| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Topic | Generalised Binomial Theorem |
| Type | Form (1+bx)^n expansion |
| Difficulty | Standard +0.3 This question combines binomial expansion with the quadratic formula in a straightforward way. Part (i) is routine application of the binomial series for fractional powers. Part (ii) requires recognizing that the quadratic formula involves $(1-4a)^{1/2}$ and substituting the expansion, then simplifying—this is a standard 'hence' question testing algebraic manipulation rather than deep insight. The multi-step nature and connection between parts elevates it slightly above average, but it remains a textbook-style exercise. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((1-4a)^2 = 1 + \frac{1}{2}(-4a) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(-4a)^2\) | M1, A1, M1 | Attempt first two terms. State \(1 - 2a\). Attempt third term |
| \(= 1 - 2a - 2a^2\) | A1 | Obtain \(-2a^2\) |
| (ii) \(x = \frac{1 \pm \sqrt{1-4a}}{2}\) | B1 | State \(\frac{1 \pm \sqrt{1-4a}}{2}\) |
| \(x = \frac{1 \pm (1 - 2a - 2a^2)}{2}\) | M1 | Substitute answer to (i) for discriminant |
| \(x = 1 - a - a^2\) | A1 | Obtain \(x = 1 - a - a^2\) AG |
| \(x = a + a^2\) | A1 | [8] |
(i) $(1-4a)^2 = 1 + \frac{1}{2}(-4a) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(-4a)^2$ | M1, A1, M1 | Attempt first two terms. State $1 - 2a$. Attempt third term
$= 1 - 2a - 2a^2$ | A1 | Obtain $-2a^2$
(ii) $x = \frac{1 \pm \sqrt{1-4a}}{2}$ | B1 | State $\frac{1 \pm \sqrt{1-4a}}{2}$
$x = \frac{1 \pm (1 - 2a - 2a^2)}{2}$ | M1 | Substitute answer to (i) for discriminant
$x = 1 - a - a^2$ | A1 | Obtain $x = 1 - a - a^2$ AG
$x = a + a^2$ | A1 | [8] | Obtain $x = a + a^2$ AG\begin{enumerate}[label=(\roman*)]
\item Find and simplify the first three terms in the expansion of $(1 - 4a)^{\frac{1}{2}}$ in ascending powers of $a$, where $|a| < \frac{1}{4}$. [4]
\item Hence show that the roots of the quadratic equation $x^2 - x + a = 0$ are approximately $1 - a - a^2$ and $a + a^2$, where $a$ is small. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q8 [8]}}