CP-8 Hyperbolic functions — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-8. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-8.1 · Understand the definitions of hyperbolic functions sinh x, cosh x and tanh x, including their domains and ranges, and be able to sketch their graphs.

Explanation

The hyperbolic functions are $\sinh x=\dfrac{e^x-e^{-x}}2$ , $\cosh x=\dfrac{e^x+e^{-x}}2$ and $\tanh x=\dfrac{\sinh x}{\cosh x}$ . Each has domain $\mathbb R$ .
The ranges are respectively $\mathbb R$ , $[1,\infty)$ and $(-1,1)$ .
Both $\sinh x$ and $\tanh x$ are odd, pass through the origin and increase; $\cosh x$ is even, with minimum $(0,1)$ .
The graph of $\tanh x$ approaches the horizontal asymptotes $y=\pm1$ but never reaches them.
Examiners expect sketches to show symmetry, intercepts, turning points and asymptotes, and exponential definitions may be used to calculate exact values or prove identities.

The characteristic shapes, intercepts and asymptotes of the three hyperbolic functions.

Worked example

Find the exact values of $\sinh(\ln4)$ , $\cosh(\ln4)$ and $\tanh(\ln4)$ .

1. $e^{\ln4}=4$ and $e^{-\ln4}=\dfrac14$ .
2. $\sinh(\ln4)=\dfrac12(4-\tfrac14)=\dfrac{15}{8}$ and $\cosh(\ln4)=\dfrac12(4+\tfrac14)=\dfrac{17}{8}$ .
3. $\tanh(\ln4)=\dfrac{15/8}{17/8}=\dfrac{15}{17}$ .

Answer: $\sinh(\ln4)=\dfrac{15}{8}$ , $\cosh(\ln4)=\dfrac{17}{8}$ and $\tanh(\ln4)=\dfrac{15}{17}$ .

Common mistakes

Don't give $\tanh x$ the closed range $[-1,1]$ even though neither asymptote is reached.
Don't sketch $\cosh x$ through the origin instead of through its minimum $(0,1)$ .
Don't treat $\cosh x$ as odd and lose its symmetry about the $y$ -axis.

Exam tip

A graph-sketch answer should label $(0,0)$ or $(0,1)$ and both asymptotes where applicable.

Tier 1 · Easy

1.
State the domain, range and $y$ -intercept of $y=\cosh x$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use the exponential definitions to solve $\cosh x-\sinh x=7$ exactly.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Using the exponential definitions, find the exact values of $\sinh(\ln3)$ , $\cosh(\ln3)$ and $\tanh(\ln3)$ .
(4)
(Total for Question 1 is 4 marks)
2.
State the domain and range of $y=\cosh(x-\ln2)-1$ , and find the coordinates of its minimum point and of any points where it meets the axes.
(5)
(Total for Question 2 is 5 marks)
3.
Sketch $y=2\sinh(3x-\ln5)+4$ . Label its point of inflection and $y$ -intercept, and state its domain and range.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Sketch $y=2-3\tanh x$ . Label its intercept and both horizontal asymptotes, and state its range.
(5)
(Total for Question 1 is 5 marks)
2.
Solve $3\cosh x-5\sinh x=1$ exactly, and justify why there is no second real solution.
(6)
(Total for Question 2 is 6 marks)
3.
Let $f(x)=13\cosh x-5\sinh x$ . Write $f(x)$ as a single translated hyperbolic cosine, state its minimum point and range, and solve $f(x)=20$ exactly.
(6)
(Total for Question 3 is 6 marks)
4.
Prove from the exponential definitions that $\cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y$ , and hence find the exact value of $\cosh(\ln2+\ln5)$ .
(6)
(Total for Question 4 is 6 marks)
5.
Using hyperbolic identities, solve $\cosh(2x)+\sinh x=2$ for real $x$ . Give both solutions in logarithmic form.
(6)
(Total for Question 5 is 6 marks)

CP-8.2 · Differentiate and integrate hyperbolic functions.

Explanation

The fundamental derivatives are $\dfrac{d}{dx}(\sinh x)=\cosh x$ , $\dfrac{d}{dx}(\cosh x)=\sinh x$ and $\dfrac{d}{dx}(\tanh x)=\operatorname{sech}^2x$ .
Chain, product and quotient rules apply in the usual way; unlike circular cosine, differentiating $\cosh x$ introduces no minus sign.
Reverse the derivative rules for integration and divide by a constant inner derivative, so $\int\sinh(ax+b)\,dx=\dfrac1a\cosh(ax+b)+C$ .
Expressions may mix powers, products and radicals, as in the specification guidance, so algebraic simplification can be needed before choosing a rule.
Examiners expect every chain-rule factor and the arbitrary constant in an indefinite integral.

Worked example

Differentiate $f(x)=x\sinh^2(3x)$ and then evaluate $f'(0)$ .

1.Use the product rule: $f'(x)=\sinh^2(3x)+x\dfrac{d}{dx}[\sinh^2(3x)]$ .
2.The chain rule gives $\dfrac{d}{dx}[\sinh^2(3x)]=6\sinh(3x)\cosh(3x)$ .
3.Thus $f'(x)=\sinh^2(3x)+6x\sinh(3x)\cosh(3x)$ , so $f'(0)=0$ .

Answer: $f'(x)=\sinh^2(3x)+6x\sinh(3x)\cosh(3x)$ and $f'(0)=0$ .

Common mistakes

Don't write $\dfrac{d}{dx}(\cosh x)=-\sinh x$ by copying the circular-trigonometric sign.
Don't differentiate $\tanh(3x)$ as $\operatorname{sech}^2(3x)$ and omit the factor $3$ .
Don't integrate $\cosh(ax)$ to $a\sinh(ax)$ instead of dividing by $a$ .

Exam tip

For a composite hyperbolic function, display the outer derivative and the inner derivative as separate factors.

Tier 1 · Easy

1.
Differentiate $y=5\cosh(4x)$ with respect to $x$ .
(2)
(Total for Question 1 is 2 marks)
2.
Differentiate $y=\tanh 3x$ with respect to $x$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Find $\displaystyle\int(5\cosh(3x)-2\sinh x)\,dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
Evaluate $\displaystyle\int_0^{\ln2}e^x\cosh x\,dx$ exactly.
(4)
(Total for Question 2 is 4 marks)
3.
For $f(x)=x/\cosh x$ , find $f'(x)$ and the exact equation of the tangent to the curve at $x=\ln2$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The function is $f(x)=e^{-x}\cosh(2x)$ . Find the exact coordinate of its stationary point and determine its nature.
(6)
(Total for Question 1 is 6 marks)
2.
For $f(x)=\dfrac{\sinh x}{1+\cosh x}$ , prove that $f'(x)=\dfrac{1}{1+\cosh x}$ . Also evaluate $\displaystyle\int_0^{\ln3}f(x)\,dx$ exactly.
(7)
(Total for Question 2 is 7 marks)
3.
For $f(x)=\dfrac{\sinh x}{\cosh^2x}$ , find the stationary points and determine their nature, hence state the range of $f$ , and find all its points of inflection.
(7)
(Total for Question 3 is 7 marks)
4.
Use integration by parts to evaluate $\displaystyle\int_0^{\ln2}x\cosh(2x)\,dx$ exactly.
(7)
(Total for Question 4 is 7 marks)
5.
Show that $\displaystyle\int\tanh x\,dx=\ln(\cosh x)+C$ , and hence find the exact value of $\displaystyle\int_0^{\ln9}\tanh x\,dx$ .
(7)
(Total for Question 5 is 7 marks)

CP-8.3 · Understand and be able to use the definitions of the inverse hyperbolic functions and their domains and ranges.

