CAIE P2 2013 November — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2013
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind stationary points coordinates
DifficultyStandard +0.3 This is a straightforward application of the quotient rule to find dy/dx, setting it equal to zero, and solving for x. The algebra is clean (exponential factor cancels nicely), and finding the y-coordinate is direct substitution. Slightly above routine due to the exponential function, but still a standard textbook exercise with no conceptual challenges.
Spec1.06d Natural logarithm: ln(x) function and properties1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation

2 The curve \(y = \frac { \mathrm { e } ^ { 3 x - 1 } } { 2 x }\) has one stationary point. Find the coordinates of this stationary point.

AnswerMarks Guidance
Use quotient or product ruleM1
Obtain correct derivative in any formA1
Equate (numerator) of derivative to zero and solve for \(x\)DM1
Obtain \(x = \frac{1}{3}\)A1
Obtain \(y = \frac{3}{2}\)A1 [5]
Use quotient or product rule | M1 |
Obtain correct derivative in any form | A1 |
Equate (numerator) of derivative to zero and solve for $x$ | DM1 |
Obtain $x = \frac{1}{3}$ | A1 |
Obtain $y = \frac{3}{2}$ | A1 | [5]
2 The curve $y = \frac { \mathrm { e } ^ { 3 x - 1 } } { 2 x }$ has one stationary point. Find the coordinates of this stationary point.

\hfill \mbox{\textit{CAIE P2 2013 Q2 [5]}}