Notes

B1: Corrects the error for the case when \(m = 5k + 2\). The correction may be on the proof in the box or may be described in the main body of the text. May just see the \(10k\) crossed out and replaced with \(20k\) in the box or described in the main body of the work.
Sight of the quadratic \((25k^2) + 20k\ (+4)\) scores B1 and isw e.g. if they make other errors in the expansion.
If B1 is not scored in (a) then allow to score if seen correct in (b) e.g. they may attempt the case \(m = 5k + 2\) as part of their proof in (b).

Notes

Main scheme method uses \(m = 5k + 3\) and \(m = 5k + 4\)
You will need to look at both cases and mark the one which is fully correct first.
Allow a different variable to \(k\) and may be different letters for the two cases.
If a candidate attempts repeated cases e.g. \(m = 5k + 4\) and \(m = 5k - 1\) then mark both and award the higher mark of the two.
Condone use of m as a variable for the first three marks.
There should be no errors in the algebra for the A marks including invisible brackets but do not be concerned with any re-attempt at doing the case \(m = 5k + 2\)

Note that there are other allowable valid pairs of combinations covering the final two distinct cases e.g. \(m = 5k - 2\) and \(m = 5k + 4\), or \(m = 5k - 1\) and \(m = 5k + 3\)
but NOT e.g. \(m = 5k + 3\) and \(m = 5k - 2\)
Note that we are not expecting candidates to state what set of numbers k belongs to but we will condone pairs such as \(m = -5k - 1\) and \(m = -5k - 2\)
Typically candidates will show the algebraic steps as in the main scheme but for this particular pair they may justify equivalence using the results in the box for \(m = 5k + 1\) and \(m = 5k + 2\) without requiring calculations which is acceptable.

M1: Considers one valid case e.g. \(m = 5k + 3\) and attempts \(m^2 = (5k + 3)^2\) or \(m = 5k + 4\) and attempts \(m^2 = (5k + 4)^2\)
Look for expanding out the brackets and simplifying to a 3TQ. Condone slips.

A1: Achieves one correct statement which includes the case, the quadratic multiplied out and written in the required form
e.g. \(m^2 = (5k + 3)^2 = 25k^2 + 30k + 9 = 5\left(5k^2 + 6k + 2\right) - 1\) or
e.g. \(m^2 = (5k + 4)^2 = 25k^2 + 40k + 16 = 5\left(5k^2 + 8k + 3\right) + 1\)

dM1: Considers both cases for a valid pair (see first M1 for guidance). It is dependent on the previous method mark. Condone slips.

A1: Full proof with correct statements for both cases for a valid pair. Each must include the case, the quadratic multiplied out and written in the form which is not in terms of \(m\) (we do not need the “where \(n = \ldots\)” at the end of the statements - you can ignore these)
Requires a minimal overall conclusion eg. Proven, QED, tick
Condone recovery of interchanging of variables.

Ignore any additional cases that are not required to complete the proof
(and ignore replications of the ones given in the box in the question)

Notes

Alternative – expands \((x+y)^3\) before differentiating

(a) Some candidates have a spurious “\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\)” appearing as their intention to differentiate e.g.

This can be condoned for the first 3 marks in both versions.
Allow equivalent notation for the \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) e.g. \(y^{\prime}\)

M1: Award this mark for one of:

• \((x+y)^3 \rightarrow k(x+y)^2\left(\lambda + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) where \(\lambda\) is 1, \(x\) or 0 but condone missing brackets e.g. \(3(x+y)^2 1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\)
• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)

A1: Either \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe
May be implied if e.g. they collect terms to one side initially.
Do not condone missing brackets unless they are implied by subsequent work.

A1: \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) and \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (seen separately or equated)
If they collect terms to one side initially then the signs must be correct.

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 2 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), one coming from the differentiation of \((x+y)^3\) and the other coming from the differentiation of “\(-3y\)”
Note that here, 2 different terms means terms such as \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(3(x+y)^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not e.g. \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(-8\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Look for \((\ldots \pm \ldots)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) which may be implied by their working.
Condone slips provided the intention is clear.
For those candidates who had a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start, they may incorporate this in their rearrangement in which case they will have 3 terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and so score M0.
If they ignore it, then this mark is available for the condition as described above.
Note that from \(3(x+y)^2\left(1 + \dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = 6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\), candidates may expand the brackets before rearranging, in which case they would need 4 different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms coming from the appropriate places.
Note that the different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms do not have to be correct as long as the above conditions are satisfied.

A1: Fully correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow any equivalent correct forms.
Apply isw as soon as a correct expression is seen.

(a) alternative by expanding:

M1: Award this mark for one of:

• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)
• Expanding \((x+y)^3\) to obtain either an \(x^2y\) term or an \(xy^2\) term and then uses the product rule to obtain \(\ldots x^2y \rightarrow \ldots x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + \ldots xy\) or \(\ldots xy^2 \rightarrow \ldots xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + \ldots y^2\)

A1: Either \(3x^2 + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6xy + 6xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y^2 + 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
May be implied if e.g. they collect terms to one side initially.

A1: \(3x^2 + 3x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6xy + 6xy\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y^2 + 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe. (seen separately or equated) If they collect terms to one side initially then the signs must be correct.

