CAIE P1 2005 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2005
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind gradient at point
DifficultyModerate -0.5 This is a straightforward differentiation question requiring the quotient rule (or rewriting as a product) followed by substitution. It's slightly easier than average because it's a single-step application with no algebraic complications, though the quotient rule itself requires care with signs and simplification.
Spec1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
2 Find the gradient of the curve \(y = \frac { 12 } { x ^ { 2 } - 4 x }\) at the point where \(x = 3\).
\(y = \frac{12}{x^2 - 4x}\)
AnswerMarks Guidance
\(\frac{dy}{dx} = -12(x^2 - 4x)^{-2} \times (2x - 4)\)B1 M1 A1 \(\checkmark\) \(-12(x^2 - 4x)^{-2}\) correct. Use of chain rule. \(\checkmark\) for B0 attempts. Quotient or product rule ok (M1A2,1)
If \(x = 3\), \(\frac{dy}{dx} = \frac{8}{3}\)A1 [4] CAO Uncancelled ok. 
$y = \frac{12}{x^2 - 4x}$

$\frac{dy}{dx} = -12(x^2 - 4x)^{-2} \times (2x - 4)$ | B1 M1 A1 $\checkmark$ | $-12(x^2 - 4x)^{-2}$ correct. Use of chain rule. $\checkmark$ for B0 attempts. Quotient or product rule ok (M1A2,1)

If $x = 3$, $\frac{dy}{dx} = \frac{8}{3}$ | A1 [4] | CAO Uncancelled ok.
2 Find the gradient of the curve $y = \frac { 12 } { x ^ { 2 } - 4 x }$ at the point where $x = 3$.

\hfill \mbox{\textit{CAIE P1 2005 Q2 [4]}}
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