CAIE P1 2006 June — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2006
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypePoints with specific gradient
DifficultyEasy -1.8 This is a straightforward differentiation and substitution problem requiring only basic knowledge that d/dx(k/x) = -k/x². Students simply differentiate, substitute x=2 and gradient=-3, then solve for k. It's simpler than typical A-level questions as it involves a single-step differentiation of a basic function with immediate substitution.
Spec1.07l Derivative of ln(x): and related functions
1 A curve has equation \(y = \frac { k } { x }\). Given that the gradient of the curve is - 3 when \(x = 2\), find the value of the constant \(k\).
AnswerMarks Guidance
\(\frac{dy}{dx} = -kx^{-2}\)B1 Negative power ok.
Subs \(x = 2, m = -3\) into \(\frac{dy}{dx}\)M1 Subs \(x=2\) into his dy/dx.
\(k = 12\)A1 co.
Total: [3]
$\frac{dy}{dx} = -kx^{-2}$ | B1 | Negative power ok.
Subs $x = 2, m = -3$ into $\frac{dy}{dx}$ | M1 | Subs $x=2$ into his dy/dx.
$k = 12$ | A1 | co.

**Total: [3]**

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1 A curve has equation $y = \frac { k } { x }$. Given that the gradient of the curve is - 3 when $x = 2$, find the value of the constant $k$.

\hfill \mbox{\textit{CAIE P1 2006 Q1 [3]}}
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