Notes

May use \(t\) or another variable which is acceptable.

M1: Uses the model with \(v = 0\) and proceeds to \(\mathrm{e}^{0.2T} = 15\) Do not be concerned by the use of an inequality sign instead of an equals.
May be implied by awrt 13.5

A1: \((T =)\) awrt 13.5 units are not required but if given they must be seconds (or e.g. secs or s)

Notes

If no attempt is seen for (b) then allow differentiation seen in (a) to score in (b)

M1: Attempts to use the product rule to differentiate \(t\,\mathrm{e}^{0.2t}\) achieving the form \(A\mathrm{e}^{0.2t} + Bt\mathrm{e}^{0.2t}\) (\(A\) and \(B\) both non zero but may be 1) which may be unsimplified. It is likely to be part of an expression.

A1: \(\left(\dfrac{\mathrm{d}v}{\mathrm{d}t} =\right) 15 - \left(0.2t\,\mathrm{e}^{0.2t} + \mathrm{e}^{0.2t}\right)\) o.e. which may be unsimplified. (Condone a missing trailing bracket)
Do not allow recovery of signs to score this mark if they initially write
e.g. \(15 - 0.2t\,\mathrm{e}^{0.2t} + \mathrm{e}^{0.2t} = 0\) and on a later line correct this. e.g. \(15 - 0.2t\,\mathrm{e}^{0.2t} - \mathrm{e}^{0.2t} = 0\)

dM1: Sets \(15 \pm A\mathrm{e}^{0.2t} \pm Bt\mathrm{e}^{0.2t} = 0\) (the =0 may be implied), attempts to make \(\mathrm{e}^{\pm 0.2t}\) (or \(C\mathrm{e}^{\pm 0.2t}\)) the subject and proceeds to the form
\(C\mathrm{e}^{0.2t} = \dfrac{D}{E + Ft}\) or \(C\mathrm{e}^{-0.2t} = \dfrac{E + Ft}{D}\) (where \(C\) can be 1 and \(A, B, C, D, E, F \neq 0\))
It is dependent on the previous method mark.
May see \(15 - \text{``}0.2t\,\mathrm{e}^{0.2t} - \mathrm{e}^{0.2t}\text{''} = 0 \Rightarrow \mathrm{e}^{0.2t} = \dfrac{15}{0.2t + 1}\) which scores dM1
They must take out a factor of \(\mathrm{e}^{\pm 0.2t}\) (or \(C\mathrm{e}^{\pm 0.2t}\)) and divide by their bracket. Condone sign slips in their rearrangement, however, if they take logs of both sides first, the rearrangement must be correct (with no sign slips).
Allow invisible brackets to be implied by further work which is not the given answer.

A1*: Achieves the given answer with no errors seen including use of invisible brackets (but condone a missing trailing bracket) All previous marks in (b) must have been scored. The = 0 must have been seen somewhere in their solution. Do not allow this mark to be scored for proceeding directly from
\(\mathrm{e}^{0.2t}(0.2t + 1) = 15 \Rightarrow t = 5\ln\left(\dfrac{75}{t + 5}\right)\) which is A0*
We must see either \(\mathrm{e}^{0.2t} = \dfrac{15}{0.2t + 1}\) o.e. or an unsimplified expression for \(t\)
e.g. \(t = 5\ln\left(\dfrac{15}{0.2t + 1}\right)\) before achieving the given answer. Condone \(t = 5\ln\dfrac{75}{t + 5}\)

Notes

(c)(i) Check by the question. If there is a contradiction between answers, the answer in the main body of the script takes precedence.

M1: Attempts to use the iteration formula at least once. \(t_2 = 5\ln\left(\dfrac{75}{8 + 5}\right)\). May be implied by awrt 8.76 or awrt 8.48 or awrt 8.58. It is not implied by awrt 8.55

A1: awrt 8.478 (on its own can score M1A1)

(c)(ii) This mark can only be scored provided in (c)(i) M1 has been scored so M0A0A1 is not a possible mark profile.

A1: awrt 8.55 seconds (e.g. s or secs) Requires units.
If the candidate lists their iterations but does not select an answer then take the final value, which still requires units to be stated (which in many cases is likely to be omitted)
Note that awrt 8.55 does not imply M1.

Notes

(a)(i)

B1: Correct derivative in any form. "\(\mathrm{f}^{\prime}(x) =\)" is not required. Apply isw if necessary.

(ii)

B1: Correct derivative in any form. "\(\mathrm{g}^{\prime}(x) =\)" is not required. Apply isw if necessary.

Notes

M1: Eliminates e by setting their \(\mathrm{f}^{\prime}(x) =\) their \(\mathrm{g}^{\prime}(x)\) where \(\mathrm{f}^{\prime}(x) = Ax\mathrm{e}^{4x^2-1}\) oe and \(\mathrm{g}^{\prime}(x) = \dfrac{B}{x}\) oe with \(A \times B \gt 0\) and proceeds via \(\mathrm{e}^{4x^2-1} = \dfrac{\ldots}{x^2}\) or equivalent work (see below) to obtain \(4x^2 - 1 = \ln\dfrac{\ldots}{x^2}\) oe e.g. \(\ln x + 4x^2 - 1 = \ln\dfrac{1}{x}\)
Allow if they use \(\alpha\) for \(x\).
Note that there are various alternatives for this mark but the derivatives must be of the form defined above and the processing must be correct with coefficient/sign slips only.

Examples of equivalent work:

A1*: Obtains the printed answer with sufficient working and no errors.
Sufficient work would require the “e” eliminated before the given answer.
Must follow correct derivatives in part (a).
Condone \(4x^2 + 2\ln|x| - 1 = 0\) and condone \(4\alpha^2 + 2\ln\alpha - 1 = 0\) or \(4\alpha^2 + 2\ln|\alpha| - 1 = 0\)

Note that if both derivatives in (a) are correct we will allow fully correct work using the equation in (b) to work backwards to verify that \(p\mathrm{f}^{\prime}(x) = q\mathrm{g}^{\prime}(x)\) for M1 then obtains \(\mathrm{f}^{\prime}(x) = \mathrm{g}^{\prime}(x)\) with a minimal conclusion for A1
If either derivative in (a) is incorrect or missing, candidates who work backwards score no marks in (b).

