Question 8 - A-Level Maths

AQA Further AS Paper 1 2020 June — Question 8 8 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2020
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Prove inverse hyperbolic logarithmic form
Difficulty	Standard +0.3 Part (a) is a standard Further Maths derivation requiring algebraic manipulation of the definition of tanh^{-1}x (5 marks suggests routine steps). Part (b) requires setting sinh x = cosh x and showing no solutions exist, which is straightforward using definitions or the identity cosh²x - sinh²x = 1. Both are textbook-style proofs with clear methods and no novel insight required, making this easier than average even for Further Maths.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials 4.07f Inverse hyperbolic: logarithmic forms

Prove that $$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$$ [5 marks]
Prove that the graphs of $$y = \sinh x \quad \text{and} \quad y = \cosh x$$ do not intersect. [3 marks]

Show mark scheme Show mark scheme source

Question 8:

Answer	Marks	Guidance
8(a)	Writes or in exponential form.	1.2

− 1

𝑦𝑦 = tanh 𝑥𝑥

𝑥𝑥 = tanh𝑦𝑦

𝑦𝑦 −𝑦𝑦

(𝑒𝑒 −𝑒𝑒 )

2 𝑥𝑥 = 𝑦𝑦 −𝑦𝑦

(𝑒𝑒 −𝑒𝑒 )

2 2𝑦𝑦

𝑒𝑒 −1

= 2𝑦𝑦

𝑒𝑒 +1

2𝑦𝑦 2𝑦𝑦

𝑥𝑥(e +1) = e −1

2y 2𝑦𝑦

𝑥𝑥e +𝑥𝑥 = e −1

2𝑦𝑦 2𝑦𝑦

1+𝑥𝑥 = e −𝑥𝑥e

2𝑦𝑦

1+𝑥𝑥 = e (1− 𝑥𝑥)

1+𝑥𝑥 2𝑦𝑦

= e

1−𝑥𝑥

1+𝑥𝑥

2𝑦𝑦 = ln� �

1−𝑥𝑥

1 1+𝑥𝑥

𝑦𝑦 = ln � �

2 1−𝑥𝑥

−1 1 1+𝑥𝑥

tanh 𝑥𝑥 = ln� �

2 1−𝑥𝑥

tanh𝑥𝑥 tanh𝑦𝑦

Answer	Marks	Guidance
Selects a method by forming an equation of the form .	3.1a	M1

𝑥𝑥 = tanh𝑦𝑦

Forms an equation in .

Answer	Marks	Guidance
2𝑦𝑦	1.1a	M1

𝑒𝑒

Isolates the terms in .

Answer	Marks	Guidance
2𝑦𝑦	1.1a	M1

𝑒𝑒

Completes a rigorous argument to show that

−1 1 1+𝑥𝑥

tanh 𝑥𝑥 = ln� �

Answer	Marks	Guidance
2 1−𝑥𝑥	2.1	R1

Answer	Marks	Guidance
8(b)	Selects a method by assuming, for contradiction, that the graphs do
meet and forms the equation .	3.1a	M1

sinh𝑥𝑥 = cosh𝑥𝑥

1 𝑥𝑥 −𝑥𝑥 1 𝑥𝑥 − 𝑥𝑥

2(e −e )= 2(e +e )

𝑥𝑥 −𝑥𝑥 𝑥𝑥 −𝑥𝑥

e −e = e +e

−𝑥𝑥

bu0t = 2e

− 𝑥𝑥

the graphs of eand > 0 do not intersect

sinh𝑥𝑥 = cosh𝑥𝑥

Deduces that or that .

Answer	Marks	Guidance
−𝑥𝑥	2.2a	A1

tanh𝑥𝑥 = 1 2e = 0

Completes a rigorous argument to prove the required result that the

graphs of and do not intersect.

Answer	Marks	Guidance
𝑦𝑦 = sinh𝑥𝑥 𝑦𝑦 = cosh𝑥𝑥	2.1	R1
Total	8	∴ 𝑦𝑦 = sinh𝑥𝑥 𝑦𝑦 = cosh𝑥𝑥
Q	Marking instructions	AO

Question 8:
--- 8(a) ---
8(a) | Writes or in exponential form. | 1.2 | B1 | Let
− 1
𝑦𝑦 = tanh 𝑥𝑥
𝑥𝑥 = tanh𝑦𝑦
𝑦𝑦 −𝑦𝑦
(𝑒𝑒 −𝑒𝑒 )
2
𝑥𝑥 = 𝑦𝑦 −𝑦𝑦
(𝑒𝑒 −𝑒𝑒 )
2
2𝑦𝑦
𝑒𝑒 −1
= 2𝑦𝑦
𝑒𝑒 +1
2𝑦𝑦 2𝑦𝑦
𝑥𝑥(e +1) = e −1
2y 2𝑦𝑦
𝑥𝑥e +𝑥𝑥 = e −1
2𝑦𝑦 2𝑦𝑦
1+𝑥𝑥 = e −𝑥𝑥e
2𝑦𝑦
1+𝑥𝑥 = e (1− 𝑥𝑥)
1+𝑥𝑥 2𝑦𝑦
= e
1−𝑥𝑥
1+𝑥𝑥
2𝑦𝑦 = ln� �
1−𝑥𝑥
1 1+𝑥𝑥
𝑦𝑦 = ln � �
2 1−𝑥𝑥
−1 1 1+𝑥𝑥
tanh 𝑥𝑥 = ln� �
2 1−𝑥𝑥
tanh𝑥𝑥 tanh𝑦𝑦
Selects a method by forming an equation of the form . | 3.1a | M1
𝑥𝑥 = tanh𝑦𝑦
Forms an equation in .
2𝑦𝑦 | 1.1a | M1
𝑒𝑒
Isolates the terms in .
2𝑦𝑦 | 1.1a | M1
𝑒𝑒
Completes a rigorous argument to show that
−1 1 1+𝑥𝑥
tanh 𝑥𝑥 = ln� �
2 1−𝑥𝑥 | 2.1 | R1
--- 8(b) ---
8(b) | Selects a method by assuming, for contradiction, that the graphs do
meet and forms the equation . | 3.1a | M1 | If the graphs meet, then
sinh𝑥𝑥 = cosh𝑥𝑥
1 𝑥𝑥 −𝑥𝑥 1 𝑥𝑥 − 𝑥𝑥
2(e −e )= 2(e +e )
𝑥𝑥 −𝑥𝑥 𝑥𝑥 −𝑥𝑥
e −e = e +e
−𝑥𝑥
bu0t = 2e
− 𝑥𝑥
the graphs of eand > 0 do not intersect
sinh𝑥𝑥 = cosh𝑥𝑥
Deduces that or that .
−𝑥𝑥 | 2.2a | A1
tanh𝑥𝑥 = 1 2e = 0
Completes a rigorous argument to prove the required result that the
graphs of and do not intersect.
𝑦𝑦 = sinh𝑥𝑥 𝑦𝑦 = cosh𝑥𝑥 | 2.1 | R1
Total | 8 | ∴ 𝑦𝑦 = sinh𝑥𝑥 𝑦𝑦 = cosh𝑥𝑥
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item Prove that
$$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$$
[5 marks]

\item Prove that the graphs of
$$y = \sinh x \quad \text{and} \quad y = \cosh x$$
do not intersect.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q8 [8]}}

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