| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard techniques: finding a tangent line using differentiation (chain rule) and calculating an area between a curve and line using definite integration. Both are routine P1 procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \left[3(3-2x)^2\right] \times \left[-2\right]\) | B1B1 | OR \(-54 + 72x - 24x^2\), \(B2, 1, 0\) |
| At \(x = \frac{1}{2}\) \(\frac{dy}{dx} = -24\) | M1 | |
| \(y - 8 = -24(x - \frac{1}{2})\) | DM1 | |
| \(y = -24x + 20\) | A1 | [5] |
| (ii) Area under curve \(= \left[\frac{(3-2x)^3}{4}\right] \times \left[-\frac{1}{2}\right]\) | B1B1 | OR \(27x - 27x^2 + 12x^3 - 2x^4\), \(B2, 1, 0\) |
| \(-2 - (-\frac{81}{8})\) | M1 | Limits 0→ ½ applied to integral with intention of subtraction shown or area trap =\(\frac{1}{2}(20 + 8) \times \frac{1}{2}\) |
| Area under tangent \(= \int(-24x + 20)\) | M1 | |
| \(= \left | -12x^2 + 20x\right | \) or 7 (from trap) |
| \(\frac{9}{8}\) or 1.125 | A1 | Dep on both M marks |
(i) $\frac{dy}{dx} = \left[3(3-2x)^2\right] \times \left[-2\right]$ | B1B1 | OR $-54 + 72x - 24x^2$, $B2, 1, 0$
At $x = \frac{1}{2}$ $\frac{dy}{dx} = -24$ | M1 |
$y - 8 = -24(x - \frac{1}{2})$ | DM1 |
$y = -24x + 20$ | A1 | [5]
(ii) Area under curve $= \left[\frac{(3-2x)^3}{4}\right] \times \left[-\frac{1}{2}\right]$ | B1B1 | OR $27x - 27x^2 + 12x^3 - 2x^4$, $B2, 1, 0$
$-2 - (-\frac{81}{8})$ | M1 | Limits 0→ ½ applied to integral with intention of subtraction shown or area trap =$\frac{1}{2}(20 + 8) \times \frac{1}{2}$
Area under tangent $= \int(-24x + 20)$ | M1 |
$= \left|-12x^2 + 20x\right|$ or 7 (from trap) | A1 | Could be implied
$\frac{9}{8}$ or 1.125 | A1 | Dep on both M marks | [6]10\\
\includegraphics[max width=\textwidth, alt={}, center]{02da6b6a-6db1-4bc3-ad4e-537e4f61dcac-4_654_974_614_587}
The diagram shows the curve $y = ( 3 - 2 x ) ^ { 3 }$ and the tangent to the curve at the point $\left( \frac { 1 } { 2 } , 8 \right)$.\\
(i) Find the equation of this tangent, giving your answer in the form $y = m x + c$.\\
(ii) Find the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2013 Q10 [11]}}