Notes

(c)(i)/(ii)

M1: Attempts to use the iterative formula with \(x_1 = 0.6\)
Award this mark for e.g. \((x_2 =)\sqrt{\dfrac{1 - 2\ln 0.6}{4}}\) or may be implied by awrt 0.71 provided no incorrect working is seen.
Candidates sometimes find \(x_3\) (or possibly subsequent terms) rather than \(x_2\) in which case the M1 can be implied. (See table below for first few iterations)

A1: \((x_2 =)\) awrt 0.7109
Sight of \((x_2 =)\) awrt 0.7109 scores M1A1

B1(A1 on ePEN): \((\alpha =)\) 0.6706 (4dp)
Must be this value and not awrt 0.6706

For reference:

Notes

M1: Attempts both f(3.6) and f(3.7) or a narrower interval that contains the root 3.672… (which may be implied by sight of f(3.6) = … and f(3.7) = … with at least one correct) and obtains at least one correct to 1 significant figure (rounded or truncated) for their interval and considers their signs. Use of degrees is M0.

Some examples for consideration of sign (which are also sufficient for the change of sign part of reasoning for the A1):

• \(\mathrm{f}(3.6) = -0.6 \lt 0\) and \(\mathrm{f}(3.7) = 0.2 \gt 0\)
• \(\mathrm{f}(3.6) \times \mathrm{f}(3.7) \lt 0\)
• \(\mathrm{f}(3.6) = -0.7,\ \mathrm{f}(3.7) = 0.2\) “change in sign”

For reference \(\mathrm{f}(3.6) = -0.68626\ldots\) and \(\mathrm{f}(3.7) = 0.21194\ldots\)

A1*: This mark requires:

• both f(3.6) and f(3.7) correct to 1 significant figure (rounded or truncated) (or their values correct to 1 significant figure if using a narrower interval)
• a reference to sign change
• a reference to continuity {of \(\mathrm{f}(x)\)}
• a (minimal) conclusion, e.g. “hence root”, “proved”, \(\checkmark\), #, QED, \(3.6 \lt \alpha \lt 3.7\)

Accept as a minimum, "change of sign, continuous, root".
Do not condone “change in sign therefore continuous” or other incorrect statements such as “\(x\) is continuous”, “the interval is continuous” – these score A0.
Condone “the graph is continuous”.
Condone reference to \(x\) in place of \(\alpha\) in their conclusion, e.g. “hence \(x\) lies in the interval”.
Condone statements such as “there is at least one root” in place of their conclusion.

Notes

Note: Their answer to (b) may be seen in part (c) provided that they have not clearly attempted part (b) incorrectly, e.g., an attempt at \(\mathrm{f}^{-1}(x)\) in (b).

M1: For \(\tan\left(\dfrac{1}{2}x\right) \rightarrow \ldots\sec^2\left(\dfrac{1}{2}x\right)\) o.e. The brackets are not required. You may see attempts at the quotient rule but the method should be correct and they should reach something equivalent to \(\ldots\sec^2\left(\dfrac{1}{2}x\right)\).

e.g. \(\tan\left(\dfrac{1}{2}x\right) = \dfrac{\sin\left(\dfrac{1}{2}x\right)}{\cos\left(\dfrac{1}{2}x\right)} \rightarrow \dfrac{k\cos\left(\dfrac{1}{2}x\right)\cos\left(\dfrac{1}{2}x\right) - -k\sin\left(\dfrac{1}{2}x\right)\sin\left(\dfrac{1}{2}x\right)}{\cos^2\left(\dfrac{1}{2}x\right)}\) where \(k\) is a positive constant scores M1. If the formula is seen it must be correct.

A1: \(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec^2\left(\dfrac{1}{2}x\right)\) o.e. which may be unsimplified and apply isw.

The brackets are not required. There is no need for \(\mathrm{f}^{\prime}(x) =\) just look for the expression.

Note that \(\{\mathrm{f}^{\prime}(x)=\}\ \dfrac{3}{2} + \dfrac{1}{2}\tan^2\left(\dfrac{1}{2}x\right)\) is correct and appears occasionally.

\(\{\mathrm{f}^{\prime}(x)=\}\ 1 + \dfrac{1}{2}\sec\dfrac{1}{2}x^2\) is condoned for M1A0 only but \(1 + \dfrac{1}{2}\left(\sec\dfrac{1}{2}x\right)^2\) scores M1A1.

Notes

M1: Attempts \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)}\) and obtains a value following through on their \(\mathrm{f}^{\prime}(x)\) as long as it is a “changed” function in terms of \(x\).

Just stating \(3.7 - \dfrac{\mathrm{f}(3.7)}{\mathrm{f}^{\prime}(3.7)} = \ldots\) without evidence of use of 3.7 in \(\mathrm{f}(x)\) (note that this evidence might come from part (a)) and in their \(\mathrm{f}^{\prime}(x)\) is M0 unless implied by a correct value for both \(\mathrm{f}(x)\) and \(\mathrm{f}^{\prime}(x)\) or by their final answer.
Must be a correct N-R formula used – you may need to check their values – accuracy of at least 3s.f. rounded or truncated required.
Allow if attempted in degrees. For reference in degrees \(\mathrm{f}(3.7) = 3.73\ldots\) and \(\mathrm{f}^{\prime}(3.7) = 1.50\ldots\) and gives \(\alpha = 1.21\ldots\)
Note that the full N-R accuracy is 3.672051617.
For reference, the value of \(\alpha\) is approximately 3.673194406… and scores M0A0 without other valid work.

A1: For awrt 3.672  Ignore any subsequent iterations.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Stretch in the \(y\)-direction by scale factor 3 and

followed by translation \(\begin{bmatrix}4\\0\end{bmatrix}\)

Typical solution

Typical solution

There could be more than one solution between the two points.

The two points are either side of the asymptote.

Typical solution

(a)(i) 7.2155

(a)(ii)

Typical solution

The curve is convex, so the trapezium rule gives an over-estimate.

