| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.5 Part (a) is straightforward composition requiring only the rule that ln(e^x) = x to get gf(x) = 3x. Part (b) requires setting 3x = e^(3ln x) = x^3 and solving x^3 - 3x = 0, then verifying only one solution is valid for the domain—this involves some algebraic manipulation and domain checking but remains a fairly standard composite function exercise with no novel insight required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{gf}(x) = 3\ln e^x\) | M1 | Apply functions in correct order |
| \(= 3x, \quad (x \in \mathbb{R})\) | A1 | Must be simplified to \(3x\); not left as \(3\ln e^x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{gf}(x) = \text{fg}(x) \Rightarrow 3x = x^3\) | M1 | Allow if \(e^{3\ln x}\) used as their \(3x\) |
| \(\Rightarrow x^3 - 3x = 0 \Rightarrow x =\) | M1 | Solve cubic; obtain at least one solution |
| \(\Rightarrow x = (+)\sqrt{3}\) only, as \(\ln x\) is not defined at \(x=0\) and \(-\sqrt{3}\) | M1 | State \(x=\sqrt{3}\) only; justify exclusion of \(x=0\) and \(x=-\sqrt{3}\) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{gf}(x) = 3\ln e^x$ | M1 | Apply functions in correct order |
| $= 3x, \quad (x \in \mathbb{R})$ | A1 | Must be simplified to $3x$; not left as $3\ln e^x$ |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{gf}(x) = \text{fg}(x) \Rightarrow 3x = x^3$ | M1 | Allow if $e^{3\ln x}$ used as their $3x$ |
| $\Rightarrow x^3 - 3x = 0 \Rightarrow x =$ | M1 | Solve cubic; obtain at least one solution |
| $\Rightarrow x = (+)\sqrt{3}$ only, as $\ln x$ is not defined at $x=0$ and $-\sqrt{3}$ | M1 | State $x=\sqrt{3}$ **only**; justify exclusion of $x=0$ and $x=-\sqrt{3}$ |
---4. Given
$$\begin{aligned}
& \mathrm { f } ( x ) = \mathrm { e } ^ { x } , \quad x \in \mathbb { R } \\
& \mathrm {~g} ( x ) = 3 \ln x , \quad x > 0 , x \in \mathbb { R }
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item find an expression for $\mathrm { gf } ( x )$, simplifying your answer.
\item Show that there is only one real value of $x$ for which $\operatorname { gf } ( x ) = \operatorname { fg } ( x )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q4 [5]}}