Question 5 - A-Level Maths

Edexcel C34 2018 October — Question 5 10 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2018
Session	October
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration with Partial Fractions
Type	Partial fractions with repeated linear factor
Difficulty	Standard +0.3 This is a standard partial fractions question with a repeated linear factor, followed by routine integration and definite integral evaluation. The decomposition form is given, finding constants uses cover-up/substitution methods, and integration of each term is straightforward. Slightly easier than average due to the scaffolding provided and mechanical nature of all steps.
Spec	1.02y Partial fractions: decompose rational functions 1.08j Integration using partial fractions

5. $$f ( x ) = \frac { 4 x ^ { 2 } + 5 x + 3 } { ( x + 2 ) ( 1 - x ) ^ { 2 } } \equiv \frac { A } { ( x + 2 ) } + \frac { B } { ( 1 - x ) } + \frac { C } { ( 1 - x ) ^ { 2 } }$$

Find the values of the constants $A$, $B$ and $C$.
1. Hence find $\int \mathrm { f } ( x ) \mathrm { d } x$.
2. Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$, writing your answer in the form $p + \ln q$, where $p$ and $q$ are constants.
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  VGHV SIHIN NI III M I ION OC VIIV SIHI NI JIIIM ION OC VI4V SIHIL NI JIIYM ION OC

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Question 5:

Part (a):

Answer	Marks	Guidance
$4x^2+5x+3 = A(1-x)^2 + B(x+2)(1-x) + C(x+2)$	B1	May be implied by sight of equivalent fractions or correct constants
Sub $x=1$ or $x=-2$, attempts to find one constant	M1	Can be scored after B0; alternatively equate coefficients
$x=1 \Rightarrow C=4$; $x=-2 \Rightarrow A=1$; any two constants correct	A1
Coefficients of $x^2$: $4 = A-B \Rightarrow B=-3$; all three constants correct	A1	(4)

Part (b)(i):

Answer	Marks	Guidance
$\int\frac{A}{x+2} \rightarrow \ldots\ln(x+2)$ and $\int\frac{B}{1-x} \rightarrow \ldots\ln(1-x)$	M1
$\int\frac{C}{(1-x)^2} \rightarrow \ldots(1-x)^{-1}$	M1
$\ln(x+2) + 3\ln(1-x) + 4(1-x)^{-1}$ $(+c)$	A1ft	All three integrals correct following through on incorrect constants (not zeros); must attempt simplest form

Part (b)(ii):

Answer	Marks	Guidance
$\left[\ln(x+2)+3\ln(1-x)+4(1-x)^{-1}\right]_0^{\frac{1}{2}}$
Sub both $x=\frac{1}{2}$ and $x=0$, involves lns, subtracts	M1	Either way around
$= \left(\ln\frac{5}{2}+3\ln\frac{1}{2}+8\right)-(\ln 2+3\ln 1+4)$ Uses correct ln work to combine ln terms	M1
$= \ln\left(\frac{\frac{5}{2}\times\left(\frac{1}{2}\right)^3}{2}\right) + \ldots = 4+\ln\left(\frac{5}{32}\right)$	A1	cao; decimal equivalent $4+\ln 0.15625$ is correct (6)

# Question 5:

## Part (a):
| $4x^2+5x+3 = A(1-x)^2 + B(x+2)(1-x) + C(x+2)$ | B1 | May be implied by sight of equivalent fractions or correct constants |
| Sub $x=1$ or $x=-2$, attempts to find one constant | M1 | Can be scored after B0; alternatively equate coefficients |
| $x=1 \Rightarrow C=4$; $x=-2 \Rightarrow A=1$; any two constants correct | A1 | |
| Coefficients of $x^2$: $4 = A-B \Rightarrow B=-3$; all three constants correct | A1 | **(4)** |