Therefore, the student is incorrect.

Typical solution

Typical solution

Let \(\mathrm{f}(x) = x^3 - \mathrm{e}^{6 - 2x}\)

Hence \(\alpha\) lies between 0 and 4

Typical solution

Typical solution

(i)

(ii)

(iii)

It is not possible to evaluate \(x_2\) as ln 0 has no value

Typical solution

1

Typical solution

(i)

(ii)

All three graphs intersect at the same point.

Typical solution

Typical solution

Typical solution

9.3

Typical solution

The second graph: the increasing curve that gets steeper, starting just above \(O\) on the \(y\)-axis

Typical solution

Area \(\approx\) 5.28

Typical solution

Shaded region is below \(x\)-axis

area = \(32\ln 2 - \dfrac{33}{2}\)

Typical solution

Area of sector = \(\dfrac{1}{2}r^2\theta\)

Area of triangle = \(\dfrac{1}{2}ab\sin C\)

Hence

Typical solution

Let \(\mathrm{f}(\theta) = \theta - \sin 2\theta\)

Hence solution lies between \(\dfrac{\pi}{5}\) and \(\dfrac{2\pi}{5}\)

Typical solution

(i)

(ii)

Use more iterations

(iii)

The value is on a stationary point

Notes

B1: B1 for \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\sin x\) oe

M1*: Obtain an integral in terms of \(u\) only – must be of the form \(\displaystyle\pm\int \frac{\pm 1 \pm u}{\pm 1 \pm u^2}\,(\mathrm{d}u)\)

Allow \(\mathrm{d}u\) missing throughout but M0 if implying \(\mathrm{d}u = \mathrm{d}x\)

M1dep*: Explicit use of \(1 - u^2 = (1 + u)(1 - u)\) or \(u^2 - 1 = (u + 1)(u - 1)\) and simplify to an integral of the form \(\displaystyle\pm\int \frac{1}{\pm 1 \pm u}\,(\mathrm{d}u)\)

A1: AG www – condone \(\ln|1 - \cos x| + c\)
Allow A1 for \(\displaystyle-\int \frac{1}{1 - u}\,\mathrm{d}u = \ln(1 - u) = \ln(1 - \cos x) + c\) (must see \(\ln(1 - u)\) as AG)
However, \(\displaystyle\int \frac{1}{u - 1}\,\mathrm{d}u = \ln|\cos x - 1| = \ln(1 - \cos x) + c\) is A0 without further correct working or correct explanation/justification
Note that \(\displaystyle\int \frac{1}{u - 1}\,\mathrm{d}u = \ln(\cos x - 1) = \ln(1 - \cos x) + c\) is A0 regardless of any further explanation/justification

\(+c\) must be present in the final given answer
e.g. as a minimum \(\ln|\cos x - 1| = \ln|(-1)(1 - \cos x)| = \ln(1 - \cos x) + c\)

Notes

B1: Possibly implied by later working – condone (use of) 180 (in degrees) or \(\cos^{-1}(-1)\) (so do not need to see this evaluated to \(\pi\))

e.g. as a limit on an integral

M1*: Correct use of their upper limit and correct lower limit to form an equation in \(a\) only so must be of the form \(\ln k - \ln(1 - \cos a) = a\) following through their value of \(k \gt 0\)

Setting equal to \(a^2\) rather than \(a\) is M0

M1dep*: Correct use of log laws and correctly remove natural log to form an equation in \(\mathrm{e}^a\) from \(\ln k - \ln(1 - \cos a) = a\) following through their \(k \gt 0\) and \(k \neq 1\)

A1: AG – sufficient working must be shown and must be \(= 0\)

allow arccos for \(\cos^{-1}\)

Notes

M1: Obtain correct value for either f(1.1) or f(1.2) to at least 1 sf rot

A1: Both correct values to at least 1 sf rot, ‘change of sign’ either shown or stated + conclusion (minimum e.g. ‘root’)

Allow, for example,’ \(-0.1 \lt 0 \lt 0.03\) therefore \(1.1 \lt a \lt 1.2\)’ for A1

Notes

B1: At least the first three values of \(a_i\) stated correctly to at least 2 decimal places rot e.g. \(a_3 = 1.18\) or \(a_3 = 1.19\)

Ignore values after the first three

B1: cao (must be to exactly 2 decimal places) – final answer must be 1.18 or \(a = 1.18\) and not a term of a sequence e.g. \(a_9 = 1.18\)

Not dependent on the previous B1. A correct answer without the correct first three terms is B0 B1

Notes

B1: Correct explanation that the graph or diagram or curve (in Fig. 2) implies that the gradient (either at \(a\) or implying all points) is negative.

Must mention ‘the graph’ oe for this mark but mention of Fig. 1 is B0

B1: Correct justification that as the iterative formula \(\left(a_{n+1} = \mathrm{F}(a_n)\right)\) converged to \(a \Rightarrow |\mathrm{F}^{\prime}(a)| \lt 1\).
As a minimum must state/mention ‘convergence’ and EITHER that the gradient at \(a\) is between \(-1\) and 1 OR that the gradient at \(a\) is between \(-1\) and 0 but only following a correct explanation that the graph (oe) implies a negative gradient

Not \(\Rightarrow |\mathrm{F}^{\prime}(a_n)| \lt 1\).
For both marks to be awarded the interval of (–1,0) must be mentioned

B2 for ‘graph shows negative gradient, and there is convergence, therefore the gradient at \(a\) lies in the interval (–1,0)’

Implying that \(a\) is between –1 and 1 and not the gradient is B0 for the second B mark

Notes

(a)(i)

M1: Correct formula oe with \(\geqslant 4\) \(y\)-values correct or omitting the \(2 \times\) May see \(\frac{1}{2} \times 0.5 \times \{1.732\ldots + 2.645\ldots + 2(1.803\ldots + 2 + 2.291\ldots)\}\)

A1: All correct. May see \(\sqrt{\frac{13}{4}}\) and \(\sqrt{\frac{21}{4}}\) oe

A1: awrt 4.14
Answer only without working scores 0/3

(a)(ii)

B1: \(0.5 \times (1.7321 + 1.8023 + 2 + 2.2913)\) awrt 3.91

B1: \(0.5 \times (1.8028 + 2 + 2.2913 + 2.6458)\) awrt 4.37

Allow B1B1 for these values seen anywhere (e.g. may see lower/upper switched). Must be evaluated.

