| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| = linear (non-modulus) |
| Difficulty | Standard +0.3 This is a standard C3 multi-part question covering routine techniques: modulus function properties, function composition, solving equations with logarithms and modulus, showing a root exists via sign change, and applying an iterative formula. While it has many parts (6 marks worth), each individual step is straightforward and follows textbook methods with no novel insight required. Slightly easier than average due to the guided nature and standard techniques. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(f(x) \geq 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(= f(0) = 5\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(fg(x) = f[\ln(x+3)] = \left | 2\ln(x+3)-5\right | \) |
| \(\therefore \left | 2\ln(x+3)-5\right | = 3\) |
| \(2\ln(x+3) = 2, 8\) | M1 | |
| \(\ln(x+3) = 1, 4\) | A1 | |
| \(x = e-3,\ e^4-3\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| 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 | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(x_1 = 3.396,\ x_2 = 3.428,\ x_3 = 3.430,\ x_4 = 3.431\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(h(3.4305) = -0.000052\), \(f(3.4315) = 0.0018\) | M1 | |
| sign change, \(h(x)\) continuous \(\therefore\) root \(\therefore \alpha = x_4\) to 4sf | A1 | (14) |
| Total | (72) |
# Question 8:
## Part (i)
| Answer | Mark | Notes |
|--------|------|-------|
| $f(x) \geq 0$ | B1 | |
## Part (ii)
| Answer | Mark | Notes |
|--------|------|-------|
| $= f(0) = 5$ | M1 A1 | |
## Part (iii)
| Answer | Mark | Notes |
|--------|------|-------|
| $fg(x) = f[\ln(x+3)] = \left|2\ln(x+3)-5\right|$ | M1 | |
| $\therefore \left|2\ln(x+3)-5\right| = 3$ | | |
| $2\ln(x+3) = 2, 8$ | M1 | |
| $\ln(x+3) = 1, 4$ | A1 | |
| $x = e-3,\ e^4-3$ | M1 A1 | |
## Part (iv)
| Answer | Mark | Notes |
|--------|------|-------|
| let $h(x) = f(x) - g(x)$ | | |
| $h(3) = -0.79$, $f(4) = 1.1$ | M1 | |
| sign change, $h(x)$ continuous $\therefore$ root | A1 | |
## Part (v)
| Answer | Mark | Notes |
|--------|------|-------|
| $x_1 = 3.396,\ x_2 = 3.428,\ x_3 = 3.430,\ x_4 = 3.431$ | M1 A1 | |
## Part (vi)
| Answer | Mark | Notes |
|--------|------|-------|
| $h(3.4305) = -0.000052$, $f(3.4315) = 0.0018$ | M1 | |
| sign change, $h(x)$ continuous $\therefore$ root $\therefore \alpha = x_4$ to 4sf | A1 | **(14)** |
| | **Total** | **(72)** |\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3
\end{aligned}$$
(i) State the range of f .\\
(ii) Evaluate fg(-2).\\
(iii) Solve the equation
$$\operatorname { fg } ( x ) = 3$$
giving your answers in exact form.\\
(iv) Show that the equation
$$\mathrm { f } ( x ) = \mathrm { g } ( x )$$
has a root, $\alpha$, in the interval [3,4].\\
(v) Use the iterative 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.\\
(vi) Show that your answer for $x _ { 4 }$ is the value of $\alpha$ correct to 4 significant figures.
\hfill \mbox{\textit{OCR C3 Q8 [14]}}