CAIE P3 2018 November — Question 9 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypePartial fractions with repeated linear factor
DifficultyStandard +0.3 This is a standard partial fractions question with a repeated linear factor, requiring routine algebraic manipulation to find constants A, B, C, followed by straightforward integration of logarithmic and rational terms. While it involves multiple steps and careful arithmetic, it follows a well-practiced algorithm with no novel insight required, making it slightly easier than average.
Spec1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08j Integration using partial fractions

Let \(f(x) = \frac{6x^2 + 8x + 9}{(2 - x)(3 + 2x)^2}\).
  1. Express \(f(x)\) in partial fractions. [5]
  2. Hence, showing all necessary working, show that \(\int_{-1}^0 f(x) dx = 1 + \frac{1}{2}\ln\left(\frac{4}{3}\right)\). [5]

Question 9:

AnswerMarks
9(i)A B C
State or imply the form + +
AnswerMarks
2−x 3+2x ( 3+2x )2B1
Use a correct method to find a constantM1
Obtain one of A = 1, B = – 1, C = 3A1
Obtain a second valueA1
Obtain the third value
A Dx+E
[Mark the form + , where A = 1, D = – 2 and
2−x ( 3+2x )2
AnswerMarks
E = 0, B1M1A1A1A1 as above.]A1
5

AnswerMarks
9(ii)Integrate and obtain terms
−ln ( 2−x ) − 1 ln ( 3+2x ) – 3
AnswerMarks Guidance
2 2 ( 3+2x )B3ft The f.t is on A, B, C; or on A, D, E.
Substitute correctly in an integral with terms a ln (2 – x),
AnswerMarks
b ln (3 + 2x) and c / (3 + 2x) where abc ≠ 0M1
Obtain the given answer after full and correct working
[Correct integration of the A, D, E form gives an extra constant term if integration by
AnswerMarks
parts is used for the second partial fraction.]A1
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 9:
--- 9(i) ---
9(i) | A B C
State or imply the form + +
2−x 3+2x ( 3+2x )2 | B1
Use a correct method to find a constant | M1
Obtain one of A = 1, B = – 1, C = 3 | A1
Obtain a second value | A1
Obtain the third value
A Dx+E
[Mark the form + , where A = 1, D = – 2 and
2−x ( 3+2x )2
E = 0, B1M1A1A1A1 as above.] | A1
5
--- 9(ii) ---
9(ii) | Integrate and obtain terms
−ln ( 2−x ) − 1 ln ( 3+2x ) – 3
2 2 ( 3+2x ) | B3ft | The f.t is on A, B, C; or on A, D, E.
Substitute correctly in an integral with terms a ln (2 – x),
b ln (3 + 2x) and c / (3 + 2x) where abc ≠ 0 | M1
Obtain the given answer after full and correct working
[Correct integration of the A, D, E form gives an extra constant term if integration by
parts is used for the second partial fraction.] | A1
5
Question | Answer | Marks | Guidance
Let $f(x) = \frac{6x^2 + 8x + 9}{(2 - x)(3 + 2x)^2}$.

\begin{enumerate}[label=(\roman*)]
\item Express $f(x)$ in partial fractions. [5]

\item Hence, showing all necessary working, show that $\int_{-1}^0 f(x) dx = 1 + \frac{1}{2}\ln\left(\frac{4}{3}\right)$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2018 Q9 [10]}}