Notes

M1: M1 for a partial or incomplete expression, e.g. missing (0,2) \(\displaystyle\lim_{\delta x \to 0} \sum \left(\sqrt{x^2 + 3}\right)\delta x\). Allow any expression involving \(\lim\), \(\Sigma\), \(\sqrt{x^2 + 3}\) (or \(y\)) and \(\delta x\) for this mark.

A1: oe (must be in terms of \(x\), not \(y\)).

Notes

M1: Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc. but signs (and values if given) must be correct to 1sf – see table.

B1: cao (Allow this mark even with no/insufficient working).

A1: By showing two values between 0.785-0.795 with opposite signs
Values of iterates not needed, i.e. condone \(\lt 0\) and \(\gt 0\) etc, but if given must be correct to 1sf
e.g. \(x = 0.79 \Rightarrow y = 0.0007\) and \(x = 0.795 \Rightarrow y = -0.0009\)

(Note that exact root is 0.792…)

Notes

M1: For an attempt at differentiating using the product rule, with at least one term correct (their derivative must have two terms).

B1: May be implied by not giving a corresponding solution (provided another solution for \(x\) is reached). (NB this mark can be gained following M0 provided their derivative has \(\mathrm{e}^{1-3x}\) as a factor)

A1: Not decimal, www (i.e. must come from a correct derivative)

Notes

M1: M1 for attempting integration by parts with at least one term correct, must see limits (condone swapped limits, may appear later). Must have two terms.

A1: A1 for both terms correct after 2^nd integral performed

A1: Need not be simplified, isw any incorrect attempts to simplify

Notes

B1: State the 4 correct non-zero \(y\)-values and no others.
Exact values (including unsimplified) or decimal equivs (0, 0.1516, 0.3679, 0.5020, 0.5413), which could be truncated or rounded.
For the first value, if \(0\mathrm{e}^0 = 1\) is seen then allow credit for the unsimplified value; if however it is only ever seen as 1 then this is B0 but M1M1 could still be awarded.
B0 if other ordinates seen, unless clearly not intended to be used.

M1*: Big brackets need to be seen or implied.
Attempts at \(y\)-values must be correctly placed (but no need to see \(y = 0\) explicitly).
If no earlier evidence of \(y\)-values seen (eg in a table) then allow M1 for the correct structure with 4 of the 5 values being correct.
Condone using more than 4 intervals as long as values equally spaced between \(x = 0\) and \(x = 2\).

M1d*: Dep on previous M1.
Or using \(k = 0.5h\), with \(h\) consistent with their different number of intervals.

A1: Obtain 0.646.
Allow answers > 3sf, as long as they round to 0.646.
A0 if not using 4 strips, even if 0.646 is obtained.
No credit if no evidence of using the trapezium rule shown.

Using separate strips (a triangle and then trapezia) is an acceptable method, and marks should be awarded as per the main MS (ie \(y\)-values / structure / widths / final answer).

Notes

B1: Convincing reason.
Allow just ‘more trapezia’ or ‘narrower trapezia’.
Could refer to strips or intervals.

Notes

B1: Curve is both convex and concave.
Comment about the shape
Referring to increasing and decreasing gradients is correct, but increasing and decreasing curve is not.
Allow BOD if muddles about which part of the curve is convex and which is concave.

B1: The tops of trapezia are both above and below the curve.
Comment about the estimates
If candidates refer to ‘it’ rather than ‘trapezia’ then allow BOD.

B marks are independent.
See appendix for further examples (below).

Appendix: exemplar responses for Q1(c)

Notes

B1: Identify both \(y\)-values being positive and state ‘no sign change’ or equiv
Could also evaluate \(\mathrm{f}(0)\) as 1 and \(\mathrm{f}(1)\) as 2.9 (or better), and refer to no sign change – these are both positive so no need to include \(\gt 0\)
Could also refer to the asymptote / discontinuity within this range (\(x = 0\) to \(x = 1\))
Also allow ‘graph is not continuous in this interval’
B0 for no reference to interval
Could say that the two points chosen are not on the same part of the curve

Notes

M1: Attempt rearrangement, as far as \(kx^2 = \ldots\)
Allow sign error(s) only

A1: Obtain given answer convincingly
If \(x = \sqrt{\dfrac{1}{4} - \dfrac{1}{8}\mathrm{e}^x}\) then an additional line of working needed before given answer (eg show common denominator of 16)

Notes

B1: Correct first iterate (at least 4sf)
State 0.2851 or better

M1: Correct iterative process (at least 3 more values)
Allow M1 for 3sf – expect 0.289, 0.288 and then 0.288 or 0.289 depending whether truncating or rounding

A1: Correct root, given to 4sf, following 2 iterates that agree to 4sf
ie at least 7 iterations needed, given to at least 4sf
A0 for eg \(x_8 = 0.2885\) (implies 8^th iterate and not root)
Process self corrects so B0M1A1 possible; or B1M1A1 if error in term other than \(x_2\)

Notes

M1: Attempt differentiation using the chain rule
Obtain derivative of form \(\dfrac{kx}{2 - 8x^2}\)
Condone subscripts still present in derivative

M1: Attempt \(\mathrm{F}'(0.3)\) – not dependent on previous M1, but must follow some attempt at differentiation
M1 can be implied by correct \(-3.75\) (from correct derivative), but explicit substitution must be seen if \(\mathrm{F}'(x)\) is incorrect
Must come from differentiating \(\mathrm{F}(x)\) and not a different function

A1: Correct reasoning, following correct \(\mathrm{F}'(0.3)\)
Allow \(\mathrm{F}'(\alpha) \lt -1\), hence will not converge
Condone \(\mathrm{F}'(x)\) not \(\mathrm{F}'(\alpha)\)
No credit for just testing the given iterative formula

