C4 June 2007 Q1
1. \[\mathrm{f}(x) = (3 + 2x)^{-3}, \qquad |x| \lt \tfrac{3}{2}.\]
Find the binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), as far as the term in \(x^3\).
Give each coefficient as a simplified fraction. (5)
** represents a constant
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = (3+2x)^{-3} = \underline{(3)^{-3}}\left(1 + \dfrac{2x}{3}\right)^{-3} = \underline{\dfrac{1}{27}}\left(1 + \dfrac{2x}{3}\right)^{-3}\) | B1 |
| \(= \tfrac{1}{27}\left\{\underline{1 + (-3)(**x);} + \dfrac{(-3)(-4)}{2!}(**x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(**x)^3 + \ldots\right\}\) with \(** \ne 1\) | M1; A1ft |
| \(= \tfrac{1}{27}\left\{1 + (-3)(\tfrac{2x}{3}) + \dfrac{(-3)(-4)}{2!}(\tfrac{2x}{3})^2 + \dfrac{(-3)(-4)(-5)}{3!}(\tfrac{2x}{3})^3 + \ldots\right\}\) | |
| \(= \tfrac{1}{27}\left\{1 - 2x + \dfrac{8x^2}{3} - \dfrac{80}{27}x^3 + \ldots\right\}\) | |
| \(= \dfrac{1}{27} - \dfrac{2x}{27};\ + \dfrac{8x^2}{81} - \dfrac{80x^3}{729} + \ldots\) | A1; A1 |
| (5) | |
| (5 marks) |
Notes
B1: Takes 3 outside the bracket to give any of \((3)^{-3}\) or \(\tfrac{1}{27}\). See note below.
M1: Expands \((1 + **x)^{-3}\) to give a simplified or an un-simplified \(1 + (-3)(**x)\);
A1ft: A correct simplified or an un-simplified \(\{\ldots\ldots\}\) expansion with candidate’s followed thro’ \((**x)\).
A1: Anything that cancels to \(\dfrac{1}{27} - \dfrac{2x}{27}\);
A1: Simplified \(\dfrac{8x^2}{81} - \dfrac{80x^3}{729}\)
Note: You would award: B1M1A0 for \[= \tfrac{1}{27}\left\{1 + (-3)(\tfrac{2x}{3}) + \dfrac{(-3)(-4)}{2!}(2x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(2x)^3 + \ldots\right\}\] because \(**\) is not consistent.
Special Case: If you see the constant \(\tfrac{1}{27}\) in a candidate’s final binomial expression, then you can award B1
Aliter: Way 2
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = (3+2x)^{-3}\) | |
| \(= \left\{\underline{(3)^{-3} + (-3)(3)^{-4}(**x);} + \dfrac{(-3)(-4)}{2!}(3)^{-5}(**x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(3)^{-6}(**x)^3 + \ldots\right\}\) with \(** \ne 1\) | B1 M1 A1ft |
| \(= \left\{(3)^{-3} + (-3)(3)^{-4}(2x); + \dfrac{(-3)(-4)}{2!}(3)^{-5}(2x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(3)^{-6}(2x)^3 + \ldots\right\}\) | |
| \(= \left\{\tfrac{1}{27} + (-3)(\tfrac{1}{81})(2x); + (6)(\tfrac{1}{243})(4x^2) + (-10)(\tfrac{1}{729})(8x^3) + \ldots\right\}\) | |
| \(= \dfrac{1}{27} - \dfrac{2x}{27};\ + \dfrac{8x^2}{81} - \dfrac{80x^3}{729} + \ldots\) | A1; A1 |
| (5) | |
| (5 marks) |
B1: \(\tfrac{1}{27}\) or \((3)^{-3}\) (See note ↓)
M1: Expands \((3 + 2x)^{-3}\) to give an un-simplified or simplified \((3)^{-3} + (-3)(3)^{-4}(**x)\);
A1ft: A correct un-simplified or simplified \(\{\ldots\ldots\}\) expansion with candidate’s followed thro’ \((**x)\)
A1: Anything that cancels to \(\dfrac{1}{27} - \dfrac{2x}{27}\); A1: Simplified \(\dfrac{8x^2}{81} - \dfrac{80x^3}{729}\)
Attempts using Maclaurin expansions need to be escalated up to your team leader.
If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader.
Special Case: If you see the constant \(\tfrac{1}{27}\) in a candidate’s final binomial expression, then you can award B1