| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.8 This is a straightforward stationary points question requiring differentiation of polynomial and reciprocal terms (standard A-level technique), solving a simple equation, and applying the second derivative test. Part (c) adds minimal complexity by asking about sign of the derivative for x>0. All steps are routine with no problem-solving insight required, making it easier than average but not trivial. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to obtain \(4x + \frac{1}{2}x^{-2}\) | B1 | OE. Condone '\(+c\)'. |
| Equate first derivative to zero and solve \(4x + \frac{K}{x^2} = 0\) as far as \(x^3 = k\), \(K\) and \(k\) non-zero | M1 | Not given if '\(+c\)' used. |
| \(x = -\frac{1}{2}\) and \(y = \frac{9}{2}\) | A1 | OE. B1 SC if no visible solution of the cubic. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate their first derivative, substitute their \(x\) value. Substitution may be implied by a correct inequality or correct value. | M1 | Must differentiate one term correctly. Expect \(4 - x^{-3} = 12\) at \(x = \frac{-1}{2}\). Alternative: substitute values of \(x\) into \(\frac{dy}{dx}\). One value \(x < -\frac{1}{2}\) and one value \(-\frac{1}{2} < x < 0\). |
| Conclude minimum | A1 | Following correct work only |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State increasing … | B1 | |
| … with clear reference to first derivative always being positive [for \(x > 0\)] | B1 | Dependent on first derivative being correct. It is not sufficient to substitute values of \(x\). |
| 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain $4x + \frac{1}{2}x^{-2}$ | B1 | OE. Condone '$+c$'. |
| Equate first derivative to zero and solve $4x + \frac{K}{x^2} = 0$ as far as $x^3 = k$, $K$ and $k$ non-zero | M1 | Not given if '$+c$' used. |
| $x = -\frac{1}{2}$ and $y = \frac{9}{2}$ | A1 | OE. **B1 SC** if no visible solution of the cubic. |
| | **3** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate their first derivative, substitute their $x$ value. Substitution may be implied by a correct inequality or correct value. | M1 | Must differentiate one term correctly. Expect $4 - x^{-3} = 12$ at $x = \frac{-1}{2}$. Alternative: substitute values of $x$ into $\frac{dy}{dx}$. One value $x < -\frac{1}{2}$ and one value $-\frac{1}{2} < x < 0$. |
| Conclude minimum | A1 | Following correct work only |
| | **2** | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State increasing … | B1 | |
| … with clear reference to first derivative always being positive [for $x > 0$] | B1 | Dependent on first derivative being correct. It is not sufficient to substitute values of $x$. |
| | **2** | |5 The equation of a curve is $y = 2 x ^ { 2 } - \frac { 1 } { 2 x } + 3$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the stationary point.
\item Determine the nature of the stationary point.
\item For positive values of $x$, determine whether the curve shows a function that is increasing, decreasing or neither. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q5 [7]}}