CAIE P1 2013 June — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeTangent with given gradient/property
DifficultyModerate -0.3 This is a straightforward tangent problem requiring differentiation of a simple rational function, equating the derivative to m, and solving simultaneous equations. While it involves multiple steps (differentiate, find gradient, use tangent equation, solve for coordinates), each step is routine and the question type is standard practice for AS-level. Slightly easier than average due to the simple function and clear structure.
Spec1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations
3 The straight line \(y = m x + 14\) is a tangent to the curve \(y = \frac { 12 } { x } + 2\) at the point \(P\). Find the value of the constant \(m\) and the coordinates of \(P\).
\(mx + 14 = \frac{12}{x} + 2 \rightarrow mx^2 + 12x - 12 = 0\)
Uses \(b^2 = 4ac\) or \(m = -3\)
AnswerMarks Guidance
\(-3x^2 + 12x - 12 = 0 \rightarrow P(2, 8)\)M1 M1 A1 DM1 A1 Eliminates \(x\) (or \(y\)). Any use of discriminant. Any valid method
[Or \(m = -12x^{-2}\) M1 Sub M1 \(x = 2\) A1] → [\(m = -3\) and \(y = 8\) M1 A1][5] 
$mx + 14 = \frac{12}{x} + 2 \rightarrow mx^2 + 12x - 12 = 0$

Uses $b^2 = 4ac$ or $m = -3$

$-3x^2 + 12x - 12 = 0 \rightarrow P(2, 8)$ | M1 M1 A1 DM1 A1 | Eliminates $x$ (or $y$). Any use of discriminant. Any valid method

[Or $m = -12x^{-2}$ M1 Sub M1 $x = 2$ A1] → [$m = -3$ and $y = 8$ M1 A1] | [5] |

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3 The straight line $y = m x + 14$ is a tangent to the curve $y = \frac { 12 } { x } + 2$ at the point $P$. Find the value of the constant $m$ and the coordinates of $P$.

\hfill \mbox{\textit{CAIE P1 2013 Q3 [5]}}
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