Notes

B1: Correct first derivative in any form, may square first and use implicit differentiation

M1: Differentiates again applying the quotient or product rule correctly to achieve the correct form or uses implicit differentiation to achieve a correct form

A1*: Correct proof with no errors. Must show sufficient working.
If uses the product rule they must show a common denominator
If uses the quotient rule there must be evidence to show the multiplication of \(\left(15 + \mathrm{e}^{2x}\right)^{\frac{1}{2}}\) top and bottom
If uses implicit differentiation we need to see the substitution for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(y\) and rearranging.

Notes

M1: A full method to obtain the required expansion. I.e. finds the values up to the second derivative and uses a correct formula

A1: Correct simplified expansion with \(y =\) or \(\mathrm{f}(x) =\) seen somewhere in their solution

Notes

B1: Deduces the correct simplified expansion, ignore any extra terms, whether correct or not

Notes

M1: A rigorous argument using their answers to (b) and (c), showing the cancelling constant term and the division by \(x\) in the numerator and denominator to establish the limiting behaviour.

A1*: Correct proof with no errors including correct limiting notation seen in the final stage cso

Note: Using L’Hospital’s Rule with the functions \(\sqrt{\left(15 + \mathrm{e}^{2x}\right)} - 4\) and \(\sin 3x\) is M0A0

Using L’Hospital’s Rule with their expansions \(\dfrac{4 + \frac{x}{4} + \frac{31}{128}x^2 - 4}{3x - \frac{9}{2}x^3}\) leading to \(\dfrac{\frac{1}{4} + \frac{31}{64}x}{3 - \frac{27}{2}x^2}\) could score M1A1

Notes

M1: Attempts to differentiate including an attempt at the product rule to achieve the correct form in any order

A1: Correct differentiation

M1: Continues to differentiate using the product rule again to reach the 4^th derivative to achieve the correct form \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = A\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + Bx\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + C\dfrac{\mathrm{d}y}{\mathrm{d}x} + D\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in any order

A1: Completes the process to obtain the correct expression (\(a = 4\), \(b = 2\), \(c = -3\)) with no errors seen and correct notation throughout and \(= 0\) where applicable cso

Notes

B1: Deduces the correct value for \(y^{\prime\prime}(2)\)

M1: Finds the values of the derivatives at \(x = 2\). There must be evidence of the correct values of \(x = 2, y = 1, \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) being used at least once

M1: Uses the correct Taylor series expansion with their derivatives. Must be using \(y = 1, \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) in their series expansion. If no method is shown they must have all terms correct for their derivatives

A1: Correct expansion must include \(y =\) or \(\mathrm{f}(x) =\) somewhere

Notes

M1: A complete method to find the value for \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\). Substitutes into the differential equation \(x = \dfrac{\pi}{4},\ y = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and rearranges to find a value for \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\).

A1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -1 \lt 0\) therefore a maximum. Must achieve the correct value, or at least a correct expression and give reason “\(\lt 0\)” (oe) and conclusion and no contradictory statements.

Notes

M1: Differentiates to the form \(\pm\sin x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \cos x\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + \alpha y\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{1\text{ or }2} + y^2\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} \pm \cos x = 0\).
Alternatively, divides through by \(\cos x\) first and differentiates to achieve the form shown in scheme.

A1: Correct differentiation, any alternative form is acceptable.

M1: Substitutes in \(x = \dfrac{\pi}{4},\ y = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), their value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \text{‘}-1\text{’}\) as appropriate for their expression, and rearranges to find a value for \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) (if not already done so).

A1*: Shows that \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \sqrt{2} - 2\) from correct working with at least one suitable intermediate line – being either an unsimplified \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\) or \(\alpha\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\), before the final answer. Must have come from a correct derivative expression. Note that in the Alt the correct form comes out much easier, so look for at least one unsimplified line before the final answer.

Notes

M1: Uses the Taylor’s series expansion with the correct values for \(y_{\frac{\pi}{4}} = 1,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0,\ \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \sqrt{2} - 2\) (or their \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) if they had a different answer to (b)) and their value of \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\). Must include the 1 at the start.

