Question 7 - A-Level Maths

Edexcel C2 2014 January — Question 7 13 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2014
Session	January
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Areas Between Curves
Type	Tangent or Normal Bounded Area
Difficulty	Standard +0.8 This C2 question requires finding a normal equation (routine differentiation and perpendicular gradient), then calculating area between a cubic curve and a normal line using integration. The area calculation involves finding intersection points, setting up the correct integral with two regions, and working with algebraic expressions. While conceptually straightforward, the multi-step nature, algebraic manipulation, and need to carefully handle the bounded region make it moderately challenging for C2 level.
Spec	1.07m Tangents and normals: gradient and equations 1.08d Evaluate definite integrals: between limits 1.08e Area between curve and x-axis: using definite integrals

7. \begin{figure}[h]

\captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of part of the curve $C$ with equation $$y = x ^ { 3 } - 6 x ^ { 2 } + 9 x + 5$$ The point $P ( 4,9 )$ lies on $C$.

Show that the normal to $C$ at the point $P$ has equation $$x + 9 y = 85$$ The region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the $y$-axis and the normal to $C$ at $P$.
Showing all your working, calculate the exact area of $R$.

Show mark scheme Show mark scheme source

Question 7(a):

$y = x^3 - 6x^2 + 9x + 5$

Answer	Marks	Guidance
$\frac{dy}{dx} = 3x^2 - 12x + 9$	M1A1	M1: $x^n \to x^{n-1}$; A1: Correct derivative
$f'(4) = 3(4)^2 - 12(4) + 9 = 9$	M1	Finds $f'(4)$
$m_N = -\frac{1}{9}$	dM1	Perpendicular gradient rule applied to their $f'(4)$; dependent on previous M
$y - 9 = -\frac{1}{9}(x-4)$	ddM1	Correct straight line method; depends on both previous M marks
$x + 9y = 85$	A1*	Correct completion to printed answer

Question 7(b) Way 1:

Answer	Marks	Guidance
$x=0 \Rightarrow y = \frac{85}{9}$	—	—
Area trapezium $= \frac{1}{2} \times 4 \times \left(9 + \frac{85}{9}\right) = \frac{332}{9}$	M1A1	M1: Correct method for trapezium; A1: Correct numerical expression
$\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x$	M1A1	M1: $x^n \to x^{n+1}$; A1: Correct integration
$\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4 = \frac{4^4}{4} - 2\times4^3 + \frac{9\times4^2}{2} + 5\times4(-0)$	M1	Use of limits 0 and 4; subtracts (either way round)
$R = \frac{332}{9} - 28 = \frac{80}{9}$	M1A1	M1: Trapezium − Integral or Integral − Trapezium; A1: cso

Question 7(b) Way 2 (Appendix):

Answer	Marks	Guidance
Area trapezium $= \int_0^4 \left(\frac{85-x}{9}\right)dx = \left[85x - \frac{x^2}{2\cdot 9}\right]$	M1	Correct method including limits; attempt to integrate rearrangement of normal
$= \frac{1}{9}\left(85(4) - \frac{4^2}{2}\right)(-0)$	A1	Correct numerical expression
$\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x$	M1A1	M1: $x^n \to x^{n+1}$; A1: Correct integration
$\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4$	M1	Use of limits 0 and 4; subtracts
$R = \frac{332}{9} - 28 = \frac{80}{9}$	M1A1	M1: Trapezium − Integral or Integral − Trapezium; A1: cso

Question 7(b) Way 3 (Appendix):

Answer	Marks	Guidance
Line $-$ Curve $= \frac{85-x}{9} - (x^3 - 6x^2 + 9x + 5)$	M1A1	Allow (Curve − Line) for both marks; A1 not awarded if brackets missing unless correct expression implied by later work
$\int(y_1 - y_2)\,dx = \frac{1}{9}\int(40 - 9x^3 + 54x^2 - 82x)\,dx$	—	—
$= \frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]$	M2A1	M2: $x^n \to x^{n+1}$ on both line and curve; A1: Correct integration
$\frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]_0^4$	M1	Use of limits 0 and 4; subtracts
$= \frac{80}{9}$ or any exact equivalent	A1cso	Correct area

