CAIE P2 2020 March — Question 3 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionMarch
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind constant from definite integral
DifficultyStandard +0.3 This is a straightforward reverse chain rule integration problem requiring students to integrate 1/(2x-5), apply limits, use logarithm properties to simplify, then solve a simple equation for a. While it involves multiple steps (integration, substitution of limits, log manipulation, solving), each step is routine and the question follows a standard template for this topic with no conceptual challenges.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)
3 It is given that \(\int _ { a } ^ { 3 a } \frac { 2 } { 2 x - 5 } \mathrm {~d} x = \ln \frac { 7 } { 2 }\).
Find the value of the positive constant \(a\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Integrate to obtain \(k\ln(2x-5)\)\*M1 For non-zero constant \(k\)
Apply limits to obtain \(\ln(6a-5) - \ln(2a-5) = \ln\frac{7}{2}\)A1
Apply subtraction law for logarithms\*M1 OE
Obtain equation \(\frac{6a-5}{2a-5} = \frac{7}{2}\)A1 OE without logarithms
Solve equation for \(a\)DM1
Obtain \(a = \frac{25}{2}\)A1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain $k\ln(2x-5)$ | \*M1 | For non-zero constant $k$ |
| Apply limits to obtain $\ln(6a-5) - \ln(2a-5) = \ln\frac{7}{2}$ | A1 | |
| Apply subtraction law for logarithms | \*M1 | OE |
| Obtain equation $\frac{6a-5}{2a-5} = \frac{7}{2}$ | A1 | OE without logarithms |
| Solve equation for $a$ | DM1 | |
| Obtain $a = \frac{25}{2}$ | A1 | |

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3 It is given that $\int _ { a } ^ { 3 a } \frac { 2 } { 2 x - 5 } \mathrm {~d} x = \ln \frac { 7 } { 2 }$.\\
Find the value of the positive constant $a$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q3 [6]}}
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