| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find constant using stationary point |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard calculus techniques: setting derivative to zero to find a stationary point (solving a simple equation), using the second derivative test, and integrating with a boundary condition. All steps are routine applications of well-practiced methods with no conceptual challenges or novel problem-solving required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(a+3)^{\frac{1}{2}} - a = 0\) | M1 | SOI. Set \(\frac{dy}{dx}=0\) when \(x=a\). Can be implied by an answer in terms of \(a\) |
| \(4(a+3) = a^2 \to a^2-4a-12=0\) | M1 | Take \(a\) to RHS and square. Form 3-term quadratic |
| \((a-6)(a+2) \to a=6\) | A1 | Must show factors, or formula or completing the square. Ignore \(a=-2\). SC If \(a\) is never used maximum of M1A1 for \(x=6\), with visible solution |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = (x+3)^{-\frac{1}{2}}-1\) | B1 | |
| Sub *their* \(a \to \frac{d^2y}{dx^2} = \frac{1}{3}-1 = -\frac{2}{3}\) (or \(< 0\)) \(\to\) MAX | M1A1 | A mark only if completely correct. If the second differential is not \(-\frac{2}{3}\) correct conclusion must be drawn to award the M1 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((y=)\,\frac{2(x+3)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{1}{2}x^2\,(+c)\) | B1B1 | |
| Sub \(x = \text{their}\,a\) and \(y=14 \to 14 = \frac{4}{3}(9)^{\frac{3}{2}}-18+c\) | M1 | Substitute into an integrated expression. \(c\) must be present. Expect \(c=-4\) |
| \(y = \frac{4}{3}(x+3)^{\frac{3}{2}} - \frac{1}{2}x^2 - 4\) | A1 | Allow \(f(x) = \ldots\) |
| Total: 4 |
## Question 10(a):
| $2(a+3)^{\frac{1}{2}} - a = 0$ | M1 | SOI. Set $\frac{dy}{dx}=0$ when $x=a$. Can be implied by an answer in terms of $a$ |
|---|---|---|
| $4(a+3) = a^2 \to a^2-4a-12=0$ | M1 | Take $a$ to RHS and square. Form 3-term quadratic |
| $(a-6)(a+2) \to a=6$ | A1 | Must show factors, or formula or completing the square. Ignore $a=-2$. SC If $a$ is never used maximum of M1A1 for $x=6$, with visible solution |
| **Total: 3** | | |
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## Question 10(b):
| $\frac{d^2y}{dx^2} = (x+3)^{-\frac{1}{2}}-1$ | B1 | |
|---|---|---|
| Sub *their* $a \to \frac{d^2y}{dx^2} = \frac{1}{3}-1 = -\frac{2}{3}$ (or $< 0$) $\to$ MAX | M1A1 | A mark only if completely correct. If the second differential is not $-\frac{2}{3}$ correct conclusion must be drawn to award the M1 |
| **Total: 3** | | |
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## Question 10(c):
| $(y=)\,\frac{2(x+3)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{1}{2}x^2\,(+c)$ | B1B1 | |
|---|---|---|
| Sub $x = \text{their}\,a$ and $y=14 \to 14 = \frac{4}{3}(9)^{\frac{3}{2}}-18+c$ | M1 | Substitute into an integrated expression. $c$ must be present. Expect $c=-4$ |
| $y = \frac{4}{3}(x+3)^{\frac{3}{2}} - \frac{1}{2}x^2 - 4$ | A1 | Allow $f(x) = \ldots$ |
| **Total: 4** | | |10 The gradient of a curve at the point $( x , y )$ is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( x + 3 ) ^ { \frac { 1 } { 2 } } - x$. The curve has a stationary point at $( a , 14 )$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item Determine the nature of the stationary point.
\item Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q10 [10]}}