| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C3 topics (sketching logarithmic and modulus functions, composite functions, and differentiation). Part (c) requires evaluating fg(k/4) which involves substituting into the modulus function then the logarithm - straightforward but requires care with the algebra. Part (d) involves differentiating ln(x+k) and setting equal to a given gradient, which is routine. Slightly easier than average due to being mostly procedural with clear steps. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions |
| Answer | Marks |
|---|---|
| Log graph: Shape | B1 |
| Intersection with \(-ve\) \(x\)-axis | dB1 |
| \((0, \ln k)\), \((1-k, 0)\) | B1 |
| Mod graph: V shape, vertex on \(+ve\) \(x\)-axis | B1 |
| \((0, k)\) and \(\left(\frac{k}{2}, 0\right)\) | B1 (5) |
| Answer | Marks |
|---|---|
| \(f(x) \in \mathbb{R}\), \(-\infty < f(x) < \infty\), \(-\infty < y < \infty\) | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(fg\left(\frac{k}{4}\right) = \ln\left\{k + \left | \frac{2k}{4} - k\right | \right\}\) or \(f\left(\left |
| \(= \ln\left(\frac{3k}{2}\right)\) | A1 (2) |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{1}{x+k}\) | B1 |
| Equating (with \(x=3\)) to grad. of line: \(\frac{1}{3+k} = \frac{2}{9}\) | M1; A1 |
| \(k = 1\frac{1}{2}\) | A1\(\sqrt{}\) (4) |
# Question 7:
## Part (a)
Log graph: Shape | B1 |
Intersection with $-ve$ $x$-axis | dB1 |
$(0, \ln k)$, $(1-k, 0)$ | B1 |
Mod graph: V shape, vertex on $+ve$ $x$-axis | B1 |
$(0, k)$ and $\left(\frac{k}{2}, 0\right)$ | B1 (5) |
## Part (b)
$f(x) \in \mathbb{R}$, $-\infty < f(x) < \infty$, $-\infty < y < \infty$ | B1 (1) |
## Part (c)
$fg\left(\frac{k}{4}\right) = \ln\left\{k + \left|\frac{2k}{4} - k\right|\right\}$ or $f\left(\left|-\frac{k}{2}\right|\right)$ | M1 |
$= \ln\left(\frac{3k}{2}\right)$ | A1 (2) |
## Part (d)
$\frac{dy}{dx} = \frac{1}{x+k}$ | B1 |
Equating (with $x=3$) to grad. of line: $\frac{1}{3+k} = \frac{2}{9}$ | M1; A1 |
$k = 1\frac{1}{2}$ | A1$\sqrt{}$ (4) |7. For the constant $k$, where $k > 1$, the functions f and g are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto \ln ( x + k ) , \quad x > - k , \\
& \mathrm {~g} : x \mapsto | 2 x - k | , \quad x \in \mathbb { R } .
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item On separate axes, sketch the graph of f and the graph of g .
On each sketch state, in terms of $k$, the coordinates of points where the graph meets the coordinate axes.
\item Write down the range of f.
\item Find $\mathrm { fg } \left( \frac { k } { 4 } \right)$ in terms of $k$, giving your answer in its simplest form.
The curve $C$ has equation $y = \mathrm { f } ( x )$. The tangent to $C$ at the point with $x$-coordinate 3 is parallel to the line with equation $9 y = 2 x + 1$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2006 Q7 [12]}}