| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward C3 inverse function question requiring standard techniques: identifying range from exponential properties, finding inverse by swapping and rearranging (involving natural log), and evaluating composite functions. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 properties |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(f(x) > 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(y = 3e^{x-1}\), so \(x - 1 = \ln\frac{y}{3}\) | M1 | |
| \(x = 1 + \ln\frac{y}{3}\) | ||
| \(f^{-1}(x) = 1 + \ln\frac{x}{3}\), \(x \in \mathbb{R}\), \(x > 0\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(f(\ln 2) = 3e^{\ln 2 - 1} = 3e^{-1}e^{\ln 2} = 6e^{-1}\) | M1 A1 | |
| \(gf(\ln 2) = g(6e^{-1}) = 30e^{-1} - 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(f^{-1}g(x) = f^{-1}(5x-2) = 1 + \ln\frac{5x-2}{3}\) | M1 A1 | |
| \(1 + \ln\frac{5x-2}{3} = 4\), so \(\frac{5x-2}{3} = e^3\) | M1 | |
| \(x = \frac{1}{5}(3e^3 + 2)\) | A1 | (11) |
# Question 7:
## Part (i):
| Working/Answer | Mark | Notes |
|---|---|---|
| $f(x) > 0$ | B1 | |
## Part (ii):
| Working/Answer | Mark | Notes |
|---|---|---|
| $y = 3e^{x-1}$, so $x - 1 = \ln\frac{y}{3}$ | M1 | |
| $x = 1 + \ln\frac{y}{3}$ | | |
| $f^{-1}(x) = 1 + \ln\frac{x}{3}$, $x \in \mathbb{R}$, $x > 0$ | A2 | |
## Part (iii):
| Working/Answer | Mark | Notes |
|---|---|---|
| $f(\ln 2) = 3e^{\ln 2 - 1} = 3e^{-1}e^{\ln 2} = 6e^{-1}$ | M1 A1 | |
| $gf(\ln 2) = g(6e^{-1}) = 30e^{-1} - 2$ | A1 | |
## Part (iv):
| Working/Answer | Mark | Notes |
|---|---|---|
| $f^{-1}g(x) = f^{-1}(5x-2) = 1 + \ln\frac{5x-2}{3}$ | M1 A1 | |
| $1 + \ln\frac{5x-2}{3} = 4$, so $\frac{5x-2}{3} = e^3$ | M1 | |
| $x = \frac{1}{5}(3e^3 + 2)$ | A1 | **(11)** |
---7. The function $f$ is defined by
$$\mathrm { f } : x \rightarrow 3 \mathrm { e } ^ { x - 1 } , \quad x \in \mathbb { R }$$
(i) State the range of f .\\
(ii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.
The function $g$ is defined by
$$g : x \rightarrow 5 x - 2 , \quad x \in \mathbb { R }$$
Find, in terms of e,\\
(iii) the value of $\mathrm { gf } ( \ln 2 )$,\\
(iv) the solution of the equation
$$\mathrm { f } ^ { - 1 } \mathrm {~g} ( x ) = 4$$
\hfill \mbox{\textit{OCR C3 Q7 [11]}}