CAIE P1 2012 November — Question 5 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeVerify stationary point location
DifficultyModerate -0.3 This is a straightforward application of the quotient/chain rule to find dy/dx, substitute x=2 to verify dy/dx=0, then use the second derivative test. The algebra is simple with small numbers, and 'verify' means students know the answer in advance. Slightly easier than average due to the verification format removing problem-solving uncertainty.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation

5 A curve has equation \(y = 2 x + \frac { 1 } { ( x - 1 ) ^ { 2 } }\). Verify that the curve has a stationary point at \(x = 2\) and determine its nature.

AnswerMarks Guidance
\(\frac{dy}{dx} = 2 - 2(x-1)^{-3}\)B2,1,0 \(-1\) each error in 2, \(-2\), \((x-1)^{-3}\)
Sub \(x = 2 \to \frac{dy}{dx} = 2 - 2 = 0 \Rightarrow\) stat value at \(x = 2\)B1 AG
\(\frac{d^2y}{dx^2} = 6(x-1)^{-4}\) (and sub \(x = 2\))M1 Reasonable attempt to diff form \((x-1)^{-n}\)
(At \(x = 2, \frac{d^2y}{dx^2} = 6\)) \(> 0 \Rightarrow\) MinimumA1 Correct \(\frac{d^2y}{dx^2}\) and 'minimum' is required. Or other valid method for last 2 marks
$\frac{dy}{dx} = 2 - 2(x-1)^{-3}$ | B2,1,0 | $-1$ each error in 2, $-2$, $(x-1)^{-3}$
Sub $x = 2 \to \frac{dy}{dx} = 2 - 2 = 0 \Rightarrow$ stat value at $x = 2$ | B1 | AG
$\frac{d^2y}{dx^2} = 6(x-1)^{-4}$ (and sub $x = 2$) | M1 | Reasonable attempt to diff form $(x-1)^{-n}$
(At $x = 2, \frac{d^2y}{dx^2} = 6$) $> 0 \Rightarrow$ Minimum | A1 | Correct $\frac{d^2y}{dx^2}$ and 'minimum' is required. Or other valid method for last 2 marks | [5]
5 A curve has equation $y = 2 x + \frac { 1 } { ( x - 1 ) ^ { 2 } }$. Verify that the curve has a stationary point at $x = 2$ and determine its nature.

\hfill \mbox{\textit{CAIE P1 2012 Q5 [5]}}