Explanation

An inverse hyperbolic function reverses a suitable one-to-one hyperbolic function. Thus $y=\operatorname{arsinh}x$ means $x=\sinh y$ and has domain and range $\mathbb R$ .
The function $\cosh x$ must first be restricted to $x\geq0$ ; consequently $\operatorname{arcosh}x$ has domain $[1,\infty)$ and range $[0,\infty)$ .
Finally, $y=\operatorname{artanh}x$ means $x=\tanh y$ , with domain $(-1,1)$ and range $\mathbb R$ .
Applying a function and its inverse is valid only on these domains.
Examiners expect domain checks when solving equations and the principal, non-negative value from $\operatorname{arcosh}$ , not both signs from the even function $\cosh$ .

Worked example

Solve $\operatorname{arcosh}(2x-1)=\ln3$ exactly.

1.Apply $\cosh$ to both sides: $2x-1=\cosh(\ln3)$ .
2. $\cosh(\ln3)=\dfrac12(3+\tfrac13)=\dfrac53$ .
3.Hence $2x=\dfrac83$ and $x=\dfrac43$ ; also $2x-1=\dfrac53\geq1$ , so the input is in the domain.

Answer: $x=\dfrac43$ .

Common mistakes

Don't give $\operatorname{arcosh}x$ range $\mathbb R$ even though the principal inverse is non-negative.
Don't accept an $\operatorname{artanh}$ input with absolute value at least $1$ .
Don't introduce a plus-or-minus sign after applying $\cosh$ to an $\operatorname{arcosh}$ equation.

Exam tip

State the inverse function's domain beside any algebraic solution that lies near an endpoint.

Tier 1 · Easy

1.
State the domain and range of $y=\operatorname{arcosh}x$ .
(2)
(Total for Question 1 is 2 marks)
2.
Find the exact values of $\cosh(\operatorname{arcosh}(7/4))$ and $\operatorname{arcosh}(\cosh(-\ln2))$ . Explain the sign in the second result.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Solve $\operatorname{arsinh}(2x-1)=\ln2$ exactly.
(3)
(Total for Question 1 is 3 marks)
2.
Solve $\operatorname{artanh}(3x+1)=\operatorname{arcosh}(5/3)$ exactly, checking the inverse-tanh domain.
(5)
(Total for Question 2 is 5 marks)
3.
Solve $\operatorname{arcosh}(x^2-1)=\operatorname{arsinh}(35/12)$ exactly, checking the domain of $\operatorname{arcosh}$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Let $u=\operatorname{artanh}(1/3)$ . Find the exact value of $x$ satisfying $\operatorname{arcosh}x=2u$ , without using decimal approximations.
(5)
(Total for Question 1 is 5 marks)
2.
Find every real solution of $\operatorname{arcosh}(1+2x^2)=\operatorname{arsinh}(3x)$ , with all domain restrictions made explicit.
(7)
(Total for Question 2 is 7 marks)
3.
Solve $\operatorname{artanh}\left(\tfrac12\operatorname{arcosh}x\right)=\operatorname{arsinh}(3/4)$ exactly. State the domain restrictions on $x$ , and hence find $\operatorname{arsinh}(\sqrt{x^2-1})$ .
(7)
(Total for Question 3 is 7 marks)
4.
Solve $\operatorname{arcosh}(\cosh(2x-3))=x+1$ for real $x$ .
(7)
(Total for Question 4 is 7 marks)
5.
For $f(x)=\operatorname{arcosh}(2-x)+\operatorname{arsinh}x$ , state its domain, find its stationary point and determine its nature. Also find $f(1)$ .
(7)
(Total for Question 5 is 7 marks)

CP-8.4 · Derive and use the logarithmic forms of the inverse hyperbolic functions.

Explanation

The logarithmic forms are $\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$ , $\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})$ for $x\geq1$ , and $\operatorname{artanh}x=\dfrac12\ln\!\left(\dfrac{1+x}{1-x}\right)$ for $|x|<1$ .
They follow by replacing the hyperbolic function by exponentials, setting $u=e^y>0$ , and solving the resulting quadratic for $u$ .
The positive exponential condition rejects the negative root.
These forms turn inverse-hyperbolic values into exact logarithms and can prove identities, provided every square root and logarithm is properly defined.
Examiners expect the domain restriction to accompany a derived form and any simplification of $\ln(z^2)$ to justify the sign of $z$ .

Worked example

Derive the logarithmic form of $\operatorname{arsinh}x$ .

1.Let $y=\operatorname{arsinh}x$ , so $2x=e^y-e^{-y}$ .
2.Set $u=e^y>0$ and multiply by $u$ : $u^2-2xu-1=0$ .
3.The roots are $u=x\pm\sqrt{x^2+1}$ ; only $x+\sqrt{x^2+1}$ is positive.
4.Therefore $e^y=x+\sqrt{x^2+1}$ and $y=\ln(x+\sqrt{x^2+1})$ .

Answer: $\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})$ for $x\in\mathbb R$ .

Common mistakes

Don't keep the negative quadratic root even though $e^y$ must be positive.
Don't use the $\operatorname{arcosh}$ logarithmic form when $x<1$ .
Don't omit the factor $\tfrac12$ from the logarithmic form of $\operatorname{artanh}x$ .

Exam tip

In a derivation, write $e^y>0$ before selecting the quadratic root; that line justifies the rejection.

Tier 1 · Easy

1.
Use a logarithmic form to evaluate $\operatorname{arsinh}(3/4)$ exactly.
(2)
(Total for Question 1 is 2 marks)
2.
Use the logarithmic form of $\operatorname{artanh}$ to solve $\operatorname{artanh}(2x)=\ln2$ exactly.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Solve $\operatorname{arsinh}x=\ln5-\ln2$ , giving $x$ exactly.
(4)
(Total for Question 1 is 4 marks)
2.
Show that $\operatorname{artanh}x=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)$ for $|x|<1$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use logarithmic forms to evaluate $\operatorname{arcosh}(25/7)+\operatorname{artanh}(5/13)$ as a single logarithm.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Using logarithmic forms, prove that $\operatorname{artanh}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\operatorname{arsinh}x$ for every real $x$ .
(6)
(Total for Question 1 is 6 marks)
2.
For $u>0$ , prove that $\operatorname{arcosh}\left(\dfrac{u+u^{-1}}2\right)=|\ln u|$ . Your proof must account for all possible sizes of $u$ .
(7)
(Total for Question 2 is 7 marks)
3.
For $0<x<1$ and $0<y<1$ , use logarithmic forms to prove $\operatorname{artanh}x+\operatorname{artanh}y=\operatorname{artanh}\left(\dfrac{x+y}{1+xy}\right)$ . Hence solve $\operatorname{artanh}(1/3)+\operatorname{artanh}z=\operatorname{artanh}(3/4)$ .
(7)
(Total for Question 3 is 7 marks)
4.
Use logarithmic forms to solve $\operatorname{artanh}x+\ln(3/2)\leq\operatorname{arsinh}(20/21)$ .
(7)
(Total for Question 4 is 7 marks)

CP-8.5 · Integrate functions of the form (x^2 + a^2)^(-1/2) and (x^2 - a^2)^(-1/2) and be able to choose substitutions to integrate associated functions.