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 4 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), 3 coming from the differentiation of \((x+y)^3\) and the other coming from the differentiation of “\(-3y\)”
Note that here, 4 different terms means terms such as \(x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(6xy\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not e.g. \(3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(-8\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Look for \((\ldots \pm \ldots \pm \ldots \pm \ldots)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) which may be implied by their working.
Condone slips provided the intention is clear.
For those candidates who had a spurious \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) at the start, they may incorporate this in their rearrangement in which case they will have 5 terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and so score M0.
If they ignore it, then this mark is available for the condition as described above.
Note that the different \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) terms do not have to be correct as long as the above conditions are satisfied. E.g. if they have an incorrect term such as \(6x\dfrac{\mathrm{d}y}{\mathrm{d}x}\), this mark is still available.

A1: Fully correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Allow any equivalent correct forms.
Condone e.g. \(3x2y\) for \(6xy\).
Apply isw as soon as a correct expression is seen.

Alternative making \(y\) the subject in (a):

Score as follows:

M1: Cube roots both sides and makes \(x + y\) or \(y\) the subject then award for

• \(\left(3x^2 - 3y - 2\right)^{\frac{1}{3}} \rightarrow \ldots\left(3x^2 - 3y - 2\right)^{-\frac{2}{3}}\) or
• \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x}\) but condone \(3x^2 - 3y - 2 \rightarrow \ldots x + \ldots\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\)

A1: For the \(\dfrac{1}{3}\left(3x^2 - 3y - 2\right)^{-\frac{2}{3}}\) or \(6x - 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Fully correct

M1: A valid attempt to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject with exactly 2 different terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Correct expression

Using partial derivatives in (a):

Score as follows:

M1: Correct structure for either partial derivative:
Doesn’t expand: \(\dfrac{\partial \mathrm{f}}{\partial x} = \ldots(x+y)^2 + \ldots x\) or \(\dfrac{\partial \mathrm{f}}{\partial y} = \ldots(x+y)^2 + \ldots\)
or
Expands: \(\dfrac{\partial \mathrm{f}}{\partial x} = \ldots x^2 + \ldots xy + \ldots y^2 + \ldots x\) or \(\dfrac{\partial \mathrm{f}}{\partial y} = \ldots x^2 + \ldots xy + \ldots y^2 + \ldots\)
Where “…” are non-zero constants

A1: Correct \(\dfrac{\partial \mathrm{f}}{\partial x}\) or correct \(\dfrac{\partial \mathrm{f}}{\partial y}\)

A1: Correct \(\dfrac{\partial \mathrm{f}}{\partial x}\) and correct \(\dfrac{\partial \mathrm{f}}{\partial y}\)

M1: Attempts \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{\partial \mathrm{f}}{\partial x} \div \dfrac{\partial \mathrm{f}}{\partial y}\)

A1: Correct expression

Notes

(b) Note that the gradient of \(\frac{1}{2}\) could have been deduced from the given equation so you will need to check their solution carefully.

M1: Substitutes \(x = 1\) and \(y = 0\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain the tangent gradient and then uses the negative reciprocal and \(x = 1\) and \(y = 0\) in a correct straight line method to obtain the normal equation with \(x = 1\) and \(y = 0\) correctly placed.
Note that when finding the normal gradient, they may find the negative reciprocal of their expression from part (a) and then substitute \(x = 1\) and \(y = 0\) which is fine.
If using \(y = mx + c\) they must proceed as far as finding a value for \(c\).
If no substitution of \(x = 1\) and \(y = 0\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is seen you will need to check their value. If they just state a value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) then it must follow their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) with \(x = 1\) and \(y = 0\)

A1*: Correct equation with no errors following a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from part (a) (unless they start again which is unlikely)
Be aware that some incorrect expressions for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from part (a) may fortuitously give \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\) and would generally score A0
In general A1* must follow the final A1 in (a) or correct differentiation in (a)

Notes

M1: Uses the equation from part (a) and substitutes \(y = \pm 2x \pm 2\) or \(x = \dfrac{\pm 2 \pm y}{2}\) to obtain an equation in one variable (usually \(x\)) (not necessarily a cubic equation). Allow slips in rearranging to obtain \(x\) in terms of \(y\) (or \(y\) in terms of \(x\)) as long as the intention is clear.

A1: Correct cubic equation with terms collected and “\(= 0\)” seen or implied.
Note that both \(-x^3 + 3x^2 - 18x + 16 = 0\) and \(-y^3 - 60y = 0\) are correct equations.

To access any of the following marks, candidates must attempt to use either the factor of \((x - 1)\) with their cubic in \(x\) or the factor of \(y\) in their cubic in \(y\) to obtain a quadratic expression in \(x\) or \(y\).
Attempts that just use a calculator to solve the cubic equation score no more marks in this part.

dM1: Uses the fact that \((x - 1)\) or \(y\) is a factor in an attempt to establish the quadratic factor.
For the cubic in \(x\), it must be of the form \(ax^3 + bx^2 + cx + d = 0\ \ a, b, c, d \neq 0\)
For the cubic in \(y\), it must be of the form \(ay^3 + by = 0\ \ a, b \neq 0\)
For the cubic in \(x\), the attempt at the quadratic factor using \((x - 1)\) may be via inspection or e.g. long division to obtain a 3 term quadratic expression. There may or may not be a remainder but they must obtain 3 terms.
For the cubic in \(y\), they would need to take out a factor of \(y\) (or divide through by \(y\)) to obtain a factor of the form \(k\left(y^2 + \alpha\right)\)

ddM1: This mark requires:

• a correct cubic equation in \(x\) or \(y\)
• the correct quadratic factor or a multiple of it e.g. \(k\left(x^2 - 2x + 16\right)\) or \(k\left(y^2 + 60\right)\)
• an attempt to show that the quadratic factor has no real roots

For the quadratic in \(x\) this could be:

Attempts discriminant: e.g. \(b^2 - 4ac = 4 - 4 \times 1 \times 16\) (may be embedded in the quadratic formula)
Attempts to complete the square: e.g. \(x^2 - 2x + 16 = (x - 1)^2 - 1 + 16\)
Uses calculus to find the turning point: e.g. \(\dfrac{\mathrm{d}\left(x^2 - 2x + 16\right)}{\mathrm{d}x} = 2x - 2 = 0 \Rightarrow x = 1 \Rightarrow y = \ldots\)
Attempts to solve: e.g. \(x^2 - 2x + 16 = 0 \Rightarrow x = \dfrac{2 \pm \sqrt{4 - 4 \times 16}}{2}\) or from a calculator \(x = 1 \pm \sqrt{15}\mathrm{i}\)

For the quadratic in \(y\) this is likely to be:

Attempts to solve: e.g. \(y^2 + 60 = 0 \Rightarrow y^2 = -60 \Rightarrow \ldots\)

A1: Fully correct argument that requires:

• fully correct work
• a justification depending on their strategy
• a conclusion depending on their strategy

Via discriminant: \(4 - 4 \times 1 \times 16 \lt 0\) so no real roots so they do not meet again
Via completing the square: \(\rightarrow (x - 1)^2 + 15\) which has a minimum value of 15 so no real roots so they do not meet again.
Via calculus: \(x = 1 \Rightarrow y = 15\) is the minimum so no real roots so they do not meet again.
Via solving: \(x = 1 \pm \sqrt{15}\mathrm{i}\) or e.g. \(x = 1 \pm 3.87\mathrm{i}\) or e.g. \(x = 1 \pm \sqrt{-15}\) so math error so they do not meet again.
For \(y\) it is likely to be more straightforward e.g. \(y^2 \neq -60\) which cannot be solved so they do not meet again.
Allow equivalent statements for “they do not meet again” e.g. so they only meet once.

(But do not condone incorrect statements such as “therefore \(P\) does not meet \(C\) again”)

A minimum justification could be:

\(x^2 - 2x + 16 = 0 \rightarrow b^2 - 4ac = (-2)^2 - 4 \times 1 \times 16\) ddM1
\(4 - 4 \times 1 \times 16 \lt 0\) so no more roots so no more intersections A1

Do not allow e.g.
“\(x^2 - 2x + 16 = 0\) gives a math error so they do not meet again”
as there has been no attempt to show why the “math error” occurs – this scores M0A0

Alternative to (c) by showing the cubic is strictly increasing (or decreasing):

M1A1: As in the main scheme then

Hence there can only be one intersection (at \(x = 1\)) so the normal and curve do not intersect again.

dM1: Differentiates their cubic of the form \(ax^3 + bx^2 + cx + d = 0\ \ a, b, c, d \neq 0\) to obtain a 3 term quadratic expression with only coefficient errors on the non-constant terms.

ddM1: This mark requires:

• a correct cubic equation in \(x\)
• the correct derivative or a multiple of it
• an attempt to show that the quadratic expression is always positive (or negative)

A1: Fully correct concluding argument e.g. that as the derivative is always positive (or always negative) the function is strictly increasing (or decreasing) and therefore there can only be one intersection (at \(x = 1\)) so the normal and curve do not meet again.

Notes

Note: Using e.g. \(x\) throughout is acceptable for both marks.

M1: Starts the process of showing the given expression is positive.
Completing the square requires \((k-2)^2 \pm \ldots\)
Differentiation requires them to differentiate to a linear expression in \(k\), set = 0 (which may be implied), solve for \(k\) and substitute their \(k\) into \(k^2 - 4k + 5\) to reach a value. Ignore e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Discriminant requires a calculation of \(b^2 - 4ac = (\pm 4)^2 - 4 \times 1 \times 5\ \{= -4\}\) and might be seen embedded in the quadratic formula.
Sketches on their own are insufficient without working to find the minimum point algebraically.
Stating the minimum is (2, 1) without any evidence is M0.

A1*: Completes the proof with no errors and correct reasoning.
Accept e.g. “hence proved” in place of “so \(k^2 - 4k + 5\) is always positive” as long as there is sufficient justification for it being “proved”.
As a minimum expect to see e.g.:

• \(k^2 - 4k + 5 = (k-2)^2 + 1\) which is always positive as \(\underline{(k-2)^2 \geqslant 0}\) (or as squares are always positive or zero). Must be a correct statement. Do not condone e.g. \((k-2)^2 \gt 0\)
• \(k^2 - 4k + 5 = (k-2)^2 + 1\) so (2, 1) is the minimum (point) so \(k^2 - 4k + 5\) is always positive.
• \(k^2 - 4k + 5 = (k-2)^2 + 1\) so \(\underline{k = 2 \pm \sqrt{-1}}\) hence \(k^2 - 4k + 5\) has no real roots and as k² has a positive coefficient (condone e.g. positive \(k^2\) or positive quadratic or \(a \gt 0\)), hence proved.
• \(2k - 4 = 0 \Rightarrow k = 2 \Rightarrow k^2 - 4k + 5 = 1\) is the minimum value so \(k^2 - 4k + 5\) is always positive.
• \(b^2 - 4ac = 16 - 20 \underline{\lt 0}\) so \(k^2 - 4k + 5 = 0\) has no real roots and as k² has a positive coefficient (condone e.g. positive \(k^2\) or positive quadratic or \(a \gt 0\)), hence proved.

Note that stating \(k^2 - 4k + 5 = (k-2)^2 + 1 \gt 0\) alone is not sufficient for the A1*
Similarly, stating \(k^2 - 4k + 5 = (k-2)^2 + 1 \gt 0\) because \((k-2)^2 \gt 0\) (or \((k-2)^2\) is positive) is incorrect and scores A0.

