CAIE P3 2005 November — Question 9 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2005
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions then binomial expansion
DifficultyStandard +0.3 This is a standard two-part question combining routine partial fractions decomposition with binomial expansion. Part (i) requires setting up and solving for constants A, B, C in the form A/(x+2) + (Bx+C)/(x²+1), which is textbook procedure. Part (ii) involves expanding each fraction using (1+u)^(-1) ≈ 1-u+u²-u³, then collecting terms—mechanical but requiring care with algebra. Slightly above average difficulty due to the quadratic factor and multi-step nature, but no novel insight required.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1

9
  1. Express \(\frac { 3 x ^ { 2 } + x } { ( x + 2 ) \left( x ^ { 2 } + 1 \right) }\) in partial fractions.
  2. Hence obtain the expansion of \(\frac { 3 x ^ { 2 } + x } { ( x + 2 ) \left( x ^ { 2 } + 1 \right) }\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).

(i)
AnswerMarks
State or imply partial fractions are of the form \(\frac{A}{x+2} + \frac{Bx+C}{x^2+1}\)B1
Use any relevant method to obtain a constantM1
Obtain \(A = 2\)A1
Obtain \(B = 1\)A1
Obtain \(C = -1\)A1
Total for (i): [5]
(ii)
AnswerMarks
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 complete unsimplified expansions of the fractions, e.g. \(2.\frac{1}{2}(1-\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^3)\); \((x-1)(1-x^2)\)A1√ + A1√
Carry out multiplication of expansion of \((1+x^2)^{-1}\) by \((x-1)\)M1(dep*)
Obtain answer \(\frac{1}{2}x + \frac{3}{4}x^2 - \frac{3}{8}x^3\)A1
[Binomial coefficients involving \(-1\), such as \(\binom{-1}{1}\), are not sufficient for the first M1.]
[f.t. is on \(A, B, C\).]
[Apply this scheme to attempts to expand \((3x^2 + x)(x + 2)^{-1}(1 + x^2)^{-1}\), giving M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.]
Total for (ii): [5]
**(i)**
State or imply partial fractions are of the form $\frac{A}{x+2} + \frac{Bx+C}{x^2+1}$ | B1 |
Use any relevant method to obtain a constant | M1 |
Obtain $A = 2$ | A1 |
Obtain $B = 1$ | A1 |
Obtain $C = -1$ | A1 |

**Total for (i): [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 complete unsimplified expansions of the fractions, e.g. $2.\frac{1}{2}(1-\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^3)$; $(x-1)(1-x^2)$ | A1√ + A1√ |
Carry out multiplication of expansion of $(1+x^2)^{-1}$ by $(x-1)$ | M1(dep*) |
Obtain answer $\frac{1}{2}x + \frac{3}{4}x^2 - \frac{3}{8}x^3$ | A1 |

[Binomial coefficients involving $-1$, such as $\binom{-1}{1}$, are not sufficient for the first M1.]

[f.t. is on $A, B, C$.]

[Apply this scheme to attempts to expand $(3x^2 + x)(x + 2)^{-1}(1 + x^2)^{-1}$, giving M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.]

**Total for (ii): [5]**

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9 (i) Express $\frac { 3 x ^ { 2 } + x } { ( x + 2 ) \left( x ^ { 2 } + 1 \right) }$ in partial fractions.\\
(ii) Hence obtain the expansion of $\frac { 3 x ^ { 2 } + x } { ( x + 2 ) \left( x ^ { 2 } + 1 \right) }$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.

\hfill \mbox{\textit{CAIE P3 2005 Q9 [10]}}