| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.3 This question requires routine application of chain rule for differentiation, standard integration techniques (reverse chain rule), and finding a minimum by solving dy/dx = 0. While multi-part with several steps, each technique is standard P1 material with no novel problem-solving insight required. The integration and solving for M are slightly above average routine exercises, placing it just above typical difficulty. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = [8] \times [(3-2x)^{-3}] + [-1]\) \(\left(= \frac{8}{(3-2x)^3} - 1\right)\) | B2, 1, 0 | B2 for all three elements correct, B1 for two elements correct, B0 for only one or no elements correct |
| \(\frac{d^2y}{dx^2} = -3 \times 8 \times (3-2x)^{-4} \times (-2)\) \(\left(= \frac{48}{(3-2x)^4}\right)\) | B1 FT | FT providing *their* bracket is to a negative power |
| \(\int y\,dx = [(3-2x)^{-1}][2 \div (-1 \times -2)][-\frac{1}{2}x^2](+c)\) \(\left(= \frac{1}{3-2x} - \frac{1}{2}x^2 + c\right)\) | B1 B1 B1 | Simplification not needed, B1 for each correct element |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 0 \rightarrow (3-2x)^3 = 8 \rightarrow 3-2x = k \rightarrow x =\) | M1 | Setting *their* 2-term differential to 0 and attempts to solve as far as \(x =\) |
| \(\frac{1}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 0 \rightarrow \frac{2}{(3-2x)^2} - x = 0 \rightarrow (x-2)(2x-1)^2 = 0 \rightarrow x =\) | M1 | Setting \(y\) to 0 and attempts to solve a cubic as far as \(x =\) (3 factors needed) |
| \(\frac{1}{2}\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area under curve \(= their\left[\frac{1}{3-2\times\left(\frac{1}{2}\right)} - \frac{\left(\frac{1}{2}\right)^2}{2}\right] - \left[\frac{1}{3-2\times 0} - 0\right]\) | M1 | Using *their* integral, *their* positive \(x\) limit from part (b) and 0 correctly |
| \(\frac{1}{24}\) | A1 | |
| 2 |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = [8] \times [(3-2x)^{-3}] + [-1]$ $\left(= \frac{8}{(3-2x)^3} - 1\right)$ | B2, 1, 0 | B2 for all three elements correct, B1 for two elements correct, B0 for only one or no elements correct |
| $\frac{d^2y}{dx^2} = -3 \times 8 \times (3-2x)^{-4} \times (-2)$ $\left(= \frac{48}{(3-2x)^4}\right)$ | B1 FT | FT providing *their* bracket is to a negative power |
| $\int y\,dx = [(3-2x)^{-1}][2 \div (-1 \times -2)][-\frac{1}{2}x^2](+c)$ $\left(= \frac{1}{3-2x} - \frac{1}{2}x^2 + c\right)$ | B1 B1 B1 | Simplification not needed, B1 for each correct element |
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 0 \rightarrow (3-2x)^3 = 8 \rightarrow 3-2x = k \rightarrow x =$ | M1 | Setting *their* 2-term differential to 0 and attempts to solve as far as $x =$ |
| $\frac{1}{2}$ | A1 | |
**Alternative method for 10(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 0 \rightarrow \frac{2}{(3-2x)^2} - x = 0 \rightarrow (x-2)(2x-1)^2 = 0 \rightarrow x =$ | M1 | Setting $y$ to 0 and attempts to solve a cubic as far as $x =$ (3 factors needed) |
| $\frac{1}{2}$ | A1 | |
| | **2** | |
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area under curve $= their\left[\frac{1}{3-2\times\left(\frac{1}{2}\right)} - \frac{\left(\frac{1}{2}\right)^2}{2}\right] - \left[\frac{1}{3-2\times 0} - 0\right]$ | M1 | Using *their* integral, *their* positive $x$ limit from **part (b)** and 0 correctly |
| $\frac{1}{24}$ | A1 | |
| | **2** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{6bcc553c-4938-46ef-bba4-97391b4d58d4-14_378_666_264_737}
The diagram shows part of the curve $y = \frac { 2 } { ( 3 - 2 x ) ^ { 2 } } - x$ and its minimum point $M$, which lies on the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Find expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and $\int y \mathrm {~d} x$.
\item Find, by calculation, the $x$-coordinate of $M$.
\item Find the area of the shaded region bounded by the curve and the coordinate axes.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q10 [10]}}