C4 January 2012 Q3
3.
Given that the binomial expansion of \(\dfrac{2 + kx}{(2 - 5x)^2}\), \(|x| \lt \dfrac{2}{5}\), is\[\frac{1}{2} + \frac{7}{4}x + Ax^2 + \ldots\]
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{(2 - 5x)^2} = (2 - 5x)^{-2} = \underline{(2)^{-2}}\left(1 - \dfrac{5x}{2}\right)^{-2} = \underline{\dfrac{1}{4}}\left(1 - \dfrac{5x}{2}\right)^{-2}\) \(\underline{(2)^{-2}}\) or \(\underline{\dfrac{1}{4}}\) | B1 |
| \(= \left\{\dfrac{1}{4}\right\}\left[1 + (-2)(**\,x) + \dfrac{(-2)(-3)}{2!}(**\,x)^2 + \ldots\right]\) see notes | M1 A1ft |
| \(= \left\{\dfrac{1}{4}\right\}\left[\underline{1 + (-2)\left(-\dfrac{5x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(-\dfrac{5x}{2}\right)^2 + \ldots}\right]\) | |
| \(= \dfrac{1}{4}\left[1 + 5x;\ + \dfrac{75}{4}x^2 + \ldots\right]\) See notes below! | |
| \(= \dfrac{1}{4} + \dfrac{5}{4}x;\ + \dfrac{75}{16}x^2 + \ldots\) | A1; A1 |
| (5) |
Notes
B1: \(\underline{(2)^{-2}}\) or \(\underline{\dfrac{1}{4}}\) outside brackets or \(\dfrac{1}{4}\) as candidate’s constant term in their binomial expansion.
M1: Expands to give a simplified or an un-simplified,
\(1 + (-2)(**\,x)\) or \((-2)(**\,x) + \dfrac{(-2)(-3)}{2!}(**\,x)^2\) or \(1 + \ldots\ldots + \dfrac{(-2)(-3)}{2!}(**\,x)^2\), where \(** \neq 1\).
A1: A correct simplified or an un-simplified \(1 + (-2)(**\,x) + \dfrac{(-2)(-3)}{2!}(**\,x)^2\) expansion with candidate’s follow through \((**\,x)\). Note that \((**\,x)\) must be consistent.
You would award B1M1A0 for \(= \dfrac{1}{4}\left[\underline{1 + (-2)\left(-\dfrac{5x}{2}\right) + \dfrac{(-2)(-3)}{2!}(-5x)^2 + \ldots}\right]\) because ** is not consistent.
Invisible brackets \(\left\{\dfrac{1}{4}\right\}\left[\underline{1 + (-2)\left(-\dfrac{5x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(-\dfrac{5x^2}{2}\right) + \ldots}\right]\) is M1A0 unless recovered.
A1: For \(\dfrac{1}{4} + \dfrac{5}{4}x\) (simplified fractions) or Also allow \(0.25 + 1.25x\) or \(\dfrac{1}{4} + 1\dfrac{1}{4}x\).
Allow Special Case A1 for either SC: \(\dfrac{1}{4}[1 + 5x;\ \ldots]\) or SC: \(K\left[1 + 5x + \dfrac{75}{4}x^2 + \ldots\right]\).
A1: Accept only \(\dfrac{75}{16}x^2\) or \(4\dfrac{11}{16}x^2\) or \(4.6875x^2\)
Alternative method for part (a)
| Scheme | Marks |
|---|---|
| \((2 - 5x)^{-2} = (2)^{-2} + (-2)(2)^{-3}(-5x);\ + \dfrac{(-2)(-3)}{2!}(2)^{-4}(-5x)^2\) |
B1: \(\dfrac{1}{4}\) or \((2)^{-2}\),
M1: Any two of three (un-simplified) terms correct.
A1: All three (un-simplified) terms correct.
