Notes

B1: Recalls the exponential definitions for \(\sinh x\) and at least one of \(\cosh x\) or \(\tanh x\), which may be embedded in their workings.

M1: Substitutes the correct exponential definitions to form an equation leading to a quartic equation for \(\mathrm{e}^x\). Any identities used such as \(\sinh 2x = 2\sinh x\cosh x\) must be correct

A1*: Correct equation following at least one intermediate stage, with no errors seen. They cannot go directly from their substitution to the given answer. cso

Notes

B1: Substitutes \(\mathrm{e}^x = 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.

OR
Substitutes \(x = \ln 3\) into each term of the equation, shows \(= 0\) and states therefore a solution or writes \(\left(\mathrm{e}^x - 3\right)\) is a factor. Allow a tick, box, QED or appropriate conclusion.

OR
Factorises their equation:
\(15\mathrm{e}^{4x} - 48\mathrm{e}^{3x} + 32\mathrm{e}^x - 15 = 0\)
\(\Rightarrow \left(\mathrm{e}^x - 3\right)\left(15\mathrm{e}^{3x} - 3\mathrm{e}^{2x} - 9\mathrm{e}^x + 5\right) = 0\)
and states hence \(\mathrm{e}^x = 3\) is a solution

Do not award this mark for simply solving a quartic such as \(15y^4 - 48y^3 + 32y - 15 = 0\) on the calculator and stating \(y = 3\), hence \(\mathrm{e}^x = 3\) is a solution.

Notes

B1: States the correct exact coordinates. Allow \(x = \ldots,\ y = \ldots\) Do not accept decimals.

Notes

M1: Use the substitution \(u = \pm\dfrac{1}{x}\) to obtain an integral of the form \(\displaystyle\int \lambda\mathrm{e}^u\,\mathrm{d}u\) oe and integrates to \(\lambda\mathrm{e}^u\)
Award for a sight of \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\) as the answer to their integral.
Alternatively applies the reverse of the chain rule \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{\frac{1}{x}}\right) = \pm\dfrac{1}{x^2}e^{\frac{1}{x}}\) and integrates to reach \(\pm\lambda\mathrm{e}^{\frac{1}{x}}\)
Answers using integration by parts are unlikely to lead to a correct solution.

A1: Correct integration. Award for correct answer with minimal workings as this can be done by inspection using the reverse of the chain rule. However, withhold this mark for incorrect workings.

M1: Uses correct notation to write the integral as the limit as \(t \to 0\), with the limits of \(t\) (or any such variable) and \(-4\). Applies correctly the limits of \(-4\) and \(t\) with correct limit notation, with limits seen substituted the right way round.
Withhold this mark if there is no evidence of using the limiting process. We must see as a minimum \(\displaystyle\lim_{t \to 0}\) oe at some stage in their work.

A1*: Produces an argument that includes an upper limit that approaches 0.
e.g. States that \(t \to 0 \Rightarrow \mathrm{e}^{\frac{1}{t}} \to 0\) and reaches a value of \(\mathrm{e}^{-\frac{1}{4}}\)

Alternative method: changing limits using \(u = \dfrac{1}{x}\)

Notes

M1: Uses the product rule to find the correct form for at least the first derivative.

A1: Achieves correct first and second derivatives. Need not be simplified.

M1: Uses the product rule to find the correct form for the second, third and fourth derivatives.

A1: Correct (unsimplified) expressions for the third and fourth derivatives.

A1: Correct fourth derivative in terms of \(y\) from correct work. Note do not accept just \(k = -4\), we must see the \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -4y\)

8(a) Alt

There may be attempts via Leibnitz theorem.

M1: Attempts first four derivatives for each of \(\cos x\) and \(\sinh x\)

A1: Correct derivatives.

M1: Applies Leibnitz theorem correctly with their derivatives.

A1: Correct unsimplified expression.

A1: Correct fourth derivative in terms of \(y\).

There may be rare cases where exponential forms are used in part (a). For reference here are the derivatives they should get.
\(y = \dfrac{1}{2}\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\cos x - \dfrac{1}{2}\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\sin x = \dfrac{1}{2}\mathrm{e}^x(\cos x - \sin x) + \dfrac{1}{2}\mathrm{e}^{-x}(\cos x + \sin x)\)
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x\ \left[= \dfrac{1}{2}\mathrm{e}^x(\cos x - \sin x) + \dfrac{1}{2}\mathrm{e}^x(-\sin x - \cos x) - \dfrac{1}{2}\mathrm{e}^{-x}(\cos x + \sin x) + \dfrac{1}{2}\mathrm{e}^{-x}(-\sin x + \cos x)\right]\)
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\sin x - \left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\cos x\)
\(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4} = -\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x - \left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x - \left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x + \left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\sin x = -2\left(\mathrm{e}^x - \mathrm{e}^{-x}\right)\cos x = -4\sinh x\cos x\)

NB Some are misreading the function as \(y = \cosh x\sinh x\) or \(y = \cos x\sin x\). These can be score method marks as long as work is equivalent (ie via product rule), but reduction of the problem to triviality via an identity is less demand and so will score M0. Allow SC A1 for the first A if the equivalent form is reached via the misread to that shown in the scheme.

Notes

B1ft: Correct fifth derivative or value for the fifth derivative deduced following through on their \(k\) from part (a). If they have stated \(y^{\prime}\) incorrectly allow for \(y^{(5)} = \text{``}{-4}\text{''} \times \text{their } y^{\prime}\) provided their \(y^{\prime}\) is non-zero

M1: Attempts the evaluation of all the derivatives at \(x = 0\) and applies the Maclaurin formula correctly with their values to achieve at least three non-zero terms.

A1: Correct simplified expansion. Accept with \(\mathrm{f}(x) = \ldots\) or with nothing before – mark the expression. Accept as a list of terms.

Notes

M1: Uses the identity \(\sinh 2x = 2\sinh x\cosh x\) and proceeds to find a value for \(\sinh x\) or \(\cosh x\)

B1: \(x = 0\)

dM1: Uses the correct formula for \(\operatorname{arcosh} x\) with their value of \(\cosh x\) to find a value for \(x\) as a natural logarithm or alternatively gives awrt to 3sf. e.g. 0.962, \(-0.962\). You may need to check their workings.

Alternatively uses the exponential definition for \(\cosh x\), forms and solves a quadratic for \(\mathrm{e}^x\) leading to find a value for \(x\) as a natural logarithm. Usual rules apply for solving a quadratic by any means including using a calculator. You may need to check their workings. If correct answers are given without workings, you may award this mark.

A1: Deduces one correct exact value for \(x\). Accept exact unsimplified equivalents.

A1: Deduces both correct exact values for \(x\). Accept exact simplified equivalents. isw
Accept, for example \(x = \pm\ln\left(\dfrac{3}{2} + \dfrac{\sqrt{5}}{2}\right)\) and \(x = \pm\ln\left(\dfrac{3}{2} + \sqrt{\dfrac{5}{4}}\right)\)
Also accept equivalent answers which you may need to check; their may be several alternatives.
e.g. Accept \(x = \ln\left(\dfrac{2}{3 + \sqrt{5}}\right)\) for \(x = \ln\left(\dfrac{3 - \sqrt{5}}{2}\right)\)

Alternative method use of exponentials

M1: Uses correct identities such as \(\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) and \(\sinh 2x = \dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2}\) to form a quartic equation for \(\mathrm{e}^x\)

B1: \(x = 0\)

dM1: Solve their quartic equation to find a value for \(x\) (other than \(x = 0\)) via a correct method. If correct answers are given from their quartic equation without workings, you may award this mark. You may need to check their workings.

A1: Deduces one correct exact value for \(x\), which may be unsimplified.