Notes

M1*: Setting equation for \(a\) equal to zero and removing \(t^2\) correctly from the denominator e.g. \(8t - \frac{1}{2}(7 + 4t^2) = 0\) to obtain the equivalent of a 3TQ in \(t\) only
This mark can be implied by a correct 3TQ in \(t\)

M1dep*: Correct method for solving their 3TQ in \(t\)
If factorising: \(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) so note that \(4t^2 - 16t + 7 = (t - 0.5)(t - 3.5)\) is M0 but e.g. \((-2t + 7)(2t - 1)\) is M1 bod
If using the formula: must apply the correct formula for their three-term quadratic in \(t\) (no errors)
If completing the square: The M mark is not awarded until correctly getting to the stage \(t - 2 = \pm\sqrt{\dfrac{9}{4}}\) for their 3TQ in \(t\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)
Must see the method – the correct answers do not imply this mark therefore \(4t^2 - 16t + 7 = 0 \Rightarrow t = 0.5\) and 3.5 scores M1 M0 B1
As a minimum must see (if correct) \(\dfrac{16 \pm \sqrt{144}}{8}\)

B1: This mark is not dependent on the previous M mark(s)
So M1 M0 B1 is common or M0 M0 B1 if no working seen

Notes

B1*: Stating the correct differential equation
Possibly implied by correct separation of variables

B1dep*: Correct lhs

B1dep*: Correct rhs – not multiplied by \(v\) or 5 (or any other constant)
Condone lack of \(+c\) for both B marks

M1*: Uses correct initial conditions to find \(c\) from an equation of the form \(k_1\ln v = k_2\ln(7 + 4t^2) + k_3t + c\) (note that e.g. \(+\,c\) may appear on the lhs)
With non-zero values of \(k_i\) - accept any equivalent form e.g. \(v = A(7 + 4t^2)\mathrm{e}^{-0.5t}\) and then use initial conditions to find \(A\) (if correct then \(A = 2.5\))

M1dep*: Uses \(t = T\), \(v = 5\) to obtain an equation in \(T\) only – dependent on previous M mark
Condone use of \(t\) for \(T\) throughout the remainder of the question

A1: AG so at least one step of intermediate working from substitution of \(t = T\) and \(v = 5\)
Condone \(T = 2\ln\left|\dfrac{7 + 4T^2}{2}\right|\)

Notes

B1: Uses given result and given starting value to obtain correct \(T_1\) and \(T_2\) (so the first two iterations after the initial value of 11.25) to at least 4 sf (rot) – but all stated values in these two terms must be correct

B1: Must be stated to 4 sf only – not dependent on the first B mark – can be awarded if either of \(T_2\) and/or \(T_3\) incorrect (assume that the iterative process corrected itself or a slip in the candidate writing down an earlier value)
Must be clear that \(T\) is 11.01 (and not the final term shown in the iterative process) – this mark can be awarded from using alternative iterative methods e.g. Newton-Raphson

Notes

B1: awrt 7.51
No follow through from incorrect earlier values

Notes

B1: Correct (possibly unsimplified) area of \(OMB\)
Could use other than \(r\) for the radius
Could set their variable equal to \(OM\), giving a radius that is double this
eg \(OM = x\) so area \(= x^2\sin\theta\)

M1: Attempt to use ratio on two correct areas
Using two of \(OMB\) \(\left(\tfrac{1}{4}r^2\sin\theta\right)\), \(MAB\) \(\left(\tfrac{1}{2}r^2\theta - \tfrac{1}{4}r^2\sin\theta\right)\) and \(OAB\) \(\left(\tfrac{1}{2}r^2\theta\right)\) oe with their variable
Must be two correct areas
Must be using the correct ratio for their two areas ie 2:3 if using \(OMB\) and \(MAB\), 2:5 if using \(OMB\) and \(OAB\) or 3:5 if using \(MAB\) and \(OAB\)
Allow ratio to be used the wrong way around eg \(2OMB = 3MAB\)

A1: Any correct statement linking the two areas
Could use other than \(r\) for the radius
Or \(2x^2\theta - x^2\sin\theta\)

A1: Simplify to given answer
At least one line of working once ratio used

Notes

B1: Obtain correct first iterate
3sf or better – more accurate answer is 0.599281923...
Condone truncating if more sig fig given

M1: Attempt correct iterative process to find at least 2 more values
M1 is for the correct process for finding \(\theta_3\) and \(\theta_4\), but these may be incorrect
M0 if working in degrees

A1: Obtain 1.13
Possibly following B0 if first iterate is wrong but process then self corrects
Must follow M1 ie a clear attempt to use the correct iterative process
Must be 3sf
Once M1 is awarded, allow A1 for 1.13 even if an incorrect iterate seen, as process will recover

Notes

B1*: Draw straight line, starting at the origin which intersects the graph
Allow point of intersection to be greater than \(\theta = \frac{1}{2}\pi\)
Ignore incorrect labels, such as \(y = x\)

B1dep*: Vertically into the curve, then horizontally into the straight line, as far as the root
Initial value should be before root
Needs point of intersection to be before \(\theta = \frac{1}{2}\pi\)

B1: Mark independently from other parts of question, including an incorrect diagram, as staircase can be deduced from the iterates in (b)

Notes

B1*: Just need correct shape for \(y = \sin^{-1}k\theta\) graph – a one to one function that starts at the origin (ignore any \(\theta \lt 0\)) and has increasing gradient for all \(\theta\)

B1dep*: Straight line from the origin to intersect their graph
Diagram is sufficient for B1 – no comment or explanation required

Notes

M1*: Attempt use of quotient rule or equivalent (e.g. product rule). Condone one incorrect term only (of the five terms) but must be subtraction in the numerator (but allow subtraction the wrong way round); condone absence of brackets; no denominator (if using quotient rule) is M0
By the five terms we mean the four in the numerator and the fifth is the term in the denominator

A1: cao must include brackets as necessary
Any correct equivalent form

M1dep*: Sets their derivative (in any form) equal to 2 (M0 if equating to normal gradient)
May equate at any stage (even after incorrect manipulation of their derivative)

