CAIE P2 2020 June — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind stationary points - polynomial/exponential products
DifficultyModerate -0.3 This is a straightforward application of the product rule to differentiate xe^(x/2), followed by solving a simple equation to find where dy/dx = 0. The algebra is clean and the exponential never equals zero, making this slightly easier than an average A-level question but still requiring proper technique.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation
2 Find the exact coordinates of the stationary point on the curve with equation \(y = 5 x \mathrm { e } ^ { \frac { 1 } { 2 } x }\).
Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Differentiate using product rule to obtain \(ae^{\frac{1}{2}x} + bxe^{\frac{1}{2}x}\)\*M1
Obtain correct \(5e^{\frac{1}{2}x} + \dfrac{5}{2}xe^{\frac{1}{2}x}\)A1 OE
Equate first derivative to zero and solve for \(x\)DM1
Obtain \(x\)-coordinate \(-2\)A1
Obtain \(y\)-coordinate \(-10e^{-1}\)A1
Total5 
**Question 2:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate using product rule to obtain $ae^{\frac{1}{2}x} + bxe^{\frac{1}{2}x}$ | \*M1 | |
| Obtain correct $5e^{\frac{1}{2}x} + \dfrac{5}{2}xe^{\frac{1}{2}x}$ | A1 | OE |
| Equate first derivative to zero and solve for $x$ | DM1 | |
| Obtain $x$-coordinate $-2$ | A1 | |
| Obtain $y$-coordinate $-10e^{-1}$ | A1 | |
| **Total** | **5** | |
2 Find the exact coordinates of the stationary point on the curve with equation $y = 5 x \mathrm { e } ^ { \frac { 1 } { 2 } x }$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q2 [5]}}
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Q1 3 Q2 5 Q3 5 Q4 5 Q5 5 Q6 8 Q7 9 Q8 10