| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Expansion up to x^2 term |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question combining the remainder theorem with standard binomial expansion. Part (a) requires substituting x=5/4 into f(x) and solving a simple equation. Part (b) is routine binomial expansion for the first three terms. Part (c) requires differentiating the expansion term-by-term and evaluating at x=0, which is a standard technique. While it tests multiple concepts, each step follows textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.04a Binomial expansion: (a+b)^n for positive integer n1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f\!\left(\tfrac{5}{4}\right)=\left(2-k\times\tfrac{5}{4}\right)^5=\dfrac{243}{32} \Rightarrow \left(2-k\times\tfrac{5}{4}\right)=\sqrt[5]{\dfrac{243}{32}} \Rightarrow k=\ldots\) | M1 | Substitutes \(x=\tfrac{5}{4}\), equates to \(\tfrac{243}{32}\), takes 5th root of both sides |
| \(\dfrac{3}{2} \Rightarrow \dfrac{5k}{4}=\dfrac{1}{2} \Rightarrow k=\dfrac{2}{5}\) | A1* | No errors; sufficient working shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k=\tfrac{2}{5},\ x=\tfrac{5}{4} \Rightarrow \left(2-\tfrac{2}{5}\times\tfrac{5}{4}\right)^5=\left(\tfrac{3}{2}\right)^5=\dfrac{243}{32}\) | M1 | Substitutes both values and raises evaluated \(2-kx\) to power 5 |
| Hence \(k=\tfrac{2}{5}\) | A1 | Fully correct with minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm{}^5C_1\times 2^4\times\left(\pm\tfrac{2}{5}x\right)\) or \(\pm{}^5C_2\times 2^3\times\left(\pm\tfrac{2}{5}x\right)^2\) | M1 | Correct structure for \(x\) or \(x^2\) term: correct binomial coefficient, correct power of 2, correct power of \(\pm\tfrac{2}{5}x\); brackets may be missing for \(x^2\) term |
| \(-32x\) | A1 | Correct simplified \(x\) term |
| \(+\dfrac{64}{5}x^2\) | A1 | Correct simplified \(x^2\) term (allow 12.8) |
| Final answer: \(32-32x+\dfrac{64}{5}x^2\) | Ignore extra terms; mark final answer; if no simplification in (b) do not allow simplified terms in (c) as recovery |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(2-\tfrac{2}{5}x\right)^5=2^5\!\left(1-\tfrac{1}{5}x\right)^5=2^5\!\left(1-5\cdot\tfrac{1}{5}x+\tfrac{5\times4}{2}\left(\tfrac{1}{5}x\right)^2+\ldots\right)\) | M1 | Score for \(2^5\!\left(\ldots\pm5\times\tfrac{1}{5}x\pm\ldots\right)\) or \(2^5\!\left(\ldots\pm\tfrac{5\times4}{2}\left(\tfrac{1}{5}x\right)^2+\ldots\right)\) |
| A marks as above | A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x)=-32+\dfrac{128}{5}x+\ldots \Rightarrow f'(0)=\ldots\) | M1 | Differentiates expansion AND substitutes \(x=0\); requires \(x^n\to x^{n-1}\) at least once (including \(k\to0\) or \(kx\to k\)) |
| \(f'(0)=-32\) | A1ft | Follow through on their expansion in (b) provided it was of form \(p+qx+rx^2\), \(p,q,r\neq0\), and differentiation is correct for their expansion |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(\tfrac{5}{4}\right)=\left(2-k\times\tfrac{5}{4}\right)^5=\dfrac{243}{32} \Rightarrow \left(2-k\times\tfrac{5}{4}\right)=\sqrt[5]{\dfrac{243}{32}} \Rightarrow k=\ldots$ | M1 | Substitutes $x=\tfrac{5}{4}$, equates to $\tfrac{243}{32}$, takes 5th root of both sides |
| $\dfrac{3}{2} \Rightarrow \dfrac{5k}{4}=\dfrac{1}{2} \Rightarrow k=\dfrac{2}{5}$ | A1* | No errors; sufficient working shown |
**Alternative by verification:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k=\tfrac{2}{5},\ x=\tfrac{5}{4} \Rightarrow \left(2-\tfrac{2}{5}\times\tfrac{5}{4}\right)^5=\left(\tfrac{3}{2}\right)^5=\dfrac{243}{32}$ | M1 | Substitutes both values and raises **evaluated** $2-kx$ to power 5 |
| Hence $k=\tfrac{2}{5}$ | A1 | Fully correct with minimal conclusion |
---
# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm{}^5C_1\times 2^4\times\left(\pm\tfrac{2}{5}x\right)$ or $\pm{}^5C_2\times 2^3\times\left(\pm\tfrac{2}{5}x\right)^2$ | M1 | Correct structure for $x$ or $x^2$ term: correct binomial coefficient, correct power of 2, correct power of $\pm\tfrac{2}{5}x$; brackets may be missing for $x^2$ term |
| $-32x$ | A1 | Correct simplified $x$ term |
| $+\dfrac{64}{5}x^2$ | A1 | Correct simplified $x^2$ term (allow 12.8) |
| Final answer: $32-32x+\dfrac{64}{5}x^2$ | | Ignore extra terms; mark final answer; if no simplification in (b) do not allow simplified terms in (c) as recovery |
**Alternative (factor out power of 2):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(2-\tfrac{2}{5}x\right)^5=2^5\!\left(1-\tfrac{1}{5}x\right)^5=2^5\!\left(1-5\cdot\tfrac{1}{5}x+\tfrac{5\times4}{2}\left(\tfrac{1}{5}x\right)^2+\ldots\right)$ | M1 | Score for $2^5\!\left(\ldots\pm5\times\tfrac{1}{5}x\pm\ldots\right)$ or $2^5\!\left(\ldots\pm\tfrac{5\times4}{2}\left(\tfrac{1}{5}x\right)^2+\ldots\right)$ |
| A marks as above | A1A1 | |
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# Question 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x)=-32+\dfrac{128}{5}x+\ldots \Rightarrow f'(0)=\ldots$ | M1 | Differentiates expansion AND substitutes $x=0$; requires $x^n\to x^{n-1}$ at least once (including $k\to0$ or $kx\to k$) |
| $f'(0)=-32$ | A1ft | Follow through on their expansion in (b) provided it was of form $p+qx+rx^2$, $p,q,r\neq0$, and differentiation is correct for their expansion |
---\begin{enumerate}
\item A curve $C$ has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = ( 2 - k x ) ^ { 5 }$$
and $k$ is a constant.\\
Given that when $\mathrm { f } ( x )$ is divided by $( 4 x - 5 )$ the remainder is $\frac { 243 } { 32 }$\\
(a) show that $k = \frac { 2 } { 5 }$\\
(b) Find the first three terms, in ascending powers of $x$, of the binomial expansion of
$$\left( 2 - \frac { 2 } { 5 } x \right) ^ { 5 }$$
giving each term in simplest form.
Using the solution to part (b) and making your method clear,\\
(c) find the gradient of $C$ at the point where $x = 0$
\hfill \mbox{\textit{Edexcel P2 2022 Q2 [7]}}