M1: Multiply both sides by \((4x^2 + 1)^2\) and simplify (so combining like terms) to obtain a quartic equation (must be expanded with at least three terms – condone lack of = 0 if all terms on the same side) – allow sign errors/minor slips but the expansion of \((4x^2 + 1)^2\) must be three terms of the form \(16x^4 + ax^2 + 1\) where \(a = \pm 4, \pm 8\)
Dependent on both previous M marks

A1: AG with explicit rejection of \(x = 0\) – as a minimum must indicate that \(x\) cannot equal 0
Just cancelling \(x\) is A0

Notes

M1: Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen
Just stating that \(\mathrm{f}(0.5) \lt 0\) and \(\mathrm{f}(1) \gt 0\) is M0

A1: Correct values together with explanation (change of sign) and correct conclusion (as a minimum ‘root’ oe)

Alternative

M1: Must be using the correct derivative. Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen
Just stating that \(\mathrm{g}(0.5) \gt 2\) and \(\mathrm{g}(1) \lt 2\) is M0

A1: Correct values together with explanation (values either side of 2) and correct conclusion (as a minimum ‘root’ oe)

Notes

B1*: Calculates correct derivative of rhs of given iterative formula

B1dep*: Correct explanation that any value in the given interval gives a gradient which is less than \(-1\)
No marks for just showing that the iteration doesn’t converge using different starting values

Notes

B1: Correct derivative (possibly seen in N-R formula)
Condone \(x\) for \(x_n\) oe

M1: Correct N-R formula seen with correct \(\mathrm{f}(x_n)\) and their \(\mathrm{f}'(x_n)\) substituted
Condone \(x\) for \(x_n\) oe

A1: First two iterations correctly stated to at least 5 decimal places (or exact) (truncated or rounded)
The correct first two iterations can imply B1 M1

A1: Independent of previous A mark (but must have scored B1 M1) – must be stated to exactly 5 decimal places
This A mark does not imply the previous A mark

B1: Independent of all previous marks – must be stated to exactly 5 decimal places
The correct answers with no evidence of N-R (e.g. no iterations stated and no N-R formula) then B0M0A0A0B1 max.

Notes

B1: State the 4 correct non-zero \(y\)-values and no others.
Exact values (including unsimplified) or decimal equivs (0, 1.12, 1.73, 2.29, 2.83) – 3sf or better
B0 if other ordinates seen unless clearly not intended to be used

M1*: Big brackets need to be seen or implied
\(y\)-values must be correctly placed
Must be using attempts for at least 4 \(y\)-values (but no need to see \(y = 0\) explicitly)
Condone using other than 4 intervals as long as values equally spaced between \(x = 1\) and \(x = 3\)

M1d*: Dep on previous M1
Or using \(k = 0.5h\), \(h\) consistent with their different number of intervals

A1: Obtain 3.28, or better
Allow answers to > 3sf, as long as they round to 3.28

Notes

B1: Under-estimate, with any valid explanation
Condone just ‘trapezia under curve’
Or curve is concave / decreasing gradient (not decreasing function)
Accept explanation on diagrams
Allow comparing to true value (3.36)
B0 if any additional incorrect or contradictory statements

Notes

B1: Convincing reason
Condone just ‘more trapezia’ or ‘narrower trapezia’
Could refer to strips or intervals

Notes

M1: Attempt differentiation using product rule
May expand first to give \(2x\ln x + \dfrac{x^2}{x} - \dfrac{2}{x}\)
(allow middle term as just \(x\))

A1: Equate to 0 and obtain given answer
Must be equated to 0 before clearing the fractions
Must be equation ie … = 0

Notes

B1: Correct derivative seen
Allow simplified middle term of \(2x\)

M1: Use correct Newton-Raphson formula, with numerator correct and their derivative in the denominator
Allow fractional term without subscripts
SC Condone use of N-R on \((x^2 - 2)\ln x\)

M1: Attempt rearrangement into single fraction with brackets expanded
Allow without subscripts
N-R not necessarily correct, but must be recognisable attempt
SC Rearrange their N-R on \((x^2 - 2)\ln x\)

A1: Obtain given answer, with no errors seen
Subscripts needed on RHS at least one step before AG
LHS needs \(x_{n+1}\) seen

Notes

B1: Condone 1.21, or better, for \(x_3\)
\(x_3 = 1.207515437\ldots\)

Notes

B1: Correct \(x\)-coordinate
B1: Correct \(y\)-coordinate

Must be 3dp or better
Could be given as single coordinate or \(x = 1.206\), \(y = -0.102\)
Allow BOD if 1.206 given but not identified as \(x\)-value

Notes

M1: With at least one correct value – values should be given to at least 2 sf (rot)
Allow degrees for M1 only: \(\mathrm{f}(0.6) = 68.8617\ldots\)

A1: Correct values together with explanation in words (change of sign) and conclusion

Notes

M1: Correct order of operations to get \(2x - 1 = \sin\left(kx^2\right)\)
\(k \neq 0\)

A1: \(p = \frac{1}{2},\ q = \frac{1}{2}\) and \(r = \frac{1}{6}\) (oe)

Notes

M1: Uses their iterative formula with correct starting value to produce terms up to at least \(x_2\) to at least 4 significant figures
Allow degrees for M1 only:
For reference:
\(x_1 = 0.5003636\ldots\)
\(x_2 = 0.5003641\ldots\)
\(x_3 = 0.5003641\ldots\)

A1: Must be stated to exactly 4 significant figures

Notes

B1 for "The graph touches the \(x\)-axis" or “repeated root” or “It is a stationary point”

Ignore all else, eg “inflection”

Notes

B1: Correct derivative of \(\mathrm{e}^{x^2}\) seen

M1: Uses quotient rule with their \(\frac{\mathrm{d}u}{\mathrm{d}x}\) and \(\frac{\mathrm{d}v}{\mathrm{d}x}\) oe

A1: Fully correct. Any form

Also allow use of the product rule with \((x+1)^{-1}\) giving \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x\mathrm{e}^{x^2}}{x+1} - \dfrac{\mathrm{e}^{x^2}}{(x+1)^2}\)