A1: Correct simplified expansion. The “\(y =\)” may be missing. Accept with \(0\left(x - \dfrac{\pi}{4}\right)\) as a second term. ISW after a correct simplified expression if further attempts to simplify are made.

Notes

M1: Attempts differentiation of the given function by applying both the chain rule and the product rule. Look for the correct form but they may make slips with signs or coefficients. Alternatively, applies the sum law for logs and differentiates using the chain rule.

A1*: Correct proof.

Notes

B1: Correct second derivative.

M1: Continues the differentiation using the chain rule and product rule to reach the 4^th derivative. Look for correct forms, allowing for slips in signs or coefficients only when differentiating. May be given in terms of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}, \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) etc for this mark.

A1: Correct 4^th derivative in terms of \(x\). Accept alternatives. Like terms should be gathered and coefficients simplified but isw after a correct suitable answer.

Notes

M1: Attempts the values of the derivatives at \(x = 0\) up to at least the third derivative. If substitution not seen at least two values should be correct for their derivatives, with values for others.

M1: Substitutes the values of their derivatives into the correct Maclaurin expansion formula up to at least the third non-zero value for their derivatives.

A1cso: Must have obtained a correct 4^th derivative in (b) (though need not have been in terms of \(x\)). Correct expansion with simplified coefficients. May be missing the “\(y =\)” and allow “\(\mathrm{f}(x) =\)”. Note the correct series will be obtained from various incorrect 4^th derivatives due to some terms evaluating to 0, but use of a clearly incorrect 4^th derivative score A0 here. However, an initially correct 4^th derivative, with incorrect simplification, can score A1 bod if no incorrect work is shown (ie correct values stated with no substitution shown, as they may have used the initially correct answer).

Notes

B1: Deduces the correct expansion in any form.

Notes

M1: Applies the subtraction law of logarithms and then substitutes their expansions.

M1: Realises that the \(x\) terms will determine the required value of \(k\) and so collects the \(x\) terms and sets the coefficient to zero and solves for \(k\) (may be implied).

A1: For \(k = 2\)

Notes

Notes: ignore extra terms throughout.

M1: Differentiates \(\mathrm{f}(x) = \ln x\) twice and finds \(\mathrm{f}(1)\), \(\mathrm{f}^{\prime}(1)\) and \(\mathrm{f}^{\prime\prime}(1)\)

A1: Correct differentiation and values for \(\mathrm{f}(1)\), \(\mathrm{f}^{\prime}(1)\) and \(\mathrm{f}^{\prime\prime}(1)\).

M1: Uses correct mathematical notation to find the Taylor series for \(\ln x\) in powers of \((x - 1)\) up to \((x - 1)^2\)

A1: Correct expansion with simplified coefficients. Do not be concerned with the left hand side.

(Corrected from the printed mark scheme: the third line of working is garbled in the printed scheme; it is shown here as \(\mathrm{f}^{\prime\prime}(x) = -\dfrac{1}{x^2} \Rightarrow \mathrm{f}^{\prime\prime}(1) = -1\), which the expansion uses.)

Notes

Question says “hence” so the result of (a) must be used. No marks for l’Hosptial’s rule on the original functions (send to review if attempted with their part (a)).

M1: Substitutes their Taylor series for \(\ln x\) in powers of \((x - 1)\) up to \((x - 1)^2\) into the limit and cancels a factor \((x - 1)\) from each term. Allow for the cancelling seeing a relevant strikethrough in all \(x - 1\) terms.

A1*: \(\lim\limits_{x \to 1}\left(\dfrac{\ln x}{x - 1}\right) = \lim\limits_{x \to 1}\left(1 - \dfrac{1}{2}(x - 1) + \ldots\right) = 1\) cso Must have come from a correct expansion. Must see the \(1 - \dfrac{1}{2}(x - 1)\)

Notes

M1: Writes the fraction in an indeterminate form \(\dfrac{f(x)}{g(x)}\) where \(\dfrac{\boldsymbol{f}(0)}{\boldsymbol{g}(0)} = \dfrac{0}{0}\) or \(\dfrac{\boldsymbol{f}(0)}{\boldsymbol{g}(0)} = \dfrac{\text{“}\infty\text{”}}{\text{“}\infty\text{”}}\)

M1: Differentiates numerator and denominator using appropriate rules, ie product rule for a product etc. Allow slips in coefficients but the form should be correct. This mark is available as long as written as \(\dfrac{f(x)}{g(x)}\) even if not an indeterminate form. May be seen written as separate from the fraction, ie \(\mathrm{f}^{\prime}(x) = \ldots\) and \(\mathrm{g}^{\prime}(x) = \ldots\)

A1: Depend on both Ms. Correct differentiation for derivatives that lead to a limit. Must have a derivative for which \(\mathrm{g}^{\prime}(0) \neq 0\) and is finite.