Explanation

Hyperbolic substitutions exploit $\cosh^2u-\sinh^2u=1$ . For $\sqrt{x^2+a^2}$ with $a>0$ , take $x=a\sinh u$ , giving $dx=a\cosh u\,du$ and $\sqrt{x^2+a^2}=a\cosh u$ .
For $\sqrt{x^2-a^2}$ on the branch $x\geq a$ , take $x=a\cosh u$ , giving $dx=a\sinh u\,du$ and $\sqrt{x^2-a^2}=a\sinh u$ .
Hence the standard reciprocal-root integrals produce $\operatorname{arsinh}(x/a)+C$ and $\operatorname{arcosh}(x/a)+C$ on the appropriate domains.
Associated integrals may require simplifying powers of $\sinh u$ and $\cosh u$ before reversing the substitution into an exact expression in $x$ .
Examiners expect $dx$ , the radical and definite limits all to be transformed before integration.

Worked example

Use $x=2\cosh u$ to evaluate $\displaystyle\int_2^4\dfrac{x}{\sqrt{x^2-4}}\,dx$ .

1. $dx=2\sinh u\,du$ and $\sqrt{x^2-4}=2\sinh u$ .
2.The limits are $u=0$ and $u=\operatorname{arcosh}2$ .
3.The integral becomes $\int_0^{\operatorname{arcosh}2}2\cosh u\,du=2\sinh(\operatorname{arcosh}2)$ .
4.If $\cosh u=2$ , then $\sinh u=\sqrt{2^2-1}=\sqrt3$ .

Answer: The integral is $2\sqrt3$ .

Common mistakes

Don't transform the square root but leave $dx$ in terms of $x$ .
Don't use $x=a\sinh u$ for $\sqrt{x^2-a^2}$ and fail to simplify the radical.
Don't keep the original $x$ -limits after changing the variable to $u$ .

Exam tip

Write a three-line substitution block for $x$ , $dx$ and the radical before changing a definite integral.

Tier 1 · Easy

1.
Find $\displaystyle\int\dfrac{1}{\sqrt{x^2+16}}\,dx$ .
(2)
(Total for Question 1 is 2 marks)
2.
For $x\geq5$ , choose a hyperbolic substitution for $\displaystyle\int\dfrac{\sqrt{x^2-25}}{x}\,dx$ , and state the resulting expressions for $dx$ and $\sqrt{x^2-25}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Using $x=3\sinh u$ , evaluate $\displaystyle\int_0^4\sqrt{x^2+9}\,dx$ exactly.
(6)
(Total for Question 1 is 6 marks)
2.
Use the substitution $2x=\sinh u$ to evaluate $\displaystyle\int_0^{3/2}\dfrac{dx}{\sqrt{4x^2+1}}$ exactly.
(4)
(Total for Question 2 is 4 marks)
3.
The positive constant $c$ satisfies $\displaystyle\int_0^c\dfrac{dx}{(x^2+4)^{3/2}}=\dfrac3{20}$ . Use a hyperbolic substitution to find $c$ exactly.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Use a hyperbolic substitution to evaluate $\displaystyle\int_3^5\dfrac{2x^2-9}{\sqrt{x^2-9}}\,dx$ exactly.
(6)
(Total for Question 1 is 6 marks)
2.
Use a suitable hyperbolic substitution, including transformed limits, to evaluate $\displaystyle\int_2^3\dfrac{x^3}{\sqrt{x^2-4}}\,dx$ exactly.
(7)
(Total for Question 2 is 7 marks)
3.
The curve $C$ has equation $y=\dfrac{\sqrt{x^2-4}}{x^2}$ . The finite region $R$ is bounded by $C$ , the $x$ -axis and the lines with equations $x=4$ and $x=6$ . Use a hyperbolic substitution to find the exact area of $R$ .
(7)
(Total for Question 3 is 7 marks)
4.
The finite region $R$ is bounded by $y=\operatorname{arcosh}(x/5)$ , the $x$ -axis and the line $x=13$ . Find the exact area of $R$ .
(7)
(Total for Question 4 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-8.1 · Understand the definitions of hyperbolic functions sinh x, cosh x and tanh x, including their domains and ranges, and be able to sketch their graphs.

Tier 1 · Easy

Mark scheme for CP-8.1 Tier 1 · Easy
Question	Scheme	Marks
1	Domain $\mathbb R$ ; range $[1,\infty)$ ; $y$ -intercept $(0,1)$ .	3
(3 marks)		3
Notes
The exponential definition is valid for every real $x$ . Also $\cosh x\geq1$ , with equality at $x=0$ , so the graph crosses the $y$ -axis at $(0,1)$ .
2	$\cosh x-\sinh x=\tfrac12(e^x+e^{-x})-\tfrac12(e^x-e^{-x})=e^{-x}$ . $e^{-x}=7$ , so $-x=\ln7$ . $x=-\ln7$ ; this is unique because the exponential is one-to-one.	3
(3 marks)		3
Notes
$\cosh x-\sinh x=\tfrac12(e^x+e^{-x})-\tfrac12(e^x-e^{-x})=e^{-x}$ . Hence $e^{-x}=7$ , so $-x=\ln7$ and $x=-\ln7$ . The exponential is one-to-one, so this is the only real solution.

Tier 2 · Standard

Mark scheme for CP-8.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\sinh(\ln3)=\dfrac43$ , $\cosh(\ln3)=\dfrac53$ , $\tanh(\ln3)=\dfrac45$ .	4
(4 marks)		4
Notes
Since $e^{\ln3}=3$ and $e^{-\ln3}=1/3$ , $\sinh(\ln3)=\frac12(3-1/3)=4/3$ and $\cosh(\ln3)=\frac12(3+1/3)=5/3$ . Their quotient is $\tanh(\ln3)=4/5$ .
2	Domain $\mathbb R$ . Since $\cosh(x-\ln2)\geq1$ , the range is $[0,\infty)$ . Equality occurs at $x=\ln2$ , so the minimum point is $(\ln2,0)$ . The graph meets the $x$ -axis only at $(\ln2,0)$ . At $x=0$ , $y=\cosh(\ln2)-1=5/4-1=1/4$ , so the $y$ -intercept is $(0,1/4)$ .	5
(5 marks)		5
Notes
Translate $y=\cosh x$ right by $\ln2$ and down by $1$ . Since $\cosh u\geq1$ , equality occurs only at $u=0$ , so the minimum is reached only when $x=\ln2$ . Therefore the curve touches the $x$ -axis there, has no other $x$ -intercept, and has $y$ -intercept $\cosh(\ln2)-1=\tfrac54-1=\tfrac14$ .
3	It is an increasing transformed $\sinh$ curve. Its point of inflection is $(\tfrac13\ln5,4)$ . Since $\sinh(-\ln5)=-12/5$ , the $y$ -intercept is $(0,-4/5)$ . The domain is $\mathbb R$ . The range is $\mathbb R$ .	5
(5 marks)		5
Notes
The inner expression is zero at $x=\tfrac13\ln5$ . Also $y''=18\sinh(3x-\ln5)$ changes sign there, so the point is an inflection point with ordinate $4$ . At $x=0$ , $\sinh(-\ln5)=\tfrac12(1/5-5)=-12/5$ , giving $y=-24/5+4=-4/5$ . The transformations do not restrict the domain or range of $\sinh$ .