A1: Deduces both correct exact values for \(x\) isw

Note: Inexact answers for the quartic equation \(\mathrm{e}^x = 2.618,\ 0.3819\) leading to awrt 0.962, \(-0.962\) scores maximum M1 B1 dM1 A0 A0

Notes

Alternative 1 for part (a)

Alternative 2 for part (a)

Alternative 3 for part (a)

Notes

(a) Do not allow misreads of \(\tan x\) for \(\tanh x\) in this question; this would be a maximum of M1A1 for part (a)

M1: Attempts to use the quotient (or product) rule on \(\dfrac{3 - x}{6 + x}\) to obtain an expression of the form

Alternatively, candidates may also write \(\dfrac{3 - x}{6 + x}\) as \(-1 + \dfrac{9}{6 + x}\) and then differentiate to find an expression of the form \(\dfrac{E}{(6 + x)^2}\) which may be seen embedded in their working.
They may also write \(\dfrac{3 - x}{6 + x}\) as \(\dfrac{3}{6 + x} - \dfrac{x}{6 + x}\) and then differentiate to find an expression of the form \(\dfrac{E}{(6 + x)^2}\) oe

A1: Correct expression in any form.

dM1: A complete method to find the derivative using the chain rule to obtain

This is dependent on the first M mark.

A1*: Reaches the printed answer with sufficient working and no errors, with at least one intermediate line.

Alternative 1:
M1: see main scheme
A1: see main scheme

dM1: A complete method to find the derivative using the chain rule:
Rearranges the equation and uses implicit differentiation. Must proceed from \(\operatorname{sech}^2 y\) or equivalent to \(1 - \tanh^2 y\) and then substitute for \(\tanh y\)
This is dependent on the first M mark.
A1*: see main scheme

Alternative 2:
M1: Uses the logarithmic form of artanh to obtain \(k\ln\left(\dfrac{1 + \dfrac{3 - x}{6 + x}}{1 - \dfrac{3 - x}{6 + x}}\right)\)

A1: Correct simplified expression

dM1: A complete method to find the derivative to obtain
\(\mathrm{f}^{\prime}(x) = k \times \left(1 \div \left(\text{their } \dfrac{9}{3 + 2x}\right)\right) \times \left(\text{their } \dfrac{9}{3 + 2x}\right)^2\)
Alternatively uses log rules to partition their expression and then differentiates each term.
This is dependent on the first M mark.

A1*: Reaches the printed answer with sufficient working and no errors.

Alternative 3:
M1: Takes tanh of both sides to obtain \(\tanh y\) in terms of \(x\) and expresses \(\tanh y\) correctly in terms of exponentials and makes \(\mathrm{e}^{2y}\) the subject.
A1: A correct expression for \(\mathrm{e}^{2y}\)

dM1: Rearranges their function to the form \(k\ln\left(\dfrac{a}{bx + c}\right)\) and uses the chain rule to find a derivative of the form \(\mathrm{f}^{\prime}(x) = \dfrac{1}{2} \times \dfrac{bx + c}{a} \times \dfrac{k}{(bx + c)^2}\) or uses log rules to partition their expression and then differentiates each term.
This is dependent on the first M mark.

A1*: Reaches the printed answer with sufficient working and no errors

Notes

B1: Correct second derivative in any form such as e.g. \(\dfrac{2}{(4x^2 + 12x + 9)}\) or \(2(2x + 3)^{-2}\)

Notes

M1: Attempts at least two of the values of f(0), f′(0) and f″(0)

M1: Correct application of the Maclaurin series for all of \(\mathrm{f}(0)\), \(\mathrm{f}^{\prime}(0)\) and \(\mathrm{f}^{\prime\prime}(0)\) where all are non-zero values. Substitutes their values into a correct expression. This mark is not dependent.

A1: Correct expansion.
Award if written as a single expression, or \(p\), \(q\) and \(r\) written separately.
Ignore any extra terms written as powers of \(x^3\) and above, isw when a correct answer is seen.

Notes

M1: Begins the proof by expressing \(\sinh x\) correctly in terms of exponentials, substitutes and makes progress in cubing the bracket.
Award for obtaining an expression of the form \(A\mathrm{e}^{3x} + B\mathrm{e}^{x} + C\mathrm{e}^{-x} + D\mathrm{e}^{-3x}\) but terms do not need to be collected.
Note that \((\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})(\mathrm{e}^x - \mathrm{e}^{-x}) = \mathrm{e}^{3x} - \mathrm{e}^{x} - 2\mathrm{e}^{x} + 2\mathrm{e}^{-x} + \mathrm{e}^{-x} - \mathrm{e}^{-3x}\)

A1*: Fully correct proof with no errors. Must see \(= \sinh 3x\) or e.g. LHS = RHS

Notes

M1: Uses part (a), collects terms and attempts to factorise or cancel \(\sinh x\).
This can be implied if they go straight from their cubic to writing all correct answers for \(\sinh x\) including zero from their calculator.

B1: Deduces the root \(x = 0\), allow ln1 but not \(\ln\left[0 + \ln\sqrt{0 + 1}\right]\)

M1: Proceeds to \(\sinh x = \alpha\) and uses the correct logarithmic form of arsinh to obtain at least one exact value for \(x\)
Alternatively, candidates proceed from \(\sinh x = \alpha\) to substitute the exponential form and then solve a 3TQ in \(\mathrm{e}^x\) to obtain at least one exact value for \(x\)

A1: One correct non-zero solution

A1: Both correct non-zero solutions and no incorrect other solutions, but isw if they go on to evaluate these answers incorrectly.
Allow \(x = \ln\left(\sqrt{5} \pm 2\right)\) for listing both solutions

Alternative if candidates substitute the exponential form at the start:

M1: Candidates substitutes the exponential form for each term and proceeds to find a four term cubic equation in \(\mathrm{e}^{2x} = 0\), which may not be correct.

B1: Deduces the root \(x = 0\), allow ln1 but not \(\ln\left[0 + \ln\sqrt{0 + 1}\right]\)

M1: They factorise their cubic, proceed to obtain exact values for \(\mathrm{e}^{2x}\), then take logs to obtain at least one exact value for \(x\)
If they go directly to decimal answers this will usually score M0 unless they recover to exact form.

A1: One correct non-zero solution

A1: Both correct non-zero solutions and no incorrect other solutions, but isw if they go on to evaluate these answers incorrectly.

This is how this would be awarded:

(corrected from the printed mark scheme: the line \(\mathrm{e}^{6x} - 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} - 1 = 0\) is printed as \(\mathrm{e}^{6x} + 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} - 1 = 0\))

Notes

Alternative using exponential definitions

Notes

M1: Differentiates to a correct form

A1: Correct derivative and reaches appropriate form to deduce the result is true for \(n = 1\), minimum \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}\left(\dfrac{4}{2}\sinh x + \dfrac{2}{2}\cosh x\right)\)

M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Allow slips in coefficients but must be evidence of the use of the product rule. Must be of the form

dM1: Dependent on the previous method mark. Factors out the exponential and gathers the \(\sinh x\) and \(\cosh x\) terms. Accept either form shown or equivalent.

A1: Reaches the correct form from correct work. Must have the “\(k\)+1” showing. Depends on the previous two method marks.

A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.

Alternative: using exponentials
M1: Uses the exponential definition of \(\sinh x\) and differentiates to a correct form
A1: Correct derivative and uses the exponential definitions of \(\sinh x\) and \(\cosh x\) to deduce the result is true for \(n = 1\)
M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Uses the exponential definition of \(\sinh x\) first and then differentiates to a correct form
dM1: Dependent on the previous method mark. Collects exponential terms and simplifies
A1: Reaches the \(k + 1\) th derivative. Uses the exponential definitions of \(\sinh x\) and \(\cosh x\) in the result, simplifies and achieves the correct result
A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.

