*\) is not 1
dM1: Fully correct strategy that is dependent on the previous TWO method marks.
There must be some attempt to use their values of \(B\) and \(C\)
A1: Correct expression or correct values for \(p,\ q\) and \(r\).
(b)(ii) B1: Correct range. Allow also other forms, for example \(-\dfrac{2}{5} \lt x \lt \dfrac{2}{5}\) or \(x \in \left(-\dfrac{2}{5},\ \dfrac{2}{5}\right)\)
Do not allow multiple answers here. The correct answer must be chosen if two answers are offered