AQA C4 2014 June — Question 8 11 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypePartial fractions for differential equations
DifficultyStandard +0.3 This is a standard two-part question combining routine partial fractions decomposition with a separable differential equation. Part (a) is textbook practice with a repeated linear factor. Part (b) requires separation of variables and integration using the partial fractions result, but follows a predictable template. The initial condition application is straightforward. Slightly above average difficulty due to the combination of techniques and algebraic manipulation required, but no novel insight needed.
Spec1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y)
8
  1. Express \(\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }\) in the form \(\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }\).
  2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\).
    Give your answer in the form \(\mathrm { f } ( y ) = \mathrm { g } ( x )\).
    [0pt] [7 marks]
Question 1
(a) Find the gradient at the point on the curve where \(t = 2\). [3 marks]
M1: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
A1: \(\frac{dx}{dt} = t\), \(\frac{dy}{dt} = -\frac{4}{t^2}\)
M1: Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
A1: Substitute \(t = 2\) to get gradient \(= -\frac{1}{2}\)
(b) Find a Cartesian equation of the curve. [2 marks]
M1: Eliminate parameter \(t\) using \(x = \frac{t^2}{2} + 1\)
A1: \(y = \frac{8}{(2x-2)^2}\) or equivalent form
Question 2
(a) Find the values of constants \(A\) and \(B\). [3 marks]
M1: Multiply both sides by \(2x^2 - x + 2\)
M1: Compare coefficients or substitute values
A1: \(A = 2\), \(B = -1\)
(b) Find the equation of the curve. [4 marks]
M1: Integrate \(\frac{dy}{dx} = Ax + \frac{B(4x-1)}{2x^2 - x + 2}\)
M1: \(\int 2x \, dx = x^2\)
M1: \(\int \frac{-1(4x-1)}{2x^2 - x + 2} \, dx = -\frac{1}{2}\ln(2x^2 - x + 2)\)
M1: Use initial condition \((-1, 2)\) to find constant of integration
A1: \(y = x^2 - \frac{1}{2}\ln(2x^2 - x + 2) + \frac{1}{2}\ln(3) + 2\)
Question 3
(a) Find the binomial expansion of \((1-4x)^{1/4}\) up to and including the term in \(x^2\). [2 marks]
M1: Use binomial series with \(n = \frac{1}{4}\), \(r = -4x\)
A1: \(1 - x - 2x^2 + \ldots\)
(b) Find the binomial expansion of \((2+3x)^{-3}\) up to and including the term in \(x^2\). [3 marks]
M1: Factor out \(2^{-3}\) and use binomial series
M1: Use binomial series with appropriate coefficients
