| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve equation involving |f(x)| and g(x) |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 topics (modulus functions, exponentials, and numerical methods). Part (a) requires routine substitution and solving; part (b) is a straightforward 'show that' involving algebraic manipulation of the modulus equation; parts (c) and (d) are standard iteration and interval verification. While it requires multiple techniques, each step follows predictable patterns with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06a Exponential function: a^x and e^x graphs and properties1.06d 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/Working | Mark | Guidance |
| (i) \(y\)-intercept \(= 21\) | B1 | Sight of 21. Accept \((0, 21)\). Do not accept just \( |
| (ii) \(4e^{2x} - 25 = 0 \Rightarrow e^{2x} = \frac{25}{4} \Rightarrow 2x = \ln\left(\frac{25}{4}\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{25}{4}\right) \Rightarrow x = \ln\left(\frac{5}{2}\right)\) | M1A1, A1 | M1: Sets \(4e^{2x} - 25 = 0\) and proceeds via \(e^{2x} = \frac{25}{4}\) or \(e^x = \frac{5}{2}\) to \(x = \ldots\). Alternatively via \((2e^x - 5)(2e^x + 5) = 0\). A1: \(\frac{1}{2}\ln\left(\frac{25}{4}\right)\) or awrt 0.92. A1 cao: \(\ln\left(\frac{5}{2}\right)\) or \(\ln 5 - \ln 2\). Accept \(\left(\ln\left(\frac{5}{2}\right), 0\right)\) |
| (iii) \(k = 25\) | B1 | Accept also 25 or \(y = 25\). Do not accept just \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4e^{2x} - 25 = 2x + 43 \Rightarrow e^{2x} = \frac{1}{2}x + 17\) | M1 | |
| \(\Rightarrow 2x = \ln\left(\frac{1}{2}x + 17\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{1}{2}x + 17\right)\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = \frac{1}{2}\ln\left(\frac{1}{2} \times 1.4 + 17\right) =\) awrt \(1.44\) | M1 | |
| awrt \(x_1 = 1.4368\), \(x_2 = 1.4373\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Defines suitable interval: \([1.4365,\ 1.4375]\) | M1 | |
| Substitutes into suitable function e.g. \(4e^{2x} - 2x - 68\), obtains correct values with both a reason and conclusion | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(4e^{2x} - 25 = 2x + 43\) and makes \(e^{2x}\) the subject, giving \(e^{2x} = \frac{1}{4}(2x + 43 + 25)\) | M1 | Condone sign slips; condone \( |
| Proceeds correctly without errors to correct solution; bracketing must be correct throughout | A1* | Allow \(e^{2x} = \frac{1}{4}(2x+43+25)\) going to \(x = \frac{1}{2}\ln\left(\frac{1}{2}x+17\right)\) without explanation; allow \(\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)\) appearing as \(\frac{1}{2}\log_e\left(\frac{1}{2}x+17\right)\) but NOT as \(\frac{1}{2}\log\left(\frac{1}{2}x+17\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes 1.4 into iterative formula: \(x_1 = \frac{1}{2}\ln\left(\frac{1}{2}\times 1.4 + 17\right) = \frac{1}{2}\ln(17.7)\) or awrt 1.44 | M1 | Attempt to find \(x_1\) |
| \(x_1 = 1.4368\), \(x_2 = 1.4373\) | A1 | Subscripts not important, mark in order given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses interval [1.4365, 1.4375] (or smaller interval spanning root = 1.4373) | M1 | Continued iteration scores M0 |
| Substitutes both values into a suitable function and calculates both values correctly to 1 sig fig | A1 | Suitable functions: \(\pm(4e^{2x}-2x-68)\), \(\pm\left(x-\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)\right)\), \(\pm\left(2x-\ln\left(\frac{1}{2}x+17\right)\right)\) |
| Using \(4e^{2x}-2x-68\): \(f(1.4365) = -0.1\), \(f(1.4375) = +0.02\) or \(+0.03\) | ||
| Using \(2e^{2x}-x-34\): \(f(1.4365) = -0.05/-0.06\), \(f(1.4375) = +0.01\) | ||
| Using \(x-\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)\): \(f(1.4365) = -0.0007\) or \(-0.0008\), \(f(1.4375) = +0.0001\) or \(+0.0002\) | ||
| States reason (e.g. change of sign) and gives minimal conclusion (e.g. root or tick) | A1 | Must state reason AND conclusion |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $y$-intercept $= 21$ | B1 | Sight of 21. Accept $(0, 21)$. Do not accept just $|4-25|$ or $(21,0)$ |
