| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Improper fraction partial fractions |
| Difficulty | Standard +0.3 This is a standard two-part partial fractions question requiring decomposition of an improper fraction (part a) followed by binomial expansion (part b). The improper fraction adds one extra step compared to typical questions, but the techniques are routine: equating coefficients or substitution for part (a), then standard binomial expansion of (1+x)^{-1} and (1-x/3)^{-1} for part (b). No novel insight required, just careful algebraic manipulation. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 3\) | B1 | Must be clearly identified as the value of \(A\). May be implied by their partial fractions |
| \(3x^2 + 4x - 7 = A(x+1)(x-3) + B(x-3) + C(x+1)\) then expands/compares coefficients or substitutes values of \(x\) Or \(3x^2 + 4x - 7 \div (x+1)(x-3) = 3 + \frac{10x+2}{(x+1)(x-3)}\) \(\Rightarrow 10x + 2 = B(x-3) + C(x+1)\) | M1 | Correct method implied by correct values with no incorrect work seen |
| \(B = 2\) or \(C = 8\) | A1 | One of \(B\) or \(C\) correct |
| \(B = 2\) and \(C = 8\) | A1 | Both \(B\) and \(C\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{x+1} = (1+x)^{-1} = (1 - x + x^2...)\) | M1 | Attempts to expand \((1+x)^{-1}\). Look for \(1 +\) a correct simplified or unsimplified second or third term |
| \(\frac{1}{x-3} = -(3-x)^{-1} = -\frac{1}{3}\left(1 - \frac{1}{3}x\right)^{-1}\) | B1 | \(\frac{1}{x-3} = -\frac{1}{3}\left(1-\frac{1}{3}x\right)^{-1}\). Takes out a correct factor including the minus sign and a correct bracket |
| \(\left(1 - \frac{1}{3}x\right)^{-1} = 1 + \frac{1}{3}x + \frac{1}{9}x^2...\) | M1 | Attempts to expand \(\left(1 \pm \frac{1}{3}x\right)^{-1}\). Look for \(1 +\) a correct simplified or unsimplified second or third term |
| \(\frac{3x^2+4x-7}{(x+1)(x-3)} \approx (3+)2(1-x+x^2) - \frac{8}{3}\left(1+\frac{1}{3}x+\frac{1}{9}x^2\right)\) | M1 | Combines using their expansions and at least their \(B\) and \(C\) (allow if they forget/don't add their \(A\)) |
| \(= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2\) | A1 | Any 2 correct terms |
| \(= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2\) | A1 | All terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+x)^{-1} = (1 - x + x^2...)\) | M1 | Attempts to expand \((1+x)^{-1}\). Look for \(1 +\) a correct simplified or unsimplified second or third term |
| \(\frac{1}{x-3} = -(3-x)^{-1} = -\frac{1}{3}\left(1-\frac{1}{3}x\right)^{-1}\) | B1 | Takes out a correct factor including the minus sign |
| \(\left(1-\frac{1}{3}x\right)^{-1} = 1 + \frac{1}{3}x + \frac{1}{9}x^2...\) | M1 | Attempts to expand \(\left(1\pm\frac{1}{3}x\right)^{-1}\) |
| \(\frac{3x^2+4x-7}{(x+1)(x-3)} \approx (3x^2+4x-7)\left(-\frac{1}{3}\right)\left(1+\frac{1}{3}x+\frac{1}{9}x^2\right)(1-x+x^2)\) | M1 | Attempts to multiply out all 3 brackets |
| \(= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2\) | A1 | Any 2 correct terms |
| \(= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2\) | A1 | All terms correct |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 3$ | B1 | Must be clearly identified as the value of $A$. May be implied by their partial fractions |
| $3x^2 + 4x - 7 = A(x+1)(x-3) + B(x-3) + C(x+1)$ then expands/compares coefficients or substitutes values of $x$ **Or** $3x^2 + 4x - 7 \div (x+1)(x-3) = 3 + \frac{10x+2}{(x+1)(x-3)}$ $\Rightarrow 10x + 2 = B(x-3) + C(x+1)$ | M1 | Correct method implied by correct values with no incorrect work seen |
| $B = 2$ or $C = 8$ | A1 | One of $B$ or $C$ correct |
| $B = 2$ and $C = 8$ | A1 | Both $B$ and $C$ correct |
### Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{x+1} = (1+x)^{-1} = (1 - x + x^2...)$ | M1 | Attempts to expand $(1+x)^{-1}$. Look for $1 +$ a correct simplified or unsimplified second or third term |
| $\frac{1}{x-3} = -(3-x)^{-1} = -\frac{1}{3}\left(1 - \frac{1}{3}x\right)^{-1}$ | B1 | $\frac{1}{x-3} = -\frac{1}{3}\left(1-\frac{1}{3}x\right)^{-1}$. Takes out a correct factor including the minus sign **and** a correct bracket |
| $\left(1 - \frac{1}{3}x\right)^{-1} = 1 + \frac{1}{3}x + \frac{1}{9}x^2...$ | M1 | Attempts to expand $\left(1 \pm \frac{1}{3}x\right)^{-1}$. Look for $1 +$ a correct simplified or unsimplified second or third term |
| $\frac{3x^2+4x-7}{(x+1)(x-3)} \approx (3+)2(1-x+x^2) - \frac{8}{3}\left(1+\frac{1}{3}x+\frac{1}{9}x^2\right)$ | M1 | Combines using their expansions and at least their $B$ and $C$ (allow if they forget/don't add their $A$) |
| $= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2$ | A1 | Any 2 correct terms |
| $= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2$ | A1 | All terms correct |
Allow $2\frac{1}{3}$ for $\frac{7}{3}$, $-2\frac{8}{9}$ for $-\frac{26}{9}$, $1\frac{19}{27}$ for $\frac{46}{27}$
### Part (b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+x)^{-1} = (1 - x + x^2...)$ | M1 | Attempts to expand $(1+x)^{-1}$. Look for $1 +$ a correct simplified or unsimplified second or third term |
| $\frac{1}{x-3} = -(3-x)^{-1} = -\frac{1}{3}\left(1-\frac{1}{3}x\right)^{-1}$ | B1 | Takes out a correct factor including the minus sign |
| $\left(1-\frac{1}{3}x\right)^{-1} = 1 + \frac{1}{3}x + \frac{1}{9}x^2...$ | M1 | Attempts to expand $\left(1\pm\frac{1}{3}x\right)^{-1}$ |
| $\frac{3x^2+4x-7}{(x+1)(x-3)} \approx (3x^2+4x-7)\left(-\frac{1}{3}\right)\left(1+\frac{1}{3}x+\frac{1}{9}x^2\right)(1-x+x^2)$ | M1 | Attempts to multiply out all 3 brackets |
| $= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2$ | A1 | Any 2 correct terms |
| $= \frac{7}{3} - \frac{26}{9}x + \frac{46}{27}x^2$ | A1 | All terms correct |
---2. Given that
$$\frac { 3 x ^ { 2 } + 4 x - 7 } { ( x + 1 ) ( x - 3 ) } \equiv A + \frac { B } { x + 1 } + \frac { C } { x - 3 }$$
\begin{enumerate}[label=(\alph*)]
\item find the values of the constants $A , B$ and $C$.
\item Hence, or otherwise, find the series expansion of
$$\frac { 3 x ^ { 2 } + 4 x - 7 } { ( x + 1 ) ( x - 3 ) } \quad | x | < 1$$
in ascending powers of $x$, up to and including the term in $x ^ { 2 }$\\
Give each coefficient as a simplified fraction.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q2 [10]}}