| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 at a stationary point to find a constant, using second derivative test, integrating to find the curve equation, and solving a quadratic. All steps are routine applications of basic differentiation/integration rules with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(6a^2-30a+6a=0 \Rightarrow 6a(a-4)=0\) | B1 | Sub \(x=a\) into \(\frac{dy}{dx}=0\). May see \(a^2-5a+a=0\) |
| \(a=4\) only | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2}=12x-30\) or correct values of \(\frac{dy}{dx}\) either side of \(x=4\) | M1 | Differentiate \(\frac{dy}{dx}\) (mult. by power or dec. power by 1). M0 if no values of \(\frac{dy}{dx}\), only signs |
| At \(x=4\), \(\frac{d^2y}{dx^2}>0\ \therefore\) minimum or \(\frac{d^2y}{dx^2}=18\ \therefore\) minimum, or concludes minimum from \(\frac{dy}{dx}\) values | A1 | WWW A0 XP if \(a=4\) obtained incorrectly in (a). Must see 'minimum'. If M0, SC B1 for 'minimum' from \(\frac{dy}{dx}\) sign diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y=]\ \frac{6}{3}x^3 - \frac{30}{2}x^2 + 6(\textit{their}\ a)x[+c]\) | B1 FT | Expect \(2x^3-15x^2+24x[+c]\). B1 poss. even if uses \('a'\) – no value in (a) – max 1/3 |
| \(-15 = 2(\textit{their}\ "4")^3 - 15(\textit{their}\ "4")^2 + 6(\textit{their}\ "4")^2 + c\) | M1 | Sub \(x=\textit{their}\ "4"\), \(y=-15\) into integral (must incl \(+c\)). Look for \(-15=128-240+96+c \Rightarrow c=1\) |
| \(y = 2x^3-15x^2+24x+1\) | A1 | Coefficients must be correct and simplified. Need to see \('y='\) or \('f(x)='\) in the working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=6x^2-30x+6(\textit{their}\ "4")[=0]\); If correct, \([6](x-1)(x-4)[=0]\) or \(\frac{30\pm\sqrt{(-30)^2-4(6)(24)}}{12}\) | M1 | OE Forming a 3-term quadratic using the given \(\frac{dy}{dx}\) and solving by factorisation, formula or completing the square. Check for working in (b) |
| Coordinates \((1, 12)\) | A1 | Allow \(x=1\), \(y=12\) (ignore \(x=4\) if present). If M0, award SC B1 for \((1,12)\) |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6a^2-30a+6a=0 \Rightarrow 6a(a-4)=0$ | B1 | Sub $x=a$ into $\frac{dy}{dx}=0$. May see $a^2-5a+a=0$ |
| $a=4$ only | B1 | |
**Total: 2 marks**
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2}=12x-30$ or correct values of $\frac{dy}{dx}$ either side of $x=4$ | M1 | Differentiate $\frac{dy}{dx}$ (mult. by power or dec. power by 1). M0 if no values of $\frac{dy}{dx}$, only signs |
| At $x=4$, $\frac{d^2y}{dx^2}>0\ \therefore$ minimum or $\frac{d^2y}{dx^2}=18\ \therefore$ minimum, or concludes minimum from $\frac{dy}{dx}$ values | A1 | WWW A0 XP if $a=4$ obtained incorrectly in (a). Must see 'minimum'. If M0, **SC B1** for 'minimum' from $\frac{dy}{dx}$ sign diagram |
**Total: 2 marks**
---
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y=]\ \frac{6}{3}x^3 - \frac{30}{2}x^2 + 6(\textit{their}\ a)x[+c]$ | B1 FT | Expect $2x^3-15x^2+24x[+c]$. B1 poss. even if uses $'a'$ – no value in (a) – max 1/3 |
| $-15 = 2(\textit{their}\ "4")^3 - 15(\textit{their}\ "4")^2 + 6(\textit{their}\ "4")^2 + c$ | M1 | Sub $x=\textit{their}\ "4"$, $y=-15$ into integral (must incl $+c$). Look for $-15=128-240+96+c \Rightarrow c=1$ |
| $y = 2x^3-15x^2+24x+1$ | A1 | Coefficients must be correct and simplified. Need to see $'y='$ or $'f(x)='$ in the working |
**Total: 3 marks**
---
## Question 11(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=6x^2-30x+6(\textit{their}\ "4")[=0]$; If correct, $[6](x-1)(x-4)[=0]$ or $\frac{30\pm\sqrt{(-30)^2-4(6)(24)}}{12}$ | M1 | OE Forming a 3-term quadratic using the given $\frac{dy}{dx}$ and solving by factorisation, formula or completing the square. Check for working in **(b)** |
| Coordinates $(1, 12)$ | A1 | Allow $x=1$, $y=12$ (ignore $x=4$ if present). If M0, award **SC B1** for $(1,12)$ |
**Total: 2 marks**
---11 The equation of a curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { 2 } - 30 x + 6 a$, where $a$ is a positive constant. The curve has a stationary point at $( a , - 15 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item Determine the nature of this stationary point.
\item Find the equation of the curve.
\item Find the coordinates of any other stationary points on the curve.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q11 [9]}}