| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Composite transformation sketch |
| Difficulty | Moderate -0.3 This is a standard C3 transformation question requiring sketches of |f(x)| and f^(-1)(x) with labeled intercepts, plus routine inverse function work. The transformations are textbook examples (reflection in x-axis for negatives, reflection in y=x for inverse), and finding the inverse of e^(2x)-k is straightforward. Slightly easier than average due to the structured parts guiding students through each step. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Curve retains shape when \(x > \frac{1}{2}\ln k\) | B1 | |
| Curve reflects through the \(x\)-axis when \(x < \frac{1}{2}\ln k\) | B1 | |
| \((0, k-1)\) and \((\frac{1}{2}\ln k, 0)\) marked in correct positions | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape of curve, contained in quadrants 1, 2 and 3 | B1 | Ignore asymptote |
| \((1-k, 0)\) and \((0, \frac{1}{2}\ln k)\) marked | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range of f: \(f(x) > -k\) or \(y > -k\) or \((-k, \infty)\) | B1 | Either \(f(x)>-k\) or \(y>-k\) or \((-k,\infty)\) or \(f>-k\) or Range \(> -k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = e^{2x} - k \Rightarrow y + k = e^{2x}\) | M1 | Attempt to make \(x\) (or swapped \(y\)) the subject |
| \(\ln(y+k) = 2x\) | M1 | Makes \(e^{2x}\) the subject and takes ln of both sides |
| \(f^{-1}(x) = \frac{1}{2}\ln(x+k)\) | A1 | \(\frac{1}{2}\ln(x+k)\) or \(\ln\sqrt{(x+k)}\), cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f^{-1}(x)\): Domain: \(x > -k\) or \((-k, \infty)\) | B1\(\checkmark\) | Either \(x>-k\) or \((-k,\infty)\) or Domain \(>-k\) or \(x\) "ft one sided inequality" their part (c) RANGE answer |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Curve retains shape when $x > \frac{1}{2}\ln k$ | B1 | |
| Curve reflects through the $x$-axis when $x < \frac{1}{2}\ln k$ | B1 | |
| $(0, k-1)$ and $(\frac{1}{2}\ln k, 0)$ marked in correct positions | B1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape of curve, contained in quadrants 1, 2 and 3 | B1 | Ignore asymptote |
| $(1-k, 0)$ and $(0, \frac{1}{2}\ln k)$ marked | B1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range of f: $f(x) > -k$ or $y > -k$ or $(-k, \infty)$ | B1 | Either $f(x)>-k$ or $y>-k$ or $(-k,\infty)$ or $f>-k$ or Range $> -k$ |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{2x} - k \Rightarrow y + k = e^{2x}$ | M1 | Attempt to make $x$ (or swapped $y$) the subject |
| $\ln(y+k) = 2x$ | M1 | Makes $e^{2x}$ the subject and takes ln of both sides |
| $f^{-1}(x) = \frac{1}{2}\ln(x+k)$ | A1 | $\frac{1}{2}\ln(x+k)$ or $\ln\sqrt{(x+k)}$, cao |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f^{-1}(x)$: Domain: $x > -k$ or $(-k, \infty)$ | B1$\checkmark$ | Either $x>-k$ or $(-k,\infty)$ or Domain $>-k$ or $x$ "ft one sided inequality" their part (c) RANGE answer |
---5.
\begin{tikzpicture}[>=latex, thick]
% Draw horizontal x-axis
\draw[->] (-5.5, 0) -- (6.5, 0) node[below] {$x$};
% Draw vertical y-axis
\draw[->] (0, -2.5) -- (0, 5.5) node[left] {$y$};
% Origin label
\node[below left] at (0, 0) {$O$};
% Plot the curve
\draw[domain=-5.3:4.4, smooth, samples=100] plot (\x, {exp(0.45*\x) - 2});
% Label Point A (y-intercept)
\node[below right=1pt] at (0, -1) {$A$};
% Label Point B (x-intercept)
\node[below right=1pt] at ({ln(2)/0.45}, 0) {$B$};
\end{tikzpicture}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$.\\
The curve meets the coordinate axes at the points $A ( 0,1 - k )$ and $B \left( \frac { 1 } { 2 } \ln k , 0 \right)$, where $k$ is a constant and $k > 1$, as shown in Figure 2.
On separate diagrams, sketch the curve with equation
\begin{enumerate}[label=(\alph*)]
\item $y = | f ( x ) |$,
\item $y = \mathrm { f } ^ { - 1 } ( x )$.
Show on each sketch the coordinates, in terms of $k$, of each point at which the curve meets or cuts the axes.
Given that $\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - k$,
\item state the range of f ,
\item find $\mathrm { f } ^ { - 1 } ( x )$,
\item write down the domain of $\mathrm { f } ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2009 Q5 [10]}}