| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch absolute value of function |
| Difficulty | Standard +0.3 This is a multi-part question involving standard C3 transformations (absolute value and composite transformations), finding intercepts, and finding an inverse function. Part (a) requires understanding of graph transformations which is routine at this level. Parts (b) and (c) involve straightforward algebraic manipulation. While it has multiple parts (7 marks typical), each component is a standard textbook exercise requiring no novel insight—slightly easier than average. |
| Spec | 1.02s Modulus graphs: sketch graph of |ax+b|1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| Graph (i): \(y=2\) asymptote, passes through \((0,q)\) and \((p,0)\), \(x=1\) asymptote | M1 A1 | |
| Graph (ii): \(y=4\) asymptote, passes through \((p-1, 0)\), \(x=0\) asymptote | M2 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(y = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2}\) \(\therefore p = \frac{1}{2}\) | M1 A1 | |
| \(x = 0 \Rightarrow y = 1\) \(\therefore q = 1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(y = \frac{2x-1}{x-1}\), \(y(x-1) = 2x-1\), \(x(y-2) = y-1\) | M1 | |
| \(x = \frac{y-1}{y-2}\) | ||
| \(\therefore f^{-1}(x) = \frac{x-1}{x-2}\) | M1 A1 | (11) |
# Question 9:
## Part (a)(i) and (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| Graph (i): $y=2$ asymptote, passes through $(0,q)$ and $(p,0)$, $x=1$ asymptote | M1 A1 | |
| Graph (ii): $y=4$ asymptote, passes through $(p-1, 0)$, $x=0$ asymptote | M2 A1 | |
## Part (b):
| Answer/Working | Mark | Notes |
|---|---|---|
| $y = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ $\therefore p = \frac{1}{2}$ | M1 A1 | |
| $x = 0 \Rightarrow y = 1$ $\therefore q = 1$ | B1 | |
## Part (c):
| Answer/Working | Mark | Notes |
|---|---|---|
| $y = \frac{2x-1}{x-1}$, $y(x-1) = 2x-1$, $x(y-2) = y-1$ | M1 | |
| $x = \frac{y-1}{y-2}$ | | |
| $\therefore f^{-1}(x) = \frac{x-1}{x-2}$ | M1 A1 | **(11)** |
**Total: (72)**9.\\
\includegraphics[max width=\textwidth, alt={}, center]{5e6a37a1-c51f-4637-aaae-48da6ab3eca0-3_727_1022_244_342}
The diagram shows the curve with equation $y = \mathrm { f } ( x )$. The curve crosses the axes at $( p , 0 )$ and $( 0 , q )$ and the lines $x = 1$ and $y = 2$ are asymptotes of the curve.
\begin{enumerate}[label=(\alph*)]
\item Showing the coordinates of any points of intersection with the axes and the equations of any asymptotes, sketch on separate diagrams the graphs of
\begin{enumerate}[label=(\roman*)]
\item $y = | \mathrm { f } ( x ) |$,
\item $y = 2 \mathrm { f } ( x + 1 )$.
Given also that
$$\mathrm { f } ( x ) \equiv \frac { 2 x - 1 } { x - 1 } , \quad x \in \mathbb { R } , \quad x \neq 1$$
\end{enumerate}\item find the values of $p$ and $q$,
\item find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q9 [11]}}