| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Standard +0.3 This is a multi-part question combining standard differentiation (finding normal equation), solving a cubic equation, and a straightforward connected rates problem using the chain rule dy/dt = (dy/dx)(dx/dt). The connected rates component is routine application of a formula with given values, slightly above average due to the multiple parts and algebraic manipulation required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = \left[-\frac{1}{2}(4x-3)^{-2}\right] \times [4]\) | B1B1 | Can gain this in part (b)(ii) |
| When \(x=1\), \(m=-2\) | B1FT | Ft from their \(\dfrac{dy}{dx}\) |
| Normal is \(y - \frac{1}{2} = \frac{1}{2}(x-1)\) | M1 | Line with gradient \(-1/m\) and through \(A\) |
| \(y = \frac{1}{2}x\) soi | A1 | Can score in part (b) |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{2(4x-3)} = \dfrac{x}{2} \to 2x(4x-3) = 2 \to (2)(4x^2-3x-1)(=0)\) | M1A1 | \(x/2\) seen on RHS of equation can score *previous* A1 |
| \(x = -1/4\) | A1 | Ignore \(x=1\) seen in addition |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of chain rule: \(\dfrac{dy}{dt} = (\text{their} -2) \times (\pm 0.3) = 0.6\) | M1A1 | Allow \(+0.3\) or \(-0.3\) for M1 |
| 2 |
## Question 10(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = \left[-\frac{1}{2}(4x-3)^{-2}\right] \times [4]$ | **B1B1** | Can gain this in part **(b)(ii)** |
| When $x=1$, $m=-2$ | **B1FT** | Ft from their $\dfrac{dy}{dx}$ |
| Normal is $y - \frac{1}{2} = \frac{1}{2}(x-1)$ | **M1** | Line with gradient $-1/m$ and through $A$ |
| $y = \frac{1}{2}x$ soi | **A1** | Can score in part **(b)** |
| | **5** | |
---
## Question 10(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{2(4x-3)} = \dfrac{x}{2} \to 2x(4x-3) = 2 \to (2)(4x^2-3x-1)(=0)$ | **M1A1** | $x/2$ seen on RHS of equation can score *previous* A1 |
| $x = -1/4$ | **A1** | Ignore $x=1$ seen in addition |
| | **3** | |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of chain rule: $\dfrac{dy}{dt} = (\text{their} -2) \times (\pm 0.3) = 0.6$ | **M1A1** | Allow $+0.3$ or $-0.3$ for M1 |
| | **2** | |
---10 A curve has equation $y = \frac { 1 } { 2 } ( 4 x - 3 ) ^ { - 1 }$. The point $A$ on the curve has coordinates $\left( 1 , \frac { 1 } { 2 } \right)$.
\begin{enumerate}[label=(\roman*)]
\item (a) Find and simplify the equation of the normal through $A$.\\
(b) Find the $x$-coordinate of the point where this normal meets the curve again.
\item A point is moving along the curve in such a way that as it passes through $A$ its $x$-coordinate is decreasing at the rate of 0.3 units per second. Find the rate of change of its $y$-coordinate at $A$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2018 Q10 [10]}}