Edexcel IAL P4 January 2025 question paper and mark scheme (WMA14/01)

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1. Vectors · 7 marks

   Line $L_1$ has vector equation $\mathbf{r_1} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ and line $L_2$ has equation $\mathbf{r_2} = \begin{pmatrix} 5 \\ -3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}$.

   1. Show that $L_1$ and $L_2$ intersect.
   2. Find the exact coordinates of the point of intersection.
   Worked answer

   1. M1 for equating the parametric equations: $1 + 2\lambda = 5 - \mu$, $\lambda = -3 + 2\mu$, and $1 + \lambda = \mu$.
      M1 for taking two equations to solve for the parameters, e.g., $\lambda = \mu - 1$ into $2\lambda + \mu = 4$.
      M1 for $2(\mu - 1) + \mu = 4 \implies 3\mu - 2 = 4 \implies 3\mu = 6$.
      A1 for finding $\mu = 2$ and $\lambda = 1$.
      B1 for verifying with the unused equation: $\lambda - 2\mu = 1 - 4 = -3$, which matches the second component equation, showing the lines intersect.
   2. M1 for substituting $\lambda = 1$ into $L_1$ (or $\mu = 2$ into $L_2$).
      A1 for the intersection point $(3, 1, 2)$.

2. Parametric Equations · 8 marks

   A curve $C$ has parametric equations $x = t^2$, $y = t^3 - 3t$.
   (a) Find $\dfrac{dy}{dx}$ in terms of $t$. (3)
   (b) Find the coordinates of the points on $C$ where the tangent is horizontal. (3)
   (c) Find the Cartesian equation of the tangent at $t = 2$. (2)

   Worked answer

   (a) $\tfrac{dx}{dt} = 2t$, $\tfrac{dy}{dt} = 3t^2 - 3$   M1 A1. $\tfrac{dy}{dx} = \tfrac{3t^2 - 3}{2t} = \tfrac{3(t^2 - 1)}{2t}$   A1.
   (b) Horizontal iff $\tfrac{dy}{dt} = 0$: $3t^2 = 3 \Rightarrow t = \pm 1$   M1. $t = 1$: $(1, -2)$; $t = -1$: $(1, 2)$   A1 A1.
   (c) At $t = 2$: $(x, y) = (4, 2)$; gradient $= \tfrac{3 \cdot 3}{4} = \tfrac{9}{4}$; equation: $y - 2 = \tfrac{9}{4}(x - 4)$, i.e. $y = \tfrac{9}{4}x - 7$   M1 A1.

3. Differential Equations · 4 marks

   Use the method of separation of variables to find the general solution to the differential equation: \[\frac{dy}{dx} = \frac{x}{y}\] Express your answer in the form \(y^2 = f(x)\).

   Worked answer

   Separate the variables by multiplying both sides by \(y\) and \(dx\): M1
   \(y \, dy = x \, dx\).
   Integrate both sides: \(\int y \, dy = \int x \, dx\). M1
   \(\frac{y^2}{2} = \frac{x^2}{2} + C\). A1
   Multiply entirely by 2 to isolate \(y^2\):
   \(y^2 = x^2 + 2C\). (Note: \(2C\) can just be written as a new constant \(K\)). A1

More on these topics: Vectors · Parametric Equations · Differential Equations

P4 topics this paper can test

Every P4 paper can test any of the 9 P4 topics in the specification.

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• Vectors
• Partial Fractions
• Parametric Equations
• Differential Equations
• Proof
• The Binomial Expansion
• Coordinate Geometry (parametric)
• Differentiation
• Integration

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What the examiners said about this paper

Summarised in our own words from the board's examiner report (Examiner Insights).

• WMA14/01: Binomial (any index) (Q3); Proof by contradiction (Q6); Connected rates of change (Q4). Official examiner report ↗

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Formula booklet

Mathematical Formulae and Statistical Tables (International AS/A Level Mathematics) · Pearson Edexcel · For the 2018 IAL specification (current)