| (ii) $4e^{2x} - 25 = 0 \Rightarrow e^{2x} = \frac{25}{4} \Rightarrow 2x = \ln\left(\frac{25}{4}\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{25}{4}\right) \Rightarrow x = \ln\left(\frac{5}{2}\right)$ | M1A1, A1 | M1: Sets $4e^{2x} - 25 = 0$ and proceeds via $e^{2x} = \frac{25}{4}$ or $e^x = \frac{5}{2}$ to $x = \ldots$. Alternatively via $(2e^x - 5)(2e^x + 5) = 0$. A1: $\frac{1}{2}\ln\left(\frac{25}{4}\right)$ or awrt 0.92. A1 cao: $\ln\left(\frac{5}{2}\right)$ or $\ln 5 - \ln 2$. Accept $\left(\ln\left(\frac{5}{2}\right), 0\right)$ |
| (iii) $k = 25$ | B1 | Accept also 25 or $y = 25$. Do not accept just $|-25|$ or $x = 25$ or $y = \pm 25$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4e^{2x} - 25 = 2x + 43 \Rightarrow e^{2x} = \frac{1}{2}x + 17$ | M1 | |
| $\Rightarrow 2x = \ln\left(\frac{1}{2}x + 17\right) \Rightarrow x = \frac{1}{2}\ln\left(\frac{1}{2}x + 17\right)$ | A1* | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \frac{1}{2}\ln\left(\frac{1}{2} \times 1.4 + 17\right) =$ awrt $1.44$ | M1 | |
| awrt $x_1 = 1.4368$, $x_2 = 1.4373$ | A1 | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Defines suitable interval: $[1.4365,\ 1.4375]$ | M1 | |
| Substitutes into suitable function e.g. $4e^{2x} - 2x - 68$, obtains correct values with both a reason and conclusion | A1 | |
# Question (b) [Iterative/Logarithm equation]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $4e^{2x} - 25 = 2x + 43$ and makes $e^{2x}$ the subject, giving $e^{2x} = \frac{1}{4}(2x + 43 + 25)$ | M1 | Condone sign slips; condone $|4e^{2x} - 25| = 2x + 43$; condone solution with $x = a/\alpha$ |
| Proceeds correctly without errors to correct solution; bracketing must be correct throughout | A1* | Allow $e^{2x} = \frac{1}{4}(2x+43+25)$ going to $x = \frac{1}{2}\ln\left(\frac{1}{2}x+17\right)$ without explanation; allow $\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)$ appearing as $\frac{1}{2}\log_e\left(\frac{1}{2}x+17\right)$ but NOT as $\frac{1}{2}\log\left(\frac{1}{2}x+17\right)$ |
---
# Question (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes 1.4 into iterative formula: $x_1 = \frac{1}{2}\ln\left(\frac{1}{2}\times 1.4 + 17\right) = \frac{1}{2}\ln(17.7)$ or awrt 1.44 | M1 | Attempt to find $x_1$ |
| $x_1 = 1.4368$, $x_2 = 1.4373$ | A1 | Subscripts not important, mark in order given |
---
# Question (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses interval [1.4365, 1.4375] (or smaller interval spanning root = 1.4373) | M1 | Continued iteration scores M0 |
| Substitutes both values into a suitable function and calculates both values correctly to 1 sig fig | A1 | Suitable functions: $\pm(4e^{2x}-2x-68)$, $\pm\left(x-\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)\right)$, $\pm\left(2x-\ln\left(\frac{1}{2}x+17\right)\right)$ |
| Using $4e^{2x}-2x-68$: $f(1.4365) = -0.1$, $f(1.4375) = +0.02$ or $+0.03$ | | |
| Using $2e^{2x}-x-34$: $f(1.4365) = -0.05/-0.06$, $f(1.4375) = +0.01$ | | |
| Using $x-\frac{1}{2}\ln\left(\frac{1}{2}x+17\right)$: $f(1.4365) = -0.0007$ or $-0.0008$, $f(1.4375) = +0.0001$ or $+0.0002$ | | |
| States reason (e.g. change of sign) and gives minimal conclusion (e.g. root or tick) | A1 | Must state reason AND conclusion |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d3ba2776-eedb-48f0-834f-41aa454afba3-06_675_1118_205_406}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = g ( x )$, where
$$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$
The curve cuts the $y$-axis at the point $A$ and meets the $x$-axis at the point $B$. The curve has an asymptote $y = k$, where $k$ is a constant, as shown in Figure 1
\begin{enumerate}[label=(\alph*)]
\item Find, giving each answer in its simplest form,
\begin{enumerate}[label=(\roman*)]
\item the $y$ coordinate of the point $A$,
\item the exact $x$ coordinate of the point $B$,
\item the value of the constant $k$.
The equation $\mathrm { g } ( x ) = 2 x + 43$ has a positive root at $x = \alpha$
\end{enumerate}\item Show that $\alpha$ is a solution of $x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)$
The iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$
can be used to find an approximation for $\alpha$
\item Taking $x _ { 0 } = 1.4$ find the values of $x _ { 1 }$ and $x _ { 2 }$
Give each answer to 4 decimal places.
\item By choosing a suitable interval, show that $\alpha = 1.437$ to 3 decimal places.
\includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2016 Q4 [11]}}