| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| 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 using the quotient rule (or chain rule), finding the gradient of the normal, and solving a linear equation for part (i), followed by standard stationary point analysis in part (ii). While it requires multiple steps, each step is routine and follows standard procedures taught in P1 with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dy}{dx} = -(x-1)^{-2} + 9(x-5)^{-2}\) | M1A1 | May be seen in part (ii) |
| \(m_{\text{tangent}} = -\dfrac{1}{4} + \dfrac{9}{4} = 2\) | B1 | |
| Equation of normal is \(y - 5 = -\frac{1}{2}(x-3)\) | M1 | Through \((3,5)\) and with \(m = -1/m_{\text{tangent}}\) |
| \(x = 13\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-5)^2 = 9(x-1)^2\) | B1 | Set \(\dfrac{dy}{dx} = 0\) and simplify |
| \(x - 5 = (\pm)3(x-1)\) or \((8)(x^2 - x - 2) = 0\) | M1 | Simplify further and attempt solution |
| \(x = -1\) or \(2\) | A1 | |
| \(\dfrac{d^2y}{dx^2} = 2(x-1)^{-3} - 18(x-5)^{-3}\) | B1 | If change of sign used, \(x\) values close to the roots must be used and all must be correct |
| When \(x = -1\), \(\dfrac{d^2y}{dx^2} = -\dfrac{1}{6} < 0\) MAX | B1 | |
| When \(x = 2\), \(\dfrac{d^2y}{dx^2} = \dfrac{8}{3} > 0\) MIN | B1 [6] |
## Question 11:
### Part (i):
$\dfrac{dy}{dx} = -(x-1)^{-2} + 9(x-5)^{-2}$ | **M1A1** | May be seen in part (ii)
$m_{\text{tangent}} = -\dfrac{1}{4} + \dfrac{9}{4} = 2$ | **B1** |
Equation of normal is $y - 5 = -\frac{1}{2}(x-3)$ | **M1** | Through $(3,5)$ and with $m = -1/m_{\text{tangent}}$
$x = 13$ | **A1** [5] |
### Part (ii):
$(x-5)^2 = 9(x-1)^2$ | **B1** | Set $\dfrac{dy}{dx} = 0$ and simplify
$x - 5 = (\pm)3(x-1)$ or $(8)(x^2 - x - 2) = 0$ | **M1** | Simplify further and attempt solution
$x = -1$ or $2$ | **A1** |
$\dfrac{d^2y}{dx^2} = 2(x-1)^{-3} - 18(x-5)^{-3}$ | **B1** | If change of sign used, $x$ values close to the roots must be used and all must be correct
When $x = -1$, $\dfrac{d^2y}{dx^2} = -\dfrac{1}{6} < 0$ MAX | **B1** |
When $x = 2$, $\dfrac{d^2y}{dx^2} = \dfrac{8}{3} > 0$ MIN | **B1** [6] |11 The point $P ( 3,5 )$ lies on the curve $y = \frac { 1 } { x - 1 } - \frac { 9 } { x - 5 }$.\\
(i) Find the $x$-coordinate of the point where the normal to the curve at $P$ intersects the $x$-axis.\\
(ii) Find the $x$-coordinate of each of the stationary points on the curve and determine the nature of each stationary point, justifying your answers.
{www.cie.org.uk} after the live examination series.
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\hfill \mbox{\textit{CAIE P1 2016 Q11 [11]}}