Edexcel C4 — Question 5 12 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeLinear combination of vectors
DifficultyModerate -0.8 This is a straightforward vector arithmetic question requiring students to express one vector as a linear combination of two others by solving a simple system of linear equations. It involves only basic algebraic manipulation with no geometric insight or problem-solving required, making it easier than the average A-level question.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms

& \mathbf { r } = \left( \begin{array} { l } a
6
3 \end{array} \right)
AnswerMarks Guidance
(a) \(2x - 4y - 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0\)M1 A2
\(\frac{dy}{dx} = \frac{2x - 4y}{4x - 4y} = \frac{x - 2y}{2x - 2y}\)M1 A1
(b) \(\text{grad} = \frac{3}{2}\)M1
\(\therefore y - 2 = \frac{3}{2}(x-1)\)M1
\(2y - 4 = 3x - 3\)
\(3x - 2y + 1 = 0\)A1
(c) \(\frac{x - 2y}{2x - 2y} = \frac{3}{2}\)M1
\(2(x-2y) = 3(2x-2y), y = 2x\)A1
sub. \(\Rightarrow x^2 - 8x^2 + 8x^2 = 1\)M1
\(x^2 = 1, x = 1\) (at \(P\)) or \(-1\)
\(\therefore Q(-1, -2)\)A1 (12) 
**(a)** $2x - 4y - 4x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0$ | M1 A2 |
$\frac{dy}{dx} = \frac{2x - 4y}{4x - 4y} = \frac{x - 2y}{2x - 2y}$ | M1 A1 |

**(b)** $\text{grad} = \frac{3}{2}$ | M1 |
$\therefore y - 2 = \frac{3}{2}(x-1)$ | M1 |
$2y - 4 = 3x - 3$ | |
$3x - 2y + 1 = 0$ | A1 |

**(c)** $\frac{x - 2y}{2x - 2y} = \frac{3}{2}$ | M1 |
$2(x-2y) = 3(2x-2y), y = 2x$ | A1 |
sub. $\Rightarrow x^2 - 8x^2 + 8x^2 = 1$ | M1 |
$x^2 = 1, x = 1$ (at $P$) or $-1$ | |
$\therefore Q(-1, -2)$ | A1 | (12)

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\\
& \mathbf { r } = \left( \begin{array} { l } 
a \\
6 \\
3
\end{array} \right)

\hfill \mbox{\textit{Edexcel C4  Q5 [12]}}
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Q1 6 Q2 7 Q3 11 Q4 11 Q5 12 Q7 15 Q14