| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Standard +0.8 This question requires finding an inverse involving exponentials and logarithms (standard C3), solving an equation combining the inverse with ln x (requiring algebraic manipulation), and most challengingly, using calculus to find when f(t) = ke^t has exactly one solution (requiring insight that this means tangency). The final part elevates this above routine inverse function questions. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06c Logarithm definition: log_a(x) as inverse of a^x1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(y = 3 - 2e^{-x}\), then \(2e^{-x} = 3-y\) | M1 | Puts \(y = f(x)\) and makes \(e^{-x}\) term subject; allow \(f(x)\) instead of \(y\) |
| \(-x = \ln\frac{3-y}{2}\) and \(x = -\ln\frac{3-y}{2}\) | M1 | Uses ln to get \(x\); allow sign errors and weak log work |
| (correct unsimplified answer for \(x\)) | A1 | Completely correct log work giving correct unsimplified answer |
| \(f^{-1}(x) = \ln\frac{2}{3-x}\) or \(-\ln\frac{3-x}{2}\) o.e. | A1 | Variable must be \(x\) not \(y\) |
| Domain is \(x < 3\) | B1 | Allow \((-\infty, 3)\); \(x \leq 3\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ln\frac{2}{3-x} = \ln x \rightarrow 2 = (3-x)x\) | M1 | Removes ln correctly on both sides and multiplies across |
| \(x^2 - 3x + 2 = 0\) | A1 | Expands bracket to give three-term quadratic |
| \(x = 2\) or \(x = 1\) | M1 A1 | Solves quadratic; need both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3 - 2e^{-t} = ke^t \rightarrow ke^{2t} - 3e^t + 2 = 0\) o.e. | M1 A1 | Sets \(3 - 2e^{-t} = ke^t\) and attempts to multiply all terms by \(e^t\) or \(e^{-t}\) |
| Use \(b^2 - 4ac = 0\) or \(b^2 = 4ac\) or attempts \(e^t = \frac{3 \pm \sqrt{9-8k}}{2k}\) | dM1 | Uses condition for equal roots; dependent on M1 |
| \(k = 1.125\) o.e. e.g. \(\frac{9}{8}\) or \(1\frac{1}{8}\) | A1 |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $y = 3 - 2e^{-x}$, then $2e^{-x} = 3-y$ | M1 | Puts $y = f(x)$ and makes $e^{-x}$ term subject; allow $f(x)$ instead of $y$ |
| $-x = \ln\frac{3-y}{2}$ and $x = -\ln\frac{3-y}{2}$ | M1 | Uses ln to get $x$; allow sign errors and weak log work |
| (correct unsimplified answer for $x$) | A1 | Completely correct log work giving correct unsimplified answer |
| $f^{-1}(x) = \ln\frac{2}{3-x}$ or $-\ln\frac{3-x}{2}$ o.e. | A1 | Variable must be $x$ not $y$ |
| Domain is $x < 3$ | B1 | Allow $(-\infty, 3)$; $x \leq 3$ is B0 | (5) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln\frac{2}{3-x} = \ln x \rightarrow 2 = (3-x)x$ | M1 | Removes ln correctly on both sides and multiplies across |
| $x^2 - 3x + 2 = 0$ | A1 | Expands bracket to give three-term quadratic |
| $x = 2$ or $x = 1$ | M1 A1 | Solves quadratic; need both correct answers | (4) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 - 2e^{-t} = ke^t \rightarrow ke^{2t} - 3e^t + 2 = 0$ o.e. | M1 A1 | Sets $3 - 2e^{-t} = ke^t$ and attempts to multiply all terms by $e^t$ or $e^{-t}$ |
| Use $b^2 - 4ac = 0$ or $b^2 = 4ac$ or attempts $e^t = \frac{3 \pm \sqrt{9-8k}}{2k}$ | dM1 | Uses condition for equal roots; dependent on M1 |
| $k = 1.125$ o.e. e.g. $\frac{9}{8}$ or $1\frac{1}{8}$ | A1 | | (4) [13] |8. The function $f$ is defined by
$$\mathrm { f } : x \rightarrow 3 - 2 \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Find the inverse function, $\mathrm { f } ^ { - 1 } ( x )$ and give its domain.
\item Solve the equation $\mathrm { f } ^ { - 1 } ( x ) = \ln x$.
The equation $\mathrm { f } ( t ) = k \mathrm { e } ^ { t }$, where $k$ is a positive constant, has exactly one real solution.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q8 [13]}}