| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration and differentiation question. Parts (a)-(c) involve routine integration twice with boundary conditions, testing the second derivative, and solving dy/dx=0. Part (d) is a standard related rates problem using the chain rule. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07f Convexity/concavity: points of inflection1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2}=6(-1)^2-\frac{4}{(-1)^3}>0\) \(\therefore\) minimum or \(\frac{d^2y}{dx^2}=10\) \(\therefore\) minimum | B1 | Sub \(x=-1\) into \(\frac{d^2y}{dx^2}\), correct conclusion. WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=2x^3+\frac{2}{x^2}[+c]\) | *M1 | Integrating \(\frac{d^2y}{dx^2}\) (at least one term correct). |
| \(0=-2+2+c\) leading to \(c=0\) | DM1 | Substituting \(x=-1,\ \frac{dy}{dx}=0\) (need to see) to evaluate \(c\). DM0 if simply state \(c=0\) or omit \(+c\). |
| \(y=\frac{1}{2}x^4-\frac{2}{x}+(\text{their }c)x+k\) | A1 FT | Integrated. FT *their* non-zero value of \(c\) if DM1 awarded. |
| \(\frac{9}{2}=\frac{1}{2}+2+k\) leading to \(k=2\) | DM1 | Substituting \(x=-1,\ y=\frac{9}{2}\) to evaluate \(k\) (dep on *M1). |
| \(y=\frac{1}{2}x^4-\frac{2}{x}+2\) | A1 | OE e.g. \(2x^{-1}\) or \(\frac{4}{2}\). A0 (wrong process) if \(c\) not evaluated but correct answer obtained. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=2x^3+\frac{2}{x^2}=0\) | M1 | *Their* \(\frac{dy}{dx}=0\). |
| Leading to \(x^5=-1\) | M1 | Reaching equation of the form \(x^5=a\). |
| Only stationary point is when \(x=-1\) | A1 | \(x=-1\) and stating e.g. 'only' or 'no other solutions'. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(x=1\), \(\frac{dy}{dx}=4\) | *M1 | Substituting \(x=1\) into *their* \(\frac{dy}{dx}\). |
| \(\frac{dx}{dt}=\frac{dx}{dy}\times\frac{dy}{dt}=\frac{1}{4}\times 5\) | DM1 | OE Using chain rule correctly SOI. |
| \(\frac{5}{4}\) | A1 | OE e.g. \(1.25\). |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2}=6(-1)^2-\frac{4}{(-1)^3}>0$ $\therefore$ minimum **or** $\frac{d^2y}{dx^2}=10$ $\therefore$ minimum | B1 | Sub $x=-1$ into $\frac{d^2y}{dx^2}$, correct conclusion. WWW |
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## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=2x^3+\frac{2}{x^2}[+c]$ | *M1 | Integrating $\frac{d^2y}{dx^2}$ (at least one term correct). |
| $0=-2+2+c$ leading to $c=0$ | DM1 | Substituting $x=-1,\ \frac{dy}{dx}=0$ (need to see) to evaluate $c$. DM0 if simply state $c=0$ or omit $+c$. |
| $y=\frac{1}{2}x^4-\frac{2}{x}+(\text{their }c)x+k$ | A1 FT | Integrated. FT *their* non-zero value of $c$ if DM1 awarded. |
| $\frac{9}{2}=\frac{1}{2}+2+k$ leading to $k=2$ | DM1 | Substituting $x=-1,\ y=\frac{9}{2}$ to evaluate $k$ (dep on *M1). |
| $y=\frac{1}{2}x^4-\frac{2}{x}+2$ | A1 | OE e.g. $2x^{-1}$ or $\frac{4}{2}$. A0 (wrong process) if $c$ not evaluated but correct answer obtained. |
---
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=2x^3+\frac{2}{x^2}=0$ | M1 | *Their* $\frac{dy}{dx}=0$. |
| Leading to $x^5=-1$ | M1 | Reaching equation of the form $x^5=a$. |
| Only stationary point is when $x=-1$ | A1 | $x=-1$ and stating e.g. 'only' or 'no other solutions'. |
---
## Question 10(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $x=1$, $\frac{dy}{dx}=4$ | *M1 | Substituting $x=1$ into *their* $\frac{dy}{dx}$. |
| $\frac{dx}{dt}=\frac{dx}{dy}\times\frac{dy}{dt}=\frac{1}{4}\times 5$ | DM1 | OE Using chain rule correctly SOI. |
| $\frac{5}{4}$ | A1 | OE e.g. $1.25$. |10 The equation of a curve is such that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 x ^ { 2 } - \frac { 4 } { x ^ { 3 } }$. The curve has a stationary point at $\left( - 1 , \frac { 9 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Determine the nature of the stationary point at $\left( - 1 , \frac { 9 } { 2 } \right)$.
\item Find the equation of the curve.
\item Show that the curve has no other stationary points.
\item A point $A$ is moving along the curve and the $y$-coordinate of $A$ is increasing at a rate of 5 units per second.
Find the rate of increase of the $x$-coordinate of $A$ at the point where $x = 1$.\\
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 2022 Q10 [12]}}