| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Two Curves Intersection Area |
| Difficulty | Standard +0.3 This is a standard A-level integration question involving finding intersection points by solving a quadratic in x³, computing area between curves, and finding where tangents are parallel. All techniques are routine for P1 level with clear structure and no novel insights required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02q Use intersection points: of graphs to solve equations1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9 - x^3 = \frac{8}{x^3}\) | M1 | Together with attempt to multiply by \(x^3\) |
| \(x^6 - 9x^3 + 8 = 0\) | A1 | AG completely correct working |
| \((X-1)(X-8) = 0 \rightarrow X = 1\) or \(8\) | M1 | Attempt to solve quadratic in \(X\) or \(x^3\) |
| \(a = 1,\ b = 2\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_1^2\left[(9-x^3) - \frac{8}{x^3}\right]dx\) | M1 | Intention to integrate the difference \(y_1 - y_2\), not \(\pi(y_1 - y_2)\) |
| \(\left[9x - \frac{x^4}{4}\right]\cdot\left[\frac{-4}{x^2}\right]\) | B1, B1 | |
| \(18 - 4 + 1 - \left(9 - \frac{1}{4} + 4\right)\) | M1 | Correct use of their limits once |
| \(2\frac{1}{4}\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-24}{x^4},\ \frac{dy}{dx} = -3x^2\) | B1, B1 | cao |
| \(\frac{-24}{c^4} = -3c^2\) | ||
| \(c^6 = 8\) | M1 | Equating and solution |
| \(c = \sqrt{2}\) or \(8^{1/6}\) or \(1.41(4\ldots)\) | A1 | Accept \(x\) or \(c\) |
| [4] |
## Question 11:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9 - x^3 = \frac{8}{x^3}$ | M1 | Together with attempt to multiply by $x^3$ |
| $x^6 - 9x^3 + 8 = 0$ | A1 | **AG** completely correct working |
| $(X-1)(X-8) = 0 \rightarrow X = 1$ or $8$ | M1 | Attempt to solve quadratic in $X$ or $x^3$ |
| $a = 1,\ b = 2$ | A1 | |
| **[4]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_1^2\left[(9-x^3) - \frac{8}{x^3}\right]dx$ | M1 | Intention to integrate the difference $y_1 - y_2$, not $\pi(y_1 - y_2)$ |
| $\left[9x - \frac{x^4}{4}\right]\cdot\left[\frac{-4}{x^2}\right]$ | B1, B1 | |
| $18 - 4 + 1 - \left(9 - \frac{1}{4} + 4\right)$ | M1 | Correct use of their limits once |
| $2\frac{1}{4}$ | A1 | |
| **[5]** | | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-24}{x^4},\ \frac{dy}{dx} = -3x^2$ | B1, B1 | cao |
| $\frac{-24}{c^4} = -3c^2$ | | |
| $c^6 = 8$ | M1 | Equating and solution |
| $c = \sqrt{2}$ or $8^{1/6}$ or $1.41(4\ldots)$ | A1 | Accept $x$ or $c$ |
| **[4]** | | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{32a57386-2696-4fda-a3cb-ca0c5c3be432-5_710_931_255_607}
The diagram shows parts of the curves $y = 9 - x ^ { 3 }$ and $y = \frac { 8 } { x ^ { 3 } }$ and their points of intersection $P$ and $Q$. The $x$-coordinates of $P$ and $Q$ are $a$ and $b$ respectively.\\
(i) Show that $x = a$ and $x = b$ are roots of the equation $x ^ { 6 } - 9 x ^ { 3 } + 8 = 0$. Solve this equation and hence state the value of $a$ and the value of $b$.\\
(ii) Find the area of the shaded region between the two curves.\\
(iii) The tangents to the two curves at $x = c$ (where $a < c < b$ ) are parallel to each other. Find the value of $c$.
\hfill \mbox{\textit{CAIE P1 2010 Q11 [13]}}