SC: Attempts at completing the square for \(k = 2n\) and/or \(k = 2n + 1\) (or \(k = 2n - 1\)) can score M1 provided at least one is attempted as far as \((\ldots)^2 + \ldots\) e.g. to \((2n-2)^2 + \ldots\) or e.g. \(4(n-1)^2 + \ldots\)
Candidates are unlikely to complete the argument to score A1 (it is not sufficient to just consider odd/even) but if you think they might deserve the A1 then send to Review.

Notes

General Note: Throughout this question we are condoning if candidates do not reference the following two points which are deemed fairly trivial and acceptable to be omitted at A-level:

• As \(x\) and \(y\) are integers then both \(3x + 2y\) and \(2x - 5y\) are integers.
• As \(x\) and \(y\) are positive then \(3x + 2y \gt 0\)

As such, we only require candidates to prove that there are no positive integer solutions \(x\) and \(y\) to simultaneous equations that have a positive RHS.

Note: Any attempt to solve the given simultaneous equations does not score any marks on its own.
Any attempts that use substitutions such as \(x = 2n\) etc. are unlikely to score any marks.
There are other methods to eliminate some pairs of simultaneous equations e.g. by showing that the only solution to \(3x + 2y = 7\) is (1, 2) which is not on \(2x - 5y = 4\). Use Review in such cases.

M1: Attempt to solve any one of the other 11 possible cases (labelled A-K) below to find a value for \(x\) or a value for \(y\). The attempt may be implied by a value for \(x\) or \(y\) which need not be correct.

dM1: Attempts to solve any two of the other 11 possible cases below to find a value for \(x\) or a value for \(y\) with at least one correct and correctly rejected. It is not necessary to find both values of \(x\) and \(y\) unless the correct value found does not cause a contradiction (see cases D and G).

ddM1: Attempts to solve all 5 pairs (A-E) of the relevant simultaneous equations with positive RHS or attempts to solve cases A and B and justifies that these are the only cases that need checking using either:

• \(3x + 2y \geqslant 5\) (because \(x\) and \(y\) are positive) or
• \(3x + 2y \gt 2x - 5y\) (because \(x\) and \(y\) are positive) [cases F, G and H do not need checking using this approach because of the general note]

Their values for \(x\) and \(y\) do not need to be correct for this mark as long as the dM1 is scored.

A1: cso Shows that all necessary cases are impossible with correct values, correct reasons, and a minimal conclusion e.g. “hence proved”. There is no need to say “contradiction” or restate the objective (you can also ignore any inaccurate attempt to restate the objective).
All 5 pairs of simultaneous may be rejected in one go if the rejection is sufficiently clear.
The mechanics of solving the simultaneous equations does not need to be shown.

Notes

(a) Way 2

(a) Condone a complete proof entirely in \(x\) (or another variable) instead of \(\theta\)
Condone “=” for “\(\equiv\)”
Note that we are marking this as B1M1A1 not M1M1A1

B1: Adds the fractions to obtain a correct single fraction (not fractions over fractions) in any form. Condone missing brackets when they combine their fractions as long as they are recovered to give a correct fraction.
This can be done in a variety of ways but when combined, the fraction must be correct e.g.
\(\dfrac{\operatorname{cosec}\theta + 1 + \operatorname{cosec}\theta - 1}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\) or \(\dfrac{2\operatorname{cosec}\theta}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\) or \(\dfrac{2\operatorname{cosec}\theta}{\operatorname{cosec}^2\theta - 1}\)
or e.g. \(\left(\dfrac{1}{\frac{1}{\sin\theta} - 1} + \dfrac{1}{\frac{1}{\sin\theta} + 1} = \dfrac{\sin\theta}{1 - \sin\theta} + \dfrac{\sin\theta}{1 + \sin\theta}\right) = \underline{\dfrac{2\sin\theta}{1 - \sin^2\theta}}\) etc.

M1: Uses a correct Pythagorean identity anywhere in their attempt e.g.
\(\operatorname{cosec}^2\theta - 1 = \cot^2\theta,\ \sin^2\theta + \cos^2\theta = 1\) etc. or equivalent

A1*: Correct work with all necessary steps shown leading to the given answer. See scheme for the necessary steps. They need to proceed via sine and cosine to the given answer. There should be no notational or bracketing errors and no mixed or missing variables. E.g. we would consider \(\cos^2\theta\) written as \(\cos\theta^2\) a notational error.
Condone reaching \(2\sec\theta\tan\theta\ *\)

Way 2 (Meet in the middle)

B1: See Way 1

M1: See Way 1

A1*: Correct work on the RHS with all necessary steps shown leading to showing the equivalence with the LHS. See scheme for the necessary steps. There should be no notational or bracketing errors and no mixed or missing variables.
For this approach there must be a (minimal) conclusion e.g. “= LHS”, “QED”, “Hence proven” etc.

It is possible to start with the rhs e.g.:

B1: Correctly reaches \(2\tan\theta\sec\theta = \dfrac{2\operatorname{cosec}\theta}{(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1)}\)

M1: Uses a correct Pythagorean identity anywhere in their attempt e.g.
\(\operatorname{cosec}^2\theta - 1 = \cot^2\theta,\ \sin^2\theta + \cos^2\theta = 1\) etc. or equivalent

A1*: Correct work with all necessary steps shown leading to the lhs. See scheme for the necessary steps. There should be no notational or bracketing errors and no mixed or missing variables.

