| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard C3/C4 composite and inverse functions question covering routine techniques: composition (with exponential/log simplification), solving an exponential equation, finding an inverse function, and sketching. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 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 properties1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(fg(x) = e^{-2\ln x}+2\) | M1 | |
| \(= e^{\ln x^{-2}}+2 = x^{-2}+2 = \left(\frac{1}{x^2}+2\right)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{-(2x+3)}+2=6 \Rightarrow e^{-(2x+3)}=4\) | M1 A1 | |
| \(\Rightarrow -(2x+3)=\ln 4\) | ||
| \(\Rightarrow x=\frac{-3-\ln 4}{2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(y=e^{-x}+2 \Rightarrow y-2=e^{-x} \Rightarrow \ln(y-2)=-x\) | M1 | |
| \(\Rightarrow x=-\ln(y-2)\) | ||
| \(f^{-1}(x)=-\ln(x-2),\quad x>2\) | A1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct shape for \(f(x)\) | B1 | |
| \((0,3)\) marked | B1 | |
| Correct shape for \(f^{-1}(x)\) | B1 | |
| \((3,0)\) marked | B1 |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $fg(x) = e^{-2\ln x}+2$ | M1 | |
| $= e^{\ln x^{-2}}+2 = x^{-2}+2 = \left(\frac{1}{x^2}+2\right)$ | M1 A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{-(2x+3)}+2=6 \Rightarrow e^{-(2x+3)}=4$ | M1 A1 | |
| $\Rightarrow -(2x+3)=\ln 4$ | | |
| $\Rightarrow x=\frac{-3-\ln 4}{2}$ | M1 A1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $y=e^{-x}+2 \Rightarrow y-2=e^{-x} \Rightarrow \ln(y-2)=-x$ | M1 | |
| $\Rightarrow x=-\ln(y-2)$ | | |
| $f^{-1}(x)=-\ln(x-2),\quad x>2$ | A1 B1 | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape for $f(x)$ | B1 | |
| $(0,3)$ marked | B1 | |
| Correct shape for $f^{-1}(x)$ | B1 | |
| $(3,0)$ marked | B1 | |
---\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{array} { l l }
\mathrm { f } : x \mapsto \mathrm { e } ^ { - x } + 2 , & x \in \mathbb { R } \\
\mathrm {~g} : x \mapsto 2 \ln x , & x > 0
\end{array}$$
(a) Find $\mathrm { fg } ( x )$, giving your answer in its simplest form.\\
(b) Find the exact value of $x$ for which $\mathrm { f } ( 2 x + 3 ) = 6$\\
(c) Find $\mathrm { f } ^ { - 1 }$, stating its domain.\\
(d) On the same axes, sketch the curves with equation $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, giving the coordinates of all the points where the curves cross the axes.\\
\hfill \mbox{\textit{Edexcel C34 Q5 [14]}}