Question 14 - A-Level Maths

SPS SPS FM 2024 November — Question 14

Exam Board	SPS
Module	SPS FM (SPS FM)
Year	2024
Session	November
Topic	Stationary points and optimisation
Type	Find stationary points coordinates
Difficulty	Moderate -0.3 This is a straightforward stationary points question requiring differentiation, setting the derivative to zero, and using the second derivative test. It's a standard A-level technique with no novel problem-solving required, making it slightly easier than average but still requiring proper execution of multiple steps.

14 November 2024 Instructions

Answer all the questions.
Use black or blue ink. Pencil may be used for graphs and diagrams only.
There are blank pages at the end of the paper for additional working. You must clearly indicate when you have moved onto additional pages on the question itself. Make sure to include the question number.
You are permitted to use a scientific or graphical calculator in this paper.
Where appropriate, your answer should be supported with working. Marks might be given for using a correct method, even if your answer is wrong.
Give non-exact numerical answers correct to 3 significant figures unless a different degree of accuracy is specified in the question.
The acceleration due to gravity is denoted by $g \mathrm {~ms} ^ { - 2 }$. When a numerical value is needed use $g = 9.8$ unless a different value is specified in the question.

Information

The total mark for this paper is $\mathbf { 6 1 }$ marks.
The marks for each question are shown in brackets.
You are reminded of the need for clear presentation in your answers.
You have $\mathbf { 6 0 }$ minutes for this paper.

\section*{Arithmetic series} $S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}$ \section*{Geometric series} $S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ $S _ { \infty } = \frac { a } { 1 - r }$ for $| r | < 1$ \section*{Binomial series} $( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )$,
where ${ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }$ $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$ \section*{Series} $\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 ) , \sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$ \section*{Maclaurin series} $\mathrm { f } ( x ) = \mathrm { f } ( 0 ) + \mathrm { f } ^ { \prime } ( 0 ) x + \frac { \mathrm { f } ^ { \prime \prime } ( 0 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { \mathrm { f } ^ { ( r ) } ( 0 ) } { r ! } x ^ { r } + \ldots$ $\mathrm { e } ^ { x } = \exp ( x ) = 1 + x + \frac { x ^ { 2 } } { 2 ! } + \ldots + \frac { x ^ { r } } { r ! } + \ldots$ for all $x$ $\ln ( 1 + x ) = x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \ldots + ( - 1 ) ^ { r + 1 } \frac { x ^ { r } } { r } + \ldots ( - 1 < x \leq 1 )$ $\sin x = x - \frac { x ^ { 3 } } { 3 ! } + \frac { x ^ { 5 } } { 5 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r + 1 } } { ( 2 r + 1 ) ! } + \ldots$ for all $x$ $\cos x = 1 - \frac { x ^ { 2 } } { 2 ! } + \frac { x ^ { 4 } } { 4 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r } } { ( 2 r ) ! } + \ldots$ for all $x$ $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$ \section*{Differentiation}

$\mathrm { f } ( x )$	$\mathrm { f } ^ { \prime } ( x )$
$\tan k x$	$k \sec ^ { 2 } k x$
$\sec x$	$\sec x \tan x$
$\cot x$	$- \operatorname { cosec } ^ { 2 } x$
$\operatorname { cosec } x$	$- \operatorname { cosec } x \cot x$
$\arcsin x$ or $\sin ^ { - 1 } x$	$\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$
$\arccos x$ or $\cos ^ { - 1 } x$	$- \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$
$\arctan x$ or $\tan ^ { - 1 } x$	$\frac { 1 } { 1 + x ^ { 2 } }$

Quotient rule $y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }$ \section*{Differentiation from first principles} $\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }$ \section*{Integration} $\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c$ $\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c$ Integration by parts $\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x$ The mean value of $\mathrm { f } ( x )$ on the interval $[ a , b ]$ is $\frac { 1 } { b - a } \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x$ Area of sector enclosed by polar curve is $\frac { 1 } { 2 } \int r ^ { 2 } \mathrm {~d} \theta$

