Hyperbolic Functions Exam Questions: Further Maths

A student can know every definition on the formula sheet and still hesitate when a hyperbolic functions question appears. The difficulty is rarely one isolated fact. It is recognising whether to use exponentials, an identity, an inverse function or a substitution.

These hyperbolic functions exam questions A-Level Further Maths students meet are designed to test that choice. This guide follows a hypothetical student, Maya, through three exam-style questions. Each solution includes indicative method marks so that you can see which lines create credit, not merely which answer finishes the question.

The allocations below are teaching guides rather than official mark schemes. An examiner's precise notation and permitted alternatives depend on the paper. You can compare the approach with an official hyperbolic functions past paper question and its published hyperbolic functions mark scheme.

The method checklist

Before beginning a hyperbolic functions question, ask:

• Can the equation be simplified using ${\cosh }^{2}x-{\sinh }^{2}x\equiv 1$?
• Would the exponential definitions turn it into a quadratic in $e^x$?
• Does an inverse function require a domain restriction?
• Is a hyperbolic substitution suggested by $\sqrt{x^2+a^2}$ or $\sqrt{x^2-a^2}$?
• Have chain-rule constants been included in differentiation or integration?
• Does the final answer need exact logarithmic form?

Maya writes these prompts at the top of her revision page. They slow her down for perhaps ten seconds, but they prevent several minutes of pursuing the wrong method.

Question one: solving with exponential definitions

Question: Solve

$$3\cosh x-2\sinh x=4,$$

giving exact values of $x$.

Maya first looks for the identity ${\cosh }^{2}x-{\sinh }^{2}x\equiv 1$. It does not naturally fit a linear combination, so she changes both functions into exponentials:

$$\cosh x=\frac{e^x+e^{-x}}{2},\sinh x=\frac{e^x-e^{-x}}{2}.$$

Substitution gives

$$\frac{3}{2}(e^x+e^{-x})-(e^x-e^{-x})=4.$$

Collecting terms,

$$\frac{1}{2}{e}^x+\frac{5}{2}{e}^{-x}=4.$$

Multiplying by $2e^x$ produces

$$e^{2x}-8e^x+5=0.$$

Now let $y=e^x$. Since $e^x>0$, Maya knows that any non-positive quadratic root would have to be rejected:

$$y^2-8y+5=0.$$

Therefore,

$$y=\frac{8\pm \sqrt{64-20}}{2}=4\pm \sqrt{11}.$$

Both values are positive, so both are valid:

$$\boxed{x=\ln (4+\sqrt{11})orx=\ln (4-\sqrt{11})}.$$

Indicative method marks

• M1: Uses the correct exponential definitions.
• A1: Obtains $e^{2x}-8e^x+5=0$.
• M1: Makes a valid substitution such as $y=e^x$ and solves the quadratic.
• A1: Finds $e^x=4\pm \sqrt{11}$.
• A1: Gives both exact logarithmic solutions.

The important line is not the final logarithm. It is the conversion to a quadratic. Even if arithmetic later goes wrong, that transformation may secure the main method mark.

If you want this recognition process organised by topic, use Maths Genie's Lessons and Question Bank alongside the A Level formula booklet before attempting full questions.

Question two: deriving an inverse hyperbolic function

Question: Given that $y=\mathrm{arcosh}x$, derive a logarithmic expression for $y$, stating the restriction on $x$.

Maya begins with what the inverse statement means:

$$x=\cosh y=\frac{e^y+e^{-y}}{2}.$$

Because $\mathrm{arcosh}x$ uses the restricted branch of $\cosh y$, she takes $y\ge 0$. Also, $\cosh y\ge 1$, so $x\ge 1$.

Multiplying by $2e^y$ gives

$$2x{e}^y=e^{2y}+1.$$

Rearranging,

$$(e^y{)}^2-2x{e}^y+1=0.$$

Treating this as a quadratic in $e^y$,

$$e^y=x\pm \sqrt{x^2-1}.$$

Since $y\ge 0$, it follows that $e^y\ge 1$. The positive branch consistent with this principal inverse is

$$e^y=x+\sqrt{x^2-1}.$$

Taking logarithms,

$$\boxed{\mathrm{arcosh}x=\ln \left(x+\sqrt{x^2-1}\right),x\ge 1}.$$

Indicative method marks

• M1: Writes $x=\frac{1}{2}(e^y+e^{-y})$.
• M1: Forms a quadratic in $e^y$.
• A1: Obtains $e^y=x\pm \sqrt{x^2-1}$.
• M1: Selects the correct branch using the range of the inverse.
• A1: States the logarithmic result and $x\ge 1$.