A1: \(\frac{1}{8} - \frac{9}{16}x + \frac{81}{128}x^2 + \ldots\)
(c) Find the binomial expansion of \(\frac{(1-4x)^{1/4}}{(2+3x)^3}\) up to and including the term in \(x^2\). [2 marks]
M1: Multiply the two expansions
A1: \(1 - \frac{17}{16}x - \frac{263}{256}x^2 + \ldots\)
Question 4
(a) Write down the value of \(A\). [1 mark]
A1: \(A = 5000\)
(b) Use the model:
(i) Show that \(p^{10} = 5\). [1 mark]
M1: Substitute \(V = 25000\), \(t = 10\) into \(V = Ap^t\)
(ii) Use logarithms to find the year in which the painting will be valued at £75000. [4 marks]
M1: Use \(75000 = 5000p^t\)
M1: Simplify to \(15 = p^t\)
M1: Take logarithms: \(t = \frac{\ln(15)}{\ln(p)}\)
M1: Use \(\ln(p) = \frac{\ln(5)}{10}\)
A1: \(t \approx 15.48\) years, so year is 2016 (or 2017)
(c)
(i) Show that \(T = \frac{\ln(5/2)}{\ln(p/q)}\). [4 marks]
M1: Set \(V = W\): \(5000p^{t-10} = 2500q^t\)
M1: Simplify to \(2p^{t-10} = q^t\)
M1: Expand to \(2p^{-10}p^t = q^t\)
M1: Rearrange to \(t\ln(p/q) = \ln(5/2)\)
A1: \(T = \frac{\ln(5/2)}{\ln(p/q)}\)
(ii) Given that \(p = 1.029q\), find the year. [1 mark]
M1: Substitute and calculate \(T\)
A1: Year 2010
Question 5
(a)
(i) Express \(3\sin x + 4\cos x\) in the form \(R\sin(x+a)\). [3 marks]
M1: \(R = \sqrt{3^2 + 4^2} = 5\)
M1: \(\tan a = \frac{4}{3}\)
A1: \(5\sin(x + 53.1°)\) or \(5\sin(x + 53.13°)\)
(ii) Hence solve \(3\sin 2y + 4\cos 2y = 5\) for \(0° < y < 360°\). [3 marks]
M1: Rewrite as \(5\sin(2y + 53.1°) = 5\)
M1: \(\sin(2y + 53.1°) = 1\), so \(2y + 53.1° = 90°\)
M1: Solve for \(y\) considering periodicity
A1: \(y = 18.4°, 198.4°\)
(b)
(i) Show that \(\tan 2y \tan y = 2\) can be written as \(2\tan^2 y = 1\). [2 marks]
M1: Use \(\tan 2y = \frac{2\tan y}{1 - \tan^2 y}\)
M1: Substitute and simplify
A1: Shown
(ii) Hence solve in the interval \(0° \leq y \leq 180°\). [2 marks]
M1: \(\tan y = \pm \frac{1}{\sqrt{2}}\)
A1: \(y = 35.3°, 144.7°\)
(c)
(i) Show that \(2x - 1\) is a factor of \(8x^3 - 4x + 1\). [1 mark]
M1: Use Factor Theorem: substitute \(x = \frac{1}{2}\) to get \(0\)
(ii) Show that \(4\cos 2y \cos y + 1\) can be written as \(8x^3 - 4x + 1\) where \(x = \cos y\). [1 mark]
M1: Use \(\cos 2y = 2\cos^2 y - 1\) and substitute \(x = \cos y\)
(iii) Show that \(\cos 72° = \frac{\sqrt{5}-1}{p}\) where \(p\) is an integer. [3 marks]
M1: From \(4\cos 2y \cos y + 1 = 0\) and \(2x - 1\) factor, find other factor \(4x^2 + 2x - 1\)
M1: Use quadratic formula on \(4x^2 + 2x - 1 = 0\)
M1: $x = \frac{-2 \pm \sqrt{4 + 16}
# Question 1

