| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring students to find a constant using given conditions, integrate a simple power function, apply a boundary condition, find and classify a stationary point. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{3}{2} = \frac{1}{2} + k\) leading to \(k = -2\) | B1 | AG. Need to see \(4^{\frac{1}{2}}\) evaluated as \(\frac{1}{\frac{1}{4^2}}\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| \([y] = 2x^{\frac{1}{2}} - 2x\ [+c]\) | M1 A1 | Allow \(\frac{x^{\frac{1}{2}}}{\frac{1}{2}} - 2x\) |
| \(-1 = 4 - 8 + c\) | M1 | Substitute \(x = 4\), \(y = -1\) (\(c\) present). Expect \(c = 3\) |
| \(y = 2x^{\frac{1}{2}} - 2x + 3\) or \(y = 2\sqrt{x} - 2x + 3\) | A1 | Allow if \(f(x) =\) or \(y =\) anywhere in the solution |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^{-1/2} - 2 = 0\) | M1 | Set *their* \(\frac{dy}{dx}\) to zero |
| \(x = \frac{1}{4}\) | A1 | If \(\left(\frac{1}{2}\right)^2 = \pm\frac{1}{4}\), max of M1A1 if \(\left(\frac{1}{4}, 3\frac{1}{2}\right)\) seen |
| \(\left(\frac{1}{4},\ 3\frac{1}{2}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-\frac{3}{2}}\) | B1 | |
| \(< 0\) (or \(-4\)) hence Maximum | DB1 | WWW. Ignore extra solutions from \(x = -\frac{1}{4}\) |
## Question 10:
**Part 10(a):**
| $-\frac{3}{2} = \frac{1}{2} + k$ leading to $k = -2$ | B1 | AG. Need to see $4^{\frac{1}{2}}$ evaluated as $\frac{1}{\frac{1}{4^2}}$ or better |
**Part 10(b):**
| $[y] = 2x^{\frac{1}{2}} - 2x\ [+c]$ | M1 A1 | Allow $\frac{x^{\frac{1}{2}}}{\frac{1}{2}} - 2x$ |
| $-1 = 4 - 8 + c$ | M1 | Substitute $x = 4$, $y = -1$ ($c$ present). Expect $c = 3$ |
| $y = 2x^{\frac{1}{2}} - 2x + 3$ or $y = 2\sqrt{x} - 2x + 3$ | A1 | Allow if $f(x) =$ or $y =$ anywhere in the solution |
**Part 10(c):**
| $x^{-1/2} - 2 = 0$ | M1 | Set *their* $\frac{dy}{dx}$ to zero |
| $x = \frac{1}{4}$ | A1 | If $\left(\frac{1}{2}\right)^2 = \pm\frac{1}{4}$, max of M1A1 if $\left(\frac{1}{4}, 3\frac{1}{2}\right)$ seen |
| $\left(\frac{1}{4},\ 3\frac{1}{2}\right)$ | A1 | |
**Part 10(d):**
| $\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-\frac{3}{2}}$ | B1 | |
| $< 0$ (or $-4$) hence Maximum | DB1 | WWW. Ignore extra solutions from $x = -\frac{1}{4}$ |
---10 At the point $( 4 , - 1 )$ on a curve, the gradient of the curve is $- \frac { 3 } { 2 }$. It is given that $\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { - \frac { 1 } { 2 } } + k$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = - 2$.
\item Find the equation of the curve.
\item Find the coordinates of the stationary point.
\item Determine the nature of the stationary point.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q10 [10]}}