Notes

B1: Correct completed square form. Allow \(3^2\) for 9.

Notes

M1: Achieves a correct form for the integration for their \(p\) and \(q\) from part (a):

\(\operatorname{arcosh}\left(\dfrac{x + p}{\sqrt{q}}\right)(+c)\) or \(\ln\left(x + p + \sqrt{(x + p)^2 - q}\right)(+c)\) or e.g. \(\ln\left(\dfrac{x + p}{\sqrt{q}} + \sqrt{\left(\dfrac{x + p}{\sqrt{q}}\right)^2 - 1}\right)(+c)\)
where \(p \neq 0,\ q \neq 1\)

Allow \(\cosh^{-1}\) for arcosh
Allow attempts that use substitution following an attempt to complete the square but must be an appropriate substitution e.g. \(x + p = \sqrt{q}\cosh u\) leading to a correct form as above.

A1: Correct integration. The “\(+ c\)” is not required. Apply isw once a correct expression is seen.
Note that \(\ln\left(\dfrac{x + 2}{3} + \sqrt{\left(\dfrac{x + 2}{3}\right)^2 - 1}\right)(+c)\) is also correct

Notes

B1: Recalls the definition of a mean function accurately. \(\dfrac{1}{13 - 3}\displaystyle\int_3^{13} \frac{1}{\sqrt{x^2 + 4x - 5}}\,\mathrm{d}x\) seen or implied.
Note that the \(\dfrac{1}{13 - 3}\) may appear at the end. \(\dfrac{1}{13 - 3}\displaystyle\int_3^{13} \mathrm{f}(x)\,\mathrm{d}x\) is sufficient as f(\(x\)) is defined in the question. Also allow it to be implied by e.g. \(\dfrac{1}{10}\left[\mathrm{g}(x)\right]_3^{13}\) where g(\(x\)) is their integrated function.

M1: Applies the correct limits the right way round to whatever they think the answer to part (b) is.
This can be awarded if the \(\dfrac{1}{10}\) is present or not.

A1: Correct answer in correct form. Allow equivalents e.g. \(\dfrac{1}{10}\ln\left(\dfrac{5}{3} + \dfrac{2\sqrt{6}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49}{9} + \dfrac{20\sqrt{6}}{9}\right)\)
And allow if the surd is not simplified e.g. \(\dfrac{1}{10}\ln\left(\dfrac{5 + \sqrt{24}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49 + \sqrt{2400}}{9}\right)\)
Apply isw once a correct answer is seen.
The brackets must be present in forms such as \(\dfrac{1}{10}\ln\left(\dfrac{5}{3} + \dfrac{2\sqrt{6}}{3}\right)\), \(\dfrac{1}{20}\ln\left(\dfrac{49}{9} + \dfrac{20\sqrt{6}}{9}\right)\) but not in e.g. \(\dfrac{1}{10}\ln\dfrac{5 + \sqrt{24}}{3}\)
If extra values are offered then score A0

Notes

B1: Correct area formula applied, including the \(\dfrac{1}{2}\) and correct limits, may be seen later, \(\mathrm{d}\theta\) may be implied.

M1: Attempts the integration \(\sinh\theta \rightarrow \pm\cosh\theta\) and \(\cosh\theta \rightarrow \pm\sinh\theta\) or in terms of exponentials \(\displaystyle\int \mathrm{e}^{\lambda\theta}\,\mathrm{d}\theta = \frac{1}{\lambda}\mathrm{e}^{\lambda\theta}\)

M1: Applies their limits to the integral, subtracts (there must be an attempt to integrate) and uses exponential definitions to achieve answer in suitable form. Condone the inclusion of i or a missing ½ from the definitions. This can be implied.

A1: Correct exact answer, no i

Notes

B1: For a suitable explanation. Technically this should refer to the interval being unbounded, but this is unlikely to be seen. Accept “Because the upper limit is infinity”, but not “because it is infinity” without reference to what “it” is. Do not accept “the upper limit tends to infinity” or “the integral is unbounded”.

Notes

B1: Writes the integral in terms of a limit as \(t \to \infty\) (or other variable) with limits 0 and “\(t\)”, or implies the integral is a limit by subsequent working by correct language.

M1: Integrates \(\cosh x\) correctly either as \(\sinh x\) or in terms of exponentials and applies correctly the limits of 0 and “\(t\)”. The bottom limit zero may be implied. No need for the \(\lim\limits_{t \to \infty}\) for this mark but substitution of \(\infty\) is M0.

A1: cso States that (as \(t \to \infty\)) \(\sinh t \to \infty\) or \(\mathrm{e}^t \to \infty\) and \(\mathrm{e}^{-t} \to 0\) therefore divergent (or not convergent), or equivalent working. Accept \(\sinh t\) is undefined as \(t \to \infty\)

Notes

M1: Divides through by \(\cosh x\) to find an expression involving \(\tanh x\)
Alternative: uses the correct exponential definitions and finds an expression for \(\mathrm{e}^{2x}\) or solves a quadratic in \(\mathrm{e}^{2x}\)

A1: Deduces the correct inequality for \(p\). Note \(|p| \lt 4\) is a correct inequality for \(p\).

Notes

M1: Uses the chain rule and product rule to find the first and second derivatives which must be of the required form, condone sign slips
Alternatively uses the exponential definition and uses the chain rule and product rule to find the first and second derivatives which must be of the required form.

A1: Correct unsimplified first and second derivatives, may be in exponential form.

M1: Uses the identity \(\pm\cosh^2 x \pm \sinh^2 x = 1\)

A1*: Achieves the printed answer with no errors or omissions e.g. missing \(x\)’s

Notes

(The printed M1 line is cut off at the edge of the page after “\(-\ldots\cosh\)”; its last term is completed here to match the A1 line.)

M1: Uses the chain rule and product rule to find the third and fourth derivatives which must be of the required form, condone sign slips

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 1

M1: Using \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2y - n(n - 1)\cosh^{n-2} x\) to find the third and fourth derivatives which must be of the required form, condone sign slips

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 2

M1: Using \(y = \cosh^n x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\)
\(y = \cosh^{n-2} x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\cosh^{n-2} x - \ldots\cosh^{n-4} x\) leading to
\(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\left[n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\right] - n(n - 1)\left[\text{their } \dfrac{\mathrm{d}\left(\cosh^{n-2} x\right)}{\mathrm{d}x}\right]\)

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 3

(Corrected from the printed mark scheme: in Alternative 3 the printed scheme gives the power of the second term of \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) as \(n - 2\) in both lines; it should be \(n\), i.e. \(n^3\cosh^n x\).)

M1: Uses the exponential definition and uses the chain rule and product rule to find the third and fourth derivatives which must be of the required form.

A1: Correct fourth derivative, does not need to be simplified ISW

Notes

M1: Attempts the evaluation of all four of their derivatives at \(x = 0\) and applies the Maclaurin formula with their values. Note that \(y^{(1)}(0) = 0\) and \(y^{(3)}(0) = 0\) may be implied as they will have a multiple of \(\sinh 0\). If their \(y^{(3)}(0) \neq 0\) they allow this mark for their first 3 non-zero terms

A1: Correct simplified expansion from correct derivatives cso

Notes

M1: Solves the quadratic equation for \(\cosh^2 x\) by any valid means. If by calculator accept for reaching the positive value for \(\cosh^2 x\) (negative may be omitted or incorrect) but do not allow for going directly to a value for \(\cosh x\). Alternatively score a correct process leading to a value for \(\sinh 2x\) or its square (Alt 1) or use of correct exponential form for \(\cosh x\) to form and expand to an equation in \(\mathrm{e}^{4x}\) and \(\mathrm{e}^{2x}\) (Alt 2)

A1: Correct value for \(\cosh^2 x\) (ignore negative or incorrect extra roots.). In Alt 1 score for a correct value for \(\sinh^2 2x\) or \(\sinh 2x\). In Alt 2 score for a correct simplified equation in \(\mathrm{e}^{4x}\).

