| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| 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: substituting a point into a derivative to find a constant, integrating to find the curve equation, finding the second derivative, and using it to classify a stationary point. All parts follow routine procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07d Second derivatives: d^2y/dx^2 notation1.07n 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 |
| At stationary point \(\frac{dy}{dx} = 0\) so \(6(3\times2-5)^3 - k\times2^2 = 0\) | M1 | Setting \(\frac{dy}{dx} = 0\) and substituting \(x = 2\) |
| \([k =] \frac{3}{2}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y =] \frac{6}{4\times3}(3x-5)^4 - \frac{1}{3}kx^3\ [+c]\) | \*M1 A1FT | Integrating (increase of power by 1 in at least one term). Expect \(\frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3\). FT their non zero \(k\) |
| \(-\frac{7}{2} = \frac{1}{2}(3\times2-5)^4 - \frac{1}{3}\times\frac{3}{2}\times2^3 + c\) leading to \(-3.5 + c = -3.5\) | DM1 | Using \((2, -3.5)\) in integrated expression. \(+c\) needed. Simply stating \(c = 0\) is DM0 |
| \(y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3\) | A1 | \(y =\) or \(f(x) =\) must be seen somewhere in solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y =] \frac{81}{2}x^4 - \frac{541}{2}x^3 + 675x^2 - 750x(+c)\) or \(-270x^3 - k\frac{x^3}{3}\) | \*M1 A1FT | From \(\frac{dy}{dx} = 162x^3 - 810x^2 - kx^2 - 1350x - 750\). FT their \(k\) |
| \(-\frac{7}{2} = \frac{81}{2}\times2^4 - \frac{541}{2}\times2^3 + 675\times2^2 - 750\times2 + c\) | DM1 | Using \((2,-3.5)\) in integrated expression. \(+c\) needed |
| \(y = \frac{81}{2}x^4 - \frac{541}{2}x^3 + 675x^2 - 750x + \frac{625}{2}\) | A1 | \(y =\) or \(f(x) =\) must be seen somewhere in solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([3\times][18(3x-5)^2][-2kx]\) | B2,1,0 FT | FT their \(k\). Square brackets indicate each required component. B2 fully correct, B1 for one error or one missing component, B0 for 2 or more errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(486x^2 - 1623x + 1350\) or \(-1620x - 2kx\) | B2,1,0 FT | FT their \(k\). B2 fully correct, B1 for one error, B0 for 2 or more errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{At } x=2]\ \left[\frac{d^2y}{dx^2} =\right] 54(3\times2-5)^2 - 4k\) or \(48\) | M1 | OE. Substituting \(x = 2\) into their second differential or other valid method |
| \([> 0]\) Minimum | A1 | WWW |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At stationary point $\frac{dy}{dx} = 0$ so $6(3\times2-5)^3 - k\times2^2 = 0$ | M1 | Setting $\frac{dy}{dx} = 0$ and substituting $x = 2$ |
| $[k =] \frac{3}{2}$ | A1 | OE |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y =] \frac{6}{4\times3}(3x-5)^4 - \frac{1}{3}kx^3\ [+c]$ | \*M1 A1FT | Integrating (increase of power by 1 in at least one term). Expect $\frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3$. FT their non zero $k$ |
| $-\frac{7}{2} = \frac{1}{2}(3\times2-5)^4 - \frac{1}{3}\times\frac{3}{2}\times2^3 + c$ leading to $-3.5 + c = -3.5$ | DM1 | Using $(2, -3.5)$ in integrated expression. $+c$ needed. Simply stating $c = 0$ is DM0 |
| $y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3$ | A1 | $y =$ or $f(x) =$ must be seen somewhere in solution |
**Alternative method for 11(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y =] \frac{81}{2}x^4 - \frac{541}{2}x^3 + 675x^2 - 750x(+c)$ or $-270x^3 - k\frac{x^3}{3}$ | \*M1 A1FT | From $\frac{dy}{dx} = 162x^3 - 810x^2 - kx^2 - 1350x - 750$. FT their $k$ |
| $-\frac{7}{2} = \frac{81}{2}\times2^4 - \frac{541}{2}\times2^3 + 675\times2^2 - 750\times2 + c$ | DM1 | Using $(2,-3.5)$ in integrated expression. $+c$ needed |
| $y = \frac{81}{2}x^4 - \frac{541}{2}x^3 + 675x^2 - 750x + \frac{625}{2}$ | A1 | $y =$ or $f(x) =$ must be seen somewhere in solution |
---
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[3\times][18(3x-5)^2][-2kx]$ | B2,1,0 FT | FT their $k$. Square brackets indicate each required component. B2 fully correct, B1 for one error or one missing component, B0 for 2 or more errors |
**Alternative for 11(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $486x^2 - 1623x + 1350$ or $-1620x - 2kx$ | B2,1,0 FT | FT their $k$. B2 fully correct, B1 for one error, B0 for 2 or more errors |
---
## Question 11(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{At } x=2]\ \left[\frac{d^2y}{dx^2} =\right] 54(3\times2-5)^2 - 4k$ or $48$ | M1 | OE. Substituting $x = 2$ into their second differential or other valid method |
| $[> 0]$ Minimum | A1 | WWW |
---11 The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 ( 3 x - 5 ) ^ { 3 } - k x ^ { 2 }$, where $k$ is a constant. The curve has a stationary point at $( 2 , - 3.5 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.\\
................................................................................................................................................. . .
\item Find the equation of the curve.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Determine the nature of the stationary point at $( 2 , - 3.5 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q11 [10]}}