Notes

M1: Uses the chain rule to differentiate to the correct form. (\(\lambda\) can be 1). Condone a missing bracket on \(3x^2\) for this mark.

A1: Correct answer. Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - (3x)^2}}\) oe

Alt

Alternatively

M1: Uses implicit differentiation and trig identities to achieve the correct form for \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) in terms of \(x\)

A1: Correct answer. Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - (3x)^2}}\) oe Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\cos(\arcsin 3x)}\)

Notes

M1: Uses the chain rule and their answer to part (a) to differentiate to the correct form. May use \(\arccos y = \arcsin 3x\) – see scheme for form but must get to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)

A1: Correct simplified answer. If using the Alt they must replace the \(y\) and simplify to achieve the same answer. Must be their answer to (b). Do not accept forms in trig functions such as \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\tan(\arcsin 3x)\) as these are not simplest form.
Do not allow recovery if they simplify the sin(arcsin..) as part of their working in part (c).

Alt

Notes

M1: A complete method to find a value for at least \(x^2\). Sets their answer to part (b) equal to 4, rearranges, squares both sides to reach a quadratic in \(x\) before achieving a value for \(x\) or \(x^2\). Note that if they never simplify the \(\sin(\arcsin 3x)\) in the derivative they will not be able to solve for \(x^2\) and so will not gain this mark.

A1: Correct value for \(x^2\) from a correct derivative in (b) (may be implied).

A1: Selects the correct value of \(x\). Must have come from a correct derivative in (b).

Caution: Watch out for those who get \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{9x}{\sqrt{1 - 9x^2}}\) in (b) which will give the correct values for part (c) but should score (b) M1A0 (c) M1A0A0.

Alt

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\tan(\arcsin 3x) = 4 \Rightarrow \tan(\arcsin 3x) = -\dfrac{4}{3} \Rightarrow \sin(\arcsin 3x) = -\dfrac{4}{5} \Rightarrow 3x = \ldots\)
Uses the alt form in trig functions, set equal to 4 and proceeds to use right angle triangle or trig identities to proceed to a value for \(3x\)

A1: Correct value for \(3x\)

A1: Correct value for \(x\)

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

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

B1: Correct series (ignore terms beyond \(x^3\)).

Notes

M1: Correctly applies the exponential Maclaurin expansion at least once, either to the base exponent or in the index. Allow 2 for 2! and 6 for 3! in the cube term. Follow through on their series seen in (a)

M1: A complete attempt to use the exponential Maclaurin series to produce a cubic expression in terms of \(x\) only. Allow if the 3! is incorrect for this mark, but a polynomial in \(x\) must have been achieved. Condone a slip with one term. Follow through on their series seen in (a)

dM1: Dependent on previous method mark only. Expands the brackets and gathers terms (not necessarily fully simplified, but should have a single term for each power).

A1: Any two correct from coefficients of \(x\), \(x^2\) and \(x^3\), need not be simplified.

A1: Fully correct answer with simplified terms.

NB: Question instructs to use standard Maclaurin series, so use of differentiation scores no mark.

Special case: Using \(\mathrm{e}^{\left(\mathrm{e}^x - 1\right)} = \mathrm{e}^{\mathrm{e}^x} \times \mathrm{e}^{-1}\) can score M1M1M0A0A0 for using Maclaurin series \(\mathrm{e}^{\mathrm{e}^x}\) and then on \(\mathrm{e}^x, \mathrm{e}^{2x}, \mathrm{e}^{3x}\)

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: Finds \(x\) in terms of \(y\) and differentiates

M1: Uses the trig identity \(\sin^2 y + \cos^2 y = 1\) to express \(\cos y\) in terms of \(x\). This may be seen in their derivative or stated on the side

A1*: Correctly achieves the printed answer \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1 - x^2}}\). cso

Notes

M1: Differentiates using the chain rule to achieve the correct form, condone \(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^{2x}}}\)
Note \(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^x}}\) is B0 for incorrect form
Alternatively restart, finds \(x\) in terms of \(y\) and differentiates

A1: Correct differentiation

A1: Follows correct differentiation. States that as \(\mathrm{e}^x \neq 0\) (or \(\mathrm{e}^x \gt 0\)) or no solutions to \(\mathrm{e}^x = 0\) therefore there are no stationary points.
Alternatively, \(\mathrm{e}^x = 0\) leading to \(x = \ln 0\) which is impossible/undefined/error therefore there are no stationary points. Ignore any reference to the denominator \(= 0\)

Notes

(a)
M1: Finds the correct form for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: States or shows that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \neq 0\) and draws the required conclusion. This mark can be scored as long as the M mark has been awarded.