$\mathrm { f } ( x )$	$\int \mathrm { f } ( x ) \mathrm { d } x$
$\frac { 1 } { \sqrt { a ^ { 2 } - x ^ { 2 } } }$	$\sin ^ { - 1 } \left( \frac { x } { a } \right) \quad ( \| x \| < a )$
$\frac { 1 } { a ^ { 2 } + x ^ { 2 } }$	$\frac { 1 } { a } \tan ^ { - 1 } \left( \frac { x } { a } \right)$
$\frac { 1 } { \sqrt { a ^ { 2 } + x ^ { 2 } } }$	$\sinh ^ { - 1 } \left( \frac { x } { a } \right)$ or $\ln \left( x + \sqrt { x ^ { 2 } + a ^ { 2 } } \right)$
$\frac { 1 } { \sqrt { x ^ { 2 } - a ^ { 2 } } }$	$\cosh ^ { - 1 } \left( \frac { x } { a } \right)$ or $\ln \left( x + \sqrt { x ^ { 2 } - a ^ { 2 } } \right) \quad ( x > a )$

\section*{Numerical methods} Trapezium rule: $\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}$, where $h = \frac { b - a } { n }$ The Newton-Raphson iteration for solving $\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }$ \section*{Complex numbers} Circles: $| z - a | = k$ Half lines: $\arg ( z - a ) = \alpha$ Lines: $| z - a | = | z - b |$ \section*{Small angle approximations} $\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta$ where $\theta$ is small and measured in radians \section*{Trigonometric identities} $\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B$ $\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B$ $\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)$ \section*{Hyperbolic functions} $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ $\sinh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } + 1 \right) } \right]$ $\cosh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } - 1 \right) } \right] , x \geq 1$ $\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) , - 1 < x < 1$

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Determine the values of $x$ for which $$64 \cosh ^ { 4 } x - 64 \sinh ^ { 2 } x - 73 = 0$$ Give your answer in the form $q \ln 2$ where $q$ is rational and in simplest form.
2. (a) Prove that $$\tanh ^ { - 1 } ( x ) = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) \quad - k < x < k$$ stating the value of the constant $k$.
(b) Hence, or otherwise, solve the equation $$2 x = \tanh ( \ln \sqrt { 2 - 3 x } )$$ 3. In this question you must show detailed reasoning. The roots of the equation $x ^ { 3 } - x ^ { 2 } + k x - 2 = 0$ are $\alpha , \frac { 1 } { \alpha }$ and $\beta$.
(a) Evaluate, in exact form, the roots of the equation.
(b) Find $k$.
[0pt]
4. (a) (i) Given that $f ( x ) = \sqrt { 1 + 2 x }$, find $f ^ { \prime } ( x )$ and $f ^ { \prime \prime } ( x )$.
(ii) Hence, find the first three terms of the Maclaurin series for $\sqrt { 1 + 2 x }$.
(b) Hence, using a suitable value for $x$, show that $\sqrt { 5 } \approx \frac { 143 } { 64 }$.
[0pt]
5. In this question you must show detailed reasoning.
(i) Given that $$z _ { 1 } = 6 \left( \cos \left( \frac { \pi } { 3 } \right) + i \sin \left( \frac { \pi } { 3 } \right) \right) \quad \text { and } \quad z _ { 2 } = 6 \sqrt { 3 } \left( \cos \left( \frac { 5 \pi } { 6 } \right) + i \sin \left( \frac { 5 \pi } { 6 } \right) \right)$$ show that $$z _ { 1 } + z _ { 2 } = 12 \left( \cos \left( \frac { 2 \pi } { 3 } \right) + i \sin \left( \frac { 2 \pi } { 3 } \right) \right)$$ (ii) Given that $$\arg ( z - 5 ) = \frac { 2 \pi } { 3 }$$ determine the least value of $| \boldsymbol { z } |$ as $Z$ varies.
[0pt]
6. A curve has polar equation $r = a ( \cos \theta + 2 \sin \theta )$, where $a$ is a positive constant and $0 \leq \theta \leq \pi$.
(a) Determine the polar coordinates of the point on the curve which is furthest from the pole.
(b) (i) Show that the curve is a circle whose radius should be specified.
(ii) Write down the polar coordinates of the centre of the circle.
[0pt]
7. (a) It is conjectured that $$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = a - \frac { b } { n ! }$$ where $a$ and $b$ are constants, and $n$ is an integer such that $n \geq 2$. By considering particular cases, show that if the conjecture is correct then $$a = b = 1$$ (b) Use induction to prove that, for $n \geq 2$, the following is true $$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = 1 - \frac { 1 } { n ! }$$ 8. (a) Use standard results for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that $$\sum _ { r = 1 } ^ { n } ( 3 r - 2 ) ^ { 2 } = \frac { 1 } { 2 } n \left[ 6 n ^ { 2 } - 3 n - 1 \right]$$ for all positive integers $n$.
(b) Hence find any values of $n$ for which $$\sum _ { r = 5 } ^ { n } ( 3 r - 2 ) ^ { 2 } + 103 \sum _ { r = 1 } ^ { 28 } r \cos \left( \frac { r \pi } { 2 } \right) = 3 n ^ { 3 }$$ [0pt]
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Show LaTeX source