Tier 3 · Hard

Mark scheme for CP-8.1 Tier 3 · Hard
Question	Scheme	Marks
1	A decreasing sigmoid through $(0,2)$ with asymptotes $y=5$ as $x\to-\infty$ and $y=-1$ as $x\to\infty$ . Range $(-1,5)$ .	5
(5 marks)		5
Notes
The graph of $\tanh x$ increases from $-1$ to $1$ and passes through the origin. Multiplication by $-3$ reverses it and scales vertically; adding $2$ gives the intercept $(0,2)$ . Transforming the limiting values gives $2-3(-1)=5$ and $2-3(1)=-1$ , neither attained.
2	Let $u=e^x$ , where $u>0$ . Using the exponential definitions gives $-u+4/u=1$ . Multiplication by $u$ gives $u^2+u-4=0$ . $u=(-1\pm\sqrt{17})/2$ . The root $(-1-\sqrt{17})/2$ is negative, so it cannot equal $e^x$ . $x=\ln\left(\dfrac{\sqrt{17}-1}{2}\right)$ , and rejecting the only other root proves uniqueness.	6
(6 marks)		6
Notes
Let $u=e^x$ , so $u>0$ . Substitution of the exponential definitions gives $-u+4/u=1$ , hence $u^2+u-4=0$ . Thus $u=(-1\pm\sqrt{17})/2$ . The minus choice is negative and cannot equal $e^x$ , while the plus choice is positive. Therefore $x=\ln((\sqrt{17}-1)/2)$ , and rejecting the only other quadratic root proves uniqueness over $\mathbb R$ .
3	$13^2-5^2=144$ , so match $f(x)=12\cosh(x-a)$ . $12\cosh a=13$ and $12\sinh a=5$ , giving $a=\ln(3/2)$ . Hence $f(x)=12\cosh(x-\ln(3/2))$ . The minimum point is $(\ln(3/2),12)$ and the range is $[12,\infty)$ . $f(x)=20$ gives $\cosh(x-\ln(3/2))=5/3$ , so $x-\ln(3/2)=\pm\ln3$ . The solutions are $x=\ln(9/2)$ and $x=\ln(1/2)$ .	6
(6 marks)		6
Notes
Write $f(x)=R\cosh(x-a)$ . Coefficient comparison gives $R\cosh a=13$ and $R\sinh a=5$ , so $R^2=169-25=144$ and $R=12$ . Since $\cosh(\ln(3/2))=13/12$ and $\sinh(\ln(3/2))=5/12$ , $a=\ln(3/2)$ . The minimum follows from $\cosh u\geq1$ . For $f=20$ , $\cosh(x-a)=5/3=\cosh(\ln3)$ , so $x=a\pm\ln3$ , which gives the stated solutions.
4	$\cosh x\cosh y=\tfrac14(e^x+e^{-x})(e^y+e^{-y})$ . $\sinh x\sinh y=\tfrac14(e^x-e^{-x})(e^y-e^{-y})$ . Adding cancels the two cross terms. The sum is $\tfrac12(e^{x+y}+e^{-(x+y)})=\cosh(x+y)$ . $\cosh(\ln2+\ln5)=\cosh(\ln10)$ . Therefore the exact value is $\tfrac12(10+1/10)=101/20$ .	6
(6 marks)		6
Notes
Expand both products using the exponential definitions. On addition, the mixed exponentials cancel and the remaining pair is exactly the exponential definition of $\cosh(x+y)$ . Apply the identity with $x=\ln2$ and $y=\ln5$ , then evaluate $\cosh(\ln10)$ directly.
5	Use $\cosh(2x)=1+2\sinh^2x$ . Writing $s=\sinh x$ gives $2s^2+s-1=0$ . The factorisation is $(2s-1)(s+1)=0$ . Hence $\sinh x=1/2$ or $\sinh x=-1$ . The first solution is $x=\ln((1+\sqrt5)/2)$ . The second solution is $x=-\ln(1+\sqrt2)$ .	6
(6 marks)		6
Notes
Replace $\cosh(2x)$ by $1+2\sinh^2x$ and solve the resulting quadratic in $s=\sinh x$ . The function $\sinh$ is one-to-one, so each value of $s$ gives one real solution. Applying $\operatorname{arsinh}u=\ln(u+\sqrt{u^2+1})$ and oddness gives the stated logarithms.

CP-8.2 · Differentiate and integrate hyperbolic functions.

Tier 1 · Easy

Mark scheme for CP-8.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=20\sinh(4x)$	2
(2 marks)		2
Notes
Differentiate $\cosh(4x)$ to obtain $4\sinh(4x)$ by the chain rule, then multiply by $5$ .
2	$\dfrac{d}{dx}(\tanh 3x)=\operatorname{sech}^2(3x)\dfrac{d}{dx}(3x)$ $\dfrac{dy}{dx}=3\operatorname{sech}^2(3x)$	2
(2 marks)		2
Notes
Apply the chain rule. The derivative of $\tanh u$ is $\operatorname{sech}^2u$ , and the derivative of the inner function $3x$ is $3$ , giving $dy/dx=3\operatorname{sech}^2(3x)$ .

Tier 2 · Standard

Mark scheme for CP-8.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac53\sinh(3x)-2\cosh x+C$	3
(3 marks)		3
Notes
Reverse the chain rule: $\int5\cosh(3x)\,dx=(5/3)\sinh(3x)$ . Also $\int-2\sinh x\,dx=-2\cosh x$ . Add the arbitrary constant.
2	$\cosh x=\tfrac12(e^x+e^{-x})$ . $e^x\cosh x=\tfrac12(e^{2x}+1)$ . An antiderivative is $\tfrac14e^{2x}+\tfrac12x$ . Evaluation from $0$ to $\ln2$ gives $\dfrac34+\dfrac12\ln2$ .	4
(4 marks)		4
Notes
From $\cosh x=(e^x+e^{-x})/2$ , the integrand is $(e^{2x}+1)/2$ . An antiderivative is $e^{2x}/4+x/2$ . Evaluation between $0$ and $\ln2$ gives $(4-1)/4+(\ln2)/2=3/4+(\ln2)/2$ .
3	$f'(x)=\dfrac{\cosh x-x\sinh x}{\cosh^2x}$ . $\cosh(\ln2)=5/4$ and $\sinh(\ln2)=3/4$ . The point of contact is $(\ln2,4\ln2/5)$ . The tangent gradient is $4(5-3\ln2)/25$ . $y-\dfrac{4\ln2}{5}=\dfrac{4(5-3\ln2)}{25}(x-\ln2)$ .	5
(5 marks)		5
Notes
The quotient rule gives $f'=(\cosh x-x\sinh x)/\cosh^2x$ . From the exponential definitions, $\cosh(\ln2)=5/4$ and $\sinh(\ln2)=3/4$ . Therefore $f(\ln2)=4\ln2/5$ and $f'(\ln2)=(5/4-3\ln2/4)/(25/16)=4(5-3\ln2)/25$ . Substitute the point and gradient into point-gradient form.