M1: For a correct method to achieve at least one value for \(x\) (from \(\cosh^2 x\)). In the main scheme or Alt 1, takes positive square root (if appropriate) and uses the correct formula for \(\operatorname{arcosh} x\) or \(\operatorname{arsinh} x\) to find a value for \(x\). (No need to see negative square root rejected.) In Alt 2 it is for solving the quadratic in \(\mathrm{e}^{4x}\) and proceeding to find a value for \(x\).
Alternatively uses the exponential definition for \(\cosh x\), forms and solves a quadratic for \(\mathrm{e}^x\) leading to a value for \(x\)

A1: Deduces (both) the correct values for \(x\) and no others. Must be in the form specified.

SC Allow M0A0M1A1 for cases where a calculator was used to get the value for \(\cosh x\) with no evidence if a correct method for find both values is shown.

Alt 1

Alt 2

Notes

M1: Uses the substitution \(x = \cosh u\) fully to achieve an integral in terms of \(u\) only. Must have replaced the \(\mathrm{d}x\) but allow if the \(\mathrm{d}u\) is missing.

A1: Correct integral in terms of \(u\). (Allow if the \(\mathrm{d}u\) is missing.)

M1: Uses correct identities \(\cosh^2 u - 1 = \sinh^2 u\) and \(\cosh 2u = 2\cosh^2 u - 1\) to achieve an integrand of the required form

M1: Integrates to achieve the correct form, may be sign errors.

M1: Uses the identities \(\sinh 2u = 2\sinh u\cosh u\) and \(\cosh^2 u - 1 = \sinh^2 u\) to attempt to find \(\sinh 2u\) in terms of \(x\). If using exponentials there must be a full and complete method to attempt the correct form.

A1*: Achieves the printed answer with no errors seen, cso

NB attempts at integration by parts are not likely to make progress – to do so would need to split the integrand as \(x\dfrac{x}{\sqrt{x^2 - 1}}\). If you see any attempts that you feel merit credit, use review.

Notes

M1: Uses integration by parts the correct way around to achieve the required form.

A1: Correct integration by parts

B1ft: Deduces the integral by using the result from part (a). Follow through on their ‘\(uv\)’

dM1: Dependent on previous method mark. Uses the limits \(x = 1\) and \(x = 3\) the correct way around and subtracts

A1*cso: Achieves the printed answer with at least one intermediate step showing the evaluation of the arcosh 3, and no errors seen.

Notes

(a)
B1: Recalls the definition for \(\sinh\left(\dfrac{1}{2}\ln 3\right)\) or forms an equation for \(\operatorname{arsinh} x\)
M1: Uses logarithms to find a value for \(\alpha\) or forms and solves a correct equation without log
A1: Deduces the correct exact value for \(\alpha\)

Note using the result
\(\ln\left(\dfrac{1}{\sqrt{3}} + \sqrt{\left(\dfrac{1}{\sqrt{3}}\right)^2 + 1}\right) = \ln\left(\dfrac{1}{\sqrt{3}} + \sqrt{\dfrac{4}{3}}\right) = \ln\sqrt{3} = \dfrac{1}{2}\ln 3\) therefore \(\operatorname{arsinh}\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{1}{2}\ln 3\)
B1 for substituting in \(\alpha\) into \(\operatorname{arsinh} x\), M1 for rearranging to show \(\dfrac{1}{2}\ln 3\), A1 for conclusion

Notes

(b)
B1: Correct expression for the volume \(\pi\displaystyle\int_0^{\frac{1}{2}\ln 3}\sinh^2 y\,\mathrm{d}y\) requires integration signs, \(\mathrm{d}y\) and correct limits.
M1: Uses the exponential formula for \(\sinh y\) or the identity \(\cosh 2y = \pm 1 \pm 2\sinh^2 y\) to write in a form which can be integrated at least one term
dM1: Dependent of previous method mark, integrates.
A1: Correct integration.
M1: Correct use of the limits \(y = 0\) and \(y = \dfrac{1}{2}\ln 3\)
A1: Correct exact volume.

Notes

B1: Correct differentiation
M1: Identifies a correct approach by using a correct identity to make progress to obtain a quadratic in \(\cosh x\)
A1: Correct 3 term quadratic obtained
M1: Solves their 3TQ
A1: Correct values (may only see 4 here)
M1: Correct process to reach at least one value for \(x\) from their \(\cosh x\)
A1: Deduces the correct 2 values with no incorrect values or work involving \(\cosh x = -\dfrac{1}{8}\)

Alternative
B1: Correct differentiation
M1: Using the exponential form for \(\cosh x\), and \(\sinh x\) if required, and forms a quartic equation for \(\mathrm{e}^x\) with all terms simplified and all on one side
A1: Correct quartic equation for \(\mathrm{e}^x\)
M1: Solves their quartic equation in \(\mathrm{e}^x\)
A1: Correct values to two decimal places or exact values
M1: \(x = \ln(b)\) where \(b\) is a real exact value
A1: Deduces the correct 2 values only

(corrected from the printed mark scheme: \(\mathrm{e}^{2x} - 8\mathrm{e}^x + 1 = 0\) is printed as \(\mathrm{e}^{2x} - 8\mathrm{e}^x + 7 = 0\))

Notes

B1: Selects an appropriate substitution leading to an integrable form

M1: Demonstrates a fully correct method for the substitution that includes substituting into the function and dealing with the “\(\mathrm{d}x\)”. The substitution being substituted does not need to be “correct” for this mark but the substitution must be an attempt at \(\displaystyle\int \frac{1}{\sqrt{4\left[\mathrm{f}(u)\right]^2 + 9}} \times \mathrm{f}^{\prime}(u)\,\mathrm{d}u\) with the \(\mathrm{f}^{\prime}(u)\) correct for their substitution. E.g. if \(x = \dfrac{1}{2}u\) is used, must see \(\mathrm{d}x = \dfrac{1}{2}\mathrm{d}u\) not \(2\mathrm{d}u\).

A1: Correct simplified integral in terms of \(u\) from correct work and from a correct substitution

A1: Correct answer including “\(+ c\)”. Allow arcsinh or arsinh for \(\sinh^{-1}\) from correct work and from a correct substitution

Alternative: Way 2

(Corrected from the printed mark scheme: the last line is printed as “\(u = \dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x}{3}\right) + c\)”; it is the integral, not \(u\), that equals this.)

Alternative: Way 3

Notes

M1: Correctly applies the method for the mean value for their integration which must be of the form specified in part (a) and substitutes the limits 0 and 3 but condone omission of 0

A1: Correct exact answer (follow through their \(A\) and \(B\)). Brackets are required if appropriate.

Notes

If you come across any attempts to use calculus to prove the result – send to review

M1: Begins the proof by expressing tanh in terms of exponentials and forms an equation in exponentials.

The exponential form can be any of \(\dfrac{\left(\mathrm{e}^y - \mathrm{e}^{-y}\right)/2}{\left(\mathrm{e}^y + \mathrm{e}^{-y}\right)/2},\ \dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}},\ \dfrac{\mathrm{e}^{2y} - 1}{\mathrm{e}^{2y} + 1}\)

Allow any variables to be used but the final answer must be in terms of \(x\). Allow alternative notation for \(\tanh^{-1}x\) e.g. artanh, arctanh.

A1: Correct expression for “\(x\)” in terms of exponentials

M1: Full method to make \(\mathrm{e}^{2\text{“}y\text{”}}\) the subject of the formula. This must be correct algebra so allow sign errors only.

A1*: Completes the proof by using logs correctly and reaches the printed answer with no errors.