Notes

(b)
M1: Substitutes \(x = 1\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Finds the normal gradient and finds the equation of the normal using \(y - \dfrac{\pi}{3} = m_n(x - 1)\)
M1: Finds where their normal cuts the \(x\)-axis and the \(y\)-axis.
M1: Finds the area of the triangle \(OAB = \dfrac{1}{2} \times x_A \times -y_B\).
A1: Correct area

Special case: If finds the tangent to the curve, the \(x\) and \(y\) intercepts and the area of the triangle max score M1 M0 M1 M0 A0

Note common error
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-1}{\sqrt{1 - \frac{1}{4}x^2}}\) In part (b) this leads to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2}{\sqrt{3}}\) leading to normal gradient \(\dfrac{\sqrt{3}}{2}\) and
\(y = \dfrac{\sqrt{3}}{2}x - \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{3}\) and \(\left(0, \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}\right)\) and \(\left(1 - \dfrac{2\pi}{3\sqrt{3}}, 0\right)\) therefore area \(= \dfrac{1}{2}\left(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}\right)\left(\dfrac{2\pi}{3\sqrt{3}} - 1\right)\)
This can score M1 M1 M1 M1 A0

Notes

(a)
M1: Finds the correct form of the first three derivatives, may be unsimplified – the third may come later.
A1: Correct first three derivatives, may be unsimplified – the third may come later.
M1: Finds \(\mathrm{f}(0)\), \(\mathrm{f}^\prime(0)\), \(\mathrm{f}^{\prime\prime}(0)\) and \(\mathrm{f}^{\prime\prime\prime}(0)\) and applies to the correct formula, needs to go up to \(x^3\)
A1: \(x + \dfrac{x^3}{6}\) cso ignore any higher terms whether correct or not.

Special case: If they think that their \(\mathrm{f}^{\prime\prime}(0) \neq 0\) then maximum score M1 A0 M1 A0. M1 for correct form of the first two derivatives. M1 Correctly uses their \(\mathrm{f}(0)\), \(\mathrm{f}^\prime(0)\), \(\mathrm{f}^{\prime\prime}(0)\) and applies to the correct formula.

Note: If candidates do not find the first three derivatives but use \(\mathrm{f}(0) = 0\), \(\mathrm{f}^\prime(0) = 1\), \(\mathrm{f}^{\prime\prime}(0) = 0\), \(\mathrm{f}^{\prime\prime\prime}(0) = 1\) and use these correctly in the formula this can score M0 A0 M1 A0

Notes

(b)
M1: Substitutes \(x = \dfrac{1}{2}\) into both sides and rearranges to find \(\pi = \ldots\)
A1ft: Infers that \(\pi = \dfrac{25}{8}\) o.e. Follow through their \(6\mathrm{f}\left(\dfrac{1}{2}\right)\)

Notes

M1: Deduces the required series by using the Maclaurin series for \(\cos x\), replacing \(x\) with \(\dfrac{x}{3}\) and squares, or first applying the double angle identity (allow sign error) and then applying the series for \(\cos x\) with \(\dfrac{2x}{3}\). Attempts at finding from differentiation score M0 as the cosine series is required.

A1: Correct series

Notes

M1: Divides their series in part (a) by \(x\) and integrates to the form \(A\ln x + Bx^2 + Cx^4\)

A1ft: Correct integration, follow through on their coefficients and need not be simplified.

A1: Deduces the definite integral awrt 0.98295

Notes

B1: Correct value.

