| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Tangent to curve at given point |
| Difficulty | Moderate -0.3 This is a standard A-level tangent/normal question requiring differentiation using chain rule, finding equations of lines, distance formula, and intersection of lines. While it has multiple parts and requires careful calculation, all techniques are routine P1 content with no novel problem-solving required. Slightly easier than average due to straightforward application of standard methods. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = [6] \times [\frac{1}{3}(6x+2)^{-\frac{2}{3}}]\) | B1B1 | Independent |
| Equation of tangent is \(y - 2 = m(x-1)\) | M1 | Where \(m = \) numerical \(\frac{dy}{dx}\) |
| Equation of normal is \(y - 2 = -\frac{1}{m}(x-1)\) | M1 | Including use of \(m_1m_2 = -1\) |
| Both eqns correct with \(m = \frac{1}{2}\) cao | A1 | SC 1/3 Blatant tangent/normal reversal |
| (ii) \(B = (0, 1\frac{1}{2})\); \(C = (2, 0)\) | B1 | Both cao |
| \(BC = \sqrt{2^2 + (1\frac{1}{2})^2} = 2\frac{1}{2}\) | M1A1 | It from their \(B\) and \(C\) |
| (iii) \(BC: y - 1\frac{1}{2} = -\frac{3}{4}(x - 0)\) or \(y = -\frac{3}{4}(x-2)\) | M1 | or \(y = -\frac{3}{4}x + 1\frac{1}{2}\) |
| Intersection \((E): -\frac{3}{4}x + 1\frac{1}{2} = 2x\) | A1 | cao |
| \(x = \frac{6}{11}\), \(y = \frac{12}{11}\) | A1 | Dependent on correct \(x\) values or \(y\) values for both \(E\) and the mid-point of \(OA\) |
| Mid-point of \(OA = (\frac{1}{2}, 1) \to E\) not mid-point | B1 | [4] |
**(i)** $\frac{dy}{dx} = [6] \times [\frac{1}{3}(6x+2)^{-\frac{2}{3}}]$ | B1B1 | Independent
Equation of tangent is $y - 2 = m(x-1)$ | M1 | Where $m = $ numerical $\frac{dy}{dx}$
Equation of normal is $y - 2 = -\frac{1}{m}(x-1)$ | M1 | Including use of $m_1m_2 = -1$
Both eqns correct with $m = \frac{1}{2}$ cao | A1 | SC 1/3 Blatant tangent/normal reversal | [5]
**(ii)** $B = (0, 1\frac{1}{2})$; $C = (2, 0)$ | B1 | Both cao
$BC = \sqrt{2^2 + (1\frac{1}{2})^2} = 2\frac{1}{2}$ | M1A1 | It from their $B$ and $C$ | [3]
**(iii)** $BC: y - 1\frac{1}{2} = -\frac{3}{4}(x - 0)$ or $y = -\frac{3}{4}(x-2)$ | M1 | or $y = -\frac{3}{4}x + 1\frac{1}{2}$
Intersection $(E): -\frac{3}{4}x + 1\frac{1}{2} = 2x$ | A1 | cao
$x = \frac{6}{11}$, $y = \frac{12}{11}$ | A1 | Dependent on correct $x$ values or $y$ values for both $E$ and the mid-point of $OA$
Mid-point of $OA = (\frac{1}{2}, 1) \to E$ not mid-point | B1 | [4]11\\
\includegraphics[max width=\textwidth, alt={}, center]{e69332d0-2e45-4a86-a1f9-5d83bca1ad9b-4_885_967_255_589}
The diagram shows the curve $y = ( 6 x + 2 ) ^ { \frac { 1 } { 3 } }$ and the point $A ( 1,2 )$ which lies on the curve. The tangent to the curve at $A$ cuts the $y$-axis at $B$ and the normal to the curve at $A$ cuts the $x$-axis at $C$.\\
(i) Find the equation of the tangent $A B$ and the equation of the normal $A C$.\\
(ii) Find the distance $B C$.\\
(iii) Find the coordinates of the point of intersection, $E$, of $O A$ and $B C$, and determine whether $E$ is the mid-point of $O A$.
\hfill \mbox{\textit{CAIE P1 2012 Q11 [12]}}