| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown power and constant |
| Difficulty | Standard +0.8 This question requires applying the generalised binomial theorem with fractional powers, manipulating the expression into standard form (1+u)^n, matching coefficients across multiple terms to find three unknowns, and then computing a further term. It goes beyond routine application by requiring algebraic manipulation and systematic coefficient matching, making it moderately challenging for C3/C4 level. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{6}{\sqrt{9+Ax^2}} = 6(9+Ax^2)^{-\frac{1}{2}} = 6 \times 9^{-\frac{1}{2}}\left(1 + \frac{A}{9}x^2\right)^{-\frac{1}{2}}\) | B1 | For \(9^{-\frac{1}{2}}\) or \(\frac{1}{3}\) or 2 before bracket |
| \(= 2\times\left(1 + \left(-\frac{1}{2}\right)\left(\frac{A}{9}x^2\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{A}{9}x^2\right)^2 + \ldots\right)\) | M1 A1 | M1 for correct binomial form with \(n = -\frac{1}{2}\) and term \(\left(\frac{A}{9}x^2\right)\) |
| \(= 2 - \frac{A}{9}x^2 + \frac{A^2}{108}x^4 + \ldots\) | ||
| Compare to \(B - \frac{2}{3}x^2 + Cx^4\): \(B = 2\) | B1 | |
| \(A = 6\) | B1 | |
| \(C = \frac{1}{108} \times A^2 = \frac{1}{3}\) | dM1 A1 | Substitute numerical \(A\) into coefficient of \(x^4\) to find \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Coefficient of \(x^6 = 2 \times \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \times \left(\frac{A}{9}\right)^3 = -\frac{5}{27}\) | M1 A1 | M1 for correct unsimplified term in \(x^6\) with \(n=-\frac{1}{2}\) using their \(A\); A1 for \(-\frac{5}{27}\) or exact equivalent |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{6}{\sqrt{9+Ax^2}} = 6(9+Ax^2)^{-\frac{1}{2}} = 6 \times 9^{-\frac{1}{2}}\left(1 + \frac{A}{9}x^2\right)^{-\frac{1}{2}}$ | B1 | For $9^{-\frac{1}{2}}$ or $\frac{1}{3}$ or 2 before bracket |
| $= 2\times\left(1 + \left(-\frac{1}{2}\right)\left(\frac{A}{9}x^2\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{A}{9}x^2\right)^2 + \ldots\right)$ | M1 A1 | M1 for correct binomial form with $n = -\frac{1}{2}$ and term $\left(\frac{A}{9}x^2\right)$ |
| $= 2 - \frac{A}{9}x^2 + \frac{A^2}{108}x^4 + \ldots$ | | |
| Compare to $B - \frac{2}{3}x^2 + Cx^4$: $B = 2$ | B1 | |
| $A = 6$ | B1 | |
| $C = \frac{1}{108} \times A^2 = \frac{1}{3}$ | dM1 A1 | Substitute numerical $A$ into coefficient of $x^4$ to find $C$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Coefficient of $x^6 = 2 \times \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \times \left(\frac{A}{9}\right)^3 = -\frac{5}{27}$ | M1 A1 | M1 for correct unsimplified term in $x^6$ with $n=-\frac{1}{2}$ using their $A$; A1 for $-\frac{5}{27}$ or exact equivalent |6. Given that the binomial expansion, in ascending powers of $x$, of
$$\frac { 6 } { \sqrt { } \left( 9 + A x ^ { 2 } \right) } , \quad | x | < \frac { 3 } { \sqrt { } | A | }$$
is $\quad B - \frac { 2 } { 3 } x ^ { 2 } + C x ^ { 4 } + \ldots$
\begin{enumerate}[label=(\alph*)]
\item find the values of the constants $A , B$ and $C$.
\item Hence find the coefficient of $x ^ { 6 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q6 [9]}}