| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Composite transformation sketch |
| Difficulty | Standard +0.3 This is a multi-part question testing standard transformations (modulus, vertical stretch/translation) and function composition. Parts (a)-(b) require routine sketching with reflection and scaling. Parts (c)-(e) involve straightforward inverse function finding and composition with logarithms. All techniques are standard C3/C4 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct curve shape in first and second quadrants with cusp on \(x\)-axis | B1 | Curve must lie completely in Q1 and Q2, correct curvature on both sections |
| Asymptote \(y = 4\) | B1 | Do not award if second asymptote given |
| \(y\)-intercept \((0, 3)\) | B1 | Condone \((3,0)\) marked on correct axis; do not award if two intercepts given |
| Touches \(x\)-axis at \(\left(-\frac{1}{3}\ln 4, 0\right)\) | B1 | \(-\frac{1}{3}\ln 4\) sufficient if marked on \(x\)-axis; do not award if two intercepts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape similar to original graph | B1 | Do not judge the stretch |
| Asymptote \(y = -2\) | B1 | Do not award if second asymptote given |
| Curve passes through origin only | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) > -4\) | B1 | Accept \((-4, \infty)\); note \(f(x) \geq -4\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = e^{-3x} - 4 \Rightarrow e^{-3x} = y + 4\) | M1 | Attempt to make \(x\) (or switched \(y\)) subject; must get \(e^{-3x}\) or \(e^{3x}\) as subject |
| \(\Rightarrow -3x = \ln(y+4)\) | dM1 | Dependent on M1; undoing exp using ln; condone imaginary brackets |
| \(f^{-1}(x) = -\frac{1}{3}\ln(x+4)\) or \(\ln\dfrac{1}{(x+4)^{\frac{1}{3}}}\), \((x > -4)\) | A1 | cao; domain not required for mark but bracket is; accept \(y =\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(x) = e^{-3\ln\left(\frac{1}{x+2}\right)} - 4\) | M1 | Correct order of operations; allow \(e^{\pm 3\ln\frac{1}{x+2}} - 4\) |
| \(= (x+2)^3 - 4\) | dM1 | Dependent; uses power law then \(e^{\ln(\cdots)} = \cdots\); condone sign errors |
| \(= x^3 + 6x^2 + 12x + 4\) | A1 | Correct expansion |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct curve shape in first and second quadrants with cusp on $x$-axis | B1 | Curve must lie completely in Q1 and Q2, correct curvature on both sections |
| Asymptote $y = 4$ | B1 | Do not award if second asymptote given |
| $y$-intercept $(0, 3)$ | B1 | Condone $(3,0)$ marked on correct axis; do not award if two intercepts given |
| Touches $x$-axis at $\left(-\frac{1}{3}\ln 4, 0\right)$ | B1 | $-\frac{1}{3}\ln 4$ sufficient if marked on $x$-axis; do not award if two intercepts |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape similar to original graph | B1 | Do not judge the stretch |
| Asymptote $y = -2$ | B1 | Do not award if second asymptote given |
| Curve passes through origin only | B1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) > -4$ | B1 | Accept $(-4, \infty)$; note $f(x) \geq -4$ is B0 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{-3x} - 4 \Rightarrow e^{-3x} = y + 4$ | M1 | Attempt to make $x$ (or switched $y$) subject; must get $e^{-3x}$ or $e^{3x}$ as subject |
| $\Rightarrow -3x = \ln(y+4)$ | dM1 | Dependent on M1; undoing exp using ln; condone imaginary brackets |
| $f^{-1}(x) = -\frac{1}{3}\ln(x+4)$ or $\ln\dfrac{1}{(x+4)^{\frac{1}{3}}}$, $(x > -4)$ | A1 | cao; domain not required for mark but bracket is; accept $y =$ |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(x) = e^{-3\ln\left(\frac{1}{x+2}\right)} - 4$ | M1 | Correct order of operations; allow $e^{\pm 3\ln\frac{1}{x+2}} - 4$ |
| $= (x+2)^3 - 4$ | dM1 | Dependent; uses power law then $e^{\ln(\cdots)} = \cdots$; condone sign errors |
| $= x^3 + 6x^2 + 12x + 4$ | A1 | Correct expansion |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-16_572_1338_278_239}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x ) , \quad x \in \mathbb { R }$\\
The curve meets the coordinate axes at the points $A ( 0 , - 3 )$ and $B \left( - \frac { 1 } { 3 } \ln 4,0 \right)$ and the curve has an asymptote with equation $y = - 4$
In separate diagrams, sketch the graph with equation
\begin{enumerate}[label=(\alph*)]
\item $y = | f ( x ) |$
\item $y = 2 \mathrm { f } ( x ) + 6$
On each sketch, give the exact coordinates of the points where the curve crosses or meets the coordinate axes and the equation of any asymptote.
Given that
$$\begin{array} { l l }
\mathrm { f } ( x ) = \mathrm { e } ^ { - 3 x } - 4 , & x \in \mathbb { R } \\
\mathrm {~g} ( x ) = \ln \left( \frac { 1 } { x + 2 } \right) , & x > - 2
\end{array}$$
\item state the range of f,
\item find $\mathrm { f } ^ { - 1 } ( x )$,
\item express $f g ( x )$ as a polynomial in $x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q11 [14]}}