Question 2 - A-Level Maths

CAIE P2 2018 November — Question 2 5 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2018
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Standard Integrals and Reverse Chain Rule
Type	Definite integral with logarithmic form
Difficulty	Moderate -0.8 This is a straightforward application of the standard integral ∫(f'(x)/f(x))dx = ln\|f(x)\| + c. Students need only recognize that 6/(2x+1) can be written as 3·(2/(2x+1)), evaluate [3ln(2x+1)] from 1 to 7, and simplify 3(ln15 - ln3) = 3ln5 = ln125. It's simpler than average because it requires just one standard technique with minimal algebraic manipulation.
Spec	1.06f Laws of logarithms: addition, subtraction, power rules 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

2 Show that \(\int _ { 1 } ^ { 7 } \frac { 6 } { 2 x + 1 } \mathrm {~d} x = \ln 125\).

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Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
Integrate to obtain form \(k\ln(2x+1)\)	M1
Obtain correct \(3\ln(2x+1)\)	A1
Use subtraction law of logarithms correctly	M1	Dependent on first M1
Use power law of logarithms correctly	M1	Dependent on first M1
Confirm \(\ln 125\)	A1
Total: 5

## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain form $k\ln(2x+1)$ | M1 | |
| Obtain correct $3\ln(2x+1)$ | A1 | |
| Use subtraction law of logarithms correctly | M1 | Dependent on first M1 |
| Use power law of logarithms correctly | M1 | Dependent on first M1 |
| Confirm $\ln 125$ | A1 | |
| **Total: 5** | | |

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2 Show that $\int _ { 1 } ^ { 7 } \frac { 6 } { 2 x + 1 } \mathrm {~d} x = \ln 125$.\\

\hfill \mbox{\textit{CAIE P2 2018 Q2 [5]}}

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Q1 5 Q2 5 Q3 5 Q4 8 Q5 9 Q6 8 Q7 10