Notes

(b) Note that attempts solve an equation of the form:
\(2\tan x\sec x = \cot 2x\sec 2x\) or e.g. \(2\tan\theta\sec\theta = \cot 2\theta\sec 2\theta\) or e.g. \(2\tan\theta\sec\theta = \cot 2x\sec 2x\)
Will generally score no marks in part (b)

Condone the use of \(\theta\) instead of \(x\) here.

B1: Deduces the correct equation using the result from part (a)

M1: Factors out or cancels the \(\sec 2x\) to obtain \(\ldots\tan 2x \pm \ldots\cot 2x = 0\) oe e.g. \(\ldots\tan 2x = \pm\ldots\cot 2x\) leading to an equation of the form:
\(\tan^2 2x = \alpha\) oe e.g. \(\cot^2 2x = \dfrac{1}{\alpha}\) where \(\alpha \gt 0\)
Or uses \(\tan 2x = \dfrac{\sin 2x}{\cos 2x}\) and \(\cot 2x = \dfrac{\cos 2x}{\sin 2x}\) and then \(\sin^2 2x = \pm 1 \pm \cos^2 2x\) or \(\cos^2 2x = \pm 1 \pm \sin^2 2x\) to obtain an equation of the form \(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) oe e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) or \(\sec^2 2x = \dfrac{1}{\gamma}\) where \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\)
Or uses \(\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}\) and \(\cot 2x = \dfrac{1 - \tan^2 x}{2\tan x}\) to obtain a 3TQ in \(\tan^2 x\) (or possibly in \(\sec^2 x\))
Or uses \(\tan 2x = \dfrac{\sin 2x}{\cos 2x}\) and \(\cot 2x = \dfrac{\cos 2x}{\sin 2x}\) and then \(2\sin^2 2x = \pm 1 \pm \cos 4x\) and \(2\cos^2 2x = \pm 1 \pm \cos 4x\) to obtain an equation of the form \(\cos 4x = k,\ \ 0 \lt k \lt 1\)

M1: Correct order of operations from \(\tan^2 2x = \alpha\) or \(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) or \(\cos 4x = k\) or equivalents e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) where \(\alpha \gt 0\) or \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\) or \(0 \lt k \lt 1\) leading to at least one value for \(x\) e.g. square roots, finds inverse tan/sin/cos/cosec and divides by 2 or inverse cos and divides by 4
or from \(\tan^2 x = k,\ k \gt 0\) which follows their equation (may need to check) and then finds \(x = \tan^{-1}\sqrt{k}\)
You may need to check their value(s) (in degrees or radians) to see if the correct order of operations has been used. May be implied by e.g. \(17.6^\circ\) or \(17.7^\circ\) provided no incorrect work is seen.

A1: Correct values. Allow awrt \(17.6^\circ\) and awrt \(72.4^\circ\). The degrees symbol is not required. Ignore any values outside the range (correct or incorrect) but if there are extra angles in range score A0. Answers in radians score A0.

Note that some candidates may convert to \(\sin x\) or \(\cos x\) and then solve:

E.g.

These can be marked in a similar way.

Alternative not using part (a):

Score as:

M1: For correct work leading to one of the forms in the main scheme e.g.
\(\tan^2 2x = \alpha\) oe e.g. \(\cot^2 2x = \dfrac{1}{\alpha}\) where \(\alpha \gt 0\)
or
\(\sin^2 2x = \beta\) or \(\cos^2 2x = \gamma\) oe e.g. \(\operatorname{cosec}^2 2x = \dfrac{1}{\beta}\) or \(\sec^2 2x = \dfrac{1}{\gamma}\) where \(0 \lt \beta \lt 1\) or \(0 \lt \gamma \lt 1\)

B1: Any correct equation e.g. \(\tan^2 2x = \dfrac{1}{2},\ \cot^2 2x = 2,\ \sin^2 2x = \dfrac{1}{3}\) etc.

Then M1A1 as main scheme

Notes

Note: Throughout the question allow use of \(\delta x\) for \(h\) or any other letter e.g. \(\alpha\) if used consistently.
If \(\delta x\) is used then you can condone e.g. \(\delta^2 x\) for \(\delta x^2\) as well as condoning e.g. poorly formed \(\delta\)’s

M1: Begins the process by writing down the gradient of the chord and attempts to expand the correct squared bracket – you can condone “poor” squaring e.g. \((x+h)^2 = x^2 + h^2\) but the \(-x^2\) must be present.

A1: Reaches a correct fraction o.e. with the \(x^2\) terms cancelled out and with no algebraic errors, e.g. \(\dfrac{\cancel{x^2} + 2xh + h^2 - \cancel{x^2}}{h}\), \(2x + h\) is correct.

A1*: Completes the process by applying a limiting argument and deduces that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) with no errors seen. They must have = \(2x\) and not just \(\lim\limits_{h \to 0} 2x\) to complete the proof.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) or an equivalent e.g. \(\mathrm{f}^{\prime}(x) =\) or “Gradient =” must be evident somewhere in their working or final line. If \(\mathrm{f}^{\prime}(x)\) is used then there is no requirement to see \(\mathrm{f}(x)\) defined first. Condone e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \rightarrow 2x\) or \(\mathrm{f}^{\prime}(x) \rightarrow 2x\).

Condone missing brackets to allow e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + h = 2x\)

Do not allow \(h = 0\) if there is never a reference to \(h \to 0\).

e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + 0 = 2x\) is acceptable

but e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2xh + h^2}{h} = 2x + 0 = 2x\) is not unless \(h \to 0\) is seen.

The \(h \to 0\) does not need to be present throughout the proof e.g. appear on every line but must appear at least once.