14 November 2024
Instructions

\begin{itemize}
  \item Answer all the questions.
  \item Use black or blue ink. Pencil may be used for graphs and diagrams only.
  \item There are blank pages at the end of the paper for additional working. You must clearly indicate when you have moved onto additional pages on the question itself. Make sure to include the question number.
  \item You are permitted to use a scientific or graphical calculator in this paper.
  \item Where appropriate, your answer should be supported with working. Marks might be given for using a correct method, even if your answer is wrong.
  \item Give non-exact numerical answers correct to 3 significant figures unless a different degree of accuracy is specified in the question.
  \item The acceleration due to gravity is denoted by $g \mathrm {~ms} ^ { - 2 }$. When a numerical value is needed use $g = 9.8$ unless a different value is specified in the question.
\end{itemize}

Information

\begin{itemize}
  \item The total mark for this paper is $\mathbf { 6 1 }$ marks.
  \item The marks for each question are shown in brackets.
  \item You are reminded of the need for clear presentation in your answers.
  \item You have $\mathbf { 6 0 }$ minutes for this paper.
\end{itemize}

\section*{Arithmetic series}
$S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}$

\section*{Geometric series}
$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$\\
$S _ { \infty } = \frac { a } { 1 - r }$ for $| r | < 1$

\section*{Binomial series}
$( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )$,\\
where ${ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }$\\
$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$

\section*{Series}
$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 ) , \sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$

\section*{Maclaurin series}
$\mathrm { f } ( x ) = \mathrm { f } ( 0 ) + \mathrm { f } ^ { \prime } ( 0 ) x + \frac { \mathrm { f } ^ { \prime \prime } ( 0 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { \mathrm { f } ^ { ( r ) } ( 0 ) } { r ! } x ^ { r } + \ldots$\\
$\mathrm { e } ^ { x } = \exp ( x ) = 1 + x + \frac { x ^ { 2 } } { 2 ! } + \ldots + \frac { x ^ { r } } { r ! } + \ldots$ for all $x$\\
$\ln ( 1 + x ) = x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \ldots + ( - 1 ) ^ { r + 1 } \frac { x ^ { r } } { r } + \ldots ( - 1 < x \leq 1 )$\\
$\sin x = x - \frac { x ^ { 3 } } { 3 ! } + \frac { x ^ { 5 } } { 5 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r + 1 } } { ( 2 r + 1 ) ! } + \ldots$ for all $x$\\
$\cos x = 1 - \frac { x ^ { 2 } } { 2 ! } + \frac { x ^ { 4 } } { 4 ! } - \ldots + ( - 1 ) ^ { r } \frac { x ^ { 2 r } } { ( 2 r ) ! } + \ldots$ for all $x$\\
$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )$

\section*{Differentiation}
\begin{center}
\begin{tabular}{|l|l|}
\hline
$\mathrm { f } ( x )$ & $\mathrm { f } ^ { \prime } ( x )$ \\
\hline
$\tan k x$ & $k \sec ^ { 2 } k x$ \\
\hline
$\sec x$ & $\sec x \tan x$ \\
\hline
$\cot x$ & $- \operatorname { cosec } ^ { 2 } x$ \\
\hline
$\operatorname { cosec } x$ & $- \operatorname { cosec } x \cot x$ \\
\hline
$\arcsin x$ or $\sin ^ { - 1 } x$ & $\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$ \\
\hline
$\arccos x$ or $\cos ^ { - 1 } x$ & $- \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$ \\
\hline
$\arctan x$ or $\tan ^ { - 1 } x$ & $\frac { 1 } { 1 + x ^ { 2 } }$ \\
\hline
\end{tabular}
\end{center}

