Edexcel C1 2006 January — Question 9 12 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2006
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent parallel to given line
DifficultyEasy -1.2 This is a straightforward C1 differentiation question requiring basic polynomial expansion, differentiation using power rule, and finding a parallel tangent. All steps are routine applications of standard techniques with no problem-solving insight needed. Easier than average A-level questions.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
9. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{815e288c-0140-4c12-9e89-b0bb4fb1a8c1-12_812_1088_317_427}
\end{figure} Figure 2 shows part of the curve \(C\) with equation $$y = ( x - 1 ) \left( x ^ { 2 } - 4 \right) .$$ The curve cuts the \(x\)-axis at the points \(P , ( 1,0 )\) and \(Q\), as shown in Figure 2.
  1. Write down the \(x\)-coordinate of \(P\), and the \(x\)-coordinate of \(Q\).
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 2 x - 4\).
  3. Show that \(y = x + 7\) is an equation of the tangent to \(C\) at the point ( \(- 1,6\) ). The tangent to \(C\) at the point \(R\) is parallel to the tangent at the point ( \(- 1,6\) ).
  4. Find the exact coordinates of \(R\).
AnswerMarks Guidance
(a) \(-2\) \((P)\), \(2\) \((Q)\)B1, B1 \((\pm 2\) scores B1 B1)
(b) \(y = x^3 - x^2 - 4x + 4\) (May be seen earlier)M1 Multiply out, giving 4 terms
\(\frac{dy}{dx} = 3x^2 - 2x - 4\)M1 A1 cso
(c) At \(x = -1\): \(\frac{dy}{dx} = 3(-1)^2 - 2(-1) - 4 = 1\)M1 A1 cso
Eqn. of tangent: \(y - 6 = 1(x-(-1))\), \(y = x + 7\)M1 A1 cso
(d) \(3x^2 - 2x - 4 = 1\) (Equating to "gradient of tangent")M1
\(3x^2 - 2x - 5 = 0\) (or equiv.), \((3x-5)(x+1) = 0\), \(x = \ldots\)M1, M1
\(x = \frac{5}{3}\) or equiv.A1
\(y = \left(\frac{5}{3}-1\right)\left(\frac{25}{9}-4\right), = \frac{2}{3} \times \left(-\frac{11}{9}\right) = -\frac{22}{27}\) or equiv.M1, A1
Total: 12 marks
AnswerMarks
Guidance: (b) Alternative: Attempt to differentiate by product rule scores the second M1: \(\frac{dy}{dx} = \{(x^2-4) \times 1\} + \{(x-1) \times 2x\}\). Then multiplying out scores the first M1, with A1 if correct (cso). (c) M1 requires full method: Evaluate \(\frac{dy}{dx}\) and use in eqn. of line through \((-1,6)\), (n.b. the gradient need not be 1 for this M1). Alternative: Gradient of \(y = x + 7\) is 1, so solve \(3x^2 - 2x - 4 = 1\), as in (d)... to get \(x = -1\). [M1, A1 cso]. (d) 2nd and 3rd M marks are dependent on starting with \(3x^2 - 2x - 4 = k\), where \(k\) is a constant. 
**(a)** $-2$ $(P)$, $2$ $(Q)$ | B1, B1 | $(\pm 2$ scores B1 B1)

**(b)** $y = x^3 - x^2 - 4x + 4$ (May be seen earlier) | M1 | Multiply out, giving 4 terms

$\frac{dy}{dx} = 3x^2 - 2x - 4$ | M1 A1 cso |

**(c)** At $x = -1$: $\frac{dy}{dx} = 3(-1)^2 - 2(-1) - 4 = 1$ | M1 A1 cso |

Eqn. of tangent: $y - 6 = 1(x-(-1))$, $y = x + 7$ | M1 A1 cso |

**(d)** $3x^2 - 2x - 4 = 1$ (Equating to "gradient of tangent") | M1 |

$3x^2 - 2x - 5 = 0$ (or equiv.), $(3x-5)(x+1) = 0$, $x = \ldots$ | M1, M1 |

$x = \frac{5}{3}$ or equiv. | A1 |

$y = \left(\frac{5}{3}-1\right)\left(\frac{25}{9}-4\right), = \frac{2}{3} \times \left(-\frac{11}{9}\right) = -\frac{22}{27}$ or equiv. | M1, A1 |

**Total: 12 marks**

| Guidance: (b) Alternative: Attempt to differentiate by product rule scores the second M1: $\frac{dy}{dx} = \{(x^2-4) \times 1\} + \{(x-1) \times 2x\}$. Then multiplying out scores the first M1, with A1 if correct (cso). (c) M1 requires full method: Evaluate $\frac{dy}{dx}$ and use in eqn. of line through $(-1,6)$, (n.b. the gradient need not be 1 for this M1). Alternative: Gradient of $y = x + 7$ is 1, so solve $3x^2 - 2x - 4 = 1$, as in (d)... to get $x = -1$. [M1, A1 cso]. (d) 2nd and 3rd M marks are dependent on starting with $3x^2 - 2x - 4 = k$, where $k$ is a constant.

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9.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
  \includegraphics[alt={},max width=\textwidth]{815e288c-0140-4c12-9e89-b0bb4fb1a8c1-12_812_1088_317_427}
\end{center}
\end{figure}

Figure 2 shows part of the curve $C$ with equation

$$y = ( x - 1 ) \left( x ^ { 2 } - 4 \right) .$$

The curve cuts the $x$-axis at the points $P , ( 1,0 )$ and $Q$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Write down the $x$-coordinate of $P$, and the $x$-coordinate of $Q$.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 2 x - 4$.
\item Show that $y = x + 7$ is an equation of the tangent to $C$ at the point ( $- 1,6$ ).

The tangent to $C$ at the point $R$ is parallel to the tangent at the point ( $- 1,6$ ).
\item Find the exact coordinates of $R$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2006 Q9 [12]}}
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