They must reach \(2x + h\) at the end and not \(\dfrac{2xh + h^2}{h}\) (without the \(h\)’s cancelled) to complete the limiting argument.

Typical solution

Let \(a\) and \(b\) be rational

Hence \(c\) is rational which contradicts that \(c\) is irrational.

Therefore, if \(c\) is irrational at least one of \(a\) or \(b\) is irrational.

Typical solution

1

Typical solution

Typical solution

\(x = 90,\ x = 270\)

Typical solution

Typical solution

Typical solution

(i)

\(\cos\theta \neq 1\) as \(\operatorname{cosec}\theta\) is undefined

(ii)

Typical solution

\(\therefore (2x + 1)\) is a factor of \(\mathrm{P}(x)\)

Typical solution

Typical solution

There are two factors. Both factors are integers not equal to 1 so \(4n^3 + 8n^2 + 11n + 4\) is never prime.

Typical solution

3 is not in the interval

Typical solution

(i)

\(y\) is between 3 and \(x\) which contradicts the definition of \(x\) as the smallest value in (3, 4)

(ii)

Step 1: Assume there is a largest number in the interval \(3 \lt k \lt 4\) and let this largest number be \(x\)

Step 2: let \(y = \dfrac{x + 4}{2}\)

Step 3: \(x \lt y \lt 4\) which is a contradiction.

Step 4: Therefore, there is no largest value in \(3 \lt k \lt 4\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Assume

Since \(a \neq b\) this is a contradiction because \((a - b)^2 \gt 0\)

Hence \(\dfrac{a}{b} + \dfrac{b}{a} \gt 2\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Since 42 and \(25n\)-9 are both integers the expression is rational

Typical solution

Hence \(a^2\) must be even, which means that \(a\) must be even

Typical solution

Hence \(2b + 1\) must be even

\(2b + 1\) is an odd number which is a contradiction

Typical solution

There are no solutions to \(a^2 - 4b - 2 = 0\) where \(a\) and \(b\) are integers

Typical solution

Typical solution

1

Typical solution

Therefore, there can be no square number which has a last digit of 8

Notes

B1: Correct condition. Accept use of \(\mathbb{Z}^+\), or \(\mathbb{Z}\) with \(b \neq 0\), when describing \(a\) and \(b\)
Allow ‘\(a\) and \(b\) are integers’ only if also stated that \(b \neq 0\)

B1: Correct condition (‘not 1’ is not required)
\(a\) and \(b\) are coprime
The HCF of \(a\) and \(b\) is 1
Condone \(\dfrac{a}{b}\) is fully simplified (or in its simplest form)
Could be implied by ‘Euclid’s infinite descent’ in (c)

Notes

B1*: Identify that \(a^2\) is a multiple of 5
eg 5 is a factor of \(a^2\)
eg \(\dfrac{a^2}{5}\) is an integer

B1dep*: Explain why \(a\) must also be a multiple of 5. Could refer to prime factors or factors of square numbers
eg \(\dfrac{a \times a}{5}\) is an integer, so \(a\) must be a multiple of 5
Not just \(a^2\) is a multiple of 5 hence \(a\) is a multiple of 5 as well, as this is given in the question

Exemplar responses for Q11(b) for the B1dep*

Notes

M1: Correctly use \(a = 5k\) to obtain an equation with no brackets

A1: Simplify and state condition on \(b\). Condone statement about \(b\), with no reference to \(b^2\) first

A1: Conclude appropriately, dep on (2^nd) B1 in part (a) for ‘no common factors’ oe
If B0 for ‘no common factors’ in part (a) then this mark is A0, as no assumption to contradict
Need reference to ‘common factor’ (or simplest form), ‘contradiction’ and ‘irrational’ in conclusion

Or eg \(\dfrac{a}{b} = \dfrac{5k}{5m}\) hence not in simplest form (as alternative evidence for sharing a common factor)

Notes

Allow “=” throughout Q4

M1: M1 for expanding correctly

A1: Must see final statement www

Alternative method

A1: Must see final statement www

Notes

M1: For writing \(\tan 2\theta\) as \(\dfrac{\sin 2\theta}{\cos 2\theta}\) or \(\dfrac{2\tan\theta}{1 - \tan^2\theta}\). May be seen embedded or implied by next A1 (Allow this mark if \(\sec 2\theta\) still used)

A1: May see \(1 + 2\cos\theta\sin\theta\) for the numerator here

A1: Allow A1 for correct numerator or correct denominator seen in terms of \(\theta\), not \(2\theta\). Allow this mark even if subsequently \(\surd\) incorrectly.

A1: correctly obtained www, ie must see previous line

Alternative method

M1: or \(\dfrac{(\cos\theta + \sin\theta)^2}{\cos^2\theta - \sin^2\theta}\) seen

A1: correct numerator seen

A1: correct denominator seen in terms of \(2\theta\), not \(\theta\)

A1: correctly obtained www, ie must see previous line

Alternative method using \(t\)-formulae

A1: www

Notes

B1: soi

B1: soi (or \(a_1 + 99d\))

M1: For an attempt at the general term \(b_i\) using \(b_1\ (= a_{100})\), i.e. an expression of the form \(a_1 + kd + (i - 1)(\pm d)\) oe

M1: Equating their \(b_i\) to an expression for \(a_i + 10\)
May see \(a_1 + (i - 1)d + 10\)

A1: www (so this mark is dependent on all previous marks)
May see \(a_1\) used throughout instead of 4
Condone other letters used throughout (e.g. \(n\) for \(i\) and \(c\) for \(-d\)):

• Must recover to \(d\) in order to solve (A1A1)
• Must recover to \(i\) to demonstrate the AG (A1)