Notes

M1: Simplifies the problem to a quadratic inequality
FT their \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
Also allow for algebraic attempt to find the stationary point if used to define a range of values
Allow if their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) leads to linear or cubic inequality

B1: \(\dfrac{-1+\sqrt{3}}{2}\) or 0.366 seen

A1: oe \(-1 \lt x \lt 0.366\ldots\)
Do not allow for \(\leqslant\) used in final answer

Notes

M1: Uses 5 values in an attempt to find the total for trapezium rule with correct \(x\)-values soi
Also allow for the total area of four trapeziums attempted
See table below for values

B1: \(\frac{h}{2} = \frac{0.25}{2}\) or 0.125 soi

Notes

B1: Also allow an explanation that involves extra area under the line that is not in the region.
Condone incorrect description of the curve, if the explanation of additional area is clear.
See appendix
Allow for comparing exact value of area found BC to their estimate and stating over-estimate

Appendix: exemplar responses for Q10(d)

Notes

B1: See appendix
Or f(2) and f(3) both positive so no change of sign
Or there are two roots between 2 and 3 which will be missed
Must mention integers if they don’t explicitly state 2 and 3
ignore attempts to evaluate f(2) and f(3)
Must have either ‘no sign change’ or ‘two roots’

Appendix: exemplar responses for Q6(a)

Notes

M1: Differentiation and setting equal to zero.
Constant must differentiate to zero; sign errors condoned.
Allow for \(13 + \frac{k}{x} = 0\) where \(k\) is a non-zero constant
Condone poor/no notation

A1: Exact fraction or recurring decimal \(2.\dot{6}1538\dot{4}\) (if given as a decimal correct dot notation oe must be used)
Ignore \(y\) value if seen
isw decimals after correct answer seen

Notes

DR: This question included the instruction: In this question you must show detailed reasoning.

M1: See appendix
Allow \(1^4 - 2^1\) or 1 – 2 and \(2^4 - 2^2\) or 16 – 4
Ignore ‘= 0’
Condone \(-1\) and 12 only provided clearly linked to \(x = 1\) and \(x = 2\) e.g. as coordinates or in a table of values
Allow for using two values between 1 and 2 that are either side of 1.24 e.g. 1.2 and 1.3

A1: Correct values of function and comment on change of sign, allow e.g. \(-1 < 0\) and \(12 > 0\)
Condone no mention of function being sufficiently well-behaved (or continuous)
e.g. 1.2 and 1.3 leading to \(-0.22\) and 0.39 (2 s.f. is sufficient)

Appendix: exemplar responses for Q5(a)

REMEMBER: It is not possible to have an award of M0A1

Notes

B1: AG
Convincing rearrangement to given result
Must show one step of working and no incorrect work seen
Must not work backwards
e.g. \(4\log_2(x) - x = 0\) is B0 as not convincingly from a correct process unless \(4\log_2(x) = x\) is seen first

Notes

M1: \(x_1 =\) awrt 1.07 (= 1.07177...).
soi by any of these values: 1.204…, 1.232…, 1.237…

A1: cao
Must have shown at least one iteration e.g. \(x_1 = 1.07\)

Notes

B1: Vertical line \(x = 0.4\) drawn to meet the curve, could start line from \(y = x\)

B1: Correct completion of staircase diagram to head towards root with at least two vertical and two horizontal lines seen (lines need not extend beyond region between line and curve)
Condone good freehand lines, mark intent
SC B1B0 for using a different starting value e.g. \(x = 0\) with at least two vertical and two horizontal lines seen. Correct staircase diagram converging to root from their starting value.

Notes

B1: AG
Convincing rearrangement to given result
Must show at least one step of working and no incorrect work seen
e.g. \(\log_2(x^4) - \log_2(2^x) = 0\) is B0 as not convincingly from a correct process unless \(\log_2(x^4) = \log_2(2^x)\) is seen first
Must not work backwards

Notes

DR: This question included the instruction: In this question you must show detailed reasoning.

M1: Or \(4\log_2 1 = 0\)

A1: Condone e.g. math error or \(x_2\) is invalid or \(x_2\) has no solution provided log 0 is mentioned

Notes

M1: For their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\)

A1: Convincing completion to given answer

Additional guidance

The B1 is for differentiating and can be given for seeing x cos x or the unsimplified version sin x + x cos x – sin x.

The M1 is given for x cos x =1 and the final A1 for finishing it off convincingly. If there is no differentiation, then the M1 is not available.

Notes

M1: Differentiation (for this mark look for a power of \(x\) and a term in \(\sin x\) or \(\cos x\) with at least one term correct)

A1: oe, e.g. \(x_{n+1} = x_n - \dfrac{\left(x_n - x_n^2\cos x_n\right)}{\left(-1 + x_n^2\sin x_n\right)}\)
(The subscripts are needed)

M1: Starting values from 3.6 to 6.1 inclusive work for the given iteration but there may be other values which also give the required root and should be awarded.

A1: BC – candidates need not show intermediate iterations
If a value outside the expected starting value is used and it converges to 4.9172 then M1A1 is awarded (no need to check)
awrt 4.9172

Alternative method

M1: Differentiation

A1: oe, e.g. \(x_{n+1} = \dfrac{\left(1 - x_n^2\sin x_n\right)}{\left(\cos x_n - x_n\sin x_n\right)}\)

M1: Starting values from 3.6 to 6.1 inclusive work for the given iteration but there may be other values which also give the required root.

A1: BC – candidates need not show intermediate iterations
If a value outside the expected starting value is used and it converges to 4.9172 then M1A1 is awarded
awrt 4.9172

Additional guidance

There are two methods given.

The first M mark is for attempting to differentiate either 1/x – cos x or x cos x -1.

The first A mark is for setting up the iteration function and it must include the subscripts.

The next M mark is for choosing a suitable starting value (accept values between 3.6 and 6.1). They might use \(x_0\) or \(x_1\) or just say ‘starting value’ – but it will be their first value.