Quotient rule $y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }$

\section*{Differentiation from first principles}
$\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }$

\section*{Integration}
$\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c$\\
$\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c$\\
Integration by parts $\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x$

The mean value of $\mathrm { f } ( x )$ on the interval $[ a , b ]$ is $\frac { 1 } { b - a } \int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x$\\
Area of sector enclosed by polar curve is $\frac { 1 } { 2 } \int r ^ { 2 } \mathrm {~d} \theta$

\begin{center}
\begin{tabular}{|l|l|}
\hline
$\mathrm { f } ( x )$ & $\int \mathrm { f } ( x ) \mathrm { d } x$ \\
\hline
$\frac { 1 } { \sqrt { a ^ { 2 } - x ^ { 2 } } }$ & $\sin ^ { - 1 } \left( \frac { x } { a } \right) \quad ( | x | < a )$ \\
\hline
$\frac { 1 } { a ^ { 2 } + x ^ { 2 } }$ & $\frac { 1 } { a } \tan ^ { - 1 } \left( \frac { x } { a } \right)$ \\
\hline
$\frac { 1 } { \sqrt { a ^ { 2 } + x ^ { 2 } } }$ & $\sinh ^ { - 1 } \left( \frac { x } { a } \right)$ or $\ln \left( x + \sqrt { x ^ { 2 } + a ^ { 2 } } \right)$ \\
\hline
$\frac { 1 } { \sqrt { x ^ { 2 } - a ^ { 2 } } }$ & $\cosh ^ { - 1 } \left( \frac { x } { a } \right)$ or $\ln \left( x + \sqrt { x ^ { 2 } - a ^ { 2 } } \right) \quad ( x > a )$ \\
\hline
\end{tabular}
\end{center}

\section*{Numerical methods}
Trapezium rule: $\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}$, where $h = \frac { b - a } { n }$\\
The Newton-Raphson iteration for solving $\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }$

\section*{Complex numbers}
Circles: $| z - a | = k$\\
Half lines: $\arg ( z - a ) = \alpha$\\
Lines: $| z - a | = | z - b |$

\section*{Small angle approximations}
$\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta$ where $\theta$ is small and measured in radians

\section*{Trigonometric identities}
$\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B$\\
$\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B$\\
$\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)$

\section*{Hyperbolic functions}
$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$\\
$\sinh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } + 1 \right) } \right]$\\
$\cosh ^ { - 1 } x = \ln \left[ x + \sqrt { \left( x ^ { 2 } - 1 \right) } \right] , x \geq 1$\\
$\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) , - 1 < x < 1$

\begin{enumerate}
  \item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}

Determine the values of $x$ for which

$$64 \cosh ^ { 4 } x - 64 \sinh ^ { 2 } x - 73 = 0$$

Give your answer in the form $q \ln 2$ where $q$ is rational and in simplest form.\\
2. (a) Prove that

$$\tanh ^ { - 1 } ( x ) = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) \quad - k < x < k$$

stating the value of the constant $k$.\\
(b) Hence, or otherwise, solve the equation

$$2 x = \tanh ( \ln \sqrt { 2 - 3 x } )$$

\\
3. In this question you must show detailed reasoning.

The roots of the equation $x ^ { 3 } - x ^ { 2 } + k x - 2 = 0$ are $\alpha , \frac { 1 } { \alpha }$ and $\beta$.\\
(a) Evaluate, in exact form, the roots of the equation.\\
(b) Find $k$.\\[0pt]
\\
4. (a) (i) Given that $f ( x ) = \sqrt { 1 + 2 x }$, find $f ^ { \prime } ( x )$ and $f ^ { \prime \prime } ( x )$.\\
(ii) Hence, find the first three terms of the Maclaurin series for $\sqrt { 1 + 2 x }$.\\
(b) Hence, using a suitable value for $x$, show that $\sqrt { 5 } \approx \frac { 143 } { 64 }$.\\[0pt]
\\
5. In this question you must show detailed reasoning.\\
(i) Given that

$$z _ { 1 } = 6 \left( \cos \left( \frac { \pi } { 3 } \right) + i \sin \left( \frac { \pi } { 3 } \right) \right) \quad \text { and } \quad z _ { 2 } = 6 \sqrt { 3 } \left( \cos \left( \frac { 5 \pi } { 6 } \right) + i \sin \left( \frac { 5 \pi } { 6 } \right) \right)$$

show that

$$z _ { 1 } + z _ { 2 } = 12 \left( \cos \left( \frac { 2 \pi } { 3 } \right) + i \sin \left( \frac { 2 \pi } { 3 } \right) \right)$$