Notes

B1: soi (accept e.g. “\(i\) must be an integer”)

B1: Ignore negative values (but B0 for any additional positive values)

Notes

B1: Any product involving 2 and a prime number, evaluated and contradiction identified.
eg \(2 \times 3 = 6\), which is not odd
Condone \(2 \times 2 = 4\), which is not odd

Notes

(b)(i) B1: Correct symbol used. Condone \(\leftarrow\)

(b)(ii) B1: Correct symbol used. Condone \(\leftrightarrow\)

(b)(iii) B1: Correct symbol used. Condone \(\rightarrow\)

Notes

B1: Correct form seen for the sum of the squares of any two odd numbers.
ie two different variables

M1: Attempt to square, add and collect like terms for their two distinct odd numbers.
Their odd numbers must both be of the form \(2p \pm q\) (where \(q\) is odd)
May involve a single variable eg \((2n+1)^2 + (2n+3)^2\)
\(= 4n^2 + 4n + 1 + 4n^2 + 12n + 9\)
\(= 8n^2 + 16n + 10\)
Allow sign and/or coefficient errors only

A1FT: Show it is a multiple of 2, by taking out a common multiple or arguing that the coefficients in all terms are even.
FT on their two odd numbers
Factorising by 2 is sufficient for A1 ie no comment required
Condone dividing by 2 to show that the quotient would be an integer

A1: Not dep on previous A1, but must follow B1 M1.
Take out a common multiple from relevant terms, or argue using coefficients of terms, or take out common factor of 4 and argue that remaining factor is not an integer.
Must be from any two odd numbers (ie two different variables)
Condone dividing by 4 to show that the quotient is not an integer
Comment required – either refer to the remainder of 2 (including ‘2 more than a multiple of 4’), or that the entire quotient is not an integer, depending on method used

SC B1 for a complete worded argument about two distinct odd numbers eg odd\(^2\) = odd for both odd numbers; odd + odd = even, hence multiple of 2

Notes

M1: Setting up so that the deduction \(a^2\) is even can be made.

A1: www, must see both statements and a convincing, correct, argument oe (e.g. \(a^2 = 2(2b + 1)\))

Alternative method

M1: For setting up and stating that \(a\) is odd \(\Rightarrow a^2\) is odd
May see (not required) \(a = 2n + 1\), \(a^2 = 2(2n^2 + 2n) + 1\) Hence \(a^2\) is odd

A1: www, Must see both statements and a convincing, correct, argument

Notes

M1: Setting up (must see assumption and use of \(a\) is even)

A1: Substituting in \(a = 2n\) and correctly reaching an equation which shows a contradiction.
Accept the equivalent in words if clear and correct.
Also accept: \(4n^2 - 4b\) is a multiple of 4, Hence \(a^2 - 4b\) is a multiple of 4, which is a contradiction

A1: www, Must see both statements and a convincing, correct, argument

Alternative method (1)

A1: Must see both statements and a convincing, correct, argument www

Alternative method (2)

A1: Must see previous two lines and a convincing, correct, argument www

Alternative method (3)

M1: For setting up using part (a) and considering either case where \(b\) is odd or even (ignore any reference to the cases where \(a\) is odd as these are not required) May see (but not required) \(a = 2n\) so \(a^2 = 4n^2\)
Condone using the same letter (e.g. \(n\)) in \(a\) and \(b\) for this mark only.

A1: For correctly considering both cases either algebraically or in words.
May see (but not required) \(b = 2m + 1\), so \(a^2 - 4b = 4(n^2 - (2m + 1))\)
And \(b = 2m\), so \(a^2 - 4b = 4(n^2 - 2m)\)
Do not award this mark if same integer (e.g. \(n\)) used in both \(a\) and \(b\)

A1: A fully correct, convincing argument with conclusion, www.

Notes

NB Other correct methods may be seen

B1: \(9m^2\) alone, not as part of longer expression

M1: At least one of these expansions attempted using \(r = 1\) or 2.
Must include three (or four) terms, Allow one error

A1: At least one of these seen explicitly

A1: Must see the statement oe.
Can be seen once at end or with each separate case
Dep complete method, with all three cases seen

Alternative method 1

M1: Attempted. Must include 3 (or 4) terms, Allow one error

A1: Explicit

A1: Must see the statement oe Dep complete method

Alternative method 2

Alternative method 3

M1: Allow one arithmetical error

A1: Both correct

A1: None of the brackets is an integer

Notes

M1*: Could be numerical or algebraic or in words
If listed, must be clear which ones are selected

M1dep: M1 for \(\left(\frac{1}{3}\right)^3\) associated with at least one of these

M1dep: M1 for \(\left(\frac{1}{3}\right)^3 \times k\) where \(k = 4\), 5 or 6, associated with (0, 1, 2)

A1: Correct answer with no working: M0M0M0A0

Alternative method for 2^nd & 3^rd M1M1

M1: Allow 7 or 8

Notes

B1: Any correct counterexample, and contradiction identified
Initial inequality soi and then contradiction eg \(-3 \gt -4\) but \(9 \lt 16\) (or \(9 \ngtr 16\))

Notes

(b)(i)

B1: Any correct statement
Identifies that \(\sin x = 0.5\) could give values of \(x\) other than \(30^\circ\)
Either specific example or general statement eg ‘many to one’ function

(b)(ii)

B1: Any correct relationship
If attempting to write general solution then must be fully correct eg \(x = 30^\circ + 360n^\circ\), \(x = 150^\circ + 360n^\circ\)
Condone \(\leftarrow\) instead of \(\Leftarrow\)