Tier 3 · Hard

Mark scheme for CP-8.2 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\dfrac14\ln3$ , $f(x)=\dfrac{2}{3^{3/4}}$ ; the stationary point is a minimum.	6
(6 marks)		6
Notes
Product differentiation gives $f'(x)=e^{-x}[2\sinh(2x)-\cosh(2x)]$ . Hence $f'(x)=0$ when $\tanh(2x)=1/2$ , so $2x=\frac12\ln3$ and $x=\frac14\ln3$ . At this value, $e^{2x}=3^{1/2}$ , so $\cosh(2x)=2/\sqrt3$ and $e^{-x}=3^{-1/4}$ ; thus $f=2/3^{3/4}$ . Since $2\tanh(2x)-1$ changes from negative to positive, the point is a minimum.
2	The quotient rule gives $f'(x)=\dfrac{\cosh x(1+\cosh x)-\sinh^2x}{(1+\cosh x)^2}$ . Using $\cosh^2x-\sinh^2x=1$ , the numerator simplifies to $1+\cosh x$ . Hence $f'(x)=\dfrac{1}{1+\cosh x}$ . For the integral, let $u=1+\cosh x$ , so $du=\sinh x\,dx$ . $\displaystyle\int f(x)\,dx=\ln(1+\cosh x)+C$ . $\cosh(\ln3)=\tfrac12(3+1/3)=5/3$ . $\displaystyle\int_0^{\ln3}f(x)\,dx=\ln(8/3)-\ln2=\ln\left(\dfrac43\right)$ .	7
(7 marks)		7
Notes
The quotient rule gives $f'=[\cosh x(1+\cosh x)-\sinh^2x]/(1+\cosh x)^2$ . Using $\cosh^2x-\sinh^2x=1$ , the numerator is $1+\cosh x$ , proving the derivative. For the integral, let $u=1+\cosh x$ , so $du=\sinh x\,dx$ and $\int f(x)\,dx=\ln(1+\cosh x)$ . Since $\cosh(\ln3)=5/3$ , evaluation gives $[\ln(1+\cosh x)]_0^{\ln3}=\ln(8/3)-\ln2=\ln(4/3)$ .
3	$f'(x)=\dfrac{1-\sinh^2x}{\cosh^3x}$ . The stationary points are $(-\ln(1+\sqrt2),-1/2)$ and $(\ln(1+\sqrt2),1/2)$ . The first is a minimum and the second is a maximum. Since $f(x)\to0$ as $x\to\pm\infty$ , the range is $[-1/2,1/2]$ . $f''(x)=\dfrac{\sinh x(\sinh^2x-5)}{\cosh^4x}$ . Thus the points of inflection are $(-\operatorname{arsinh}\sqrt5,-\sqrt5/6)$ , $(0,0)$ and $(\operatorname{arsinh}\sqrt5,\sqrt5/6)$ . The denominator is positive and the numerator changes sign at each of the three stated values, confirming all three inflections.	7
(7 marks)		7
Notes
Differentiating $\sinh x\,\cosh^{-2}x$ gives $f'=(1-\sinh^2x)/\cosh^3x$ . Its sign changes show a minimum at $\sinh x=-1$ and a maximum at $\sinh x=1$ ; there $\cosh x=\sqrt2$ , so the values are $-1/2$ and $1/2$ . Together with the zero limits at both ends, these give the range. Differentiating again and using $\cosh^2x=1+\sinh^2x$ gives $f''=\sinh x(\sinh^2x-5)/\cosh^4x$ . Its numerator has simple sign-changing zeros at $\sinh x=-\sqrt5,0,\sqrt5$ , and $\cosh^2x=6$ at the outer two, giving the three stated points.
4	Take $u=x$ and $dv=\cosh(2x)\,dx$ . Then $du=dx$ and $v=\tfrac12\sinh(2x)$ . An antiderivative is $\tfrac12x\sinh(2x)-\tfrac14\cosh(2x)$ . $\sinh(2\ln2)=\sinh(\ln4)=15/8$ . $\cosh(2\ln2)=\cosh(\ln4)=17/8$ . Evaluation at the bounds gives $\dfrac{15}{16}\ln2-\dfrac{17}{32}+\dfrac14$ . The exact value is $\dfrac{15}{16}\ln2-\dfrac9{32}$ .	7
(7 marks)		7
Notes
Integration by parts gives $[x\sinh(2x)/2-\cosh(2x)/4]_0^{\ln2}$ . The exponential definitions give $\sinh(\ln4)=15/8$ and $\cosh(\ln4)=17/8$ . Substitution, including the lower-limit contribution $1/4$ , simplifies to $15\ln2/16-9/32$ .
5	$\tanh x=\sinh x/\cosh x$ . Set $u=\cosh x$ , so $du=\sinh x\,dx$ . The integral becomes $\int du/u=\ln\|u\|+C$ . Since $\cosh x>0$ , this is $\ln(\cosh x)+C$ . $\displaystyle\int_0^{\ln9}\tanh x\,dx=[\ln(\cosh x)]_0^{\ln9}$ . $\cosh(\ln9)=\tfrac12(9+1/9)=41/9$ and $\cosh0=1$ . The exact value is $\ln(41/9)$ .	7
(7 marks)		7
Notes
Write tanh as $\sinh x/\cosh x$ and substitute $u=\cosh x$ . Positivity of cosh removes the absolute value. Evaluate the resulting logarithm at the two limits using the exponential definition of cosh.

CP-8.3 · Understand and be able to use the definitions of the inverse hyperbolic functions and their domains and ranges.

Tier 1 · Easy

Mark scheme for CP-8.3 Tier 1 · Easy
Question	Scheme	Marks
1	Domain $[1,\infty)$ ; range $[0,\infty)$ .	2
(2 marks)		2
Notes
The inverse uses the one-to-one branch of $\cosh y$ for $y\geq0$ . That branch takes values from $1$ upwards.
2	$\cosh(\operatorname{arcosh}(7/4))=7/4$ because $7/4$ is in the domain of arcosh. $\cosh(-\ln2)=\cosh(\ln2)$ because cosh is even. $\operatorname{arcosh}(\cosh(-\ln2))=\ln2$ because arcosh returns the principal non-negative value.	3
(3 marks)		3
Notes
The first composition returns $7/4$ because this lies in the domain $[1,\infty)$ of arcosh. For the second, $\cosh$ is even, so $\cosh(-\ln2)=\cosh(\ln2)$ . The principal arcosh value is non-negative, and therefore it returns $\ln2$ rather than $-\ln2$ .

