| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Trapezium rule estimation |
| Difficulty | Moderate -0.3 This is a multi-part question requiring quotient rule differentiation, solving dy/dx=0 for a stationary point, substituting into a function, and applying the trapezium rule with given ordinates. All techniques are standard Core 3/4 material with no novel insights required, making it slightly easier than average but still requiring competent execution across multiple skills. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 1 | 2 | 3 |
| \(y\) | 0 | \(\frac { 3 } { 5 } \ln 5\) | \(\frac { 3 } { 10 } \ln 10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{(x^2+1)\times\frac{6x}{x^2+1} - 6x\times\ln(x^2+1)}{(x^2+1)^2}\) or \(\frac{6x\left(1-\ln(x^2+1)\right)}{(x^2+1)^2}\) | M1A1 | M1: applies quotient rule; A1: fully correct unsimplified form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 0 \Rightarrow (x^2+1)\frac{3\cdot 2x}{(x^2+1)} - 3\ln(x^2+1)(2x) = 0\) | M1 | Sets numerator \(= 0\), proceeds to form without fractional terms |
| \(\ln(x^2+1) = 1\) so \(x = \sqrt{e-1}\) | M1A1 | M1: correct work to \(\ln(x^2+1)=A\); A1: \(x=\sqrt{e-1}\) from correct numerator |
| \(y = \frac{3}{e}\) | ddM1A1 | ddM1: substitute \(x\) to find \(y\); A1: correct solution only, \(x>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{2}\ln 2\) or \(1.0397\) | B1 | Exact equivalent such as \(\frac{1}{2}\ln 8\) accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 1 \times \{\ldots\}\) | B1 oe | \(h=1\), implied by \(\frac{1}{2}\times 1\) or \(\frac{1}{2}\) outside bracket |
| \(\frac{1}{2}\times 1\times\left\{0 + \frac{3}{10}\ln 10 + 2\left(\frac{3}{2}\ln 2 + \frac{3}{5}\ln 5\right)\right\}\) | M1 | Correct bracketed expression: \(0 + \frac{3}{10}\ln 10 + 2\!\left(\frac{3}{2}\ln 2 + \frac{3}{5}\ln 5\right)\) |
| \(= 2.351\) (awrt 4 sf) | A1 |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(x^2+1)\times\frac{6x}{x^2+1} - 6x\times\ln(x^2+1)}{(x^2+1)^2}$ or $\frac{6x\left(1-\ln(x^2+1)\right)}{(x^2+1)^2}$ | M1A1 | M1: applies quotient rule; A1: fully correct unsimplified form |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 0 \Rightarrow (x^2+1)\frac{3\cdot 2x}{(x^2+1)} - 3\ln(x^2+1)(2x) = 0$ | M1 | Sets numerator $= 0$, proceeds to form without fractional terms |
| $\ln(x^2+1) = 1$ so $x = \sqrt{e-1}$ | M1A1 | M1: correct work to $\ln(x^2+1)=A$; A1: $x=\sqrt{e-1}$ from correct numerator |
| $y = \frac{3}{e}$ | ddM1A1 | ddM1: substitute $x$ to find $y$; A1: correct solution only, $x>0$ |
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{2}\ln 2$ or $1.0397$ | B1 | Exact equivalent such as $\frac{1}{2}\ln 8$ accepted |
## Question 7(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 1 \times \{\ldots\}$ | B1 oe | $h=1$, implied by $\frac{1}{2}\times 1$ or $\frac{1}{2}$ outside bracket |
| $\frac{1}{2}\times 1\times\left\{0 + \frac{3}{10}\ln 10 + 2\left(\frac{3}{2}\ln 2 + \frac{3}{5}\ln 5\right)\right\}$ | M1 | Correct bracketed expression: $0 + \frac{3}{10}\ln 10 + 2\!\left(\frac{3}{2}\ln 2 + \frac{3}{5}\ln 5\right)$ |
| $= 2.351$ (awrt 4 sf) | A1 | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-10_543_817_278_584}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows part of the curve $C$ with equation
$$y = \frac { 3 \ln \left( x ^ { 2 } + 1 \right) } { \left( x ^ { 2 } + 1 \right) } , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$
\item Using your answer to (a), find the exact coordinates of the stationary point on the curve $C$ for which $x > 0$. Write each coordinate in its simplest form.\\
(5)
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis and the line $x = 3$
\item Complete the table below with the value of $y$ corresponding to $x = 1$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$y$ & 0 & & $\frac { 3 } { 5 } \ln 5$ & $\frac { 3 } { 10 } \ln 10$ \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the $y$ values in the completed table to find an approximate value for the area of $R$, giving your answer to 4 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q7 [11]}}