A1cso: Deduces the correct limit from fully correct work.

Notes

M1: Attempts the first derivative, including use of the chain rule. May expand first. E.g. accept forms as shown.

A1: Correct first derivative, need not be simplified.

M1: Attempts second derivative using quotient rule or product rule – examples as shown, or equivalents accepted.

A1*: Correct result achieved from correct work.

Notes

M1: Applies quotient rule or product rule to achieve third derivative. If formula is quoted it must be correct, if not accept derivatives of the form shown as there may be confusion with the minus sign.

A1: Correct third derivative, any form.

Notes

M1: Find value of derivatives at \(x = 1\).

M1: Applies Taylor series expansion

A1: Correct series, may be unsimplified, isw once correct series seen. Must be using a correct third derivative.

Notes

M1: Applies the series to the limit and cancels terms in numerator to leave term in \((x - 1)^3\) and above only (may not see \(+\ldots\) for this mark)

M1: Simplifies and realises that the \((x - 1)^3\) cancels and achieves a constant \(A\)

A1: Correct limit deduced with reasoning given why the remaining terms disappear.

Notes

M1: Establishes the non-disappearing derivatives of \(x^4\). Allow slips in coefficients, but powers must decrease.

M1: Identifies the relevant derivatives for \(\sin(2x)\), up to the 8^th derivative or establishes the correct pattern. Look for alternating between sin and cos. Condone use of \(x\).

A1: Correct sizes for the coefficients, allow sign errors for this mark (may be due to incorrect signs when differentiating sin and cos) Must have angle \(2x\).

A1: All derivatives correctly established. (Note the sin terms may be omitted if the student has made clear they will disappear, but if present they must be correct).

M1: Applies Leibnitz’s theorem to get the 8^th derivative with their expressions. Binomial coefficients must be present.

M1: Evaluates their 8^th derivative at \(\pi\)

M1: Uses Taylor series – divides their value for \(\mathrm{f}^{(8)}(\pi)\) by \(8!\)

A1: Simplifies to the correct answer.

Note: If do not use Leibnitz’s theorem then maximum M0 M0 A0 A0 M0 M1 M1 A0

Notes

M1: Attempts to find the second and third derivatives:
This requires \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) followed by \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \pm 2y\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} \pm \ldots\) or \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \pm \ldots \pm 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)

A1: Correct second and third derivatives.

dM1: Continues to differentiate to reach the 5^th derivative. This is dependent on the first method mark but there is no need to check the detail and the mark can be awarded as long as the 5^th derivative is reached.

A1: Completes the process, collecting terms if necessary, to obtain the correct expression
(NB \(a = -2\), \(b = -8\), \(c = -6\))

Allow dash/dot notation for the derivatives but the final answer must be in the correct form.

Note that if \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is obtained initially, allow a full recovery in (a).

Note that (a) can be found using Leibnitz’s theorem and the following scheme should be applied:

M1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) followed by an attempt to differentiate \(y\) 3 times and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) 3 times.

A1: All correct

dM1: \(\dfrac{\mathrm{d}^{5}y}{\mathrm{d}x^{5}} = -2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 3 \times 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 3 \times 2\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2 - 2\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (correct application of Leibnitz)

A1: \(= -2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 8\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 6\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2\)

(Corrected from the printed mark scheme: the Leibnitz line is printed with the first term as \(-2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) and the last as \(-2y\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\dfrac{\mathrm{d}y}{\mathrm{d}x}\); the factors \(y\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) are the wrong way round.)

As in the main scheme, if \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is obtained initially, allow a full recovery in (a).

