| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Normal to curve at given point |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on normals to curves. Part (i) requires finding dy/dx, evaluating at x=3, finding the perpendicular gradient, and verifying the given equation—all routine steps. Parts (ii) and (iii) involve basic coordinate geometry (finding intercepts and their midpoint) and solving a simultaneous equation (linear normal with quadratic curve). While it has multiple parts worth several marks, each step uses standard AS-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 4 - 2x\); At \(x=3,\ m=-2\) | \(B1\) | co |
| Gradient of normal \(= \frac{1}{2}\) | \(M1\) | Use of \(m_1 m_2 = -1\) |
| Equation of normal: \(y - 6 = \frac{1}{2}(x-3)\) | \(M1\ A1\) | Use of \(y-k = m(x-h)\) or \(y=mx+c\) (where \(m\) is gradient of normal) |
| \(\rightarrow 2y = x + 9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Meets axes at \(\left(0,\frac{9}{2}\right)\) and \((-9, 0)\) | \(M1\) | Sets \(x\) and \(y\) to \(0\) + midpoint formula |
| Mid-point is \(\left(\frac{-9}{2}, \frac{9}{4}\right)\) | \(A1\) | co |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2y = x+9,\ y = 4x - x^2 + 3\) | ||
| \(\rightarrow 2x^2 - 7x + 3 = 0\) oe | \(M1\ A1\) | Eliminates \(x\) completely. Correct equation. |
| \(\rightarrow \left(\frac{1}{2},\ 4\frac{3}{4}\right)\) | \(M1\ A1\) | Solution of quadratic. co |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4 - 2x$; At $x=3,\ m=-2$ | $B1$ | co |
| Gradient of normal $= \frac{1}{2}$ | $M1$ | Use of $m_1 m_2 = -1$ |
| Equation of normal: $y - 6 = \frac{1}{2}(x-3)$ | $M1\ A1$ | Use of $y-k = m(x-h)$ or $y=mx+c$ (where $m$ is gradient of normal) |
| $\rightarrow 2y = x + 9$ | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Meets axes at $\left(0,\frac{9}{2}\right)$ and $(-9, 0)$ | $M1$ | Sets $x$ and $y$ to $0$ + midpoint formula |
| Mid-point is $\left(\frac{-9}{2}, \frac{9}{4}\right)$ | $A1$ | co |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2y = x+9,\ y = 4x - x^2 + 3$ | | |
| $\rightarrow 2x^2 - 7x + 3 = 0$ oe | $M1\ A1$ | Eliminates $x$ completely. Correct equation. |
| $\rightarrow \left(\frac{1}{2},\ 4\frac{3}{4}\right)$ | $M1\ A1$ | Solution of quadratic. co |10 The equation of a curve is $y = 3 + 4 x - x ^ { 2 }$.\\
(i) Show that the equation of the normal to the curve at the point $( 3,6 )$ is $2 y = x + 9$.\\
(ii) Given that the normal meets the coordinate axes at points $A$ and $B$, find the coordinates of the mid-point of $A B$.\\
(iii) Find the coordinates of the point at which the normal meets the curve again.
\hfill \mbox{\textit{CAIE P1 2010 Q10 [10]}}