Notes

B1: Makes a quantitative statement about the accuracy, so e.g. how many decimal places or significant figures it is correct to, or calculates a percentage accuracy to deduce it is reasonable. Do not accept just “underestimate” or similar without quantitative evidence. Allow for a reasonable comment as long as (b) is correct to at least 2 s.f. but (c) must be the correct value.

Notes

(a)
M1: Makes progress in establishing the derivative by taking the tan of both sides and differentiating with respect to \(y\) or implicitly with respect to \(x\)
M1: Use of the correct identity
A1*: Fully correct proof

Notes

(b)
B1: Correct derivative
M1: Uses integration by parts in the correct direction
A1: Correct expression
M1: Adopts a correct strategy for the integration by splitting into two fractions or using a substitution of \(4x = \tan u\) to get to an integrable form
A1: Correct answer

(corrected from the printed mark scheme: \(\displaystyle\frac{1}{32}\int \left(\sec^2 u - 1\right)\mathrm{d}u\) is printed as \(\displaystyle\frac{1}{32}\int \sec^2 u - u\,\mathrm{d}u\))

Notes

(c)
M1: Correctly applies the method for the mean value for their integration. The limit of zero can be implied if it comes to 0.
A1: Correct exact answer. Allow exact equivalents e.g. \(\dfrac{4\pi\sqrt{3} - 9}{72},\ \dfrac{\pi\sqrt{3}}{18} - \dfrac{1}{8}\)

Typical solution

Typical solution

Typical solution

\(1 - 2x^2\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

The expansion is only valid for

\(x = 1\) is not in this valid range

So Julia should not have substituted \(x = 1\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Let

Then

So, by l’Hôpital’s rule,

Typical solution

\(\dfrac{\pi}{2}\) and \(\dfrac{3\pi}{2}\)

Typical solution

Typical solution

By using Simpson’s rule with 5 ordinates

Typical solution

Typical solution

Typical solution

Typical solution

\(\displaystyle\lim_{x \to \infty}\left(\frac{\mathrm{e}^x}{x^5}\right)\)

Typical solution

(i)

(ii)

Typical solution

Typical solution

Let \(\mathrm{f}(x) = x,\ \mathrm{g}(x) = \mathrm{e}^x\)

Then \(x\mathrm{e}^{-x} = \dfrac{\mathrm{f}(x)}{\mathrm{g}(x)}\)

and \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) both tend to \(\infty\) as \(x \to \infty\)

\(\mathrm{f}^{\prime}(x) = 1\) and \(\mathrm{g}^{\prime}(x) = \mathrm{e}^x\)

\(= 0\) as required

Typical solution

Typical solution

Typical solution

First non-zero terms of series expansion of \(\dfrac{\sin x}{x} - \cos x\) are \(\dfrac{x^2}{3}\) and \(-\dfrac{x^4}{30}\)

First non-zero terms of series expansion of \(1 - \cos x\) are \(\dfrac{x^2}{2}\) and \(-\dfrac{x^4}{24}\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Notes

Throughout this question condone single line slips to sin or cos provided recovered.

B1: May see \(n = 1\) in addition but this mark is for \(n = 0\) only. Must see LHS = RHS \(\Rightarrow\) true for \(n = 0\) oe (but may appear later e.g. in conclusion)

M1: Sets up and uses inductive hypothesis by showing an intent to differentiate twice and using the inductive assumption at least once. Complete statement might not be seen until after first or even second differentiation.
Must have statement in terms of some other variable than \(n\) in all three places consistently.
Condone minor slips in notation e.g. \(\frac{\mathrm{d}^2\boldsymbol{f}}{\mathrm{d}x^2}\left(\frac{\mathrm{d}^{2k}\mathrm{f}}{\mathrm{d}x^{2k}}\right)\) for all but the final A mark.

M1: Differentiates using product rule. Condone sign errors, missing 2s, \(\frac{1}{2}\) instead of 2 but not \(\frac{1}{2}x^2\) as the derivative of \(x\).
May be seen separately. Could be done with exponential form but M1 awarded on correct return to hyperbolic form (which could be after later differentiation).