Allow e.g. \(\dfrac{1}{2}\ln\left(\dfrac{x + 1}{1 - x}\right),\ \dfrac{1}{2}\ln\dfrac{x + 1}{1 - x},\ \dfrac{1}{2}\ln\left|\dfrac{x + 1}{1 - x}\right|\). Need to see \(\tanh^{-1}x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) as a conclusion but allow if the proof concludes that \(y = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) with \(y\) defined as \(\tanh^{-1}x\) earlier.

B1: Correct value for \(k\) or writes \(-1 \lt x \lt 1\)

Alternative: Way 2

M1: Starts with result, takes tanh of both sides and expresses in terms of exponentials

A1: Correct expression

M1: Eliminates exponentials and logs and simplifies

A1: Correct result (i.e. \(x = x\)) with conclusion

B1: Correct value for \(k\) or writes \(-1 \lt x \lt 1\)

Notes

M1: Adopts a correct strategy by taking \(\tanh^{-1}\) of both sides

M1: Makes the link with part (a) by replacing artanh\((2x)\) with \(\dfrac{1}{2}\ln\left(\dfrac{1 + 2x}{1 - 2x}\right)\) and demonstrates the use of the power law of logs to obtain an equation with logs removed correctly.

A1: Obtains the correct 3TQ

M1: Solves their 3TQ using a correct method (see General Guidance – if no working is shown (calculator) and the roots are correct for their quadratic, allow M1)

A1: Correct value with the other solution rejected (accept rejection by omission) so \(x = \dfrac{9 \pm \sqrt{57}}{12}\) scores A0 unless the positive root is rejected

Alternative for first 2 marks of (b)

M1: Adopts a correct strategy by expressing tanh in terms of exponentials

M1: Demonstrates the use of the power law of logs to obtain an equation with logs removed correctly

Typical solution

Let \(s = \sinh x\)

Solution of inequality:

Typical solution

Typical solution

Area of \(R = A_1 + A_2\), where

Area of \(R = \frac{9}{2} + \ln 2 + 15\ln\frac{4}{3}\)

Typical solution

Let \(y = \coth^{-1}(x)\)

Then \(x = \coth y\)

Typical solution

Typical solution

Typical solution

but \(\mathrm{e}^x \gt 0\)

Typical solution

The error is in the 3rd line of working.

Geraldine uses an incorrect definition of tanh.

She should replace \(\tanh y\) with

Typical solution

Typical solution

\(\operatorname{sech}^2 x + \tanh^2 x = 1\)

Typical solution

Let \(x + 3 = \cosh\theta\)

Then

Let \(I = \displaystyle\int_{-2}^{1} \sqrt{x^2 + 6x + 8}\,\mathrm{d}x\)

When \(x = -2\), \(\cosh\theta = 1\) and \(\theta = 0\)

When \(x = 1\), \(\cosh\theta = 4\) and \(\theta = \cosh^{-1}(4)\)

When \(\cosh\theta = 4\), \(\sinh\theta = \sqrt{4^2 - 1} = \sqrt{15}\)

Typical solution

Integrating factor \(= \cosh x\)

When \(x = \ln 2\), \(\sinh x = \dfrac{3}{4}\) and \(\cosh x = \dfrac{5}{4}\)

Typical solution

Typical solution

Let \(s = \sinh x\)

Typical solution

Typical solution

Typical solution

\(0 \lt \mathrm{g}(x) \leqslant 1\)

Typical solution

\(1 + \sinh^2 x\)

Typical solution

Typical solution

(i)

(ii)

For \(x \gt 0\),

and

Typical solution

Typical solution

Typical solution

Typical solution

\(\{x : x \geqslant 4\}\)

Typical solution

\(2\sinh x\)

Typical solution

\(0\)

Typical solution

Typical solution

The roots of the quadratic are \(\sinh\theta_1\) and \(\sinh\theta_2\)

So \(\sinh\theta_1 + \sinh\theta_2 = 1\)

Typical solution

Typical solution

Let \(y = \tanh^{-1} x\)

Then \(x = \tanh y\)

as required.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

At P, \(y^2 - x^2 = 16\) and \(x^2 + y^2 = 32\)

Let \(x = 4\sinh u\)

Then \((x^2 + 16)^{\frac{1}{2}} = 4\cosh u\)

and \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 4\cosh u\)

When \(x = 2\sqrt{2}\),

\(\sinh u = \dfrac{\sqrt{2}}{2}\) and \(\cosh u = \dfrac{\sqrt{6}}{2}\)

OP makes an angle of \(\dfrac{\pi}{3}\) with the \(x\)-axis

as required

Typical solution

Typical solution

The required integral is \(S_4\) with \(a = 1\)

Typical solution

Typical solution

but \(\cosh\theta \geqslant 1 \quad \therefore \cosh\theta \neq -1\)

Typical solution

Typical solution

But \(\operatorname{sech} x \gt 0\) \(\therefore \operatorname{sech} x = \dfrac{1}{3}\)

Typical solution

Typical solution

Typical solution

Let \(y = \cosh^{-1}\left(\frac{x}{a}\right)\); then \(x = a\cosh y\)

\(\times\,\mathrm{e}^y\):

Product of roots = 1, so one root is greater than 1 and the other is less than 1.

\(y \geqslant 0\) by definition of \(\cosh^{-1} \therefore \mathrm{e}^y \geqslant 1\)

So we choose the larger root.

and

Typical solution

Typical solution

Typical solution

Let \(y = \tanh^{-1} x\)

Notes

(corrected from the printed mark scheme: the typical solution prints the denominator of the compound fraction for \(x\) as \(\dfrac{(\mathrm{e}^y - \mathrm{e}^{-y})}{2}\); it should be \(\dfrac{(\mathrm{e}^y + \mathrm{e}^{-y})}{2}\).)

Typical solution

If the graphs meet, then \(\sinh x = \cosh x\)

but \(\mathrm{e}^{-x} \gt 0\)

\(\therefore\) the graphs of \(y = \sinh x\) and \(y = \cosh x\) do not intersect

Typical solution

Typical solution

Typical solution

In the 2^nd line, \(\sinh^2 x\) should not have been replaced with \(1 - \cosh^2 x\)

Typical solution

but \(\cosh(2x) \geqslant 1\) \(\therefore \cosh(2x) = 1 + \sqrt{3}\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

but \(\tanh\theta \neq -1\) \(\therefore \tanh\theta = \frac{1}{2}\) only

Typical solution

It is an error because \(y - \sqrt{y^2 + 1} \lt 0\) and \(\mathrm{e}^x \gt 0\) so there is a contradiction.

Typical solution

Notes

B1: For both. % symbol not necessary.
\(1.865736036\ldots\)
\(4.540199101\ldots \times 10^{-3}\)

Notes

B1: For both. % symbol not necessary.
\(3.731472073\ldots\)
\(9.08039820 \times 10^{-3}\)

Notes

B1: One correct approximation soi.

M1: Correct use of approximations to derive an inequality in \(n\).
Accept equations in place of inequalities for all M marks.

M1*: Rearranging to a 3 term quadratic inequality in \(\mathrm{e}^{0.005n}\)
May see quadratic in \(\mathrm{e}^{0.005n}\).

M1dep*: Correctly solves their quadratic equation to find expression for \(\mathrm{e}^{0.005n}\)
\(17.47497913\ldots\) Or \(\left(\frac{5 + \sqrt{897}}{2}\right)\). Other root does not need to be explicitly rejected.

A1: Must see values, not just an assertion.
NB no marks for \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) without justification.

Alternative method

B1: One correct approximation soi.

M1: Correct use of both approximations to derive an inequality in sinh or cosh only. May see \(\cosh 0.01n - 5\cosh 0.005n - 9\) and use of \(\cosh x\) throughout.
Accept equations in place of inequalities for all M marks.

