Edexcel C2 — Question 7 11 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea under polynomial curve
DifficultyModerate -0.3 This is a standard C2 calculus question covering routine techniques: finding roots, differentiation for tangents and stationary points, and definite integration. All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and integration calculation make it slightly more substantial than the most basic exercises.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals

\includegraphics{figure_1} Fig. 1 shows part of the curve \(C\) with equation \(y = \frac{1}{3}x^2 - \frac{1}{4}x^3\). The curve \(C\) touches the \(x\)-axis at the origin and passes through the point \(A(p, 0)\).
  1. Show that \(p = 6\). [1]
  2. Find an equation of the tangent to \(C\) at \(A\). [4]
The curve \(C\) has a maximum at the point \(P\).
  1. Find the \(x\)-coordinate of \(P\). [2]
The shaded region \(R\), in Fig. 1, is bounded by \(C\) and the \(x\)-axis.
  1. Find the area of \(R\). [4]

AnswerMarks Guidance
(a) Solve \(\frac{3}{2}x^2 - \frac{1}{4}x^3 = 0\) to find \(p = 6\), or verify: \(\frac{3}{2} \times 6^2 - \frac{1}{4} \times 6^4 = 0\) (*)B1 (1 mark)
(b) \(\frac{dy}{dx} = 3x - \frac{3x^2}{4}\)M1, A1
\(m = -9\), \(y - 0 = -9(x - 6)\) (Any correct form)M1, A1 (4 marks)
(c) \(3x - \frac{3x^2}{4} = 0\), \(x = 4\)M1, A1ft (2 marks)
(d) \(\int(\frac{3x^2}{2} - \frac{x^4}{4})dx = \frac{x^3}{2} - \frac{x^4}{16}\) (Allow unsimplified versions)M1, A1
\([.........]_b^6 = \frac{6^3}{2} - \frac{6^4}{16} = 27\)M1, A1 M: Need 6 and 0 as limits.
**(a)** Solve $\frac{3}{2}x^2 - \frac{1}{4}x^3 = 0$ to find $p = 6$, or verify: $\frac{3}{2} \times 6^2 - \frac{1}{4} \times 6^4 = 0$ (*) | B1 | (1 mark)

**(b)** $\frac{dy}{dx} = 3x - \frac{3x^2}{4}$ | M1, A1 |
$m = -9$, $y - 0 = -9(x - 6)$ (Any correct form) | M1, A1 | (4 marks)

**(c)** $3x - \frac{3x^2}{4} = 0$, $x = 4$ | M1, A1ft | (2 marks)

**(d)** $\int(\frac{3x^2}{2} - \frac{x^4}{4})dx = \frac{x^3}{2} - \frac{x^4}{16}$ (Allow unsimplified versions) | M1, A1 |
$[.........]_b^6 = \frac{6^3}{2} - \frac{6^4}{16} = 27$ | M1, A1 | M: Need 6 and 0 as limits. | (4 marks)
\includegraphics{figure_1}

Fig. 1 shows part of the curve $C$ with equation $y = \frac{1}{3}x^2 - \frac{1}{4}x^3$.

The curve $C$ touches the $x$-axis at the origin and passes through the point $A(p, 0)$.

\begin{enumerate}[label=(\alph*)]
\item Show that $p = 6$. [1]
\item Find an equation of the tangent to $C$ at $A$. [4]
\end{enumerate}

The curve $C$ has a maximum at the point $P$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the $x$-coordinate of $P$. [2]
\end{enumerate}

The shaded region $R$, in Fig. 1, is bounded by $C$ and the $x$-axis.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the area of $R$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2  Q7 [11]}}