| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Modulus function transformations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard C3/C4 content: sketching logarithmic functions and their modulus transformations, solving modulus equations, and function composition. All parts follow routine procedures with no novel problem-solving required. Part (a) requires knowing that |f(x)| reflects negative portions above the x-axis; part (b) is a standard two-case modulus equation; parts (c-d) involve direct substitution and recognizing range from the composite function form. Slightly above average difficulty due to multiple parts and requiring careful execution, but well within typical C3/C4 expectations. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02m Graphs of functions: difference between plotting and sketching1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Logarithmic shaped curve in correct position | B1 | Shape |
| \((e^2, 0)\) | B1 | Intersection with \(x\)-axis |
| Asymptote \(x=0\) | B1 | Equation of asymptote (do not allow "\(y\)-axis") |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape (reflection in \(x\)-axis, above and below) | B1ft | Shape — must appear both above and below \(x\)-axis with correct curvature to left of intercept |
| Asymptote and coordinate \((e^2, 0)\) | B1ft | Follow through on (a)(i) coordinates and asymptote |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\ln x - 4 = 4 \Rightarrow \ln x = 4 \Rightarrow x = e^4\) | M1A1 | Sets \(2\ln x -4=4\), proceeds to \(x=e^{\ldots}\) |
| \(2\ln x - 4 = -4 \Rightarrow \ln x = 0 \Rightarrow x = 1\) | M1A1 | Sets \(-2\ln x+4=4\) or \(2\ln x+4=-4\); \(x=1\) may be by symmetry if \((1,-4)\) identified |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2\ln x-4)^2=16 \Rightarrow 4(\ln x)^2-16\ln x+16=16 \Rightarrow \ln x=\ldots\) | M1 | Squares both sides, expands, proceeds to solve for \(\ln x\) |
| Proceeds from \(\ln x=\ldots\) to at least one value of \(x\) | M1 | |
| \(x=e^4\) | A1 | |
| \(x=1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{gf}(x)=e^{2\ln x-4+5}-2 = e^1 \times e^{2\ln x}-2 = ex^2-2\) | M1, dM1A1 | M1: attempts gf correct way; look for \(e^{2\ln x\pm\ldots}\). dM1: correct processing to form \(e^k x^2-2\), \(k\neq 0\) (only allow slips on "\(-4+5\)"). A1: \(ex^2-2\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{gf}(x) > -2\) | B1 | Accept: "\(>-2\)", \(\text{gf}(x)>-2\), range \(>-2\), \(y>-2\), \(-2<\text{gf}(x)<\infty\), \((-2,\infty)\). Allow in words. Not \(x>-2\) |
| (1) |
# Question 9:
**(a)(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Logarithmic shaped curve in correct position | B1 | Shape |
| $(e^2, 0)$ | B1 | Intersection with $x$-axis |
| Asymptote $x=0$ | B1 | Equation of asymptote (do not allow "$y$-axis") |
| | **(3)** | |
**(a)(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape (reflection in $x$-axis, above and below) | B1ft | Shape — must appear both above and below $x$-axis with correct curvature to left of intercept |
| Asymptote and coordinate $(e^2, 0)$ | B1ft | Follow through on (a)(i) coordinates and asymptote |
| | **(2)** | |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\ln x - 4 = 4 \Rightarrow \ln x = 4 \Rightarrow x = e^4$ | M1A1 | Sets $2\ln x -4=4$, proceeds to $x=e^{\ldots}$ |
| $2\ln x - 4 = -4 \Rightarrow \ln x = 0 \Rightarrow x = 1$ | M1A1 | Sets $-2\ln x+4=4$ or $2\ln x+4=-4$; $x=1$ may be by symmetry if $(1,-4)$ identified |
| | **(4)** | |
**Alternative by squaring:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2\ln x-4)^2=16 \Rightarrow 4(\ln x)^2-16\ln x+16=16 \Rightarrow \ln x=\ldots$ | M1 | Squares both sides, expands, proceeds to solve for $\ln x$ |
| Proceeds from $\ln x=\ldots$ to at least one value of $x$ | M1 | |
| $x=e^4$ | A1 | |
| $x=1$ | A1 | |
**(c)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{gf}(x)=e^{2\ln x-4+5}-2 = e^1 \times e^{2\ln x}-2 = ex^2-2$ | M1, dM1A1 | M1: attempts gf correct way; look for $e^{2\ln x\pm\ldots}$. dM1: correct processing to form $e^k x^2-2$, $k\neq 0$ (only allow slips on "$-4+5$"). A1: $ex^2-2$ |
| | **(3)** | |
**(d)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{gf}(x) > -2$ | B1 | Accept: "$>-2$", $\text{gf}(x)>-2$, range $>-2$, $y>-2$, $-2<\text{gf}(x)<\infty$, $(-2,\infty)$. Allow in words. Not $x>-2$ |
| | **(1)** | |
---9.
$$\mathrm { f } ( x ) = 2 \ln ( x ) - 4 , \quad x > 0 , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch, on separate diagrams, the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = \mathrm { f } ( x )$
\item $y = | \mathrm { f } ( x ) |$
On each diagram, show the coordinates of each point at which the curve meets or cuts the axes.
On each diagram state the equation of the asymptote.
\end{enumerate}\item Find the exact solutions of the equation $| \mathrm { f } ( x ) | = 4$
$$\mathrm { g } ( x ) = \mathrm { e } ^ { x + 5 } - 2 , \quad x \in \mathbb { R }$$
\item Find $\mathrm { gf } ( x )$, giving your answer in its simplest form.
\item Hence, or otherwise, state the range of gf.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q9 [13]}}