M1*: Using an appropriate identity (e.g. \(\sinh 2x \equiv 2\sinh x\cosh x\)) and using approximation to rearrange to a 3 term quadratic inequality in sinh or cosh only.
May see quadratic in e.g. \(y = \sinh u = \sinh 0.005n\) i.e. \(2y^2 - 5y - 109 \gt 0\).

M1dep*: Correctly solves their quadratic equation to find expression for \(\cosh u\) or \(\sinh u\)
\(y \gt 8.737489\ldots\) or \(\frac{5 + \sqrt{897}}{4}\). Other root does not need to be explicitly rejected.

A1: Must see values, not just an assertion.
NB no marks for \(\mathrm{f}(572) = 99.6\ldots \lt 100\) and \(\mathrm{f}(573) = 100.96\ldots \gt 100\) without justification.

Notes

B1: Accept any equivalent (un-simplified) forms e.g. \(\dfrac{1}{3} \times \dfrac{1}{\sqrt{\left(\frac{x}{3}\right)^2 - 1}}\) or \(\dfrac{1}{\sqrt{x^2 - 3^2}}\). Invisible brackets e.g. \(\dfrac{1}{3} \times \dfrac{1}{\sqrt{\frac{x}{3}^2 - 1}}\) is B0 unless correctly recovered. ISW if correct un-simplified form is simplified incorrectly

Notes

B1: Correct normal gradient (soi) – possibly BC

B1: For correct un-simplified (or simplified) \(y\)-coordinate in logarithmic form when \(x = 5\) – ISW if simplified incorrectly

M1: Any complete correct method for the equation of a straight line with \(x = 5\), their \(y\) coordinate (which must be of the form \(\ln(a)\) where \(a \gt 0\)) and any non-zero gradient. If using \(y = mx + c\) then must explicitly find \(c\) using \(x = 5\), any non-zero \(m\) and their \(y\) which must be of the form as stated above (but allow errors in the evaluation of \(c\))

A1: oe but must be integer values e.g. \(8x + 2y = 40 + \ln 9\) and must be of the form \(ax + by = c + \ln d\) e.g. allow \(-\ln 3 - 20 = -y - 4x\)

Notes

M1: Use correct definitions of \(\cosh u\) and \(\sinh u\) in terms of exponentials. Condone omission of \(2\sinh u\cosh u\) for this mark. (Note \(2\sinh u\cosh u\) or \(LHS\) must be included for subsequent A1.) M1 can be implied for stating that \(2\sinh u\cosh u \equiv \frac{1}{2}(\mathrm{e}^{u} - \mathrm{e}^{-u})(\mathrm{e}^u + \mathrm{e}^{-u})\) (but not for the A mark). (Corrected from the printed mark scheme: the first bracket is printed as \((\mathrm{e}^{-u} - \mathrm{e}^{-u})\).)

A1: Complete proof with no errors. Allow \(LHS\) and \(RHS\) in place of \(2\sinh u\cosh u\) and \(\sinh 2u\) respectively. Minimally acceptable proof is of the form \(RHS \equiv 2\left(\dfrac{\mathrm{e}^u - \mathrm{e}^{-u}}{2}\right)\left(\dfrac{\mathrm{e}^u + \mathrm{e}^{-u}}{2}\right) \equiv \dfrac{\mathrm{e}^{2u} - \mathrm{e}^{-2u}}{2} \equiv LHS\) in either direction.
Do not condone sin for sinh or cos for cosh or \(x\) for \(u\) (unless “let \(x = u\)” stated) for A1.
If candidate shows that both \(LHS\) and \(RHS\) are equal to some third expression, then a conclusion must be clearly stated for the A1 mark.
SC B1 If candidate works with given identity and concludes, for example, that 1 = 1.

Alternative method (working left to right)

M1: Use definition of \(\sinh 2u\) in terms of exponentials. Condone omission of \(\sinh 2u\) for this mark. (Note \(\sinh 2u\) or \(RHS\) must be included for subsequent A1).

A1: Complete proof must see \(2 \times \dfrac{1}{2}(\mathrm{e}^u - \mathrm{e}^{-u}) \times \dfrac{1}{2}(\mathrm{e}^u + \mathrm{e}^{-u})\) before stating correct \(RHS\).

Notes

M1*: Differentiating to get \(\pm 16\sinh x \pm k\cosh 2x\) for some \(k \neq 0\). May be in exponential form \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\left(\mathrm{e}^x - \mathrm{e}^{-x}\right) - (\mathrm{e}^{2x} + \mathrm{e}^{-2x})\) - if using exponential form then the derivative must be of the form \(\frac{\mathrm{d}y}{\mathrm{d}x} = \pm 8\left(\mathrm{e}^x - \mathrm{e}^{-x}\right) \pm k(\mathrm{e}^{2x} + \mathrm{e}^{-2x})\) for some \(k \neq 0\).

A1: Correct second derivative, condone un-simplified. May be in exponential form. \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 2(\mathrm{e}^{2x} - \mathrm{e}^{-2x})\) (oe).

M1dep*: Correct use of the result from part (a) in their second derivative or their \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 2(\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\) using difference of two squares (oe e.g. setting equal to zero and converting to a quartic in \(\mathrm{e}^x\) which if correct is \(\mathrm{e}^{4x} - 4\mathrm{e}^{3x} - 4\mathrm{e}^x - 1 = 0\)). If converting to exponentials at any point, then correct exponential expression(s) for sinh and cosh must be used.
Or in exponential form \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(\mathrm{e}^x + \mathrm{e}^{-x})(4 - \mathrm{e}^x + \mathrm{e}^{-x})\)

A1: Must state \(\cosh x \gt 0\) or \(\cosh x \geqslant 1\) or show that there is no solution of the equation \(\cosh x = 0\) (oe e.g. ‘cosh cannot be 0’, ‘cosh is at least 1’, ‘cosh has a minimum at (0, 1)’, ‘\(\cosh^{-1}\) only valid for \(x \geqslant 1\)’ but not just ‘cosh \(x = 0\) has no solutions’) - allow stating that the only solution satisfies the equation \(\sinh x = 2\). If using exponentials then must show clearly that there is only one solution of the correct equation (which for reference is \(\ln(2 + \sqrt{5})\)).

Alternative method for second and third marks

M1dep*: Use of \(\cosh 2x = \cosh^2 x + \sinh^2 x\) in their first derivative. \(\frac{\mathrm{d}y}{\mathrm{d}x} = 8(\mathrm{e}^x - \mathrm{e}^{-x}) - \frac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x})^2 - \frac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})^2\)

A1: Correct second derivative, condone un-simplified. May use exponential form: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - (\mathrm{e}^x + \mathrm{e}^{-x})(\mathrm{e}^x - \mathrm{e}^{-x}) - (\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

Alternative method for first three marks

B1: Using the result from part (a). Or \(y = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - 0.5(\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

M1: Differentiating to get \(\pm 16\sinh x \pm k\cosh^2 x \pm k\sinh^2 x\) or \(\frac{\mathrm{d}y}{\mathrm{d}x} = \pm 8(\mathrm{e}^x - \mathrm{e}^{-x}) \pm k(\mathrm{e}^x + \mathrm{e}^{-x})^2 \pm k(\mathrm{e}^x - \mathrm{e}^{-x})^2\) for some \(k \neq 0\).

A1: Correct second derivative, condone un-simplified. May use exponential form: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(\mathrm{e}^x + \mathrm{e}^{-x}) - (\mathrm{e}^x + \mathrm{e}^{-x})(\mathrm{e}^x - \mathrm{e}^{-x}) - (\mathrm{e}^x - \mathrm{e}^{-x})(\mathrm{e}^x + \mathrm{e}^{-x})\)

Notes

B1: Condone \(\ln\left|2 + \sqrt{5}\right|\).

