CAIE P3 2020 November — Question 10 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeIntegration with differentiation context
DifficultyStandard +0.3 Part (a) requires product rule differentiation and solving for a stationary point—standard A-level calculus. Part (b) involves integration by parts of an exponential-polynomial product, which is a routine technique at this level. Both parts follow predictable methods with no novel insight required, making this slightly easier than average.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08i Integration by parts

10 \includegraphics[max width=\textwidth, alt={}, center]{5f80ae11-34c3-4d2f-89f8-71b4ac021c7d-16_426_908_262_616} The diagram shows the curve \(y = ( 2 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x }\), and its minimum point \(M\).
  1. Find the exact coordinates of \(M\).
  2. Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e.

Question 10(a):
AnswerMarks Guidance
AnswerMark Guidance
Use correct product or quotient rule\*M1 \(\frac{dy}{dx} = \left(-\frac{1}{2}\right)(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x}\); M1 requires at least one of derivatives correct
Obtain correct derivative in any formA1
Equate derivative to zero and solve for \(x\)DM1
Obtain \(x = 4\)A1 ISW
Obtain \(y = -2e^{-2}\), or exact equivalentA1
Total5
Question 10(b):
AnswerMarks Guidance
AnswerMark Guidance
Commence integration and reach \(a(2-x)e^{-\frac{1}{2}x} + b\int e^{-\frac{1}{2}x}\,dx\)\*M1 Condone omission of \(dx\); \(-2(2-x)e^{-\frac{1}{2}x} + 4e^{-\frac{1}{2}x}\) or \(2xe^{-\frac{1}{2}x}\)
Obtain \(-2(2-x)e^{-\frac{1}{2}x} - 2\int e^{-\frac{1}{2}x}\,dx\)A1 OE
Complete integration and obtain \(2xe^{-\frac{1}{2}x}\)A1 OE
Use correct limits \(x=0\) and \(x=2\), correctly, having integrated twiceDM1 Ignore omission of zeros and allow max of 1 error
Obtain answer \(4e^{-1}\), or exact equivalentA1 ISW
Total5
Alternative method for 10(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{d\left(2xe^{-\frac{1}{2}x}\right)}{dx} = 2e^{-\frac{1}{2}x} - xe^{-\frac{1}{2}x}\)\*M1 A1
\(\therefore 2xe^{-\frac{1}{2}x}\)A1
Use correct limits \(x=0\) and \(x=2\), correctly, having integrated twiceDM1 Ignore omission of zeros and allow max of 1 error
Obtain answer \(4e^{-1}\), or exact equivalentA1 ISW
Total5
## Question 10(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product or quotient rule | \*M1 | $\frac{dy}{dx} = \left(-\frac{1}{2}\right)(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x}$; M1 requires at least one of derivatives correct |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for $x$ | DM1 | |
| Obtain $x = 4$ | A1 | ISW |
| Obtain $y = -2e^{-2}$, or exact equivalent | A1 | |
| **Total** | **5** | |

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## Question 10(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $a(2-x)e^{-\frac{1}{2}x} + b\int e^{-\frac{1}{2}x}\,dx$ | \*M1 | Condone omission of $dx$; $-2(2-x)e^{-\frac{1}{2}x} + 4e^{-\frac{1}{2}x}$ or $2xe^{-\frac{1}{2}x}$ |
| Obtain $-2(2-x)e^{-\frac{1}{2}x} - 2\int e^{-\frac{1}{2}x}\,dx$ | A1 | OE |
| Complete integration and obtain $2xe^{-\frac{1}{2}x}$ | A1 | OE |
| Use correct limits $x=0$ and $x=2$, correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer $4e^{-1}$, or exact equivalent | A1 | ISW |
| **Total** | **5** | |

**Alternative method for 10(b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d\left(2xe^{-\frac{1}{2}x}\right)}{dx} = 2e^{-\frac{1}{2}x} - xe^{-\frac{1}{2}x}$ | \*M1 A1 | |
| $\therefore 2xe^{-\frac{1}{2}x}$ | A1 | |
| Use correct limits $x=0$ and $x=2$, correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer $4e^{-1}$, or exact equivalent | A1 | ISW |
| **Total** | **5** | |

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10\\
\includegraphics[max width=\textwidth, alt={}, center]{5f80ae11-34c3-4d2f-89f8-71b4ac021c7d-16_426_908_262_616}

The diagram shows the curve $y = ( 2 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x }$, and its minimum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e.
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2020 Q10 [10]}}