| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a standard C3 composite/inverse functions question with routine parts: stating range of absolute value function, evaluating a composite, solving fg(x)=3 (requires basic log manipulation), showing a root exists (sign change), and performing iteration. All techniques are textbook exercises requiring careful execution but no novel insight. Slightly easier than average due to straightforward structure. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping)1.06d Natural logarithm: ln(x) function and properties1.06g Equations with exponentials: solve a^x = b1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| \(f(x) \geq 0\) | B1 |
| Answer | Marks |
|---|---|
| \(= f(0) = 5\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(fg(x) = f[\ln(x + 3)] = | 2\ln(x + 3) - 5 | \) |
| \(\therefore | 2\ln(x + 3) - 5 | = 3\) |
| \(2\ln(x + 3) = 2, 8\) | M1 | |
| \(\ln(x + 3) = 1, 4\) | ||
| \(x = e - 3, e^4 - 3\) | A1 | |
| M1 A1 |
| Answer | Marks |
|---|---|
| Let \(h(x) = f(x) - g(x)\) | |
| \(h(3) = -0.79, f(4) = 1.1\) | M1 |
| Sign change, \(h(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks |
|---|---|
| \(x_1 = 3.396, x_2 = 3.428, x_3 = 3.430\) | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(h(3.4305) = -0.000052, f(3.4315) = 0.0018\) | M1 | |
| Sign change, \(h(x)\) continuous \(\therefore\) root \(\therefore \alpha = x_4\) to 4sf | A1 | (15 marks) |
**(a)**
$f(x) \geq 0$ | B1 |
**(b)**
$= f(0) = 5$ | M1 A1 |
**(c)**
$fg(x) = f[\ln(x + 3)] = |2\ln(x + 3) - 5|$ | M1 |
$\therefore |2\ln(x + 3) - 5| = 3$ | |
$2\ln(x + 3) = 2, 8$ | M1 |
$\ln(x + 3) = 1, 4$ | |
$x = e - 3, e^4 - 3$ | A1 |
| M1 A1 |
**(d)**
Let $h(x) = f(x) - g(x)$ | |
$h(3) = -0.79, f(4) = 1.1$ | M1 |
Sign change, $h(x)$ continuous $\therefore$ root | A1 |
**(e)**
$x_1 = 3.396, x_2 = 3.428, x_3 = 3.430$ | M1 A2 |
**(f)**
$h(3.4305) = -0.000052, f(3.4315) = 0.0018$ | M1 |
Sign change, $h(x)$ continuous $\therefore$ root $\therefore \alpha = x_4$ to 4sf | A1 | (15 marks)
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**Total: 75 marks**7. The functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3
\end{aligned}$$
(a) State the range of f .\\
(b) Evaluate fg(-2).\\
(c) Solve the equation
$$\operatorname { fg } ( x ) = 3$$
giving your answers in exact form.\\
(d) Show that the equation
$$\mathrm { f } ( x ) = \mathrm { g } ( x )$$
has a root, $\alpha$, in the interval [3,4].\\
(e) Use the iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$
with $x _ { 0 } = 3$, to find $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 significant figures.\\
(f) Show that your answer for $x _ { 4 }$ is the value of $\alpha$ correct to 4 significant figures.
\end{document}
\hfill \mbox{\textit{Edexcel C3 Q7 [15]}}