Tier 2 · Standard

Mark scheme for CP-8.3 Tier 2 · Standard
Question	Scheme	Marks
1	$x=\dfrac78$	3
(3 marks)		3
Notes
Apply $\sinh$ to both sides: $2x-1=\sinh(\ln2)=\frac12(2-1/2)=3/4$ . Hence $2x=7/4$ and $x=7/8$ .
2	Let $u=\operatorname{arcosh}(5/3)$ , so $\cosh u=5/3$ and $u\geq0$ . $\sinh u=\sqrt{(5/3)^2-1}=4/3$ . $\tanh u=(4/3)/(5/3)=4/5$ . $3x+1=4/5$ , so $x=-1/15$ . At this value the artanh input is $4/5$ , and $\|4/5\|<1$ .	5
(5 marks)		5
Notes
Let $u=\operatorname{arcosh}(5/3)$ . Then $\cosh u=5/3$ and, since $u\geq0$ , $\sinh u=\sqrt{(5/3)^2-1}=4/3$ . Hence $\tanh u=(4/3)/(5/3)=4/5$ . Applying $\tanh$ gives $3x+1=4/5$ , so $x=-1/15$ . The inverse-tanh input is $4/5$ , whose modulus is below $1$ .
3	$\operatorname{arsinh}(35/12)=\ln6$ because $\sinh(\ln6)=35/12$ . Applying $\cosh$ gives $x^2-1=\cosh(\ln6)=37/12$ . Hence $x^2=49/12$ . The candidates are $x=\pm\dfrac{7\sqrt3}{6}$ . For both candidates, $x^2-1=37/12\geq1$ , so both satisfy the arcosh domain and both are solutions.	5
(5 marks)		5
Notes
Since $\sinh(\ln6)=\tfrac12(6-1/6)=35/12$ , the right side is $\ln6$ . Applying $\cosh$ is reversible on the principal arcosh range and gives $x^2-1=\cosh(\ln6)=\tfrac12(6+1/6)=37/12$ . Thus $x^2=49/12$ , so $x=\pm7\sqrt3/6$ . For either sign the arcosh input is $37/12$ , so the required domain condition is satisfied.

Tier 3 · Hard

Mark scheme for CP-8.3 Tier 3 · Hard
Question	Scheme	Marks
1	$x=\dfrac54$	5
(5 marks)		5
Notes
Since $\tanh u=1/3$ , the double-angle identity gives $\cosh(2u)=\dfrac{1+\tanh^2u}{1-\tanh^2u}=\dfrac{1+1/9}{1-1/9}=5/4$ . Applying $\cosh$ to $\operatorname{arcosh}x=2u$ gives $x=\cosh(2u)=5/4$ , which is in the required domain $x\geq1$ .
2	The arcosh input satisfies $1+2x^2\geq1$ for every real $x$ . Since arcosh is non-negative, equality requires $\operatorname{arsinh}(3x)\geq0$ , so $x\geq0$ . Applying cosh gives $1+2x^2=\sqrt{1+9x^2}$ . Both sides are non-negative, so squaring gives $(1+2x^2)^2=1+9x^2$ . Simplification gives $x^2(4x^2-5)=0$ . Subject to $x\geq0$ , the candidates are $x=0$ and $x=\dfrac{\sqrt5}{2}$ . Both candidates satisfy the unsquared equation, so both are solutions.	7
(7 marks)		7
Notes
The left side is non-negative, so equality requires $\operatorname{arsinh}(3x)\geq0$ and hence $x\geq0$ ; its arcosh input is valid for every real $x$ . Applying $\cosh$ gives $1+2x^2=\sqrt{1+9x^2}$ . Both sides are non-negative, so squaring is reversible: $4x^4-5x^2=0$ . Thus $x=0$ or $x=\sqrt5/2$ after the condition $x\geq0$ . Direct substitution into the unsquared equation confirms both values.
3	The nested expression requires $x\geq1$ and $0\leq\operatorname{arcosh}x<2$ . $\operatorname{arsinh}(3/4)=\ln2$ because $\sinh(\ln2)=3/4$ . Applying $\tanh$ gives $\tfrac12\operatorname{arcosh}x=\tanh(\ln2)=3/5$ . Hence $\operatorname{arcosh}x=6/5$ . $x=\cosh(6/5)=\tfrac12(e^{6/5}+e^{-6/5})$ . Here $0\leq\operatorname{arcosh}x=6/5<2$ , so every domain restriction holds. Since $x=\cosh(6/5)$ with $6/5>0$ , $\sqrt{x^2-1}=\sinh(6/5)$ and the final value is $6/5$ .	7
(7 marks)		7
Notes
For the inner arcosh to exist, $x\geq1$ ; its half must also lie in $(-1,1)$ , which reduces here to $0\leq\operatorname{arcosh}x<2$ . Since $\sinh(\ln2)=3/4$ , the right side is $\ln2$ . Applying $\tanh$ gives half the arcosh value as $3/5$ , hence $x=\cosh(6/5)$ . This meets the nested domain condition. The positive identity $\sqrt{\cosh^2u-1}=\sinh u$ for $u=6/5$ then makes the final inverse equal to $6/5$ .
4	For real $u$ , $\operatorname{arcosh}(\cosh u)=\|u\|$ . The equation is therefore $\|2x-3\|=x+1$ , with $x\geq-1$ . If $2x-3\geq0$ , then $2x-3=x+1$ . This gives $x=4$ , which satisfies the branch condition. If $2x-3<0$ , then $3-2x=x+1$ . This gives $x=2/3$ , which satisfies the branch condition. Hence the real solutions are $x=4$ and $x=2/3$ .	7
(7 marks)		7
Notes
The principal inverse of cosh is non-negative, so $\operatorname{arcosh}(\cosh u)=\|u\|$ . Split the resulting absolute-value equation into the two sign branches and check each solution against its branch condition.
5	The condition $2-x\geq1$ gives the domain $x\leq1$ . $f'(x)=-1/\sqrt{(2-x)^2-1}+1/\sqrt{x^2+1}$ . Solving $f'(x)=0$ gives $(2-x)^2-1=x^2+1$ , hence $x=1/2$ . $f(1/2)=\ln((3+\sqrt5)/2)+\ln((1+\sqrt5)/2)$ . Combining gives $\ln((3+\sqrt5)(1+\sqrt5)/4)=\ln(2+\sqrt5)$ . $f'$ changes from positive to negative at $1/2$ , so the point is a maximum. $f(1)=\ln(1+\sqrt2)$ .	7
(7 marks)		7
Notes
The arcosh input fixes the domain. Differentiate the two inverse functions and equate the positive denominators to locate the only stationary point; a sign check gives its nature. Evaluate its ordinate and the endpoint value with the logarithmic inverse-hyperbolic forms, combining the two logarithms exactly.

CP-8.4 · Derive and use the logarithmic forms of the inverse hyperbolic functions.

Tier 1 · Easy

Mark scheme for CP-8.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\ln2$	2
(2 marks)		2
Notes
$\operatorname{arsinh}(3/4)=\ln(3/4+\sqrt{9/16+1})=\ln(3/4+5/4)=\ln2$ .
2	$\dfrac12\ln\left(\dfrac{1+2x}{1-2x}\right)=\ln2$ , so $\dfrac{1+2x}{1-2x}=4$ . $1+2x=4-8x$ , hence $x=3/10$ ; then $\|2x\|=3/5<1$ .	2
(2 marks)		2
Notes
Use $\operatorname{artanh}z=\tfrac12\ln((1+z)/(1-z))$ : $\tfrac12\ln((1+2x)/(1-2x))=\ln2$ . Therefore $(1+2x)/(1-2x)=4$ , so $1+2x=4-8x$ and $x=3/10$ . The resulting input $2x=3/5$ lies in $(-1,1)$ .