M1: Differentiates again. Same guidance. Or \(= 4^k(4k\cosh 2x + 4\cosh 2x + 4x\sinh 2x)\)
Allow both differentiations in one step if fully correct.

A1: Rewrites to the correct form.

A1: Clear conclusion for induction process with implication stated. Dependent on all previous M and A marks www and fully correct notation throughout.
Allow this mark if the basis case was given as \(n = 1\) only but their conclusion must match their basis case (i.e. \(n = 1 \Rightarrow n \geq 1\)).

Notes

M1: \(= 1024\) Chooses correct value of \(n\) and uses formula to find value for appropriate derivative when \(x = 0\).

M1: Using the correct rule for finding the coefficient of \(x^8\). This is independent of the previous mark and \(\mathrm{f}^{(8)}(0)\) need not be correct or even evaluated but must be \(x = 0\).

A1: Must be simplified. Do not accept eg 0.0254 etc but ISW once 8/315 seen unless it is clear that a different value is intended (eg 8 from multiplying by 315).
Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\). Answer only with no working scores 0/3.

Notes

B1: Uses exponential definition of \(\sinh x\) with \(2x\) as argument…

M1: … and given general term (or complete series) for \(\mathrm{e}^x\) to try to find (coefficient of) \(x^7\) term of \(\sinh 2x\) (or \(x^8\) term of \(x\sinh 2x\)).

A1: No conclusion necessary. Must be in same form as answer to (b) (ie explicit verification) (but allow this mark for a correct answer if no solution in (b)).
Allow SC A1 if (b) and (c) both given as eg 0.0254. Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\)

Notes

M1: Squaring the correct result for \(\ln(1 + x)\) with at least the first three terms.

B1: First two terms correct.

A1: Third term correct allow any equivalent fraction or exact decimal equivalent. ISW and ignore higher power terms.

Alternative method

M1: For \(\mathrm{f}(0) = 0\) and correct first derivative (or stating that the second derivative is zero too).

A1: Correct values for first, second and third derivatives
For reference: \(\mathrm{f}''(x) = \dfrac{2 - 2\ln(x + 1)}{(x + 1)^2}\) and \(\mathrm{f}'''(x) = \dfrac{2(-3 + 2\ln(x + 1))}{(x + 1)^3}\) and \(\mathrm{f}^{(4)}(x) = \dfrac{22 - 12\ln(x + 1)}{(x + 1)^4}\) and \(\mathrm{f}^{(4)}(0) = 22\).

A1: All terms correct (allow simplified or exact decimal equivalents for coefficients). ISW and ignore higher power terms.

Notes

M1*: Setting their expansion from part (a) equal to \(2x^3\). Their expansion must have at least three terms with non-zero quadratic, cubic and quartic terms.

M1dep*: Rearranging and factorising (allow sign slips only). Ignore higher order terms if the factor of \(x^2\) is seen – their expansion must have contained no constant or linear terms. Stating their quadratic \(1 - 3x + \frac{11}{12}x^2 = 0\) is fine for this mark. Stating either non-zero root of the correct (or their) quadratic can imply this mark.

A1: Selects the smallest non-zero root (so must see both correct roots either exact \(\dfrac{18 \pm 8\sqrt{3}}{11}\) or to at least 2 decimal places) and justifies with interval for validity being \(-1 \lt x \leqslant 1\). Allow awrt 0.38 or \(\dfrac{18 - 8\sqrt{3}}{11}\).
Condone any indication that expansion is only valid between values of \(-1\) and 1 or for values less than, or less than or equal, to 1.
Note that including \(x^5\) terms in expansion gives \(x = 0.35997\) and including \(x^5\) and \(x^6\) terms in expansion give \(x = 0.36504\). Ignore any attempt to work out corresponding \(y\)-coordinate.

Notes

B1: For \(\dfrac{1}{\sqrt{(1 - (x^2)^2}}\) seen.

M1: For \(2x \times \mathrm{f}(x)\) where \(\mathrm{f}(x) = \dfrac{1}{\sqrt{1 - (x^2)^2}}\) or \(\dfrac{1}{\sqrt{1 - x^2}}\) ONLY.