## (a) Find the gradient at the point on the curve where $t = 2$. [3 marks]

M1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

A1: $\frac{dx}{dt} = t$, $\frac{dy}{dt} = -\frac{4}{t^2}$

M1: Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

A1: Substitute $t = 2$ to get gradient $= -\frac{1}{2}$

## (b) Find a Cartesian equation of the curve. [2 marks]

M1: Eliminate parameter $t$ using $x = \frac{t^2}{2} + 1$

A1: $y = \frac{8}{(2x-2)^2}$ or equivalent form

---

# Question 2

## (a) Find the values of constants $A$ and $B$. [3 marks]

M1: Multiply both sides by $2x^2 - x + 2$

M1: Compare coefficients or substitute values

A1: $A = 2$, $B = -1$

## (b) Find the equation of the curve. [4 marks]

M1: Integrate $\frac{dy}{dx} = Ax + \frac{B(4x-1)}{2x^2 - x + 2}$

M1: $\int 2x \, dx = x^2$

M1: $\int \frac{-1(4x-1)}{2x^2 - x + 2} \, dx = -\frac{1}{2}\ln(2x^2 - x + 2)$

M1: Use initial condition $(-1, 2)$ to find constant of integration

A1: $y = x^2 - \frac{1}{2}\ln(2x^2 - x + 2) + \frac{1}{2}\ln(3) + 2$

---

# Question 3

## (a) Find the binomial expansion of $(1-4x)^{1/4}$ up to and including the term in $x^2$. [2 marks]

M1: Use binomial series with $n = \frac{1}{4}$, $r = -4x$

A1: $1 - x - 2x^2 + \ldots$

## (b) Find the binomial expansion of $(2+3x)^{-3}$ up to and including the term in $x^2$. [3 marks]

M1: Factor out $2^{-3}$ and use binomial series

M1: Use binomial series with appropriate coefficients

A1: $\frac{1}{8} - \frac{9}{16}x + \frac{81}{128}x^2 + \ldots$

## (c) Find the binomial expansion of $\frac{(1-4x)^{1/4}}{(2+3x)^3}$ up to and including the term in $x^2$. [2 marks]

M1: Multiply the two expansions

A1: $1 - \frac{17}{16}x - \frac{263}{256}x^2 + \ldots$

---

# Question 4

## (a) Write down the value of $A$. [1 mark]

A1: $A = 5000$

## (b) Use the model:

### (i) Show that $p^{10} = 5$. [1 mark]

M1: Substitute $V = 25000$, $t = 10$ into $V = Ap^t$

## (ii) Use logarithms to find the year in which the painting will be valued at £75000. [4 marks]

M1: Use $75000 = 5000p^t$

M1: Simplify to $15 = p^t$

M1: Take logarithms: $t = \frac{\ln(15)}{\ln(p)}$

M1: Use $\ln(p) = \frac{\ln(5)}{10}$

A1: $t \approx 15.48$ years, so year is 2016 (or 2017)

## (c)

### (i) Show that $T = \frac{\ln(5/2)}{\ln(p/q)}$. [4 marks]

M1: Set $V = W$: $5000p^{t-10} = 2500q^t$

M1: Simplify to $2p^{t-10} = q^t$

M1: Expand to $2p^{-10}p^t = q^t$

M1: Rearrange to $t\ln(p/q) = \ln(5/2)$

A1: $T = \frac{\ln(5/2)}{\ln(p/q)}$

### (ii) Given that $p = 1.029q$, find the year. [1 mark]

M1: Substitute and calculate $T$

A1: Year 2010

---

# Question 5

## (a)

### (i) Express $3\sin x + 4\cos x$ in the form $R\sin(x+a)$. [3 marks]

M1: $R = \sqrt{3^2 + 4^2} = 5$

M1: $\tan a = \frac{4}{3}$

A1: $5\sin(x + 53.1°)$ or $5\sin(x + 53.13°)$

### (ii) Hence solve $3\sin 2y + 4\cos 2y = 5$ for $0° < y < 360°$. [3 marks]

M1: Rewrite as $5\sin(2y + 53.1°) = 5$

M1: $\sin(2y + 53.1°) = 1$, so $2y + 53.1° = 90°$

M1: Solve for $y$ considering periodicity

A1: $y = 18.4°, 198.4°$

## (b)

### (i) Show that $\tan 2y \tan y = 2$ can be written as $2\tan^2 y = 1$. [2 marks]

M1: Use $\tan 2y = \frac{2\tan y}{1 - \tan^2 y}$

M1: Substitute and simplify

A1: Shown

### (ii) Hence solve in the interval $0° \leq y \leq 180°$. [2 marks]

M1: $\tan y = \pm \frac{1}{\sqrt{2}}$

A1: $y = 35.3°, 144.7°$

## (c)

### (i) Show that $2x - 1$ is a factor of $8x^3 - 4x + 1$. [1 mark]

M1: Use Factor Theorem: substitute $x = \frac{1}{2}$ to get $0$

### (ii) Show that $4\cos 2y \cos y + 1$ can be written as $8x^3 - 4x + 1$ where $x = \cos y$. [1 mark]

M1: Use $\cos 2y = 2\cos^2 y - 1$ and substitute $x = \cos y$

### (iii) Show that $\cos 72° = \frac{\sqrt{5}-1}{p}$ where $p$ is an integer. [3 marks]

M1: From $4\cos 2y \cos y + 1 = 0$ and $2x - 1$ factor, find other factor $4x^2 + 2x - 1$

M1: Use quadratic formula on $4x^2 + 2x - 1 = 0$

M1: $x = \frac{-2 \pm \sqrt{4 + 16}
8
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$ in the form $\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }$.
\item Solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$

where $y = 0$ when $x = 0$.\\
Give your answer in the form $\mathrm { f } ( y ) = \mathrm { g } ( x )$.\\[0pt]
[7 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA C4 2014 Q8 [11]}}
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