Alternative for (a)

M1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - 2y\left(x - y^2\right) = 1 - 2xy + 2y^3 \Rightarrow \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -2y - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Score for the second derivative form as in the main scheme and then an attempt at the third derivative with at least 2 terms correct.

A1: Fully correct

Then as main scheme.

Notes

B1: Deduces the correct value for \(y^{\prime}(0)\)

M1: Finds the values of all the other derivatives at \(x = 0\) up to 5^th. There is no need to check their values as long as there is no obvious incorrect work, but values for all the derivatives up to the 5^th must be found

A1: All values correct (as single values – e.g. do not allow unsimplified)

M1: Applies the correct Maclaurin series for their values including the factorials up to the term in \(x^5\)

A1ft: Correct expansion, follow through their values for the derivatives. This does not have to be simplified but the factorials need to be evaluated. Once a correct, or correct follow through expression is seen apply isw.

Notes

ALT 1   Divide by y before differentiating:

ALT 2   Re-arrange and divide by y before differentiating:

Notes

B1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\tan x\sec^2 x\)

M1: attempting the second derivative, inc using the product rule or \(\sec^2\theta = \tan^2\theta + 1\)   Must start from the result given in (a)

A1: a correct second derivative in any form

A1cso: for a correct result following completely correct working \(\sec^2\theta = \tan^2\theta + 1\) must be d seen or used

Notes

M1: attempting the third derivative, inc using the chain rule

A1: a correct derivative

A1: a completely correct final result

Notes

B1: \(y_{\frac{\pi}{3}} = \left(\sqrt{3}\right)^2\) or 3 and \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{3}} = 2\sqrt{3}\times\left(\dfrac{2}{1}\right)^2\) or \(8\sqrt{3}\)

M1: obtaining values for second and third derivatives at \(\dfrac{\pi}{3}\) (need not be correct but must be obtained from their derivatives)

M1: using a correct Taylor’s expansion using \(\left(x - \dfrac{\pi}{3}\right)\) and their derivatives. (2! or 2, 3! or 6 must be seen or implied by the work shown)  This mark is not dependent.

A1: for a correct final answer  Must start \(\tan^2 x = \ldots\) or \(y = \ldots\)   \(\mathrm{f}(x)\) scores A0 unless defined as \(\tan^2 x\) or \(y\) here or earlier. Accept equivalents eg awrt 610 (609.6…) \(\sqrt{371712}\) But no factorials in this final answer.

Notes

B1  \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) seen in the differentiation
M1  divide equation by \(y\) and differentiate wrt \(x\) chain and product/quotient rules needed
A1A1  -1 for each error. Ignore any simplification following the differentiation and obtaining \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \ldots\)

ALT:
B1 as above
M1 differentiating before dividing
A1A1  rearrange to a correct expression for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\),  -1 each error

Notes

M1  using values for \(x\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain a value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1  correct value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1  correct value for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\)
M1  Taylor’s series formed using their values for the differentials, accept 2! or 2 and 3! or 6.
A1  correct series, must start \(y =\)   (or end \(= y\))

(Corrected from the printed mark scheme: the decimal for \(\dfrac{37}{16}\) is printed as 2.325; \(\dfrac{37}{16} = 2.3125\).)

Notes

Q4a – Alternative 1

Accept the dash notation in this question

M1 for using the product rule to differentiate \(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\).

M1 for differentiating \(5y\) and using the product rule or chain rule to differentiate \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)

A2,1,0 for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{-5\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}}{y}\) Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error. If there are two sign errors and no other error then give A1A0.
Do NOT deduct if the two \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) terms are shown separately.

Alternative to Q4a

Can be re-arranged first and then differentiated.

