| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration question requiring: (i) finding where curve touches x-axis (simple algebra), (ii) finding tangent equation and x-intercept (standard differentiation), and (iii) calculating area between curve and line (routine integration). All techniques are standard P1 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 1/3\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = \left[\dfrac{2}{16}(3x-1)\right][3]\) | B1B1 | |
| When \(x = 3\): \(\dfrac{dy}{dx} = 3\) soi | M1 | |
| Equation of \(QR\) is \(y - 4 = 3(x - 3)\) | M1 | |
| When \(y = 0\): \(x = 5/3\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area under curve \(= \left[\dfrac{1}{16 \times 3}(3x-1)^3\right]\left[\times\dfrac{1}{3}\right]\) | B1B1 | |
| \(\dfrac{1}{16 \times 9}\left[8^3 - 0\right] = \dfrac{32}{9}\) | M1A1 | Apply limits: *their* \(\dfrac{1}{3}\) and 3 |
| Area of \(\Delta = 8/3\) | B1 | |
| Shaded area \(= \dfrac{32}{9} - \dfrac{8}{3} = \dfrac{8}{9}\) (or 0.889) | A1 [6] |
## Question 10:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1/3$ | B1 [1] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = \left[\dfrac{2}{16}(3x-1)\right][3]$ | B1B1 | |
| When $x = 3$: $\dfrac{dy}{dx} = 3$ soi | M1 | |
| Equation of $QR$ is $y - 4 = 3(x - 3)$ | M1 | |
| When $y = 0$: $x = 5/3$ | A1 [5] | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area under curve $= \left[\dfrac{1}{16 \times 3}(3x-1)^3\right]\left[\times\dfrac{1}{3}\right]$ | B1B1 | |
| $\dfrac{1}{16 \times 9}\left[8^3 - 0\right] = \dfrac{32}{9}$ | M1A1 | Apply limits: *their* $\dfrac{1}{3}$ and 3 |
| Area of $\Delta = 8/3$ | B1 | |
| Shaded area $= \dfrac{32}{9} - \dfrac{8}{3} = \dfrac{8}{9}$ (or 0.889) | A1 [6] | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-5_650_1038_260_550}
The diagram shows part of the curve $y = \frac { 1 } { 16 } ( 3 x - 1 ) ^ { 2 }$, which touches the $x$-axis at the point $P$. The point $Q ( 3,4 )$ lies on the curve and the tangent to the curve at $Q$ crosses the $x$-axis at $R$.\\
(i) State the $x$-coordinate of $P$.
Showing all necessary working, find by calculation\\
(ii) the $x$-coordinate of $R$,\\
(iii) the area of the shaded region $P Q R$.
\hfill \mbox{\textit{CAIE P1 2016 Q10 [12]}}