CAIE P1 2016 June — Question 8 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypePerpendicular bisector of segment
DifficultyStandard +0.8 This question requires multiple connected steps: finding the gradient of AB, differentiating to find dy/dx, solving a quadratic equation to find points C and D where the tangent is parallel to AB, finding the midpoint of CD, and finally determining the perpendicular bisector equation. While each individual step uses standard techniques (differentiation, coordinate geometry), the multi-stage problem-solving and coordination of several concepts makes this moderately challenging, above average difficulty but not requiring exceptional insight.
Spec1.07m Tangents and normals: gradient and equations
8 A curve has equation \(y = 3 x - \frac { 4 } { x }\) and passes through the points \(A ( 1 , - 1 )\) and \(B ( 4,11 )\). At each of the points \(C\) and \(D\) on the curve, the tangent is parallel to \(A B\). Find the equation of the perpendicular bisector of \(C D\).
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y = 3x - \frac{4}{x}\); \(\frac{dy}{dx} = 3 + \frac{4}{x^2}\)B1
\(m\) of \(AB = 4\)B1
Equate \(\rightarrow x = \pm 2\); \(\rightarrow C(2,4)\) and \(D(-2,-4)\)M1 A1 Equating + solution
\(\rightarrow M(0,0)\) or stating \(M\) is the origin; \(m\) of \(CD = 2\)B1\(\checkmark\) \(\checkmark\) on their \(C\) and \(D\)
Perpendicular gradient \(= -\frac{1}{2}\)M1 A1 Use of \(m_1m_2 = -1\), must use \(m_{CD}\) (not \(m = 4\))
\(\rightarrow y = -\frac{1}{2}x\) 
## Question 8:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 3x - \frac{4}{x}$; $\frac{dy}{dx} = 3 + \frac{4}{x^2}$ | **B1** | |
| $m$ of $AB = 4$ | **B1** | |
| Equate $\rightarrow x = \pm 2$; $\rightarrow C(2,4)$ and $D(-2,-4)$ | **M1 A1** | Equating + solution |
| $\rightarrow M(0,0)$ or stating $M$ is the origin; $m$ of $CD = 2$ | **B1$\checkmark$** | $\checkmark$ on their $C$ and $D$ |
| Perpendicular gradient $= -\frac{1}{2}$ | **M1 A1** | Use of $m_1m_2 = -1$, must use $m_{CD}$ (not $m = 4$) |
| $\rightarrow y = -\frac{1}{2}x$ | | |
8 A curve has equation $y = 3 x - \frac { 4 } { x }$ and passes through the points $A ( 1 , - 1 )$ and $B ( 4,11 )$. At each of the points $C$ and $D$ on the curve, the tangent is parallel to $A B$. Find the equation of the perpendicular bisector of $C D$.

\hfill \mbox{\textit{CAIE P1 2016 Q8 [7]}}
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