AQA Further Paper 1 2024 June — Question 9 8 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeSolve using double angle formulas
DifficultyStandard +0.8 Part (a) requires manipulating the exponential definition of sinh and algebraic manipulation with surds to prove an inverse hyperbolic identity—moderately challenging but follows a standard approach. Part (b) involves using hyperbolic identities to convert to a quadratic in sinh x, then applying the inverse sinh formula from part (a). This is a solid Further Maths question requiring multiple techniques and careful algebra, but follows recognizable patterns for students familiar with hyperbolic functions.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07f Inverse hyperbolic: logarithmic forms
  1. It is given that $$p = \ln\left(r + \sqrt{r^2 + 1}\right)$$ Starting from the exponential definition of the sinh function, show that \(\sinh p = r\) [4 marks]
  2. Solve the equation $$\cosh^2 x = 2\sinh x + 16$$ Give your answers in logarithmic form. [4 marks]
Question 9:
AnswerMarks
9(a)ep –e–p
Writes down OE
AnswerMarks Guidance
21.2 B1
s p = in h
2
(r ) ( )
+ ln 2 r + – r + 1 ln r 2 + 1
e – e
=
2
 
1 1
= r +  2 r + – 1
2 r + 2 r + 1
(  ) 2 
1 r  2 + r + 1 – 1
=
2 2 r + r + 1
(  ) 
r 2 2 + r r + 2 1 + 2 r + 1 – 1
1 
=
2 r + r 2 + 1
1  r 2 2 2 + r r + 2 1 
= 
2 2 r + r + 1
 ( ) 
r 2 2 r + r + 1
1 
=
2 2 r + r + 1
= r
Substitutes p with the given
expression and removes
e and ln to obtain
1
r + r 2 + 1 or
AnswerMarks Guidance
r + r 2 + 13.1a M1
Collects over common
denominator r + r 2 + 1
or
1
Multiplies by
r + r 2 + 1
r− r2 +1
AnswerMarks Guidance
r− r2 +11.1a M1
Completes reasoned
argument to obtain
sinh p=r
AnswerMarks Guidance
AG2.1 R1
Subtotal4
QMarking instructions AO
AnswerMarks
9(b)Uses c o s h 2 x = 1 + s i n h 2 x
to obtain
1 + s i n h 2 x = 2 s i n h x + 1 6
or
Substitutes in correct
exponential form to obtain
 e x + e – x  2  e x – e – x 
= 2 + 1 6
AnswerMarks Guidance
2 23.1a M1
s2 + 1 = 2s + 16
s2 – 2s – 15 = 0
s = –3,5
x = ln ( – 3 + 1 0 ) , x = ln ( 5 + 2 6 )
Obtains ( s i n h x = ) –3,5
or
Obtains
( )
AnswerMarks Guidance
ex = 10.099...,0.162...1.1b A1
Uses
sinh−1 x=ln (x+ x2 +1)
AnswerMarks Guidance
FT their –3 or 51.1a M1
Obtains
ln ( – 3 + 1 0 ) , ln ( 5 + 2 6 )
AnswerMarks Guidance
and no other solutions1.1b A1
Subtotal4
Question total8
QMarking instructions AO 
Question 9:
--- 9(a) ---
9(a) | ep –e–p
Writes down OE
2 | 1.2 | B1 | e p – e – p
s p = in h
2
(r ) ( )
+ ln 2 r + – r + 1 ln r 2 + 1
e – e
=
2
 
1 1
= r +  2 r + – 1
2 r + 2 r + 1
(  ) 2 
1 r  2 + r + 1 – 1
=
2 2 r + r + 1
(  ) 
r 2 2 + r r + 2 1 + 2 r + 1 – 1
1 
=
2 r + r 2 + 1
1  r 2 2 2 + r r + 2 1 
= 
2 2 r + r + 1
 ( ) 
r 2 2 r + r + 1
1 
=
2 2 r + r + 1
= r
Substitutes p with the given
expression and removes
e and ln to obtain
1
r + r 2 + 1 or
r + r 2 + 1 | 3.1a | M1
Collects over common
denominator r + r 2 + 1
or
1
Multiplies by
r + r 2 + 1
r− r2 +1
r− r2 +1 | 1.1a | M1
Completes reasoned
argument to obtain
sinh p=r
AG | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Uses c o s h 2 x = 1 + s i n h 2 x
to obtain
1 + s i n h 2 x = 2 s i n h x + 1 6
or
Substitutes in correct
exponential form to obtain
 e x + e – x  2  e x – e – x 
= 2 + 1 6
2 2 | 3.1a | M1 | Let s = sinh x
s2 + 1 = 2s + 16
s2 – 2s – 15 = 0
s = –3,5
x = ln ( – 3 + 1 0 ) , x = ln ( 5 + 2 6 )
Obtains ( s i n h x = ) –3,5
or
Obtains
( )
ex = 10.099...,0.162... | 1.1b | A1
Uses
sinh−1 x=ln (x+ x2 +1)
FT their –3 or 5 | 1.1a | M1
Obtains
ln ( – 3 + 1 0 ) , ln ( 5 + 2 6 )
and no other solutions | 1.1b | A1
Subtotal | 4
Question total | 8
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item It is given that
$$p = \ln\left(r + \sqrt{r^2 + 1}\right)$$
Starting from the exponential definition of the sinh function, show that $\sinh p = r$
[4 marks]

\item Solve the equation
$$\cosh^2 x = 2\sinh x + 16$$
Give your answers in logarithmic form.
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2024 Q9 [8]}}
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