(ii) Given that

$$\arg ( z - 5 ) = \frac { 2 \pi } { 3 }$$

determine the least value of $| \boldsymbol { z } |$ as $Z$ varies.\\[0pt]
\\
6. A curve has polar equation $r = a ( \cos \theta + 2 \sin \theta )$, where $a$ is a positive constant and $0 \leq \theta \leq \pi$.\\
(a) Determine the polar coordinates of the point on the curve which is furthest from the pole.\\
(b) (i) Show that the curve is a circle whose radius should be specified.\\
(ii) Write down the polar coordinates of the centre of the circle.\\[0pt]
\\
7. (a) It is conjectured that

$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = a - \frac { b } { n ! }$$

where $a$ and $b$ are constants, and $n$ is an integer such that $n \geq 2$. By considering particular cases, show that if the conjecture is correct then

$$a = b = 1$$

(b) Use induction to prove that, for $n \geq 2$, the following is true

$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = 1 - \frac { 1 } { n ! }$$

\\
8. (a) Use standard results for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that

$$\sum _ { r = 1 } ^ { n } ( 3 r - 2 ) ^ { 2 } = \frac { 1 } { 2 } n \left[ 6 n ^ { 2 } - 3 n - 1 \right]$$

for all positive integers $n$.\\
(b) Hence find any values of $n$ for which

$$\sum _ { r = 5 } ^ { n } ( 3 r - 2 ) ^ { 2 } + 103 \sum _ { r = 1 } ^ { 28 } r \cos \left( \frac { r \pi } { 2 } \right) = 3 n ^ { 3 }$$

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\hfill \mbox{\textit{SPS SPS FM 2024 Q14}}

This paper (8 questions)

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Q1 4 Q2 10 Q3 8 Q4 6 Q5 6 Q6 10 Q7 7 Q14

\(\mathrm { f } ( x )\)	\(\mathrm { f } ^ { \prime } ( x )\)
\(\tan k x\)	\(k \sec ^ { 2 } k x\)
\(\sec x\)	\(\sec x \tan x\)
\(\cot x\)	\(- \operatorname { cosec } ^ { 2 } x\)
\(\operatorname { cosec } x\)	\(- \operatorname { cosec } x \cot x\)
\(\arcsin x\) or \(\sin ^ { - 1 } x\)	\(\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\)
\(\arccos x\) or \(\cos ^ { - 1 } x\)	\(- \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\)
\(\arctan x\) or \(\tan ^ { - 1 } x\)	\(\frac { 1 } { 1 + x ^ { 2 } }\)

\(\mathrm { f } ( x )\)	\(\int \mathrm { f } ( x ) \mathrm { d } x\)
\(\frac { 1 } { \sqrt { a ^ { 2 } - x ^ { 2 } } }\)	\(\sin ^ { - 1 } \left( \frac { x } { a } \right) \quad ( \| x \| < a )\)
\(\frac { 1 } { a ^ { 2 } + x ^ { 2 } }\)	\(\frac { 1 } { a } \tan ^ { - 1 } \left( \frac { x } { a } \right)\)
\(\frac { 1 } { \sqrt { a ^ { 2 } + x ^ { 2 } } }\)	\(\sinh ^ { - 1 } \left( \frac { x } { a } \right)\) or \(\ln \left( x + \sqrt { x ^ { 2 } + a ^ { 2 } } \right)\)
\(\frac { 1 } { \sqrt { x ^ { 2 } - a ^ { 2 } } }\)	\(\cosh ^ { - 1 } \left( \frac { x } { a } \right)\) or \(\ln \left( x + \sqrt { x ^ { 2 } - a ^ { 2 } } \right) \quad ( x > a )\)