| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Two Curves Intersection Area |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: finding derivatives to get gradients, using arctan for angle between lines, and integrating the difference of two simple functions between given limits. All steps are routine A-level procedures 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 equations1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(y = (4x+1)^{\frac{1}{2}}\), \(\frac{dy}{dx} = \left[\frac{1}{2}(4x+1)^{-\frac{1}{2}}\right] \times [4]\) | B1B1 | |
| When \(x = 2\), gradient \(m_1 = \frac{2}{3}\) | B1✓ | Ft from *their* derivative above |
| For \(y = \frac{1}{2}x^2 + 1\), \(\frac{dy}{dx} = x \rightarrow\) gradient \(m_2 = 2\) | B1 | |
| \(\alpha = \tan^{-1}m_2 - \tan^{-1}m_1\) | M1 | |
| \(\alpha = 63.43 - 33.69 = 29.7°\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int(4x+1)^{\frac{1}{2}}\,dx = \left[\frac{(4x+1)^{\frac{3}{2}}}{2/3}\right] \div [4]\) | B1B1 | |
| \(\int\left(\frac{1}{2}x^2+1\right)dx = \frac{1}{6}x^3 + x\) | B1 | |
| \(\int_0^2(4x+1)^{\frac{1}{2}}\,dx = \frac{1}{6}[27-1]\), \(\int_0^2\left(\frac{1}{2}x^2+1\right)dx = \left[\frac{8}{6}+2\right]\) | M1 | Apply limits \(0 \to 2\) to at least the 1st integral |
| \(\frac{13}{3} - \frac{10}{3}\) | M1 | Subtract the integrals (at some stage) |
| \(1\) | A1 |
## Question 11:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $y = (4x+1)^{\frac{1}{2}}$, $\frac{dy}{dx} = \left[\frac{1}{2}(4x+1)^{-\frac{1}{2}}\right] \times [4]$ | **B1B1** | |
| When $x = 2$, gradient $m_1 = \frac{2}{3}$ | **B1**✓ | Ft from *their* derivative above |
| For $y = \frac{1}{2}x^2 + 1$, $\frac{dy}{dx} = x \rightarrow$ gradient $m_2 = 2$ | **B1** | |
| $\alpha = \tan^{-1}m_2 - \tan^{-1}m_1$ | **M1** | |
| $\alpha = 63.43 - 33.69 = 29.7°$ cao | **A1** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int(4x+1)^{\frac{1}{2}}\,dx = \left[\frac{(4x+1)^{\frac{3}{2}}}{2/3}\right] \div [4]$ | **B1B1** | |
| $\int\left(\frac{1}{2}x^2+1\right)dx = \frac{1}{6}x^3 + x$ | **B1** | |
| $\int_0^2(4x+1)^{\frac{1}{2}}\,dx = \frac{1}{6}[27-1]$, $\int_0^2\left(\frac{1}{2}x^2+1\right)dx = \left[\frac{8}{6}+2\right]$ | **M1** | Apply limits $0 \to 2$ to at least the 1st integral |
| $\frac{13}{3} - \frac{10}{3}$ | **M1** | Subtract the integrals (at some stage) |
| $1$ | **A1** | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{77543862-ed95-42bf-b788-a9a43f039a89-4_995_905_260_621}
The diagram shows parts of the curves $y = ( 4 x + 1 ) ^ { \frac { 1 } { 2 } }$ and $y = \frac { 1 } { 2 } x ^ { 2 } + 1$ intersecting at points $P ( 0,1 )$ and $Q ( 2,3 )$. The angle between the tangents to the two curves at $Q$ is $\alpha$.\\
(i) Find $\alpha$, giving your answer in degrees correct to 3 significant figures.\\
(ii) Find by integration the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2014 Q11 [12]}}