B1FT: Finding the value of \(\cosh x\) in a form not involving logs or exponentials following through their \(x\) of the form \(\ln(a + \sqrt{b})\) where \(a\) is non-zero and \(b \gt 0\) and \(a + \sqrt{b} \gt 0\) – allow un-simplified. The correct \(y\)-coordinate implies this and the next B mark. For reference: \(\cosh(\ln(a + \sqrt{b})) = \frac{1}{2}\left(a + \sqrt{b} + \dfrac{1}{a + \sqrt{b}}\right)\).
Or from using \(\cosh^2 x - \sinh^2 x \equiv 1\) with their value of \(\sinh x\) from part (b).

B1: Need not be stated as a coordinate. Accept any exact one term equivalent.

Notes

M1: Substituting correct exponential definitions of hyperbolic functions into expression

M1: Collecting terms
eg \(s\mathrm{e}^x + t\mathrm{e}^{-x}\) where \(s\), \(t\) are non-zero constants, possibly unsimplified.

A1: Answer can be embedded. ISW.

Notes

M1: DR. Formula for VoR correctly used in solution (limits and/or \(\pi\) may come later) soi. \(\mathrm{d}x\) must be seen here but can be implied later.

M1: Squaring out and using their part (a) formula

M1: Useful substitution stated or used
eg \(u = \mathrm{e}^{-x}\) or \(\mathrm{e}^x = 4\tan u\) etc

M1*: Making the substitution (including \(\mathrm{d}x\)) to reduce to integrable form. Ignore limits here.
\(\mathrm{e}^x\,\mathrm{d}x = 4\sec^2 u\,\mathrm{d}u\) and \(V = \frac{\pi}{4}\int\mathrm{d}u\) if using \(\mathrm{e}^x = 4\tan u\)

B1FT: Correct arctan integration (ignore limits etc). FT their 16 and \(\sqrt{16}\)
Could be implicit in the substitution.
(ie \(\int\mathrm{d}u = [u]\) if using \(\mathrm{e}^x = 4\tan u\))

depM1*: Dealing with limits correctly (either converting to \(u\)-space or substituting back to \(x\)-space – if the latter ignore missing “\(x =\)” in the limits if correct at resubstitution)
Could see \(\dfrac{\pi}{4}\left[\tan^{-1}\dfrac{\mathrm{e}^x}{4}\right]_0^{\ln 3}\)

A1: Formula for \(\tan(A - B)\) must be used.
(NB It is possible to find 8/19 using tan and \(\tan^{-1}\) on a calculator)
0.3129987969... alone or used to derive \(k\) and/or \(q\) gets A0

Notes

B1: Condone use of “origin” rather than “pole” in explanations.

B1: Must link domain of \(\theta\) with quadrant.

B1: Convincingly explaining why it is a single loop. \(\theta\) values/\(r = 0\) must be linked to loop.
or “There are no other solutions to \(r = 0\) (in the domain)”.

Notes

B1*: All 4 entries correct and non-zero ones correct to at least 3 sf. ie awrt 0.912, 1.59, 1.35.

M1: All points, theirs and given, plotted approximately correctly (ie on the \(\theta\) = constant lines and in the correct half between the appropriate \(r\) = constant lines)
Points can be inferred by curve

A1dep*: Smooth curve plotted between correctly plotted points (see criteria above) joining at the pole.
Single maximum \(r\) at some angle between \(\pi/4\) and \(5\pi/12\) (it might be on the wrong side of \(\theta = \pi/3\).
Curve needs to be: smooth (except at pole); curved (not line segments); closed; single loop; cusp at pole

Notes

M1: Attempting to differentiate \(r(\theta)\) using the product rule.
\(a\sinh\frac{1}{3}\theta\cos 2\theta + b\sin 2\theta\cosh\frac{1}{3}\theta\) where \(a\), \(b\) are non-zero constants

A1: All correct (any form)

M1: Setting their derivative to 0 and rearranging to a form containing tan and tanh only.
If using exponential form, must reach tan and reduce to one exponential term.
eg \(\mathrm{e}^{\frac{2}{3}\phi}(6 + \tan 2\phi) = 6 - \tan 2\phi\)

A1: AG so intermediate step must be shown.
Condone use of \(\theta\) rather than \(\phi\) but final answer must be in terms of \(\phi\) (or explicitly stating “\(\theta = \phi\)”).

Notes

B1: 0.0741 comes from using degrees.

Notes

B1: (1st) Correct condition for intersection of curves leading to a correct expression or value for an angle at a PoI

B1ft: For finding (\(\pm\)) the correct area of the sector (or half of it) either using \(\frac{1}{2}r^2\theta\) or \(\frac{1}{2}\int_{\theta_1}^{\theta_2}(5)^2\,\mathrm{d}\theta\). Accept lower limit of 0. May be embedded in a calculation for total area.
Here \(\theta_1 = -\cosh^{-1}\frac{5}{3}\), \(\theta_2 = \cosh^{-1}\frac{5}{3}\) or \((\theta_1 = -\ln 3), (\theta_2 = \ln 3)\), i.e. may later evaluate integral… for some \(k\)

M1: (1st) Correct use of area formula with \(\pm\)their value for limits. Condone lower limit of 0. Here \(\theta_1 = -\cosh^{-1}\frac{5}{3}\), \(\theta_2 = \cosh^{-1}\frac{5}{3}\). Limits can be seen later. Later doubling may be seen

*M1: Correct conversion of \(\cosh^2\theta\) into a form which can be integrated, or \(\int\cosh^2\theta\,\mathrm{d}\theta = \frac{1}{2}\int 1 + \cosh 2\theta\,\mathrm{d}\theta\)

dep*M1: Correctly integrating \(\cosh^2\theta\) and substituting their limits. Could be embedded.
Or \(\displaystyle\int_{-\ln 3}^{\ln 3}\cosh^2\theta\,\mathrm{d}\theta = \frac{1}{2}\int_{-\ln 3}^{\ln 3}1 + \cosh 2\theta\,\mathrm{d}\theta = \frac{1}{2}\left[\theta + \frac{1}{2}\sinh 2\theta\right]_{-\ln 3}^{\ln 3} = \frac{20}{9} + \ln 3\)
NB \(A_{C_2} = \dfrac{9}{2} \times \dfrac{1}{4}\left(\dfrac{80}{9} + 4\ln 3\right) = 10 + \dfrac{9}{2}\ln 3\)

A1: (1st) Accept unsimplified form. If B1B1M1M0M0 can score ScB1 both here and in the next line for correct answers

Notes

B1: For using the definition of the mean value of \(p\) wrt \(t\), correct limits and 1/0.5.

M1: For \(\int \tanh 1.44t\,\mathrm{d}t = k\ln\left|\mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\right| (+c)\) \(k\) can \(= 1\)
Or \(\int \tanh 1.44t\,\mathrm{d}t = k\ln|\cosh 1.44t| (+c)\)
May see integration by substitution, eg. \(u = \mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\) or \(u = \cosh 1.44t\). If so, award this mark for \(k\ln|u|\) seen

A1: For fully correct integration of \(\tanh 1.44t\).
Either for \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln\left|\mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\right| (+c)\)
or \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln|\cosh 1.44t| (+c)\)
or \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln|u| (+c)\) with \(u\) as above.
Condone missing modulus.

A1: cao, with clear working.

Notes

M1: AG. Use of exponential definition of sinh or cosh. Must be used on LHS or RHS

A1: AG. So an intermediate step must be shown, e.g. \(2\left(\frac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{4}\right)\). LHS must be equated to RHS. Accept full reverse argument

Notes

B1: Use of identity in (a).

