| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify shape type from coordinates |
| Difficulty | Moderate -0.5 This is a straightforward C1 coordinate geometry question requiring standard techniques: gradient calculation, perpendicular line verification (product = -1), midpoint formula, distance formula, and finding equation of a line. All parts are routine applications with no problem-solving insight needed, making it slightly easier than average, though the multi-part structure and final synthesis step (recognizing the line of symmetry passes through B and M) prevents it from being trivial. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{5+x}{(1-x)(2+x)} = \frac{A}{1-x} + \frac{B}{2+x}\), \(\Rightarrow 5 + x = A(2+x) + B(1-x)\) | M1 | Either multiplication by denominator or cover up rule attempted |
| Substitute \(x = 1\); Substitute \(x = -2\) | m1 | Either use (any) two values or equate coefficients to form and attempt to solve \(A - B = 1\) and \(2A + B = 5\) |
| \(A = 2\), \(B = 1\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-x)^{-1} = 1 + (-1)(-x) + px^2\) | M1 | \(p \neq 0\) |
| \(= 1 + x + x^2\ldots\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^{-1}\left[1 + \frac{x}{2}\right]^{-1} = \frac{1}{2}\left[1 + (-1)\left(\frac{x}{2}\right) + \frac{(-1)(-2)}{2!}\left(\frac{x}{2}\right)^2 + \ldots\right]\) | M1 | \(\left[1 + (-1)\left(\frac{x}{2}\right) + kx^2\right]\) |
| A1 | Correct expansion of \(\left(1 + \frac{x}{2}\right)^{-1}\) | |
| \(\frac{5+x}{(1-x)(2+x)} = 2(1-x)^{-1} + (2+x)^{-1}\) | M1 | Using (a) with powers \(-1\) |
| \(= 2(1 + x + x^2\ldots) + \frac{1}{2}\left(1 - \frac{x}{2} + \frac{x^2}{4} + \ldots\right)\) | m1 | Dep on prev 3Ms |
| \(= 2.5 + 1.75x + 2.125x^2 + \ldots\) | A1F | 5 |
# Question 2(a) [XMCA2]:
$\frac{5+x}{(1-x)(2+x)} = \frac{A}{1-x} + \frac{B}{2+x}$, $\Rightarrow 5 + x = A(2+x) + B(1-x)$ | M1 | Either multiplication by denominator or cover up rule attempted
Substitute $x = 1$; Substitute $x = -2$ | m1 | Either use (any) two values or equate coefficients to form and attempt to solve $A - B = 1$ and $2A + B = 5$
$A = 2$, $B = 1$ | A1 | **3** |
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# Question 2(b)(i) [XMCA2]:
$(1-x)^{-1} = 1 + (-1)(-x) + px^2$ | M1 | $p \neq 0$
$= 1 + x + x^2\ldots$ | A1 | **2** |
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# Question 2(b)(ii) [XMCA2]:
$2^{-1}\left[1 + \frac{x}{2}\right]^{-1} = \frac{1}{2}\left[1 + (-1)\left(\frac{x}{2}\right) + \frac{(-1)(-2)}{2!}\left(\frac{x}{2}\right)^2 + \ldots\right]$ | M1 | $\left[1 + (-1)\left(\frac{x}{2}\right) + kx^2\right]$
| A1 | Correct expansion of $\left(1 + \frac{x}{2}\right)^{-1}$
$\frac{5+x}{(1-x)(2+x)} = 2(1-x)^{-1} + (2+x)^{-1}$ | M1 | Using (a) with powers $-1$
$= 2(1 + x + x^2\ldots) + \frac{1}{2}\left(1 - \frac{x}{2} + \frac{x^2}{4} + \ldots\right)$ | m1 | Dep on prev 3Ms
$= 2.5 + 1.75x + 2.125x^2 + \ldots$ | A1F | **5** | Ft only on wrong integer values for $A$ and $B$, ie simplified $(A + \frac{1}{2}B) + (A - \frac{1}{4}B)x + (A + \frac{1}{8}B)\ldots$
---2 The triangle $A B C$ has vertices $A ( 1,3 ) , B ( 3,7 )$ and $C ( - 1,9 )$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item Hence show that angle $A B C$ is a right angle.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the coordinates of $M$, the mid-point of $A C$.
\item Show that the lengths of $A B$ and $B C$ are equal.
\item Hence find an equation of the line of symmetry of the triangle $A B C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q2 [12]}}