AQA Further Paper 1 2019 June — Question 6 8 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeExpress hyperbolic in exponential form
DifficultyStandard +0.8 This is a Further Maths hyperbolic functions question requiring algebraic manipulation and use of identities. Part (a) needs expansion using definitions and simplification (standard technique, ~3 steps). Part (b) requires recognizing the factorization a³-b³=(a-b)(a²+ab+b²), applying part (a), and using cosh²x-sinh²x=1, demanding good algebraic insight across multiple steps. The multi-stage reasoning and need to connect different hyperbolic identities places it moderately above average difficulty.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1
  1. Show that $$\cosh^3 x + \sinh^3 x = \frac{1}{4}e^{mx} + \frac{3}{4}e^{nx}$$ where \(m\) and \(n\) are integers. [3 marks]
  2. Hence find \(\cosh^6 x - \sinh^6 x\) in the form $$\frac{a \cosh(kx) + b}{8}$$ where \(a\), \(b\) and \(k\) are integers. [5 marks]
Question 6:
AnswerMarks
6(a)Uses correct expressions
for coshxand sinhx,and
uses them to simplify
AnswerMarks Guidance
LHS.AO1.1a M1
8
sinh3 x= 1( e3x −3ex +3e−x −e−3x )
8
1 3
cosh3 x+sinh3 x= e3x + e−x
4 4
Finds a correct,
unsimplified expansion of
the LHS in terms of
AnswerMarks Guidance
exponentials.AO1.1b A1
Completes a rigorous
argument to obtain the
correct result.
Must include clear
definitions for &
.
AnswerMarks Guidance
NMS = 0/3 cosh𝑥𝑥AO2.1 R1
AnswerMarks
6(b)sFiinndhs𝑥𝑥 in
exponentia3l form 3
PI corrceocsth ex𝑥𝑥po−nesnintihal 𝑥𝑥
AnswerMarks Guidance
expression.AO3.1a B1
3 1
cosh3 x−sinh3 x= ex + e−3x
4 4
1 3 3 1 
cosh6x−sinh6x= e3x+ e−x ex+ e−3x
 
4 4 4 4 
3 9 1 3
= e4x+ + + e−4x
16 16 16 16
3e4x+e−4x 5
=  +
8 2  8
3cosh4x+5
=
8
Uses their expressions to
find cosh6 x−sinh6 x in
AnswerMarks Guidance
exponential form.AO3.1a M1
Obtains their correct resultAO1.1b A1F
Correctly separates out
3
cosh4x or equivalent
8
AnswerMarks Guidance
from their expressionAO2.2a M1
Completes a rigorous
argument to obtain the
AnswerMarks Guidance
correct result.AO2.1 R1
Total8
QMarking Instructions AO 
Question 6:
--- 6(a) ---
6(a) | Uses correct expressions
for coshxand sinhx,and
uses them to simplify
LHS. | AO1.1a | M1 | cosh3 x= 1( e3x +3ex +3e−x +e−3x )
8
sinh3 x= 1( e3x −3ex +3e−x −e−3x )
8
1 3
cosh3 x+sinh3 x= e3x + e−x
4 4
Finds a correct,
unsimplified expansion of
the LHS in terms of
exponentials. | AO1.1b | A1
Completes a rigorous
argument to obtain the
correct result.
Must include clear
definitions for &
.
NMS = 0/3 cosh𝑥𝑥 | AO2.1 | R1
--- 6(b) ---
6(b) | sFiinndhs𝑥𝑥 in
exponentia3l form 3
PI corrceocsth ex𝑥𝑥po−nesnintihal 𝑥𝑥
expression. | AO3.1a | B1 | cosh6x−sinh6x=( cosh3x+sinh3x )( cosh3x−sinh3x )
3 1
cosh3 x−sinh3 x= ex + e−3x
4 4
1 3 3 1 
cosh6x−sinh6x= e3x+ e−x ex+ e−3x
 
4 4 4 4 
3 9 1 3
= e4x+ + + e−4x
16 16 16 16
3e4x+e−4x 5
=  +
8 2  8
3cosh4x+5
=
8
Uses their expressions to
find cosh6 x−sinh6 x in
exponential form. | AO3.1a | M1
Obtains their correct result | AO1.1b | A1F
Correctly separates out
3
cosh4x or equivalent
8
from their expression | AO2.2a | M1
Completes a rigorous
argument to obtain the
correct result. | AO2.1 | R1
Total | 8
Q | Marking Instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\cosh^3 x + \sinh^3 x = \frac{1}{4}e^{mx} + \frac{3}{4}e^{nx}$$
where $m$ and $n$ are integers.
[3 marks]

\item Hence find $\cosh^6 x - \sinh^6 x$ in the form
$$\frac{a \cosh(kx) + b}{8}$$
where $a$, $b$ and $k$ are integers.
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2019 Q6 [8]}}
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