June 2024 Paper 2 Q9
9
(a)
(i) Find the binomial expansion of \((1 + 3x)^{-1}\) up to and including the term in \(x^2\) [2 marks]
(ii) Show that the first three terms in the binomial expansion of\[\frac{1}{2 - 3x}\]form a geometric sequence and state the common ratio. [5 marks]
(b) It is given that\[\frac{36x}{(1 + 3x)(2 - 3x)} \equiv \frac{P}{(2 - 3x)} + \frac{Q}{(1 + 3x)}\]where \(P\) and \(Q\) are integers.
Find the value of \(P\) and the value of \(Q\) [3 marks]
(c)
(i) Using your answers to parts (a) and (b), find the binomial expansion of\[\frac{12x}{(1 + 3x)(2 - 3x)}\]up to and including the term in \(x^2\) [2 marks]
(ii) Find the range of values of \(x\) for which the binomial expansion of\[\frac{12x}{(1 + 3x)(2 - 3x)}\]is valid. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(1 + (-1)(3x) + \dfrac{(-1)(-2)(3x)^2}{2!}\) OE with at least two terms correct | M1 | 1.1a |
| Obtains \(1 - 3x + 9x^2\) | A1 | 1.1b |
| (2) | ||
| (ii) Writes fraction as \((2 - 3x)^{-1}\) PI by \(\dfrac{1}{2} + \dfrac{3}{4}x + \dfrac{9}{8}x^2\) | B1 | 1.1b |
| Factorises to obtain the form \(2^{-1}(1 - Ax)^{-1}\) PI by \(\dfrac{1}{2} + \dfrac{3}{4}x + \dfrac{9}{8}x^2\) | M1 | 1.1a |
| Expands \(\left(1 - \dfrac{3x}{2}\right)^{-1}\) to obtain \(1 + (-1)\left(\pm\dfrac{3x}{2}\right) + \dfrac{(-1)(-2)}{2!}\left(\pm\dfrac{3x}{2}\right)^2\) OE Condone one sign error | M1 | 1.1a |
| Completes a correct argument to show \(\dfrac{1}{2 - 3x} \approx \dfrac{1}{2} + \dfrac{3}{4}x + \dfrac{9}{8}x^2\) | R1 | 2.1 |
| States that the common ratio is \(\dfrac{3}{2}x\) | B1 | 2.2a |
| (5) |
Typical solution
(i)
\[(1 + 3x)^{-1} \approx 1 + (-1)(3x) + \frac{(-1)(-2)(3x)^2}{2!}\]\[= 1 - 3x + 9x^2\](ii)
\[\frac{1}{2 - 3x} = (2 - 3x)^{-1}\]\[= 2^{-1}\left(1 + \left(-\frac{3x}{2}\right)\right)^{-1}\]\[\approx \frac{1}{2}\left(1 + (-1)\left(-\frac{3x}{2}\right) + \frac{(-1)(-2)}{2!}\left(-\frac{3x}{2}\right)^2\right)\]\[\frac{1}{2 - 3x} \approx \frac{1}{2} + \frac{3}{4}x + \frac{9}{8}x^2\]\(\dfrac{1}{2}\), \(\dfrac{3}{4}x\) and \(\dfrac{9}{8}x^2\) form a geometric sequence with common ratio \(\dfrac{3}{2}x\)
| Scheme | Marks | AO |
|---|---|---|
| Uses a valid method to find \(P\) or \(Q\) Substitution of \(x = -\dfrac{1}{3}\) or \(x = \dfrac{2}{3}\) Or Rearranging and substitution or comparison of coefficients | M1 | 1.1a |
| Obtains \(P\) = 8 | A1 | 1.1b |
| Obtains \(Q = -4\) | A1 | 1.1b |
| (3) |
Typical solution
\[\frac{36x}{(1 + 3x)(2 - 3x)} = \frac{P}{(2 - 3x)} + \frac{Q}{(1 + 3x)}\]\[36x = P(1 + 3x) + Q(2 - 3x)\]Let \(x = -\dfrac{1}{3}\)
\[\Rightarrow -12 = 3Q\]\[Q = -4\]\[x = \frac{2}{3}\]\[\Rightarrow 24 = 3P\]\[P = 8\]| Scheme | Marks | AO |
|---|---|---|
| (i) Multiplies their \(P\) by their expansion in (a)(ii) and multiplies their \(Q\) by their expansion in (a)(i) Condone a sign error Or Multiplies their \(\dfrac{P}{3}\) by their expansion in (a)(ii) and multiplies their \(\dfrac{Q}{3}\) by their expansion in (a)(i) Condone a sign error Or Writes the product of 12\(x\) or 36\(x\) with their three-term expansion in (a)(i) and their three-term expansion in (a)(ii) Condone a sign error | M1 | 3.1a |
| Obtains \(6x - 9x^2\) | A1 | 1.1b |
| (2) | ||
| (ii) Deduces \(|x| \lt \dfrac{1}{3}\) ACF | R1 | 2.2a |
| (1) | ||
| (13 marks) |
Typical solution
(i)
\[\frac{8}{(2 - 3x)} - \frac{4}{(1 + 3x)}\]\[\approx 8\left(\frac{1}{2} + \frac{3}{4}x + \frac{9}{8}x^2\right) - 4\left(1 - 3x + 9x^2\right)\]\[= 18x - 27x^2\]\[\therefore \frac{12x}{(1 + 3x)(2 - 3x)} \approx 6x - 9x^2\](ii)
\[|x| \lt \frac{1}{3}\]