CAIE P3 2006 June — Question 9 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2006
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions with quadratic factor
DifficultyStandard +0.3 This is a standard two-part question combining partial fractions with binomial expansion. Part (i) is routine A-level technique with straightforward cover-up method. Part (ii) requires expanding two simple binomial terms and collecting coefficients, which is methodical but not conceptually demanding. The question is slightly easier than average due to its predictable structure and lack of any novel insight required.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1

  1. Express \(\frac{10}{(2-x)(1+x^2)}\) in partial fractions. [5]
  2. Hence, given that \(|x| < 1\), obtain the expansion of \(\frac{10}{(2-x)(1+x^2)}\) in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying the coefficients. [5]

(i)
AnswerMarks Guidance
State or imply partial fractions are of the form \(\frac{A}{2-x} + \frac{Bx+C}{1+x^2}\)B1
Use any relevant method to obtain a constantM1
Obtain one of the values \(A = 2, B = 2, C = 4\)A1
Obtain a second valueA1
Obtain the third valueA1 5
(ii)
AnswerMarks Guidance
Use correct method to obtain the first two terms of the expansion of \((2-x)^{-1}\) or \((1-\frac{1}{2}x)^{-1}\) or \((1+x^2)^{-1}\)M1
Obtain any correct unsimplified expansion of the partial fractions up to the terms in \(x^3\), e.g. \((2x + 4)(1)(1 + (-1)x^2)\) (deduct A1 for each incorrect expansion)A1♦ + A1♦
Carry out multiplication of expansion of \((1 + x^2)^{-1}\) by \((2x + 4)\)M1
Obtain answer \(5 + \frac{1}{2}x - \frac{13}{2}x^2 - \frac{1}{8}x^3\)A1 5
[Binomial coefficients involving \(-1\), e.g. \(\binom{-1}{1}\), are not sufficient for the M1 mark. The f.t. is on \(A, B, C\).]
[In the case of an attempt to expand \(10(2-z)^{-1}(1+x^2)^{-1}\), give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer (f.t. is on \(A, B, C\) if used).]
[Allow the use of Maclaurin, giving M1A1♦ for \(f(0) = 5\) and \(f'(0) = \frac{1}{2}\), A1♦ for \(f''(0) = -13\), A1♦ for \(f'''(0) = -\frac{45}{4}\), and A1 for obtaining the correct final answer (f.t. is on \(A, B, C\) if used).]
**(i)**

| State or imply partial fractions are of the form $\frac{A}{2-x} + \frac{Bx+C}{1+x^2}$ | B1 |
| Use any relevant method to obtain a constant | M1 |
| Obtain one of the values $A = 2, B = 2, C = 4$ | A1 |
| Obtain a second value | A1 |
| Obtain the third value | A1 | 5 |

**(ii)**

| Use correct method to obtain the first two terms of the expansion of $(2-x)^{-1}$ or $(1-\frac{1}{2}x)^{-1}$ or $(1+x^2)^{-1}$ | M1 |
| Obtain any correct unsimplified expansion of the partial fractions up to the terms in $x^3$, e.g. $(2x + 4)(1)(1 + (-1)x^2)$ (deduct A1 for each incorrect expansion) | A1♦ + A1♦ |
| Carry out multiplication of expansion of $(1 + x^2)^{-1}$ by $(2x + 4)$ | M1 |
| Obtain answer $5 + \frac{1}{2}x - \frac{13}{2}x^2 - \frac{1}{8}x^3$ | A1 | 5 |

[Binomial coefficients involving $-1$, e.g. $\binom{-1}{1}$, are not sufficient for the M1 mark. The f.t. is on $A, B, C$.] |

[In the case of an attempt to expand $10(2-z)^{-1}(1+x^2)^{-1}$, give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer (f.t. is on $A, B, C$ if used).] |

[Allow the use of Maclaurin, giving M1A1♦ for $f(0) = 5$ and $f'(0) = \frac{1}{2}$, A1♦ for $f''(0) = -13$, A1♦ for $f'''(0) = -\frac{45}{4}$, and A1 for obtaining the correct final answer (f.t. is on $A, B, C$ if used).] |
\begin{enumerate}[label=(\roman*)]
\item Express $\frac{10}{(2-x)(1+x^2)}$ in partial fractions. [5]

\item Hence, given that $|x| < 1$, obtain the expansion of $\frac{10}{(2-x)(1+x^2)}$ in ascending powers of $x$, up to and including the term in $x^3$, simplifying the coefficients. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2006 Q9 [10]}}