M1: Writing as \(= 0\) and factorising (where \(s = \sinh x\) and \(c = \cosh x\))

A1: (1st) Complete solution in any form (assume that \(\cosh^{-1}\) is multi-valued here)

A1: (3rd) Must show \(\pm\) explicitly (or have both \(\ln 2\) and \(\ln\frac{1}{2}\))

Alternative method

B1: Use of exponential definitions of \(\sinh x\), \(\cosh x\) and \(\sinh 2x\) in equation. Also award if starts with main method before using exponentials

*M1: Multiplying by \(\mathrm{e}^{2x}\) and collecting like terms to write as quartic equation in \(\mathrm{e}^x\). Could see a substitution, e.g. \(y = \mathrm{e}^x\) leading to \(6y^4 - 31y^3 + 40y^2 - y - 6 = 0\). Could use Pythagoras to derive quartic in sinh or cosh

*dep*M1: Using factor theorem to deduce that \(\mathrm{e}^x = 2\) (or 3 or \(\frac{1}{2}\)) is a solution and factorising; or \((6y^2 - y - 1)(y^2 - 5y + 6)\) seen

dep*M1: Using factor theorem to find another factor and fully factorising

A1: Must reject negative root explicitly for A1. ScB1 for correct solution after B1M1M0M0

Notes

M1: AG. Taking sinh of both sides, differentiating and using \(\cosh^2 y - \sinh^2 y = 1\)
Condone missing \(\pm\) for M1. Accept \(\frac{\mathrm{d}u}{\mathrm{d}y}\) at this stage

A1: AG so reason required (accept “always positive”); poor notation can be recovered

Alternative method

M1: AG. Using the definition of \(\sinh^{-1}\) in logarithmic form and attempting to differentiate using the chain rule on ln function.

A1: AG so some intermediate working must be seen. www

Alternative method 2

M1: Correctly integrates RHS. Condone omission of \(c\)

Notes

M1: (1st) Differentiating using chain rule or the formula booklet, giving \(\dfrac{1}{\sqrt{\frac{1}{4} + x^2}}\)

M1: (2nd) Substituting the \(x\)-value into the equation to find the \(y\) coordinate of the point

M1: (3rd) Substituting the \(x\)-value into their gradient and taking negative reciprocal

A1: oe in correct form

Notes

B1: Evidence of correct differentiation using product rule and a substitution \(n = 1\) into RHS

M1dep: Differentiate using the product rule

M1dep: Differentiate second time using the product rule

A1*: Must be equated to correct derivative form

A1dep: Dependent on previous A1 and the B1

Notes

M1: Using definitions of \(\cosh x\) and \(\sinh x\).

A1: AG. Intermediate working must be seen.

Notes

M1: Using the identity from (b) to reduce equation to 3 term quadratic

A1: Both solutions found… …and \(-1\) rejected explicitly with valid reason. E.g “\(-1\) is outside the range of \(\cosh x\)”
Could be BC

M1: Use of formula for \(\cosh^{-1}\) (or by solving quadratic in \(\mathrm{e}^x\)).

A1: \(x = \ln 5\) or \(\ln(1/5)\). Must have both solutions.

Alternative method

M1: Using the exponential definition of cosh to reduce the given equation to a quartic equation in \(\mathrm{e}^x\).

M1: Factorising or using quartic solver.

A1: Negative solution must be rejected and a valid reason given.

A1: \(x = \pm\ln 5\). Must have both solutions.

Notes

M1: Correct use of definition of \(\cosh x\) must be seen

A1: AG, must see either \(\mathrm{e}^{\ln 9}\) and \(\mathrm{e}^{\ln\frac{1}{9}}\) or \(3^2\) and \(3^{-2}\) or \(\dfrac{1}{2}\left(9 + \dfrac{1}{9}\right)\)

Notes

M1: oe, intention to integrate \(y^2\).
Condone missing \(\pi\), ignore limits.

A1: For \(+\cosh x\). Ignore any reference to \(c\)

M1: Substituting correct limits and subtracting

A1: Ignore units

Notes

M1: Diffn (Hyperbolics or exponentials)

M1: Set = 0 and use exponential forms – can change to exponentials before diffn.

A1: Conclusion with justification

Alternative method

M1: Differentiate

M1: Set = 0 and use formula for tanh

A1: Conclusion with justification

Notes

M1: Use of exponentials

M1: equation of the form \(a\mathrm{e}^{2x} + b\mathrm{e}^x + c = 0\) (for non-zero \(a\), \(b\) and \(c\))

A1: Two roots for \(\mathrm{e}^x\)

A1: One rejected plus reason

A1: Ignore inclusion of 2^nd root

Alternative method

M1: Use Pythagoras

M1: Quadratic in cosh (or sinh)

A1: Two roots

A1: One rejected plus reason

(Full working of this alternative method is taken from the appendix of the mark scheme.)

Notes

M1: Uses correct exponential form in an attempt at proof

A1: AG. Proof must be complete

(Corrected from the printed mark scheme: the second line is printed as \(2\left(\dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4}\right) - 1 = \dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2}\), with the \(-1\) missing on the right-hand side, as typed above.)

Notes

M1: (first) Use of identity in (a) to leave a three term quadratic equation in just \(\cosh x\)

M1: (second) Attempt to solve eg \(\dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times (-2)}}{2 \times 2}\) or \(2\left(\cosh x - \dfrac{3}{4}\right)^2 - \dfrac{9}{8} - 2 = 0\)

A1: (first) Or solves quadratic BC

A1: (second) Justification must be seen and must contain no incorrect statements

A1: (third) For either correct answer seen

A1: (fourth) Both correct values for \(x\)
Or \(x = -\ln\left(2 + \sqrt{3}\right)\)
Mark final answer

Notes

M1: Using minimum point of curve and knowledge of cosh graph. Could be derived by differentiation

A1: If zero scored then sc1 for k=1 www

Notes

*M1: Use of \((0, 3)\) to derive an expression for \(b\). accept \(\cosh(-b) = \frac{4}{k}\)

dep*M1: Correct numerical use of formula. Or rearranges.

M1: Use of \((2, 0)\) to derive \(b = 2a\). Could be from (a). Allow ft

Notes

B1: Condone 27/4

M1: Use of \(x = 5\) with their values of \(a\) and \(b\) to predict \(d\). Must have \(-1\)
\(a = 1.0317\ldots,\ b = 2.0634\ldots,\ d_{\mathrm{J}} = 10.067\ldots\) Condone 10.07 only if clear evidence of production

A1: From correct values

Notes

M1: Use of exponential form for \(\sinh u\)

Notes

M1: Attempt to find \(\dfrac{\mathrm{d}x}{\mathrm{d}u}\)

M1: Use double angle formulae and attempt to integrate.

A1: Ignore \(c\).

A1: \(c\) must be included here as part of \(\mathrm{f}(x)\) – allow \(a\) and \(b\) not being stated explicitly but \(\mathrm{f}(x)\) must be [the guidance ends here in the printed mark scheme]

Alternative method

M1: Use exponentials and attempt to integrate.

A1: Ignore \(c\).

A1: \(c\) must be included here as part of \(\mathrm{f}(x)\) – allow \(a\) and \(b\) not being stated explicitly but \(\mathrm{f}(x)\) must be [the guidance ends here in the printed mark scheme]

Notes

M1: Correct limits substituted and subtracted into their answer to (c) soi

A1: \(p\), \(q\), \(r\) must be stated

Notes

M1: Use of \(\cosh x\) in exponentials

Notes

M1: Substitute and use (a)

A1: any form entirely in terms of \(u\). Allow absence of \(\mathrm{d}u\)

M1: Use standard form for integral and substitute back

A1: Must include \(c\)

Alternative method

Alternative method

M1: Substitute

A1: In correct form

M1: Integrate and substitute back

A1: Must include \(c\)