This is where notation and domain knowledge become part of the method. Writing the familiar formula from memory may earn an answer mark, but deriving it coherently protects the earlier marks.

Question three: integration by hyperbolic substitution

Question: Use a hyperbolic substitution to find

$$\int \sqrt{x^2+9}\mathrm{d}x.$$

The expression $\sqrt{x^2+a^2}$ suggests $x=a\sinh u$. Maya chooses

$$x=3\sinh u.$$

Then

$$\mathrm{d}x=3\cosh u\mathrm{d}u$$

and

$$\sqrt{x^2+9}=\sqrt{9{\sinh }^{2}u+9}=3\sqrt{{\sinh }^{2}u+1}=3\cosh u.$$

Here she has used $1+{\sinh }^{2}u={\cosh }^{2}u$. The integral becomes

$$9\int {\cosh }^{2}u\mathrm{d}u.$$

Using

$${\cosh }^{2}u=\frac{1+\cosh 2u}{2},$$

we obtain

$$9\int {\cosh }^{2}u\mathrm{d}u=\frac{9}{2}\int (1+\cosh 2u)\mathrm{d}u=\frac{9}{2}u+\frac{9}{4}\sinh 2u+C.$$

Since $\sinh 2u=2\sinh u\cosh u$,

$$\frac{9}{4}\sinh 2u=\frac{9}{2}\left(\frac{x}{3}\right)\left(\frac{\sqrt{x^2+9}}{3}\right)=\frac{x}{2}\sqrt{x^2+9}.$$

Also,

$$u=\mathrm{arsinh}\left(\frac{x}{3}\right).$$

Therefore,

$$\boxed{\int \sqrt{x^2+9}\mathrm{d}x=\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\mathrm{arsinh}\left(\frac{x}{3}\right)+C}.$$

Indicative method marks

• M1: Uses $x=3\sinh u$.
• A1: Finds $\mathrm{d}x=3\cosh u\mathrm{d}u$.
• M1: Simplifies the integral to $9\int {\cosh }^{2}u\mathrm{d}u$.
• M1: Integrates using the double-angle identity.
• A1: Substitutes correctly for $\sinh u$ and $\cosh u$.
• A1: Gives a correct final expression, including $C$.

Common mistakes in hyperbolic functions questions

Copying trigonometric signs

The central identity is

$${\cosh }^{2}x-{\sinh }^{2}x\equiv 1,$$

not a sum of squares. That minus sign affects rearrangements, substitutions and proofs.

Losing the chain-rule factor

For example,

$$\frac{\mathrm{d}}{\mathrm{d}x}\sinh (4x)=4\cosh (4x),$$

while

$$\int \cosh (4x)\mathrm{d}x=\frac{1}{4}\sinh (4x)+C.$$

Ignoring domains and ranges

Inverse hyperbolic functions are not simply buttons on a calculator. For instance, $\mathrm{arcosh}x$ has real domain $x\ge 1$. A derivation should explain why a branch is selected.

Giving decimals too early

Exact answers reveal structure and are often explicitly required. Keep forms involving logarithms and square roots unless the question requests a numerical approximation.

Treating a mark scheme as a model solution

A hyperbolic functions markscheme records acceptable evidence efficiently. It may omit the connective reasoning a student needs when learning. Use the mark scheme to identify credit, then use a full solution to understand why the method works. Maths Genie provides further practice through resources such as the Edexcel Core Pure paper and an OCR Further Maths mark scheme.

Exam-style practice

Try these without looking back at the worked solutions:

1. Solve $5\cosh x-3\sinh x=6$, giving exact answers.
2. Differentiate $x\tanh (2x)$.
3. Use $x=4\cosh u$ to begin evaluating $\int \sqrt{x^2-16}\mathrm{d}x$ for $x\ge 4$.

For each question, write the identity, definition or substitution before doing any manipulation. That first line makes your intention visible and often provides the evidence needed for a method mark. Once individual methods feel secure, attempt complete papers, including Maths Genie's Further Pure Mathematics paper, under timed conditions.

Turn recognition into exam marks

Maya's progress is not a sudden leap in algebra. She learns to pause, classify the expression and write the mark-earning first step before calculating. You now have the same method, but knowing it and applying it consistently across unfamiliar questions are different tasks.

Build that consistency with Maths Genie's Lessons and Question Bank: review the matching hyperbolic functions lesson, then answer the topic questions in exam conditions.