| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of normal |
| Difficulty | Standard +0.3 This is a straightforward application of differentiation (quotient rule), finding a normal line, and calculating a distance. It requires multiple steps but uses standard techniques: finding where the curve crosses the x-axis, differentiating to find the gradient, using the perpendicular gradient for the normal, and simple coordinate geometry. Slightly above average due to the multi-step nature and quotient rule differentiation, but all steps are routine for A-level. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) is \((3, 0)\) | B1 | Anywhere – but not from given answer |
| \(\frac{dy}{dx} = 18(2x+3)^{-2} \times 2\); if \(x=3\), \(m = \frac{4}{9}\) | B1 B1 | B1 for \(18(2x+3)^{-2}\), B1 for \(\times 2\) |
| \(m\) of normal \(= -\frac{9}{4}\) | M1 | Use of \(m_1 m_2 = -1\) with \(m\) from \(dy/dx\) |
| Equation of normal \(y = -\frac{9}{4}(x-3)\) | M1 | Correct method for normal |
| \(\rightarrow 4y + 9x = 27\) | A1 [6] | co (answer was given) |
| Answer | Marks | Guidance |
|---|---|---|
| Normal meets \(y\)-axis at \((0,\ 6\frac{3}{4})\); curve meets \(y\)-axis at \((0,\ -4)\); \(\rightarrow BC = 10\frac{3}{4}\) | M1 A1 [2] | Needs to put \(x=0\) in both normal and curve. co |
## Question 7:
$y = 2 - \frac{18}{2x+3}$
**Part (i):**
$A$ is $(3, 0)$ | B1 | Anywhere – but not from given answer
$\frac{dy}{dx} = 18(2x+3)^{-2} \times 2$; if $x=3$, $m = \frac{4}{9}$ | B1 B1 | B1 for $18(2x+3)^{-2}$, B1 for $\times 2$
$m$ of normal $= -\frac{9}{4}$ | M1 | Use of $m_1 m_2 = -1$ with $m$ from $dy/dx$
Equation of normal $y = -\frac{9}{4}(x-3)$ | M1 | Correct method for normal
$\rightarrow 4y + 9x = 27$ | A1 [6] | co (answer was given)
**Part (ii):**
Normal meets $y$-axis at $(0,\ 6\frac{3}{4})$; curve meets $y$-axis at $(0,\ -4)$; $\rightarrow BC = 10\frac{3}{4}$ | M1 A1 [2] | Needs to put $x=0$ in both normal and curve. co
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-3_766_589_251_778}
The diagram shows part of the curve $y = 2 - \frac { 18 } { 2 x + 3 }$, which crosses the $x$-axis at $A$ and the $y$-axis at $B$. The normal to the curve at $A$ crosses the $y$-axis at $C$.\\
(i) Show that the equation of the line $A C$ is $9 x + 4 y = 27$.\\
(ii) Find the length of $B C$.
\hfill \mbox{\textit{CAIE P1 2010 Q7 [8]}}