Notes

M1: Differentiates numerator and denominator to the correct form, may be done separately

A1: Correct derivatives

M1: Recognises the requirement to differentiate numerator and denominator again and achieves the correct form of the derivatives may be seen separately or as a fraction \(\dfrac{A\cos 7x}{B\cos 9x + C\cos 9x + Dx\sin 9x}\)

A1*: Completes the proof, including correct limit notation seen when \(x = 0\) is substituted with as a minimum \(\dfrac{49}{9 + 9} = \dfrac{49}{18}\) or \(\lim\limits_{x \to 0}\dfrac{49\cos 7x}{18\cos 9x - 81x\sin 9x} = \dfrac{49}{18}\)

Notes

M1: Differentiates \(u = x^2\) twice, may be seen as part of the expression for the \(k\)^th derivative

M1: Uses \(v = \mathrm{e}^{3x}\) to establish the form of the derivatives \(\dfrac{\mathrm{d}^k v}{\mathrm{d}x^k} = 3^k\mathrm{e}^{3x}\). Look for multiples of \(\mathrm{e}^{3x}\) increasing by a factor of 3 each time, need at least three. This may be seen as part of the expression for the \(k\)^th derivative

M1: A correct strategy for the \(k\)^th derivative. This requires the correct derivative of \(x^2\) combined with the correct derivative of \(\mathrm{e}^{3x}\) in terms of \(k\) together with the correct binomial coefficient. Condone using \(k(k - 1)\) for the third term

A1: A correct unsimplified expression

A1: Correct expression in the required form with correct values of \(A\), \(B\) and \(C\). Condone missing trailing bracket
(NB \(A = 9\), \(B = 6\), \(C = 1\))

Repeated differentiation to find at least the third derivative
\(y = x^2\mathrm{e}^{3x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\mathrm{e}^{3x} + 3x^2\mathrm{e}^{3x} \Rightarrow \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} + 6x\mathrm{e}^{3x} + 9x^2\mathrm{e}^{3x}\)
\(\Rightarrow \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 6\mathrm{e}^{3x} + 12\mathrm{e}^{3x} + 36x\mathrm{e}^{3x} + 18x\mathrm{e}^{3x} + 27x^2\mathrm{e}^{3x}\)
they can score the first M1M1 for differentiating \(u\) twice and \(v\) three times. Then will score M0A0A0.

Notes

M1: Substitutes \(k = 9\) into their result from part (a) and solves the resulting 3TQ which leads to real roots

A1: These values only following full marks in (a)

Notes

M1: Finds the correct form of the first four derivatives of \(\mathrm{e}^{3x}\). Must all be of the form \(\alpha\mathrm{e}^{3x}\).
SC: M1 may also be awarded for at least four relevant derivatives found if they do not find all four for the \(\mathrm{e}^{3x}\) or the \(\sin x\).

A1: Correct derivatives. Allow if the general form is identified rather than individual ones given.
SC allow for at least four correct derivatives if not all are found.

M1: Finds the correct form of the first four derivatives of \(\sin x\), condone sign slips only.

A1: Correct derivatives.

M1: Applies Leibnitz’s theorem to get the 4^th derivative with their expressions. Binomial coefficients must be present and correct numerical expressions (not notational forms).

A1*: Correct simplified 4^th derivative from fully correct working, including the \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} =\) seen at some stage.

Note: If do not use Leibnitz’s theorem then up to the first 4 marks can be awarded for using the product rule to differentiate. Accept correct forms for the derivatives.

