| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown power and constant |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion formula for fractional powers. Part (a) requires matching coefficients to find k, A, and B using the standard formula, while part (b) involves a routine substitution. The question is slightly above average difficulty due to the multi-step nature and need for careful algebraic manipulation, but it follows a standard template with no novel insight required. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}k = \frac{1}{8} \Rightarrow k = \frac{1}{4}\) | M1A1 | M1: Sets \(\frac{1}{2}k = \frac{1}{8}\) or \(\frac{1}{2}kx = \frac{1}{8}x\) and proceeds to find \(k\). A1: \(k = \frac{1}{4}\) oe such as \(\frac{2}{8}\) or 0.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)}{2!}\times "k"^2 = -\frac{1}{128}\) | M1 A1 | M1: Correct attempt at 3rd or 4th term with substitution of \(k\). Award for \(A = -\frac{1}{8}\times("k")^2\). No requirement to simplify fraction |
| \(B = \frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{3!}\times "k"^3 = \frac{1}{1024}\) | A1 | Award for \(B = \frac{1}{16}\times("k")^3\). May see \(B = \frac{3}{3072}\) which is fine |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(x = 0.6 \Rightarrow \sqrt{1.15} = 1 + \frac{1}{8}\times 0.6 - \frac{1}{128}\times 0.6^2 + \frac{1}{1024}\times 0.6^3 = 1.072398\) | M1 A1 | M1: Either \(kx = 0.15\) into expansion of form \(1 \pm p\times(kx) \pm q\times(kx)^2 \pm r\times(kx)^3\), or \(x = \frac{0.15}{"k"}\) into their \(1 + \frac{1}{8}x + "A"x^2 + "B"x^3\). A1: 1.072398 must be to 6 decimal places |
# Question 1:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}k = \frac{1}{8} \Rightarrow k = \frac{1}{4}$ | M1A1 | M1: Sets $\frac{1}{2}k = \frac{1}{8}$ or $\frac{1}{2}kx = \frac{1}{8}x$ and proceeds to find $k$. A1: $k = \frac{1}{4}$ oe such as $\frac{2}{8}$ or 0.25 |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)}{2!}\times "k"^2 = -\frac{1}{128}$ | M1 A1 | M1: Correct attempt at 3rd **or** 4th term with substitution of $k$. Award for $A = -\frac{1}{8}\times("k")^2$. No requirement to simplify fraction |
| $B = \frac{\left(\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)\times\left(-\frac{3}{2}\right)}{3!}\times "k"^3 = \frac{1}{1024}$ | A1 | Award for $B = \frac{1}{16}\times("k")^3$. May see $B = \frac{3}{3072}$ which is fine |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $x = 0.6 \Rightarrow \sqrt{1.15} = 1 + \frac{1}{8}\times 0.6 - \frac{1}{128}\times 0.6^2 + \frac{1}{1024}\times 0.6^3 = 1.072398$ | M1 A1 | M1: Either $kx = 0.15$ into expansion of form $1 \pm p\times(kx) \pm q\times(kx)^2 \pm r\times(kx)^3$, or $x = \frac{0.15}{"k"}$ into their $1 + \frac{1}{8}x + "A"x^2 + "B"x^3$. A1: 1.072398 must be to 6 decimal places |
---\begin{enumerate}
\item Given that $k$ is a constant and the binomial expansion of
\end{enumerate}
$$\sqrt { 1 + k x } \quad | k x | < 1$$
in ascending powers of $x$ up to the term in $x ^ { 3 }$ is
$$1 + \frac { 1 } { 8 } x + A x ^ { 2 } + B x ^ { 3 }$$
(a) (i) find the value of $k$,\\
(ii) find the value of the constant $A$ and the constant $B$.\\
(b) Use the expansion to find an approximate value to $\sqrt { 1.15 }$
Show your working and give your answer to 6 decimal places.\\
\hfill \mbox{\textit{Edexcel P4 2021 Q1 [7]}}