Edexcel C2 — Question 8 12 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea involving fractional powers
DifficultyStandard +0.3 This is a straightforward C2 integration question requiring differentiation to find a minimum (standard quotient rule or rewriting as x^{-2}), then integration of x^{1/2} and x^{-2} using standard power rule. The arithmetic with surds is routine, and the answer is given to verify. Slightly easier than average due to being a standard two-part question with no novel problem-solving required.
Spec1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals

8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{089f5506-94ac-489f-b219-e67fa6ca834f-4_536_883_248_486} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows the curve with equation \(y = \sqrt { x } + \frac { 8 } { x ^ { 2 } } , x > 0\).
  1. Find the coordinates of the minimum point of the curve.
  2. Show that the area of the shaded region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 9\) is \(24 \frac { 4 } { 9 }\).

(a)
AnswerMarks
\(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 16x^{-3}\)M1 A2
For minimum, \(\frac{1}{2}x^{-\frac{1}{2}} - 16x^{-3} = 0\)M1
\(\frac{1}{2}x^{-3}(x^{\frac{5}{2}} - 32) = 0\)
\(x^{\frac{5}{2}} = 32\)A1
\(x = (\sqrt[5]{32})^2 = 4\)M1 A1
\(\therefore (4, \frac{5}{2})\)
(b)
AnswerMarks
\(= \int_1^9 \left(\sqrt{x} + \frac{8}{x^2}\right) dx\)
\(= \left[\frac{2}{3}x^{\frac{3}{2}} - 8x^{-1}\right]_1^9\)M1 A2
\(= (18 - \frac{8}{9}) - (\frac{2}{3} - 8)\)M1
\(= 24\frac{4}{9}\)A1
**(a)**

$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 16x^{-3}$ | M1 A2 |
For minimum, $\frac{1}{2}x^{-\frac{1}{2}} - 16x^{-3} = 0$ | M1 |
$\frac{1}{2}x^{-3}(x^{\frac{5}{2}} - 32) = 0$ | |
$x^{\frac{5}{2}} = 32$ | A1 |
$x = (\sqrt[5]{32})^2 = 4$ | M1 A1 |
$\therefore (4, \frac{5}{2})$ | |

**(b)**

$= \int_1^9 \left(\sqrt{x} + \frac{8}{x^2}\right) dx$ | |
$= \left[\frac{2}{3}x^{\frac{3}{2}} - 8x^{-1}\right]_1^9$ | M1 A2 |
$= (18 - \frac{8}{9}) - (\frac{2}{3} - 8)$ | M1 |
$= 24\frac{4}{9}$ | A1 |
8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{089f5506-94ac-489f-b219-e67fa6ca834f-4_536_883_248_486}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows the curve with equation $y = \sqrt { x } + \frac { 8 } { x ^ { 2 } } , x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the minimum point of the curve.
\item Show that the area of the shaded region bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 9$ is $24 \frac { 4 } { 9 }$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q8 [12]}}