A1: Allow any equivalent correct form e.g. \(2x(1 - x^4)^{-0.5}\).
Must be in terms of \(x\).
Condone \((x^2)^2\) for \(x^4\).
ISW once correct answer seen.

Alternative method

B1: For correctly differentiating implicitly with respect to either \(x\) or \(y\).

M1: Replacing \(\cos y\) with \(\pm\sqrt{\pm 1 \pm (x^2)^2}\) in their derivative of the form \(\pm\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) (or equivalent if differentiating with respect to \(y\)).

A1: Allow any equivalent correct form e.g. \(2x(1 - x^4)^{-0.5}\).
Must be in terms of \(x\).
Condone \((x^2)^2\) for \(x^4\).
ISW once correct answer seen.

Notes

M1: Differentiating correctly first derivative, or \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{a}{1 + at} = \left(a^{-1} + t\right)^{-1}\) etc

A1: (1st) For all; constants need not be evaluated and may be unsimplified, e.g. could see \((-1)^2\) or \((-1)^3\)

A1: (2nd) Could be \((-1)^{n+1}\) oe, or \((-1)^{n-1}(n - 1)!\left(a^{-1} + t\right)^{-n}\). Omission of factorial term can score a possible B1M1 in part (b)

Notes

*B1ft: Convincingly showing that their conjecture works for \(n = 1\) for their 1^st derivative in part (a); accept omission of working shown in 1^st line here

M1: Forming the inductive hypothesis and making it clear that the \((k + 1)\)^th derivative is the derivative of the \(k\)^th derivative

*A1: Differentiating and rewriting into correct form. Some intermediate working and/or justification must be seen. Could see substitution \(u = 1 + at\) etc

dep*A1: full correct argument

Notes

M1: Considering the seventh derivative.

A1: 5.625. Ignore attempt at units.

Notes

(a)(i)

B1: Convincingly shown

(corrected from the printed mark scheme: the printed answer says “sign of \(\frac{\mathrm{d}y}{\mathrm{d}t}\)”; the term in the equation is \(\frac{\mathrm{d}x}{\mathrm{d}t}\))

(a)(ii)

M1: Substitutes \(t = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \pm 5\)

A1: Or 6 towards \(O\)

Notes

(b)(i)

B1: First two terms of Maclaurin’s formula, possibly in generalised form, must be seen
AG Convincingly shown.

(b)(ii)

B1: AG Convincingly shown

(b)(iii)

M1: For \(\frac{\mathrm{d}}{\mathrm{d}t}(\cosh t) = \sinh t\) and product rule attempted

A1: Fully correct differentiation

M1: Substitute \(t = 0\)

A1: AG. Allow embedded answer.

Notes

(c)(i)

B1: Substitutes \(t = 0.25\) to obtain an approximation and correct conclusion

(c)(ii)

B1: Allow any correct explanation that explains/implies that for \(t \gt 1\) some of the higher power terms are large and so non-negligible.
\(\mathrm{f}(10) = 156\) is too large is not enough

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

M1: Quoting and using the Maclaurin series

A1: AG. Result with sufficient justification

Notes

B1: AG

Notes

B1: Some justification that \(t \gt 1\) is required

M1: Substituting their choice into the inequality

A1: Answer without use of inequality in part (b) scores M0A0

Alternative method

B1: Some justification that \(t \gt 1\) is required

Notes

(a)(i)

B1: Not for \(45^\circ\)

(a)(ii)

M1: Diffn – Must be seen
\(\mathrm{f}'(x) = \dfrac{1}{1 + x^2}\) is M0

(a)(iii)

M1: Diffn their \(\mathrm{f}'(x)\)

A1: oe, e.g. \(\mathrm{f}''(x) = -\dfrac{2(1 + x)}{\left(1 + (1 + x)^2\right)^2}\)

A1: \(\mathrm{f}''(0)\) must be seen. The substitution must be seen (implied by \(-\frac{2}{4}\))
AG