## Part (b)(i):
| $\int\frac{A}{x+2} \rightarrow \ldots\ln(x+2)$ and $\int\frac{B}{1-x} \rightarrow \ldots\ln(1-x)$ | M1 | |
| $\int\frac{C}{(1-x)^2} \rightarrow \ldots(1-x)^{-1}$ | M1 | |
| $\ln(x+2) + 3\ln(1-x) + 4(1-x)^{-1}$ $(+c)$ | A1ft | All three integrals correct following through on incorrect constants (not zeros); must attempt simplest form |

## Part (b)(ii):
| $\left[\ln(x+2)+3\ln(1-x)+4(1-x)^{-1}\right]_0^{\frac{1}{2}}$ | | |
| Sub both $x=\frac{1}{2}$ and $x=0$, involves lns, subtracts | M1 | Either way around |
| $= \left(\ln\frac{5}{2}+3\ln\frac{1}{2}+8\right)-(\ln 2+3\ln 1+4)$ Uses correct ln work to combine ln terms | M1 | |
| $= \ln\left(\frac{\frac{5}{2}\times\left(\frac{1}{2}\right)^3}{2}\right) + \ldots = 4+\ln\left(\frac{5}{32}\right)$ | A1 | cao; decimal equivalent $4+\ln 0.15625$ is correct **(6)** |

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5.

$$f ( x ) = \frac { 4 x ^ { 2 } + 5 x + 3 } { ( x + 2 ) ( 1 - x ) ^ { 2 } } \equiv \frac { A } { ( x + 2 ) } + \frac { B } { ( 1 - x ) } + \frac { C } { ( 1 - x ) ^ { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A$, $B$ and $C$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int \mathrm { f } ( x ) \mathrm { d } x$.
\item Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$, writing your answer in the form $p + \ln q$, where $p$ and $q$ are constants.\\
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VGHV SIHIN NI III M I ION OC & VIIV SIHI NI JIIIM ION OC & VI4V SIHIL NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel C34 2018 Q5 [10]}}

This paper (13 questions)

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Q1 8 Q2 7 Q3 6 Q4 10 Q5 10 Q6 9 Q7 8 Q8 10 Q9 9 Q10 12 Q11 10 Q12 13 Q13 13

Answer	Marks	Guidance
\(4x^2+5x+3 = A(1-x)^2 + B(x+2)(1-x) + C(x+2)\)	B1	May be implied by sight of equivalent fractions or correct constants
Sub \(x=1\) or \(x=-2\), attempts to find one constant	M1	Can be scored after B0; alternatively equate coefficients
\(x=1 \Rightarrow C=4\); \(x=-2 \Rightarrow A=1\); any two constants correct	A1
Coefficients of \(x^2\): \(4 = A-B \Rightarrow B=-3\); all three constants correct	A1	(4)

Answer	Marks	Guidance
\(\int\frac{A}{x+2} \rightarrow \ldots\ln(x+2)\) and \(\int\frac{B}{1-x} \rightarrow \ldots\ln(1-x)\)	M1
\(\int\frac{C}{(1-x)^2} \rightarrow \ldots(1-x)^{-1}\)	M1
\(\ln(x+2) + 3\ln(1-x) + 4(1-x)^{-1}\) \((+c)\)	A1ft	All three integrals correct following through on incorrect constants (not zeros); must attempt simplest form

Answer	Marks	Guidance
\(\left[\ln(x+2)+3\ln(1-x)+4(1-x)^{-1}\right]_0^{\frac{1}{2}}\)
Sub both \(x=\frac{1}{2}\) and \(x=0\), involves lns, subtracts	M1	Either way around
\(= \left(\ln\frac{5}{2}+3\ln\frac{1}{2}+8\right)-(\ln 2+3\ln 1+4)\) Uses correct ln work to combine ln terms	M1
\(= \ln\left(\frac{\frac{5}{2}\times\left(\frac{1}{2}\right)^3}{2}\right) + \ldots = 4+\ln\left(\frac{5}{32}\right)\)	A1	cao; decimal equivalent \(4+\ln 0.15625\) is correct (6)