C4 January 2007 Q1
1.
\[\mathrm{f}(x) = (2 - 5x)^{-2}, \qquad |x| \lt \tfrac{2}{5}.\]
Find the binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), as far as the term in \(x^3\), giving each coefficient as a simplified fraction. (5)
** represents a constant
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = (2 - 5x)^{-2} = \underline{(2)^{-2}}\left(1 - \dfrac{5x}{2}\right)^{-2} = \underline{\dfrac{1}{4}}\left(1 - \dfrac{5x}{2}\right)^{-2}\) | B1 |
| \(= \frac{1}{4}\left\{\underline{1 + (-2)(**x); + \dfrac{(-2)(-3)}{2!}(**x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(**x)^3 + \ldots}\right\}\) | M1 A1 |
| \(= \frac{1}{4}\left\{\underline{1 + (-2)\left(\tfrac{-5x}{2}\right); + \dfrac{(-2)(-3)}{2!}\left(\tfrac{-5x}{2}\right)^2 + \dfrac{(-2)(-3)(-4)}{3!}\left(\tfrac{-5x}{2}\right)^3 + \ldots}\right\}\) | |
| \(= \frac{1}{4}\left\{1 + 5x; + \dfrac{75x^2}{4} + \dfrac{125x^3}{2} + \ldots\right\}\) | |
| \(= \dfrac{1}{4} + \dfrac{5x}{4}; + \dfrac{75x^2}{16} + \dfrac{125x^3}{8} + \ldots\) | A1; A1 |
| \(= \dfrac{1}{4} + 1\dfrac{1}{4}x; + 4\dfrac{11}{16}x^2 + 15\dfrac{5}{8}x^3 + \ldots\) | |
| (5) | |
| (5 marks) |
Notes
B1 Takes 2 outside the bracket to give any of \((2)^{-2}\) or \(\frac{1}{4}\).
M1 Expands \((1 + **x)^{-2}\) to give an unsimplified \(1 + (-2)(**x)\);
A1 A correct unsimplified \(\{\ldots\ldots\}\) expansion with candidate’s \((**x)\)
A1; Anything that cancels to \(\frac{1}{4} + \frac{5x}{4}\);
A1 Simplified \(\frac{75x^2}{16} + \frac{125x^3}{8}\)
Aliter Way 2
| \(\mathrm{f}(x) = (2 - 5x)^{-2}\) | B1 |
| \(= \left\{\underline{(2)^{-2} + (-2)(2)^{-3}(**x); + \dfrac{(-2)(-3)}{2!}(2)^{-4}(**x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(2)^{-5}(**x)^3 + \ldots}\right\}\) | M1 A1 |
| \(= \left\{(2)^{-2} + (-2)(2)^{-3}(-5x); + \dfrac{(-2)(-3)}{2!}(2)^{-4}(-5x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(2)^{-5}(-5x)^3 + \ldots\right\}\) | |
| \(= \left\{\tfrac{1}{4} + (-2)\left(\tfrac{1}{8}\right)(-5x); + (3)\left(\tfrac{1}{16}\right)(25x^2) + (-4)\left(\tfrac{1}{32}\right)(-125x^3) + \ldots\right\}\) | |
| \(= \dfrac{1}{4} + \dfrac{5x}{4}; + \dfrac{75x^2}{16} + \dfrac{125x^3}{8} + \ldots\) | A1; A1 |
| \(= \dfrac{1}{4} + 1\dfrac{1}{4}x; + 4\dfrac{11}{16}x^2 + 15\dfrac{5}{8}x^3 + \ldots\) | |
| [5] |
B1 \(\frac{1}{4}\) or \((2)^{-2}\)
M1 Expands \((2 - 5x)^{-2}\) to give an unsimplifed \((2)^{-2} + (-2)(2)^{-3}(**x)\);
A1 A correct unsimplified \(\{\ldots\ldots\}\) expansion with candidate’s \((**x)\)
A1; Anything that cancels to \(\frac{1}{4} + \frac{5x}{4}\);
A1 Simplified \(\frac{75x^2}{16} + \frac{125x^3}{8}\)
CHECK (corrected from the printed mark scheme: in Way 2 the \(x^3\) term is printed as \((-4)\left(\frac{1}{16}\right)(-125x^3)\); since \((2)^{-5} = \frac{1}{32}\) it is \((-4)\left(\frac{1}{32}\right)(-125x^3)\), which gives the \(\frac{125x^3}{8}\) shown.)