| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.2 This is a straightforward multi-part question on standard C3 inverse function techniques. Parts (a)-(b) require routine identification of range and finding inverse of an exponential function. Parts (c)-(e) involve logarithm laws and composition—all textbook exercises with no novel problem-solving required. Slightly easier than average due to the step-by-step scaffolding. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) > k^2\) | B1 | Accept \(f > k^2\), Range \(> k^2\), \((k^2, \infty)\), \(y > k^2\). Do not accept \(f(x) \geq k^2\), \(x > k^2\), \([k^2, \infty)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = e^{2x} + k^2 \Rightarrow e^{2x} = y - k^2\) | M1 | Attempts to make \(x\) or swapped \(y\) the subject. Score for \(e^{2x} = y \pm k^2\) and proceeding to \(x = \ln\ldots\) |
| \(\Rightarrow x = \frac{1}{2}\ln(y - k^2)\) | dM1 | Dependent on previous M. Taking ln of whole RHS then dividing by 2 |
| \(f^{-1}(x) = \frac{1}{2}\ln(x-k^2),\quad x > k^2\) | A1 | Must include domain \(x > k^2\). Accept \(y = 0.5\ln |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ln 2x + \ln 2x^2 + \ln 2x^3 = 6\) | M1 | Attempts to solve by writing down the equation |
| \(\Rightarrow \ln 8x^6 = 6\) | M1 | Uses addition laws of logs to write in form \(\ln Ax^n = 6\) |
| \(\Rightarrow 8x^6 = e^6 \Rightarrow x = \ldots\) | M1 | Takes exp and proceeds to solution |
| \(x = \frac{e}{\sqrt{2}}\) | A1 | Ignore any reference to \(-\frac{e}{\sqrt{2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ln 2 + \ln x + \ln 2 + 2\ln x + \ln 2 + 3\ln x = 6 \Rightarrow 3\ln 2 + 6\ln x = 6\) | M1 | |
| \(\Rightarrow \ln x = 1 - \frac{1}{2}\ln 2\) | M1 | |
| \(\Rightarrow x = e^{1-\frac{1}{2}\ln 2} = \frac{e}{\sqrt{2}}\) | M1, A1 | Ignore reference to \(-\frac{e}{\sqrt{2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(fg(x) = e^{2\times\ln(2x)} + k^2\) | M1 | Correct order: g before f. Accept \(y = e^{\ln(2x)^2} + k^2\) |
| \(fg(x) = (2x)^2 + k^2 = 4x^2 + k^2\) | A1 | Accept \(fg(x) = (2x)^2 + k^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(fg(x) = 2k^2 \Rightarrow 4x^2 + k^2 = 2k^2 \Rightarrow 4x^2 = k^2 \Rightarrow x = \ldots\) | M1 | Sets answer to (d) in form \(Ax^2 + k^2 = 2k^2\) where \(A = 2\) or \(4\), proceeds to \(Ax^2 = k^2\) |
| \(x = \frac{k}{2}\) only | A1 | Answer \(x = \pm\frac{k}{2}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\Rightarrow \ln(2x)^2 = \ln(k^2) \Rightarrow 4x^2 = k^2 \Rightarrow x = \frac{k}{2}\) | M1, A1 |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) > k^2$ | B1 | Accept $f > k^2$, Range $> k^2$, $(k^2, \infty)$, $y > k^2$. Do not accept $f(x) \geq k^2$, $x > k^2$, $[k^2, \infty)$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = e^{2x} + k^2 \Rightarrow e^{2x} = y - k^2$ | M1 | Attempts to make $x$ or swapped $y$ the subject. Score for $e^{2x} = y \pm k^2$ and proceeding to $x = \ln\ldots$ |
| $\Rightarrow x = \frac{1}{2}\ln(y - k^2)$ | dM1 | Dependent on previous M. Taking ln of whole RHS then dividing by 2 |
| $f^{-1}(x) = \frac{1}{2}\ln(x-k^2),\quad x > k^2$ | A1 | Must include domain $x > k^2$. Accept $y = 0.5\ln|x-k^2|$, domain $(k^2, \infty)$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\ln 2x + \ln 2x^2 + \ln 2x^3 = 6$ | M1 | Attempts to solve by writing down the equation |
| $\Rightarrow \ln 8x^6 = 6$ | M1 | Uses addition laws of logs to write in form $\ln Ax^n = 6$ |
| $\Rightarrow 8x^6 = e^6 \Rightarrow x = \ldots$ | M1 | Takes exp and proceeds to solution |
| $x = \frac{e}{\sqrt{2}}$ | A1 | Ignore any reference to $-\frac{e}{\sqrt{2}}$ |
**Alt (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\ln 2 + \ln x + \ln 2 + 2\ln x + \ln 2 + 3\ln x = 6 \Rightarrow 3\ln 2 + 6\ln x = 6$ | M1 | |
| $\Rightarrow \ln x = 1 - \frac{1}{2}\ln 2$ | M1 | |
| $\Rightarrow x = e^{1-\frac{1}{2}\ln 2} = \frac{e}{\sqrt{2}}$ | M1, A1 | Ignore reference to $-\frac{e}{\sqrt{2}}$ |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $fg(x) = e^{2\times\ln(2x)} + k^2$ | M1 | Correct order: g before f. Accept $y = e^{\ln(2x)^2} + k^2$ |
| $fg(x) = (2x)^2 + k^2 = 4x^2 + k^2$ | A1 | Accept $fg(x) = (2x)^2 + k^2$ |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $fg(x) = 2k^2 \Rightarrow 4x^2 + k^2 = 2k^2 \Rightarrow 4x^2 = k^2 \Rightarrow x = \ldots$ | M1 | Sets answer to (d) in form $Ax^2 + k^2 = 2k^2$ where $A = 2$ or $4$, proceeds to $Ax^2 = k^2$ |
| $x = \frac{k}{2}$ **only** | A1 | Answer $x = \pm\frac{k}{2}$ is A0 |
**Alt (e):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\Rightarrow \ln(2x)^2 = \ln(k^2) \Rightarrow 4x^2 = k^2 \Rightarrow x = \frac{k}{2}$ | M1, A1 | |
---6. The function f is defined by
$$\mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } + k ^ { 2 } , \quad x \in \mathbb { R } , \quad k \text { is a positive constant. }$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f .
\item Find $\mathrm { f } ^ { - 1 }$ and state its domain.
The function g is defined by
$$g : x \rightarrow \ln ( 2 x ) , \quad x > 0$$
\item Solve the equation
$$\mathrm { g } ( x ) + \mathrm { g } \left( x ^ { 2 } \right) + \mathrm { g } \left( x ^ { 3 } \right) = 6$$
giving your answer in its simplest form.
\item Find $\mathrm { fg } ( x )$, giving your answer in its simplest form.
\item Find, in terms of the constant $k$, the solution of the equation
$$\mathrm { fg } ( x ) = 2 k ^ { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q6 [12]}}