| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Tangent with specified gradient |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring: (i) finding a line through two points, (ii) differentiating y=(x-1)^(1/2) using chain rule and solving dy/dx=m for x, (iii) using perpendicular distance formula. All techniques are standard AS-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of \(AB = \frac{1}{2}\) | B1 | |
| Equation of \(AB\) is \(y = \frac{1}{2}x - \frac{1}{2}\) | B1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2}(x-1)^{-\frac{1}{2}}\) | B1 | |
| \(\frac{1}{2}(x-1)^{-\frac{1}{2}} = \frac{1}{2}\). Equate their \(\frac{dy}{dx}\) to their \(\frac{1}{2}\) | *M1 | |
| \(x = 2\), \(y = 1\) | A1 | |
| \(y - 1 = \frac{1}{2}(x-2)\) (thro' their\((2,1)\) & their \(\frac{1}{2}\)) \(\rightarrow y = \frac{1}{2}x\) | DM1 A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| EITHER: \(\sin\theta = \frac{d}{1} \rightarrow d = \sin\theta\) | (M1 | Where \(\theta\) is angle between \(AB\) and the \(x\)-axis |
| gradient of \(AB = \frac{1}{2} \Rightarrow \tan\theta = \frac{1}{2} \Rightarrow \theta = 26.5(7)°\) | B1 | |
| \(d = \sin 26.5(7)° = 0.45\) (or \(\frac{1}{\sqrt{5}}\)) | A1) | |
| OR1: Perpendicular through \(O\) has equation \(y = -2x\) | (M1 | |
| Intersection with \(AB\): \(-2x = \frac{1}{2}x - \frac{1}{2} \rightarrow \left(\frac{1}{5}, \frac{-2}{5}\right)\) | A1 | |
| \(d = \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{2}{5}\right)^2} = 0.45\) (or \(\frac{1}{\sqrt{5}}\)) | A1) | |
| OR2: Perpendicular through \((2,1)\) has equation \(y = -2x + 5\) | (M1 | |
| Intersection with \(AB\): \(-2x+5 = \frac{1}{2}x - \frac{1}{2} \rightarrow \left(\frac{11}{5}, \frac{3}{5}\right)\) | A1 | |
| \(d = \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{2}{5}\right)^2} = 0.45\) (or \(\frac{1}{\sqrt{5}}\)) | A1) | |
| OR3: \(\triangle OAC\) has area \(\frac{1}{4}\) [where \(C = (0, -\frac{1}{2})\)] | (B1 | |
| \(\frac{1}{2} \times \frac{\sqrt{5}}{2} \times d = \frac{1}{4} \rightarrow d = \frac{1}{\sqrt{5}}\) | M1 A1) | |
| Total: 3 |
## Question 11(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of $AB = \frac{1}{2}$ | B1 | |
| Equation of $AB$ is $y = \frac{1}{2}x - \frac{1}{2}$ | B1 | |
| **Total: 2** | | |
## Question 11(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = \frac{1}{2}(x-1)^{-\frac{1}{2}}$ | B1 | |
| $\frac{1}{2}(x-1)^{-\frac{1}{2}} = \frac{1}{2}$. Equate their $\frac{dy}{dx}$ to their $\frac{1}{2}$ | *M1 | |
| $x = 2$, $y = 1$ | A1 | |
| $y - 1 = \frac{1}{2}(x-2)$ (thro' their$(2,1)$ & their $\frac{1}{2}$) $\rightarrow y = \frac{1}{2}x$ | DM1 A1 | |
| **Total: 5** | | |
## Question 11(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **EITHER:** $\sin\theta = \frac{d}{1} \rightarrow d = \sin\theta$ | (M1 | Where $\theta$ is angle between $AB$ and the $x$-axis |
| gradient of $AB = \frac{1}{2} \Rightarrow \tan\theta = \frac{1}{2} \Rightarrow \theta = 26.5(7)°$ | B1 | |
| $d = \sin 26.5(7)° = 0.45$ (or $\frac{1}{\sqrt{5}}$) | A1) | |
| **OR1:** Perpendicular through $O$ has equation $y = -2x$ | (M1 | |
| Intersection with $AB$: $-2x = \frac{1}{2}x - \frac{1}{2} \rightarrow \left(\frac{1}{5}, \frac{-2}{5}\right)$ | A1 | |
| $d = \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{2}{5}\right)^2} = 0.45$ (or $\frac{1}{\sqrt{5}}$) | A1) | |
| **OR2:** Perpendicular through $(2,1)$ has equation $y = -2x + 5$ | (M1 | |
| Intersection with $AB$: $-2x+5 = \frac{1}{2}x - \frac{1}{2} \rightarrow \left(\frac{11}{5}, \frac{3}{5}\right)$ | A1 | |
| $d = \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{2}{5}\right)^2} = 0.45$ (or $\frac{1}{\sqrt{5}}$) | A1) | |
| **OR3:** $\triangle OAC$ has area $\frac{1}{4}$ [where $C = (0, -\frac{1}{2})$] | (B1 | |
| $\frac{1}{2} \times \frac{\sqrt{5}}{2} \times d = \frac{1}{4} \rightarrow d = \frac{1}{\sqrt{5}}$ | M1 A1) | |
| **Total: 3** | | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{17ca6dd2-271b-4b06-8433-354493feaf06-18_428_857_260_644}
The diagram shows the curve $y = ( x - 1 ) ^ { \frac { 1 } { 2 } }$ and points $A ( 1,0 )$ and $B ( 5,2 )$ lying on the curve.\\
(i) Find the equation of the line $A B$, giving your answer in the form $y = m x + c$.\\
(ii) Find, showing all necessary working, the equation of the tangent to the curve which is parallel to $A B$.\\
(iii) Find the perpendicular distance between the line $A B$ and the tangent parallel to $A B$. Give your answer correct to 2 decimal places.\\
\hfill \mbox{\textit{CAIE P1 2017 Q11 [10]}}