A1: \(\dfrac{1}{4} + \dfrac{5}{4}x\)
A1: \(\dfrac{75}{16}x^2\)
Note: The terms in C need to be evaluated, so \({}^{-2}C_0(2)^{-2} + {}^{-2}C_1(2)^{-3}(-5x);\ + {}^{-2}C_2(2)^{-4}(-5x)^2\) without further working is B0M0A0.
| Scheme | Marks |
|---|---|
| \(\left\{\dfrac{2 + kx}{(2 - 5x)^2}\right\} = (2 + kx)\left(\dfrac{1}{4} + \dfrac{5}{4}x + \left\{\dfrac{75}{16}x^2 + \ldots\right\}\right)\) Can be implied by later work even in part (c). | M1 |
| \(x\) terms: \(\ \dfrac{2(5x)}{4} + \dfrac{kx}{4} = \dfrac{7x}{4}\) giving, \(10 + k = 7 \Rightarrow \underline{k = -3}\) \(\underline{k = -3}\) | A1 |
| (2) |
Notes
M1: Candidate writes down \((2 + kx)(\text{their part (a) answer, at least up to the term in } x.)\)
\((2 + kx)\left(\dfrac{1}{4} + \dfrac{5}{4}x + \ldots\right)\) or \((2 + kx)\left(\dfrac{1}{4} + \dfrac{5}{4}x + \dfrac{75}{16}x^2 + \ldots\right)\) are fine.
This mark can also be implied by candidate multiplying out to find two terms (or coefficients) in \(x\).
A1: \(k = -3\)
Alternative method for parts (b) and (c)
| Scheme | Marks |
|---|---|
| \((2 + kx) = (2 - 5x)^2\left(\dfrac{1}{2} + \dfrac{7}{4}x + Ax^2 + \ldots\right)\) \((2 + kx) = (4 - 20x + 25x^2)\left(\dfrac{1}{2} + \dfrac{7}{4}x + Ax^2 + \ldots\right)\) \((2 + kx) = 2 + (7x - 10x) + \left(4Ax^2 - 35x^2 + \dfrac{25}{2}x^2\right)\) | |
| Equate \(x\) terms: \(\ \underline{k = -3}\) | |
| Equate \(x^2\) terms: \(\ 0 = 4A - 35 + \dfrac{25}{2} \Rightarrow 4A = \dfrac{45}{2} \Rightarrow \underline{A = \dfrac{45}{8}}\) |
(b) M1: For \((2 + kx) = (4 \pm \lambda x + 25x^2)\left(\dfrac{1}{2} + \dfrac{7}{4}x + Ax^2 + \ldots\right)\), where \(\lambda \neq 0\)
A1: \(k = -3\)
(c) M1: Multiplies out to obtain three \(x^2\) terms/coefficients, equates to 0 and attempts to find \(A\).
A1: Either \(\dfrac{45}{8}\) or \(5\dfrac{5}{8}\) or 5.625 Note: \(\dfrac{45}{8}x^2\) is A0.
| Scheme | Marks |
|---|---|
| \(x^2\) terms: \(\ \dfrac{150x^2}{16} + \dfrac{5kx^2}{4}\) | M1 |
| So, \(A = \dfrac{75}{8} + \dfrac{5(-3)}{4} = \dfrac{75}{8} - \dfrac{15}{4} = \underline{\dfrac{45}{8}}\) \(\underline{\dfrac{45}{8}}\) or \(5\dfrac{5}{8}\) or \(\underline{5.625}\) | A1 |
| (2) | |
| (9 marks) |
Notes
M1: Multiplies out their \((2 + kx)\left(\dfrac{1}{4} + \dfrac{5}{4}x + \dfrac{75}{16}x^2 + \ldots\right)\) to give exactly two terms (or coefficients) in \(x^2\) and attempts to find \(A\) using a numerical value of \(k\).
A1: Either \(\dfrac{45}{8}\) or \(5\dfrac{5}{8}\) or 5.625 Note: \(\dfrac{45}{8}x^2\) is A0.
For the alternative method for parts (b) and (c), see the notes to part (b).