| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change question requiring basic differentiation. Part (i) involves simple geometry (triangle area formula), and part (ii) requires differentiating with respect to time using the chain rule with a given constant rate. The setup is clear, the algebra is routine, and no novel insight is needed—slightly easier than average for A-level. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| (ii) | y = 2x² , X(−2, 0) and P(p, 0) |
| Answer | Marks |
|---|---|
| dt dt dt | M1 A1 |
| Answer | Marks |
|---|---|
| [3] | Attempt at base and height in terms |
| Answer | Marks |
|---|---|
| (ii) | (1−x)2(1+2x)6 |
| Answer | Marks |
|---|---|
| → 60 – 72 + 15 = 3 | B2,1 |
| Answer | Marks |
|---|---|
| [3] | −1 each error |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 11 |
Question 2:
--- 2
(i)
(ii) ---
2
(i)
(ii) | y = 2x² , X(−2, 0) and P(p, 0)
1
A = × (2 + p) × 2p² (= 2p² + p³)
2
dA
= 4p + 3p²
dp
dA dA dp
= × = 0.02 × 20 = 0.4
dt dp dt
dA dp dp
or =4p +3p2
dt dt dt | M1 A1
[2]
B1
M1 A1
[3] | Attempt at base and height in terms
bh
of p and use of
2
cao
any correct method, cao
3
(i) (a)
(b)
(ii) | (1−x)2(1+2x)6
.
(1−x)6 = 1 – 6x + 15x²
(1+2x)6 = 1 + 12x + 60x²
Product of (a) and (b) with >1 term
→ 60 – 72 + 15 = 3 | B2,1
[2]
B2,1
[2]
M1
DM1A1
[3] | −1 each error
−1 each error
SC B1 only, in each part, for all 3
correct descending powers
SC only one penalty for omission
of the ‘1’ in each expansion
Must be 2 or more products
M1 exactly 3 products. cao,
3x2
condone
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 11\includegraphics{figure_2}
The diagram shows the curve $y = 2x^2$ and the points $X(-2, 0)$ and $P(p, 0)$. The point $Q$ lies on the curve and $PQ$ is parallel to the $y$-axis.
\begin{enumerate}[label=(\roman*)]
\item Express the area, $A$, of triangle $XPQ$ in terms of $p$. [2]
\end{enumerate}
The point $P$ moves along the $x$-axis at a constant rate of 0.02 units per second and $Q$ moves along the curve so that $PQ$ remains parallel to the $y$-axis.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the rate at which $A$ is increasing when $p = 2$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2015 Q2 [5]}}