## Question 7(a):

$y = x^3 - 6x^2 + 9x + 5$

| $\frac{dy}{dx} = 3x^2 - 12x + 9$ | M1A1 | M1: $x^n \to x^{n-1}$; A1: Correct derivative |
|---|---|---|
| $f'(4) = 3(4)^2 - 12(4) + 9 = 9$ | M1 | Finds $f'(4)$ |
| $m_N = -\frac{1}{9}$ | dM1 | Perpendicular gradient rule applied to their $f'(4)$; dependent on previous M |
| $y - 9 = -\frac{1}{9}(x-4)$ | ddM1 | Correct straight line method; depends on both previous M marks |
| $x + 9y = 85$ | A1* | Correct completion to printed answer |

## Question 7(b) Way 1:

| $x=0 \Rightarrow y = \frac{85}{9}$ | — | — |
|---|---|---|
| Area trapezium $= \frac{1}{2} \times 4 \times \left(9 + \frac{85}{9}\right) = \frac{332}{9}$ | M1A1 | M1: Correct method for trapezium; A1: Correct numerical expression |
| $\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x$ | M1A1 | M1: $x^n \to x^{n+1}$; A1: Correct integration |
| $\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4 = \frac{4^4}{4} - 2\times4^3 + \frac{9\times4^2}{2} + 5\times4(-0)$ | M1 | Use of limits 0 and 4; subtracts (either way round) |
| $R = \frac{332}{9} - 28 = \frac{80}{9}$ | M1A1 | M1: Trapezium − Integral or Integral − Trapezium; A1: cso |

## Question 7(b) Way 2 (Appendix):

| Area trapezium $= \int_0^4 \left(\frac{85-x}{9}\right)dx = \left[85x - \frac{x^2}{2\cdot 9}\right]$ | M1 | Correct method including limits; attempt to integrate rearrangement of normal |
|---|---|---|
| $= \frac{1}{9}\left(85(4) - \frac{4^2}{2}\right)(-0)$ | A1 | Correct numerical expression |
| $\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x$ | M1A1 | M1: $x^n \to x^{n+1}$; A1: Correct integration |
| $\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4$ | M1 | Use of limits 0 and 4; subtracts |
| $R = \frac{332}{9} - 28 = \frac{80}{9}$ | M1A1 | M1: Trapezium − Integral or Integral − Trapezium; A1: cso |

## Question 7(b) Way 3 (Appendix):

| Line $-$ Curve $= \frac{85-x}{9} - (x^3 - 6x^2 + 9x + 5)$ | M1A1 | Allow (Curve − Line) for both marks; A1 not awarded if brackets missing unless correct expression implied by later work |
|---|---|---|
| $\int(y_1 - y_2)\,dx = \frac{1}{9}\int(40 - 9x^3 + 54x^2 - 82x)\,dx$ | — | — |
| $= \frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]$ | M2A1 | M2: $x^n \to x^{n+1}$ on both line and curve; A1: Correct integration |
| $\frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]_0^4$ | M1 | Use of limits 0 and 4; subtracts |
| $= \frac{80}{9}$ or any exact equivalent | A1cso | Correct area |

---

Show LaTeX source

7.

\begin{figure}[h]
\begin{center}
  \begin{tikzpicture}[
    x=1.2cm, 
    y=0.8cm, 
    >=latex,
    font=\rmfamily
]

    % Define the mathematical functions
    \def\curve{\x^3 - 6*\x^2 + 9*\x + 5}
    \def\normalline{-\x/9 + 85/9}

    % 1. Shaded Region R
    \fill[gray!30]
        (0, {85/9}) --
        (4, 9) --
        plot[domain=4:0, samples=100] (\x, {\curve}) --
        cycle;

    % 2. Axes
    \draw[->] (-1.5, 0) -- (6.5, 0) node[below] {$x$};
    \draw[->] (0, -2.5) -- (0, 11) node[left] {$y$};
    \node[below left] at (0, 0) {$O$};