The final A mark is for getting the root 4.9172. The question asked for 4dps. They do not need to show all the working (BC means ‘by calculator’) to get the A1. If they choose a value outside the range and it gives 4.9172 then give M1 A1. A value outside the range giving the wrong answer will get M0 A0.

Notes

B1: Explanation referring to gradient of curve or to convergence to a different root.
Not just that it is close to another root
isw after a correct answer

Additional guidance

The comment needs to be more than just ‘it is close to another root’.

Notes

B1: Finding \(\mathrm{f}(1)\) or \(\mathrm{f}(2)\)

B1: Completion to show change of sign with explanation.

Additional guidance

The first B mark is for finding either f(1) or f(2). The second B mark is for finding the other value AND saying there is a sign change.

Notes

B1: Any value in this range. Candidates may give the range.

Additional guidance

Any value between 1.125 and 1.25 will do for this mark. Some candidates are just halving the range and using 1.0625 which scores B0.

Notes

B1: Justified by their calculations (which may not necessarily use 1.15).

Additional guidance

They must give a value for, say f(1.15) = -0.09,(or f(1.15625) = -0.0496, f(1.16) = -0.025, f(1.163) = -0.05, f(1.164) = 0.0015), and a final value of 1.2. The question asks for 1dp so only accept 1.2. 1.2 with no other working scores B0. Other values for f are possible.

Notes

E1: Incorrect maths (e.g. it implies a y-intercept) B0

Additional guidance

A change is sign is all that is required here.

Notes

E1: E.g.

• The function is undefined for \(x = \frac{2}{3}\) [and it looks as if the spreadsheet is homing in on this value]
• Accept ‘discontinuous’ or ‘asymptote’ for ‘undefined’
• Fig. 9.1 shows that there is only one root.
• Could refer to the table (e.g. \(\mathrm{f}(x)\) values diverging)

Additional guidance

We are allowing various comments. See the MS guidance. ISW after a correct answer has been seen.

Notes

B1: Correct triangles identified

B1: At least one height correctly found and related to diagram

B1: Correct completion (including base =1 either labelled on at least one triangle or stated (probably in calculation))
Dep on B2

Notes

M1: allow M1A1 for calculation of exact values using formula in parts (a) and (b)

A1: need to see 0.73373… as well as 0.7337 for A1

Notes

M1: or \((3.66866 - 0.5 + 1) \times 0.2\)

Notes

B1: FT their 0.8337(32) – 0.7337(32), dependent on award of M1 in part (b)

Notes

M1: if M0 allow SC1 for correct interval identified eg 0.77912 to 0.79162

Notes

B1: do not allow eg reduce interval
eg just ‘smaller’ rectangles – need to specify width reduction

Notes

M1: Attempt to differentiate. Both an \(x\) term and a cos term needed condone other errors.
If \(\pi^2\) is treated as a variable and incorrectly differentiated eg to \(2\pi\) then M0. Similarly with \(\pi\).

M1: At least one substitution into their expression

A1: Both correct (at least 2d.p., rounded or truncated)

E1: Can be implied by ‘sign change’ or sketch
Dependent on M2 but can be earned following M2A0

B1: Allow 2^nd derivative used at turning point found BC (\(x = 2.67\), 2^nd deriv \(= -0.356\)) or 2 relevant values eg \(-0.295\) and \(-0.383\) seen and stated as \(< 0\) therefore max

Notes

E1: Argues from change of sign that this argument is valid
Allow argument is not valid as he does not state that the function is continuous

Notes

E1: Allow for a comment that implies changes of sign are missed

Notes

(i) M1: Attempt to differentiate

A1: cao, but condone missing subscripts in the fraction

(ii) M1: Produces at least two iterations

A1: Three iterations with correct values either rounded or truncated to at least 3 decimal places

A1: Correct to at least 3 s.f. FT their values if their sequence seems to converge

(root is 2.907845109 to 10 sf)

Notes

(i) B1: Attempt to draw a tangent at \(x = 5\) as far as the \(x\)-axis

B1: Drawing the second tangent approximately at the point where \(x = 3.97\) as far as the \(x\)-axis.

(ii) B1: Conveys the idea that the stationary point or the value of the gradient causes the problem

B1: Conveys the idea that the wrong root is found

(iii) E1: Allow for a ‘starting value between 5 and 6’ oe

Notes

M1: may see \(\cos x \approx 1 - \frac{x^2}{2}\)

A1: intermediate step needed from here to earn final mark

(corrected from the printed mark scheme: the last line is printed as \(\ln(0.05) - \ln(0.01) + 0.0012 \approx \ln 5\); with \(\mathrm{F}[x] = \ln x - \frac{1}{2}x^2\) the 0.0012 is subtracted)

Notes

M1: or differentiation of \(y\) using quotient rule and use of small angle approximation

A1: from \(-\frac{1}{x^2}\) or \(-\frac{1}{x^2} - 1\) oe

Notes

E1: allow sign change between 0.947745 and 0.9477475

Notes

M1: Finding at least 2 distinct values for \(h\)
Need not be a table of values

M1: Using rectangles forming UB for area with width 0.5. Need not be drawn
Allow both M marks if only half the area considered.

Notes

M1: Using trapezium rule with 5 \(h\) values
FT incorrect \(h\) values for the M mark

A1: Allow awrt 0.22

Using the definite integral functionality of calculator gives 0.2214297… So method must be seen.

Notes

M1: Condone missing factor of 2 for method mark

A1: FT part (b)

Notes

(i): If answers given in both degrees and radians follow inst 2g

(i) B1: 180 gets 0

(ii) B1: (90, 1) or (1.57, 1) get 0

Notes

B1: Both correct
OE E,g, may use \(\operatorname{cosec} x - x\)

E1: Condone no mention of continuity
AG
Dep on B mark

Notes

(i) B1: Both correct to at least 3dp.

Notes

E1: OR same as their (c) (ii) or ‘the root between 1 and 2’ etc
Just ‘No’ gets 0
‘Yes’ with anything gets 0

Notes

B1: Starting point between min and right asymptote

B1: Initial “staircase” (≥2 horiz sections)

B1: Spirals into lower root

3 B marks all independent)