Tier 2 · Standard

Mark scheme for CP-8.4 Tier 2 · Standard
Question	Scheme	Marks
1	$x=\dfrac{21}{20}$	4
(4 marks)		4
Notes
The right side is $\ln(5/2)$ . Applying $\sinh$ gives $x=\frac12(5/2-2/5)=\frac12(21/10)=21/20$ .
2	Let $y=\operatorname{artanh}x$ , so $x=\tanh y=\dfrac{e^{2y}-1}{e^{2y}+1}$ . $x(e^{2y}+1)=e^{2y}-1$ . Hence $(1-x)e^{2y}=1+x$ , so $e^{2y}=\dfrac{1+x}{1-x}$ . Taking logarithms gives $y=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)$ . For $\|x\|<1$ , $1-x>0$ and $1+x>0$ , so the logarithm is defined.	5
(5 marks)		5
Notes
Let $y=\operatorname{artanh}x$ . Then $x=\tanh y=(e^{2y}-1)/(e^{2y}+1)$ . Rearranging $x(e^{2y}+1)=e^{2y}-1$ gives $(1-x)e^{2y}=1+x$ , and hence $e^{2y}=(1+x)/(1-x)$ . Taking logarithms yields $y=\tfrac12\ln((1+x)/(1-x))$ . When $\|x\|<1$ , both $1-x$ and $1+x$ are positive, so the quotient and its logarithm are defined.
3	$\sqrt{(25/7)^2-1}=24/7$ . $\operatorname{arcosh}(25/7)=\ln((25+24)/7)=\ln7$ . $\operatorname{artanh}(5/13)=\tfrac12\ln((18/13)/(8/13))$ . Hence $\operatorname{artanh}(5/13)=\tfrac12\ln(9/4)=\ln(3/2)$ . The required value is $\ln7+\ln(3/2)=\ln(21/2)$ .	5
(5 marks)		5
Notes
The logarithmic arcosh form gives $\ln(25/7+24/7)=\ln7$ . For the inverse tanh, $(1+5/13)/(1-5/13)=18/8=9/4$ , so the half-logarithm is $\ln(3/2)$ . The logarithm product rule combines the sum to $\ln(21/2)$ .

Tier 3 · Hard

Mark scheme for CP-8.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\operatorname{artanh}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\ln(x+\sqrt{1+x^2})=\operatorname{arsinh}x$ .	6
(6 marks)		6
Notes
Put $s=\sqrt{1+x^2}$ . Since $\|x\|<s$ , the inverse-tanh input is valid. Its logarithmic form is $\frac12\ln((s+x)/(s-x))$ . But $(s-x)(s+x)=s^2-x^2=1$ , so $(s+x)/(s-x)=(s+x)^2$ . As $s+x>0$ , this becomes $\ln(s+x)$ , which is the logarithmic form of $\operatorname{arsinh}x$ .
2	For $u>0$ , $\dfrac{u+u^{-1}}2\geq1$ , so the arcosh input is valid. $\sqrt{((u+u^{-1})/2)^2-1}=\dfrac{\|u-u^{-1}\|}{2}$ . The logarithmic form gives $\operatorname{arcosh}((u+u^{-1})/2)=\ln\left(\dfrac{u+u^{-1}+\|u-u^{-1}\|}{2}\right)$ . If $u>1$ , then $\|u-u^{-1}\|=u-u^{-1}$ , so the result is $\ln u$ . If $0<u<1$ , then $\|u-u^{-1}\|=u^{-1}-u$ , so the result is $\ln(u^{-1})=-\ln u$ . If $u=1$ , both sides equal $0$ . The three cases combine to give $\operatorname{arcosh}((u+u^{-1})/2)=\|\ln u\|$ for every $u>0$ .	7
(7 marks)		7
Notes
For $u>0$ , $(u+u^{-1})/2\geq1$ , so arcosh is defined. Its logarithmic form gives $\ln((u+u^{-1}+\|u-u^{-1}\|)/2)$ . If $u\geq1$ , the absolute value is $u-u^{-1}$ and the result is $\ln u$ . If $0<u<1$ , it is $u^{-1}-u$ and the result is $\ln(u^{-1})=-\ln u$ . At $u=1$ both sides are $0$ , so the two cases combine to $\|\ln u\|$ .
3	The left side is $\tfrac12\ln\left(\dfrac{(1+x)(1+y)}{(1-x)(1-y)}\right)$ . Let $w=(x+y)/(1+xy)$ ; then $1+w=(1+x)(1+y)/(1+xy)$ . Also $1-w=(1-x)(1-y)/(1+xy)$ . Therefore the logarithmic expression is $\tfrac12\ln((1+w)/(1-w))=\operatorname{artanh}w$ . The restrictions give $0<w<1$ , so every inverse-tanh input is valid. For the stated equation, $(1/3+z)/(1+z/3)=3/4$ . Solving gives $z=5/9$ , which lies in $(-1,1)$ .	7
(7 marks)		7
Notes
Add the two logarithmic forms and combine their arguments. With $w=(x+y)/(1+xy)$ , direct simplification gives $(1+w)/(1-w)=(1+x)(1+y)/[(1-x)(1-y)]$ . For $0<x,y<1$ , both $1-w$ and $w$ are positive, so $0<w<1$ and the form is valid. Injectivity of $\operatorname{artanh}$ changes the numerical equation to $(1/3+z)/(1+z/3)=3/4$ , whose solution is $z=5/9$ .
4	The artanh term requires $-1<x<1$ . $\operatorname{arsinh}(20/21)=\ln(7/3)$ . The inequality becomes $\operatorname{artanh}x\leq\ln(14/9)$ . Since artanh is strictly increasing, $x\leq\tanh(\ln(14/9))$ . $\tanh(\ln a)=(a^2-1)/(a^2+1)$ for $a>0$ . Thus $\tanh(\ln(14/9))=(196-81)/(196+81)=115/277$ . The solution set is $-1<x\leq115/277$ .	7
(7 marks)		7
Notes
Evaluate the fixed arsinh value by its logarithmic form and combine the remaining logarithms. Monotonicity of artanh preserves the inequality. Its inverse at $\ln(14/9)$ is evaluated with the exponential definition of tanh, then intersected with the original domain.

CP-8.5 · Integrate functions of the form (x^2 + a^2)^(-1/2) and (x^2 - a^2)^(-1/2) and be able to choose substitutions to integrate associated functions.

Tier 1 · Easy

Mark scheme for CP-8.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\operatorname{arsinh}(x/4)+C$ , equivalently $\ln(x+\sqrt{x^2+16})+C$ .	2
(2 marks)		2
Notes
Use the standard form with $a=4$ : the antiderivative is $\operatorname{arsinh}(x/4)+C$ . Its logarithmic form differs from $\ln(x+\sqrt{x^2+16})$ only by the constant $-\ln4$ .
2	Use $x=5\cosh u$ with $u\geq0$ . $dx=5\sinh u\,du$ . $\sqrt{x^2-25}=5\sinh u$ .	3
(3 marks)		3
Notes
The radical has the form $\sqrt{x^2-a^2}$ on the branch $x\geq a$ , so set $x=a\cosh u$ with $a=5$ and $u\geq0$ . Then $\cosh^2u-1=\sinh^2u$ gives the stated positive radical and differentiation gives $dx$ .