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\mathrm{e}^{3x}\cos x + A\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = C\mathrm{e}^{3x}\cos x + D\mathrm{e}^{3x}\sin x\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{3x}\cos x + 3\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 6\mathrm{e}^{3x}\cos x + 8\mathrm{e}^{3x}\sin x\)

dM1: Forms correct next two. A1: \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = 26\mathrm{e}^{3x}\cos x + 18\mathrm{e}^{3x}\sin x\) and \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = 28\mathrm{e}^{3x}\sin x + 96\mathrm{e}^{3x}\cos x\)

Notes

(b) Note this appears on epen as MAA but is being marked as BMA

B1cao: Deduces the correct value of \(R\), simplified, \(R = 100\)

M1: Attempts the value of \(\alpha\) via \(\tan\alpha = \pm\dfrac{B}{A}\) or \(\tan\alpha = \pm\dfrac{A}{B}\) or \(\sin\alpha = \pm\dfrac{B}{R}\) or \(\cos\alpha = \pm\dfrac{A}{R}\)
Note that \(\alpha =\) awrt 1.29 or \(\alpha =\) awrt \(74^\circ\) imply this mark.

A1: Correct answer, the \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) may be missing but must be given as the correct expression with \(\alpha =\) awrt 1.29 not just separate values for \(R\) and \(\alpha\)

Notes

M1: Complete method to write the function as a quotient.

dM1: Attempts differentiation of both numerator and denominator, including use of product rule with the denominator. Either numerator or denominator of the correct form. May be done separately.

A1: Fully correct differentiation of both numerator and denominator (may be separate).

ddM1: Dependent on previous method mark. Recognises the need to differentiate again and carries out the differentiation to complete the method.

M1: Clear demonstration of L’Hospital’s Rule being used to attain a limit, e.g. clear statement using limits reaching an expression that can be evaluated. Both numerator and denominator must be used appropriately. The substitution may be implied by 0/2 being reached from a suitable expression (oe if errors) but must have achieved a non-zero denominator. Not dependent, so may be scored if genuine errors in the first derivative lead to an expression that has a non-zero denominator. They may have only differentiated once.

A1*: Fully correct solution. Must see clear use of a substitution of \(x = 0\) into their derivatives. Accept as minimum e.g. \(\dfrac{0}{2 - 0}\) with each term seen evaluated or equivalent working shown. Needs to be a correct line showing substitution before reaching the printed answer with use of some limit notation. All aspects of the proof should be clear for this mark to be awarded and no errors seen.
NB Proceeding to fourth derivatives before evaluating the limit is a correct approach, and may score the final M once a limit is reached, and final A if all aspects are correct.

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: 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 differentiation of both numerator and denominator, including at least one use of the chain rule. Either numerator or denominator of the correct form. May be done separately.

A1: Numerator correct

A1: Denominator correct

M1: Applies l’Hospital’s Rule, must see clear use of a substitution of \(x = \dfrac{\pi}{2}\) into their derivatives, not the original expression. (no need to see check that limits of numerator and denominator are non-zero).

A1*: Needs to be a correct intermediate line following substitution before reaching the printed answer with use of some limit notation. All aspects of the proof should be clear for this mark to be awarded and no errors seen.

Notes

M1: Differentiates \(u = x^3\) three times. Need to see \(x^3 \to \ldots x^2 \to \ldots x \to k\)

M1: Uses \(v = \sin kx\) to establish the form of the derivatives. Need to see at least alternating \(k^{\cdots}\sin kx\) and \(k^{\cdots}\cos kx\) with increasing powers of \(k\) for at least 3 derivatives.

M1: Uses a correct formula with 2 and 3! (or 6) with terms shown to disappear after the fourth term. This needs to be a correct application of the theorem so that the correct binomial coefficients need to go with the correct pairings of their derivatives. If there is any doubt, at least 3 terms should have the correct structure. Allow equivalent notation for the binomial coefficients e.g. \(\dbinom{5}{0}, \dbinom{5}{1}\) etc. or \({}^5\mathrm{C}_0, {}^5\mathrm{C}_1\) etc.

A1: Correct expression in the required form with correct values of \(A\), \(B\) and \(C\). Apply isw if necessary e.g. if a correct expression is followed by \(A = 60\), \(B = 15\), \(C = -4\)
(NB \(A = -60\), \(B = 15\), \(C = -4\))

If there is no use Leibnitz’s theorem e.g. repeated differentiation of products, this scores no marks.