M1M1 for differentiating, product and chain rule both needed (or quotient rule as an alternative to product rule)

A2,1,0 for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{y^2}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3 - \dfrac{2}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)\) Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error

Notes

Alternative

M1 for substituting \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\) and \(y = 2\) in the equation to obtain a numerical value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)

A1 for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -7\)

A1 for obtaining the correct value, 16, for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\)

M1 for using the series \(y = \mathrm{f}(x) = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2!}\mathrm{f}^{\prime\prime}(0) + \dfrac{x^3}{3!}\mathrm{f}^{\prime\prime\prime}(0) + \ldots\) (2! or 2, 3! or 6) (The general series may be shown explicitly or implied by their substitution)

A1 for \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\) oe Must have \(y = \ldots\) and be in ascending powers of

Alternative to Q4b

M1 for setting \(y = 2 + 2x + ax^2 + bx^3\)

M1 for \(\left(2 + 2x + ax^2 + bx^3\right)(2a + 6bx) + \left(2 + 2ax + 3bx^2\ldots\right)^2 + 5\left(2 + 2x + ax^2 + bx^3\right) = 0\)

A1 for equating constant terms to get \(a = -\dfrac{7}{2}\)

A1 for equating coeffs of \(x\) to get \(b = \dfrac{8}{3}\) (corrected from the printed mark scheme, which says “coeffs of \(x^2\)”; the scheme above equates the coefficients of \(x\))

A1 for \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\)

Notes

B1 for \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = -2\) wherever seen

M1 for attempting the differentiation of the given equation. To obtain \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} \pm k\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm \cos x\ (= 0)\) oe

A1 for substituting \(x = 0\) to obtain \(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = \dfrac{1}{2}\)

M1 for using the expansion \(\left[y = \mathrm{f}(x)\right] = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2(!)}\mathrm{f}^{\prime\prime}(0) + \dfrac{x^3}{3!}\mathrm{f}^{\prime\prime\prime}(0)\) with their values for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\). Factorial can be omitted in the \(x^2\) term but must be shown explicitly in the \(x^3\) term or implied by further working eg using 6.

A1cao for \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) (Ignore any higher powers included) Exact decimals allowed. Must include \(y = \ldots\)

Alternative

(corrected from the printed mark scheme: the printed alternative omits the factor 4 in front of the bracket, which is shown in the notes below.)

B1 for \(y = \dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3 + \ldots\)

M1 for differentiating this twice to get \(y'' = 2a + 6bx + \ldots\) (may not be completely correct)

A1 for correct differentiation and using the given equation and the expansion of \(\sin x\) to get \(2a + 6bx + \ldots = \left(x - \dfrac{x^3}{6} + \ldots\right) - 4\left(\dfrac{1}{2} + \dfrac{x}{8} + \ldots\right)\) (corrected from the printed mark scheme: \(\dfrac{x^3}{3}\) is printed for \(\dfrac{x^3}{6}\))

M1 for equating coefficients to obtain a value for \(a\) or \(b\)

A1 cao for \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) (Ignore any higher powers included)

Notes

Finding second derivative and substituting into given answer acceptable

Notes

1^st M1 for differentiating second term to obtain an expression involving \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)

B1B1B1 for 4, 7, 32 seen respectively

2^nd M1 require \(\mathrm{f}(1)\) or 1, \(\mathrm{f}^{\prime}(1)\) etc and \(x - 1\) and at least first 3 terms

A1 for 4 terms following through their constants

Condone \(\mathrm{f}(x) =\) instead of \(y =\)

Notes

1^st M1 for evidence of Product Rule

1^st A1 for completely correct expression or equivalent

2^nd A1 for correct expression or \(k = 4\) stated

Notes

2^nd M1 require four terms and denominators of 2 and 6 (might be implied)

A1 follow through from their values in the final answer.

Notes

(corrected from the printed mark scheme: the line giving 80 is printed with \(\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{\frac{\pi}{4}}\) instead of \(\left(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{4}}\), and the two expansion lines are printed as \(\sec x \approx \ldots\) instead of \(\sec^2 x \approx \ldots\))

Notes

M: Use of product rule (at least once) and implicit differentiation (at least once).

Notes

M: Use of series expansion with values for the derivatives (can be allowed without the first term 1, and can also be allowed if final term uses 3 rather than 3!)

Notes

(corrected from the printed mark scheme: the \(\left(x - \tfrac{\pi}{4}\right)^3\) term of the general expansion is printed with \(\mathrm{f}'\left(\tfrac{\pi}{4}\right)\) instead of \(\mathrm{f}'''\left(\tfrac{\pi}{4}\right)\))

Notes

[Alternative (c):