Edexcel C2 — Question 6 10 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyModerate -0.8 Part (a) requires standard differentiation and solving dy/dx=0 (routine calculus). Part (b) applies the trapezium rule formula with given ordinates—a straightforward numerical method requiring only substitution and arithmetic. Both parts are below-average difficulty for A-level, being direct applications of standard C2 techniques with no problem-solving insight required.
Spec1.07n Stationary points: find maxima, minima using derivatives1.09f Trapezium rule: numerical integration

6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e44996a-4635-46f6-bd45-7799a8c49463-3_589_894_248_397} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows the curve with equation \(y = 4 x + \frac { 1 } { x } , x > 0\).
  1. Find the coordinates of the minimum point of the curve. The shaded region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\).
  2. Use the trapezium rule with three intervals of equal width to estimate the area of \(R\).

AnswerMarks Guidance
(a) \(\frac{dy}{dx} = 4 - x^{-2}\)M1 A1
For minimum, \(4 - x^{-2} = 0\)M1
\(x^2 = \frac{1}{4}\)M1
\(x > 0 \quad \therefore x = \frac{1}{2} \quad \therefore \left(\frac{1}{2}, 4\right)\)A2
(b)\(x\) \(1\)
\(4x + x^{-1}\)\(5\) \(8\frac{1}{2}\)
Area\(= \frac{1}{2} \times 1 \times [5 + 16\frac{1}{4} + 2(8\frac{1}{2} + 12\frac{1}{3})]\) B1 M1 A1
\(= 31.5\) (3sf)A1 (10 marks)
**(a)** $\frac{dy}{dx} = 4 - x^{-2}$ | M1 A1 |
For minimum, $4 - x^{-2} = 0$ | M1 |
$x^2 = \frac{1}{4}$ | M1 |
$x > 0 \quad \therefore x = \frac{1}{2} \quad \therefore \left(\frac{1}{2}, 4\right)$ | A2 |

**(b)** | $x$ | $1$ | $2$ | $3$ | $4$ |
| | $4x + x^{-1}$ | $5$ | $8\frac{1}{2}$ | $12\frac{1}{3}$ | $16\frac{1}{4}$ | B1 |
| Area | $= \frac{1}{2} \times 1 \times [5 + 16\frac{1}{4} + 2(8\frac{1}{2} + 12\frac{1}{3})]$ | B1 M1 A1 |
| | $= 31.5$ (3sf) | A1 | **(10 marks)**
6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3e44996a-4635-46f6-bd45-7799a8c49463-3_589_894_248_397}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows the curve with equation $y = 4 x + \frac { 1 } { x } , x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the minimum point of the curve.

The shaded region $R$ is bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 4$.
\item Use the trapezium rule with three intervals of equal width to estimate the area of $R$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q6 [10]}}