Question 2(a)(ii) Alternative solution

Notes

M1: Using the formula and substituting their value for f;(0)

A1: ft their values from (a)

Notes

(a)(i)

M1: Formula from the Formula Booklet and attempt differentiation. To within sign error

(a)(ii)

B1: or \(a_0 = 0\), \(a_1 = 1\) and \(a_2 = 0\). Ignore sign error in \(\mathrm{f}''(x)\)

M1: Differentiate and simplify far enough to be able to justify value 1. Either full derivative or “zero term” denoted as such

A1: Condone 3! In place of 6. Not BC. If M0 then SC1 for correct expansion

(a)(iii)

M1: Integral of their 2 term cubic with limits

A1: Could be BC

Notes

M1: Attempt integration by parts. ignore limits. Formula for parts must be correct

Notes

B1: soi

M1: Differentiating using chain rule (or rule for \(\ln(f(x))\)) and evaluating when \(x = 0\). Allow sign error in numerator

A1: Differentiating again using quotient (or product/chain) rule. NB Simplifies to \(-\dfrac{\frac{1}{2}\cos x + 1}{\left(\frac{1}{2} + \cos x\right)^2}\)

A1: www

If zero scored then SC1 for correct expansion

Notes

B1: Finding either \(\pm\pi/3\) as a root. Allow \(60^\circ\) for B1. Ignore other roots

M1: Substituting their root, in radians, into their Maclaurin series and equating (approximately) to 0.
Or equating their expression (approximately) to 0 and rearranging for \(x\): \(\ln\left(\frac{3}{2}\right) - \frac{x^2}{3} \approx 0 \Rightarrow x \approx \sqrt{3\ln\left(\frac{3}{2}\right)}\)
\(\frac{\pi}{3} \approx \sqrt{3\ln\left(\frac{3}{2}\right)} \Rightarrow \pi \approx 3\sqrt{3\ln\left(\frac{3}{2}\right)}\)

A1: Could see \(\pm\) but must be removed by final conclusion. Must use approximately equals symbol (not just equals symbol)

Notes

M1: Correct form, in terms of \(\theta\). M1 can be implied by 1.0757… BC

*A1: Integrand has been squared out. Must include limits (can be seen later)

dep*M1: Might be as result of substitution. Allow coefficient error for M1. eg \(\frac{3}{4}\left[\mathrm{e}^{u}\right]_{-\frac{2}{3}}^{\frac{2}{3}}\) or \(\frac{3}{4}\left[\mathrm{e}^{\frac{2}{3}u}\right]_{-1}^{1}\) oe

A1: isw

Notes

*M1: Attempt to differentiate using product and chain rules. Must be in the form \(uv' + u'v\) with at most one of \(u, v, u'\) or \(v'\) incorrect or omitted

dep*M1: Setting \(r'\) to zero and factorising/cancelling to produce a quadratic equation in cos and/or sin

M1: Use of \(\cos^2 + \sin^2 = 1\) to find 3 term quadratic equation in \(\cos\theta\). Or could be in \(\sin^2\theta\): \(4\sin^4\theta + 9\sin^2\theta - 9 = 0\)

*A1: Solving quadratic correctly. \(\sin^2\theta = \frac{3}{4}, -3\)

dep*A1: Explicitly rejecting root. Rejects \(\sin^2\theta = -3\) and \(\sin\theta = -\frac{\sqrt{3}}{2}\)

A1: AG. At least one intermediate step must be seen. Can be awarded even if rejection of root(s) was implicit.

Notes

M1: Attempt to differentiate using chain rule, i.e. product of 2 terms

Alternative method

M1: Make a substitution

(corrected from the printed mark scheme: the printed alternative has \((1 + \tan^2 x)\), which should be \((1 + \tan^2 y)\), and shows a total of [4] for this 2-mark part.)

Notes

M1: Get into form of (a). Ignore limits

A1: Correct form

M1: Use (a) and correct limits in correct order.

