| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This is a multi-part question testing standard integration techniques (finding y from d²y/dx²), using stationary point conditions, and basic circle geometry with tangents. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07e Second derivative: as rate of change of gradient1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=3x^2\ [+c]\) | B1 | |
| \(3\times2^2+c=0\) | M1 | Substitute \(x=2\) and \(\frac{dy}{dx}=0\) into an integral (\(c\) must be present) |
| \(\frac{dy}{dx}=3x^2-12\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=x^3-12x\ [+k]\) | B1 FT | FT on their non-zero \(c\) (dependent on \(c\) being found at some stage) |
| \(-10=2^3-12\times2+k\) | M1 | Substitute \(x=2\), \(y=-10\) (\(k\) present) |
| \(y=x^3-12x+6\) | A1 | Must be \(y=\) (unless \(y=x^3-12x+k\) stated earlier) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3x^2-12=0\) [leading to \(x=-2\)] | M1 | Set their two term \(\frac{dy}{dx}=0\). Expect \(x=-2\). Ignore \(x=2\) given in addition |
| \(y=(-2)^3-12\times(-2)+6=22\) leading to \((-2,\,22)\) | A1 | |
| When \(x=-2\), \(\frac{d^2y}{dx^2}\lt0\) (or \(-12\)) hence Maximum | A1 | Can be from correct conclusion from \(\frac{dy}{dx}\) sign diagram if \(\frac{dy}{dx}\) calculated correctly. Do not allow concave downward for final A1. Can be awarded if the only error is incorrect or missing \(y\)-coordinate |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(x=0\), \(\frac{dy}{dx}=-12\), \(y=6\) | M1 | Both required. FT on their \(\frac{dy}{dx}\) and \(y\) |
| \(y-6=-12x\) | A1 | OE |
| Total | 2 |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=3x^2\ [+c]$ | B1 | |
| $3\times2^2+c=0$ | M1 | Substitute $x=2$ and $\frac{dy}{dx}=0$ into an integral ($c$ must be present) |
| $\frac{dy}{dx}=3x^2-12$ | A1 | |
| **Total** | **3** | |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=x^3-12x\ [+k]$ | B1 FT | FT on their non-zero $c$ (dependent on $c$ being found at some stage) |
| $-10=2^3-12\times2+k$ | M1 | Substitute $x=2$, $y=-10$ ($k$ present) |
| $y=x^3-12x+6$ | A1 | Must be $y=$ (unless $y=x^3-12x+k$ stated earlier) |
| **Total** | **3** | |
---
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3x^2-12=0$ [leading to $x=-2$] | M1 | Set their two term $\frac{dy}{dx}=0$. Expect $x=-2$. Ignore $x=2$ given in addition |
| $y=(-2)^3-12\times(-2)+6=22$ leading to $(-2,\,22)$ | A1 | |
| When $x=-2$, $\frac{d^2y}{dx^2}\lt0$ (or $-12$) hence Maximum | A1 | Can be from correct conclusion from $\frac{dy}{dx}$ sign diagram if $\frac{dy}{dx}$ calculated correctly. Do not allow concave downward for final A1. Can be awarded if the only error is incorrect or missing $y$-coordinate |
| **Total** | **3** | |
---
## Question 10(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $x=0$, $\frac{dy}{dx}=-12$, $y=6$ | M1 | Both required. FT on their $\frac{dy}{dx}$ and $y$ |
| $y-6=-12x$ | A1 | OE |
| **Total** | **2** | |
---10 A curve has a stationary point at $( 2 , - 10 )$ and is such that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 x$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find the equation of the curve.
\item Find the coordinates of the other stationary point and determine its nature.
\item Find the equation of the tangent to the curve at the point where the curve crosses the $y$-axis.\\
\includegraphics[max width=\textwidth, alt={}, center]{5e3e5418-7976-4232-8550-1da6420a3fcb-18_689_828_276_646}
The diagram shows the circle with equation $( x - 4 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 40$. Parallel tangents, each with gradient 1 , touch the circle at points $A$ and $B$.\\
(a) Find the equation of the line $A B$, giving the answer in the form $y = m x + c$.\\
(b) Find the coordinates of $A$, giving each coordinate in surd form.\\
(c) Find the equation of the tangent at $A$, giving the answer in the form $y = m x + c$, where $c$ is in surd form.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q10 [11]}}