    % 3. Normal Line
    \draw (-1.5, {1.5/9 + 85/9}) -- (5.5, {-5.5/9 + 85/9});

    % 4. Curve C
    \draw[thick] plot[domain=-0.45:4.2, samples=100] (\x, {\curve}) node[right] {$C$};

    % 5. Labels and Points
    \node[above right, inner sep=2pt] at (4, 9) {$P$};
    \node[left] at (0, 5) {$5$};
    \node at (2.2, 7.5) {$R$};

\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of part of the curve $C$ with equation

$$y = x ^ { 3 } - 6 x ^ { 2 } + 9 x + 5$$

The point $P ( 4,9 )$ lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal to $C$ at the point $P$ has equation

$$x + 9 y = 85$$

The region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the $y$-axis and the normal to $C$ at $P$.
\item Showing all your working, calculate the exact area of $R$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2 2014 Q7 [13]}}

This paper (9 questions)

View full paper

Q1 5 Q2 6 Q3 11 Q4 8 Q5 7 Q6 5 Q7 13 Q8 11 Q9 9

Answer	Marks	Guidance
\(\frac{dy}{dx} = 3x^2 - 12x + 9\)	M1A1	M1: \(x^n \to x^{n-1}\); A1: Correct derivative
\(f'(4) = 3(4)^2 - 12(4) + 9 = 9\)	M1	Finds \(f'(4)\)
\(m_N = -\frac{1}{9}\)	dM1	Perpendicular gradient rule applied to their \(f'(4)\); dependent on previous M
\(y - 9 = -\frac{1}{9}(x-4)\)	ddM1	Correct straight line method; depends on both previous M marks
\(x + 9y = 85\)	A1*	Correct completion to printed answer

Answer	Marks	Guidance
\(x=0 \Rightarrow y = \frac{85}{9}\)	—	—
Area trapezium \(= \frac{1}{2} \times 4 \times \left(9 + \frac{85}{9}\right) = \frac{332}{9}\)	M1A1	M1: Correct method for trapezium; A1: Correct numerical expression
\(\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\)	M1A1	M1: \(x^n \to x^{n+1}\); A1: Correct integration
\(\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4 = \frac{4^4}{4} - 2\times4^3 + \frac{9\times4^2}{2} + 5\times4(-0)\)	M1	Use of limits 0 and 4; subtracts (either way round)
\(R = \frac{332}{9} - 28 = \frac{80}{9}\)	M1A1	M1: Trapezium − Integral or Integral − Trapezium; A1: cso

Answer	Marks	Guidance
Area trapezium \(= \int_0^4 \left(\frac{85-x}{9}\right)dx = \left[85x - \frac{x^2}{2\cdot 9}\right]\)	M1	Correct method including limits; attempt to integrate rearrangement of normal
\(= \frac{1}{9}\left(85(4) - \frac{4^2}{2}\right)(-0)\)	A1	Correct numerical expression
\(\int y\,dx = \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\)	M1A1	M1: \(x^n \to x^{n+1}\); A1: Correct integration
\(\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} + 5x\right]_0^4\)	M1	Use of limits 0 and 4; subtracts
\(R = \frac{332}{9} - 28 = \frac{80}{9}\)	M1A1	M1: Trapezium − Integral or Integral − Trapezium; A1: cso

Answer	Marks	Guidance
Line \(-\) Curve \(= \frac{85-x}{9} - (x^3 - 6x^2 + 9x + 5)\)	M1A1	Allow (Curve − Line) for both marks; A1 not awarded if brackets missing unless correct expression implied by later work
\(\int(y_1 - y_2)\,dx = \frac{1}{9}\int(40 - 9x^3 + 54x^2 - 82x)\,dx\)	—	—
\(= \frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]\)	M2A1	M2: \(x^n \to x^{n+1}\) on both line and curve; A1: Correct integration
\(\frac{1}{9}\left[40x - \frac{9x^4}{4} + \frac{54x^3}{3} - \frac{82x^2}{2}\right]_0^4\)	M1	Use of limits 0 and 4; subtracts
\(= \frac{80}{9}\) or any exact equivalent	A1cso	Correct area