A1: oe

Alternative method

M1: Make a substitution

A1: Get into correct form

M1: Use standard result with correct limits in correct order

Notes

B1: no more simplified in initial step than \(\frac{(-1)^2(0!)}{1 + x}\); condone missing brackets

M1: differentiating expression for \(\mathrm{f}^{(k)}(x)\); condone bracketing errors or a sign slip only

A1: condone \(\frac{(-1)^{(k+2)}k!}{(1 + x)^{k+1}}\). Do not condone bracketing errors. M1 step must be seen.

A1: www. \(\frac{(-1)^{(k+1)+1}((k + 1) - 1)!}{(1 + x)^{k+1}}\) must have been seen to score this mark. \(n = 1\) must have been considered. Cannot score A0A1.

Notes

M1: using result from (a) to find at least two derivatives (can include general term). May be embedded in coefficients. Must include powers of \(-1\) and factorials in each term; condone missing division by 1. Allow one slip.

M1: considering coefficient of general term and simplifying using \(\frac{(n - 1)!}{n!}\)

A1: AG complete argument, www. Must clearly be a sum of infinitely many terms. Must have considered \(\mathrm{f}(0) = 0\).

Alternative method

M1: use of Maclaurin series with result from (a) used for all terms other than \(\mathrm{f}(0)\)

A1: \(\frac{(n - 1)!}{n!} = \frac{1}{n}\) seen or used

A1: AG

Notes

B1: Must be simplified

Notes

M1: Or \(x^3 = 0.125\). Must be seen, can be implied by explicit substitution.

A1: oe correct substitution. If \(x^3 = 0.125\) used then \(0.125 - \frac{0.125^2}{2} + \frac{0.125^3}{3}\). FT their three terms. Cannot be implied.

A1: cao

Notes

M1: May be embedded

A1: See appendix. Do not award if spoilt by an incorrect statement.

Alternative method

M1: May be embedded

A1: See appendix. Do not award if spoilt by an incorrect statement.

Appendix: exemplar responses for Q10c – A1 mark only; the M1 mark is not implied by any of the responses below and must be secured separately

The following statements do not score by themselves but will not spoil an otherwise correct answer:

Notes

(a)(ii)

M1: Their \(\mathrm{f}(0)\), \(\mathrm{f}^{\prime}(0)\) and \(\mathrm{f}^{\prime\prime}(0)\) evaluated and substituted into Maclaurin

A1: Must come from correct expressions for \(\mathrm{f}^{\prime}(x)\), \(\mathrm{f}^{\prime\prime}(x)\); cannot come from binomial expansion. Ignore subsequent terms.

Notes

M1: using \(x = \frac{1}{8}\) in their expansion

A1: AG

Notes

M1: May be in exponential form

A1: AG

Notes

M1: chain rule

Notes

M1: quotient or product rule

M1: \(\cosh^2 x - \sinh^2 x = 1\) used

Notes

B1ft: soi

M1: Maclaurin expansion attempted, must see their values substituted in

Notes

A1: Allow \(-0.48\%\)

Notes

M1: substitute \(x = 1\) in quadratic

Notes

B1: \(\dfrac{k}{\sqrt{1 - 4x^2}}\)

M1: Chain rule (x 2)

A1: Fully correct \(\frac{\mathrm{d}y}{\mathrm{d}x}\)

Alternative solution

M1: For \(\frac{\mathrm{d}x}{\mathrm{d}y} = k\cos\frac{y}{6}\)

M1: Using chain rule \(\left(\times \frac{1}{6}\right)\)

A1: Or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 \times \dfrac{1}{\sqrt{1 - 4x^2}} \times 2\)

(Corrected from the printed mark scheme: the guidance for the first M1 is printed as \(\frac{\mathrm{d}x}{\mathrm{d}y} = k\cos\frac{\pi}{6}\); it should be \(k\cos\frac{y}{6}\), as typed above.)

Notes

M1: substituting

Notes

SC B1 if other methods used

NB AG

Notes

A1: accept \(f^{(n)}(x) = 2^{n-1}\cosh(2x)\)

Notes

M1: at least 3 terms correct, or good attempt from 1st principles

A1: Allow unsimplified fractions (without factorials)