| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | State domain or range |
| Difficulty | Standard +0.3 This is a standard multi-part C3 question covering routine techniques: stating range of exponential function, function composition, finding inverse (straightforward logarithm manipulation), and simple fixed-point iteration. Part (iv) requires minimal insight (just recognizing the definition of a fixed point), and the iteration in (v) is mechanical. Slightly easier than average due to straightforward algebraic steps throughout. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) < 3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= f(2) = 3 - e^4\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 3 - e^{2x}\), \(\quad e^{2x} = 3 - y\), \(\quad 2x = \ln(3-y)\), \(\quad x = \frac{1}{2}\ln(3-y)\) | M1 | |
| \(\therefore f^{-1}(x) = \frac{1}{2}\ln(3-x)\), \(x \in \mathbb{R}\), \(x < 3\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\) so they intersect on the line \(y = x\), hence \(f^{-1}(x) = f(x) \Rightarrow f^{-1}(x) = x\) | B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = 0.4581\), \(x_2 = 0.4664\), \(x_3 = 0.4648\), \(x_4 = 0.4651\), \(x_5 = 0.4651\) | M1 A1 | |
| \(\therefore \alpha = 0.465\) (3sf) | A1 | (11) |
# Question 9:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) < 3$ | B1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= f(2) = 3 - e^4$ | M1 A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 3 - e^{2x}$, $\quad e^{2x} = 3 - y$, $\quad 2x = \ln(3-y)$, $\quad x = \frac{1}{2}\ln(3-y)$ | M1 | |
| $\therefore f^{-1}(x) = \frac{1}{2}\ln(3-x)$, $x \in \mathbb{R}$, $x < 3$ | A2 | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ so they intersect on the line $y = x$, hence $f^{-1}(x) = f(x) \Rightarrow f^{-1}(x) = x$ | B2 | |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = 0.4581$, $x_2 = 0.4664$, $x_3 = 0.4648$, $x_4 = 0.4651$, $x_5 = 0.4651$ | M1 A1 | |
| $\therefore \alpha = 0.465$ (3sf) | A1 | **(11)** |
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**Total: (72)**9. $\quad f ( x ) = 3 - e ^ { 2 x } , \quad x \in \mathbb { R }$.\\
(i) State the range of f .\\
(ii) Find the exact value of $\mathrm { ff } ( 0 )$.\\
(iii) Define the inverse function $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.
Given that $\alpha$ is the solution of the equation $\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )$,\\
(iv) explain why $\alpha$ satisfies the equation
$$x = \mathrm { f } ^ { - 1 } ( x )$$
(v) use the iterative formula
$$x _ { n + 1 } = \mathrm { f } ^ { - 1 } \left( x _ { n } \right)$$
with $x _ { 0 } = 0.5$ to find $\alpha$ correct to 3 significant figures.
\hfill \mbox{\textit{OCR C3 Q9 [11]}}