Tier 2 · Standard

Mark scheme for CP-8.5 Tier 2 · Standard
Question	Scheme	Marks
1	$10+\dfrac92\ln3$	6
(6 marks)		6
Notes
With $x=3\sinh u$ , $dx=3\cosh u\,du$ and $\sqrt{x^2+9}=3\cosh u$ . The limits are $0$ and $\operatorname{arsinh}(4/3)=\ln3$ . Hence the integral is $9\int_0^{\ln3}\cosh^2u\,du=9[u/2+\sinh(2u)/4]_0^{\ln3}$ . Since $\sinh(2\ln3)=40/9$ , the value is $(9/2)\ln3+10$ .
2	$2x=\sinh u$ gives $dx=\tfrac12\cosh u\,du$ . $\sqrt{4x^2+1}=\cosh u$ , and the transformed limits are $0$ and $\operatorname{arsinh}3$ . $\displaystyle\int_0^{3/2}\dfrac{dx}{\sqrt{4x^2+1}}=\tfrac12\operatorname{arsinh}3$ . $\displaystyle\tfrac12\operatorname{arsinh}3=\tfrac12\ln(3+\sqrt{10})$ .	4
(4 marks)		4
Notes
Set $2x=\sinh u$ . Then $dx=\tfrac12\cosh u\,du$ and $\sqrt{4x^2+1}=\cosh u$ . The limits $x=0$ and $x=3/2$ become $u=0$ and $u=\operatorname{arsinh}3$ . Hence the integral is $\tfrac12\int_0^{\operatorname{arsinh}3}du=\tfrac12\operatorname{arsinh}3=\tfrac12\ln(3+\sqrt{10})$ .
3	Let $x=2\sinh u$ , so $dx=2\cosh u\,du$ . Then $(x^2+4)^{3/2}=8\cosh^3u$ . The limits are $u=0$ and $u=\operatorname{arsinh}(c/2)$ , so the integral is $\tfrac14\tanh(\operatorname{arsinh}(c/2))$ . In terms of $c$ , this is $c/(4\sqrt{c^2+4})$ , so $c/(4\sqrt{c^2+4})=3/20$ . Squaring $5c=3\sqrt{c^2+4}$ gives $16c^2=36$ . Since $c>0$ , $c=3/2$ .	6
(6 marks)		6
Notes
With $x=2\sinh u$ , the transformed integrand is $\tfrac14\operatorname{sech}^2u$ , and the upper limit is $\operatorname{arsinh}(c/2)$ . Thus the integral is one quarter of the corresponding tanh value. Since $\sinh u=c/2$ and $\cosh u=\sqrt{c^2+4}/2$ , this is $c/(4\sqrt{c^2+4})$ . Equating to $3/20$ and using $c>0$ gives $c=3/2$ .

Tier 3 · Hard

Mark scheme for CP-8.5 Tier 3 · Hard
Question	Scheme	Marks
1	$20$	6
(6 marks)		6
Notes
Let $x=3\cosh u$ . Then $dx=3\sinh u\,du$ , $\sqrt{x^2-9}=3\sinh u$ , and the limits are $u=0$ and $u=\operatorname{arcosh}(5/3)=\ln3$ . The integrand becomes $18\cosh^2u-9=9\cosh(2u)$ . Therefore the integral is $9\int_0^{\ln3}\cosh(2u)\,du=(9/2)\sinh(2\ln3)=(9/2)(40/9)=20$ .
2	Set $x=2\cosh u$ . $dx=2\sinh u\,du$ and $\sqrt{x^2-4}=2\sinh u$ . The transformed limits are $u=0$ and $u=\operatorname{arcosh}(3/2)$ . The integral becomes $8\displaystyle\int_0^{\operatorname{arcosh}(3/2)}\cosh^3u\,du$ . An antiderivative is $8(\sinh u+\tfrac13\sinh^3u)$ . At the upper limit, $\sinh u=\sqrt{(3/2)^2-1}=\sqrt5/2$ . Therefore the value is $8(\sqrt5/2+5\sqrt5/24)=\dfrac{17\sqrt5}{3}$ .	7
(7 marks)		7
Notes
Take $x=2\cosh u$ . Then $dx=2\sinh u\,du$ , $\sqrt{x^2-4}=2\sinh u$ , and the limits are $0$ and $\operatorname{arcosh}(3/2)$ . The integral becomes $8\int\cosh^3u\,du=8[\sinh u+\sinh^3u/3]$ between these limits. At the upper limit, $\sinh u=\sqrt{(3/2)^2-1}=\sqrt5/2$ , so the value is $8(\sqrt5/2+5\sqrt5/24)=17\sqrt5/3$ .
3	Put $\cosh u=x/2$ . Then $\sqrt{x^2-4}=2\sinh u$ and $dx=2\sinh u\,du$ . The limits are $u=\operatorname{arcosh}2$ and $u=\operatorname{arcosh}3$ . The integrand becomes $\tanh^2u=1-\operatorname{sech}^2u$ . An antiderivative is $u-\tanh u$ . At the limits, $\tanh u=\sqrt3/2$ and $2\sqrt2/3$ respectively. $\operatorname{arcosh}2=\ln(2+\sqrt3)$ and $\operatorname{arcosh}3=\ln(3+2\sqrt2)$ . The area is $\ln\left(\dfrac{3+2\sqrt2}{2+\sqrt3}\right)+\dfrac{\sqrt3}{2}-\dfrac{2\sqrt2}{3}$ .	7
(7 marks)		7
Notes
Since the radical is $\sqrt{x^2-2^2}$ , set $x=2\cosh u$ . Then the radical and differential are both $2\sinh u$ factors, so $y\,dx=\tanh^2u\,du$ . The bounds $x=4,6$ become $u=\operatorname{arcosh}2,\operatorname{arcosh}3$ . Evaluating $u-\tanh u$ there, using $\tanh(\operatorname{arcosh}z)=\sqrt{z^2-1}/z$ and the logarithmic arcosh form, gives the stated exact area.
4	$R$ has area $\displaystyle\int_5^{13}\operatorname{arcosh}(x/5)\,dx$ . Use integration by parts with $u=\operatorname{arcosh}(x/5)$ and $dv=dx$ . $du=dx/\sqrt{x^2-25}$ and $v=x$ . The area is $[x\operatorname{arcosh}(x/5)]_5^{13}-\displaystyle\int_5^{13}x/\sqrt{x^2-25}\,dx$ . The remaining integral is $[\sqrt{x^2-25}]_5^{13}=12$ . $\operatorname{arcosh}(13/5)=\ln5$ . Therefore the exact area is $13\ln5-12$ .	7
(7 marks)		7
Notes
The curve meets the axis at $x=5$ . Integrate its inverse-hyperbolic ordinate by parts, using the derivative of arcosh to reduce the remaining integral to the derivative of $\sqrt{x^2-25}$ . Exact evaluation at $x=13$ uses $\cosh(\ln5)=13/5$ .