Notes

B1*: Four consecutive multiples of 4 written correctly in terms of \(n\), or any other variable
Allow BOD if \(n\) not explicitly stated to be an integer
Sufficient to just list the 4 terms, rather than as a sum
Not necessarily starting on \(4n\)
Could also define \(k\) as a multiple of 4 and then have \(k\), \(k + 4\) etc

M1dep*: Correctly sum terms, and correctly take out common factor of 8
Or sum and then consider each term separately
Could be a different factor if using \(k\)

A1: Conclude appropriately
Allow BOD if \(2n + 3\) not explicitly stated to be an integer
If using \(k\)… expect \(8(0.5k + 3)\) then justify \(0.5k\) as an integer, or \(4(k + 6)\) then justify \(k + 6\) is a multiple of 2

Notes

(a)(i) B1: Justification for \(CD\)
Need to see either \(\cos y = \frac{CD}{a}\) or adj = hyp × \(\cos\theta\) before given answer

(a)(ii) B1: Use area of triangle to show given answer
Could quote general expression for area and then show clear substitution
If not, then sides being used need to be clearly identified through statement or diagram
Could also use right-angled triangle, with base as \(AD\)
Condone not being rearranged to given expression

(a)(iii)

B1: Correct \(CD\) in terms of \(b\) and \(x\)

B1: Correct area of triangle \(BCD\)
B0 B1 if correct area stated with no justification

B1: Correct area of triangle \(ABC\)

B1: Equate area of \(ABC\) to the sum of the areas of the two small triangles and complete proof convincingly
Allow alternative proofs eg using lengths

Notes

B1: Correct use of compound angle formulae
Could be implied if exact values used immediately – allow BOD for RHS
May be seen as two separate expressions, not yet equated

M1: Use exact trig values
In either equation or two expressions
Must see all 4 values, but expansions may not be fully correct

M1: Gather like terms and attempt \(\tan\alpha\)
May still have fractions in the fraction
\(\tan\alpha\) does not yet need to be the subject, but must only appear once for M1

M1: Attempt to rationalise their denominator
Clear intention seen to multiply throughout by the conjugate of their denominator

A1: Obtain given answer
With full detail, including (at least) \(3 - 2\) in denominator

Notes

or \((n + 1)(n^2 - n + 1)\)

Notes

B1: Stated. Allow \(x\) instead of \(n\)
NOT \(n^3 + 1\) can be expressed as \((n + 1)(n^2 - n + 1)\)

B1: Must see \(n \gt 2\)
Allow omission of this step

A1: Conclusion stated, from correct working seen.
Dep at least B1M1 and correct reasoning given

SC: \((n + 1) \gt 1\) or \(n + 1 \gt 3\) (or \(n^2 - n + 1 \gt 1\))   B1
\((n + 1)\) & \((n^2 - n + 1)\) are factors of \(K\)   B1

Notes

M1: Finding common denominator or attempt to rationalise denominator for at least one fraction e.g. \(\frac{1}{\sqrt{n+1}-1}\cdot\frac{\sqrt{n+1}+1}{\sqrt{n+1}+1} - \frac{1}{\sqrt{n+1}+1}\cdot\frac{\sqrt{n+1}-1}{\sqrt{n+1}-1}\) or \(\frac{\sqrt{n+1}+1}{n} - \frac{\left(\sqrt{n+1}-1\right)}{n}\) oe
Condone consistently considering any general term in the sequence e.g. \(a_{k+1}\) for all marks \(\dfrac{1}{\sqrt{k+2}-1}\cdot\dfrac{\sqrt{k+2}+1}{\sqrt{k+2}+1} - \dfrac{1}{\sqrt{k+2}+1}\cdot\dfrac{\sqrt{k+2}-1}{\sqrt{k+2}-1}\)

A1: Correct simplification
\(a_{k+1}\) leads to \(\frac{2}{k+1}\)

A1: nfww
Correct explanation of why the general term must be rational e.g. in form \(\frac{p}{q}\) oe \(p\), \(q\) integers
Condone rational as \(n\) an integer or \(k\) integer
‘\(\frac{2}{n}\) is rational’ alone is A0 - must include comment that \(n\) is an integer

Notes

M1: Must relate to \(\frac{2}{n}\), \(a_n \to 0\) is not sufficient
Allow "\((n > 0\) and\()\ \lim\limits_{n\to\infty}\frac{2}{n} = 0\)"
Condone omission of \(n > 0\)
May find the difference between 2 terms or use inequalities
Allow \(\mathrm{f}(x) = \frac{2}{x} \Rightarrow \mathrm{f}^\prime(x) = -\frac{2}{x^2} < 0\) (condone \(n\) instead of \(x\))
Condone listing terms for M1 A0
FT their \(\frac{2}{n}\) provided in the form \(\frac{b}{n+c}\) where \(b, c\) integers

A1: See appendix
Convincing conclusion e.g. difference between terms is negative, so sequence is decreasing or \(\frac{2}{n+1} < \frac{2}{n}\), so decreasing or function \(\mathrm{f}(x)\) is decreasing therefore sequence \(a_n\) decreasing
A1 is only available if their \(\frac{2}{n}\) is correct for their term e.g. \(\dfrac{2}{k+1}\) from considering \(a_{k+1}\)
SC B1 for ‘decreasing’ following M0 for a correct but incomplete argument. But not just for ‘decreasing’ unsupported.

Appendix: exemplar responses for Q11(b)

Notes

B1: Use of Pythagoras implied by \(17 + x^2\)

B1: Understanding that \(17 + x^2\) is a square number soi e.g. by equating to at least one square number \(17 + x^2 = 25\)
Allow e.g. \(n = \sqrt{17 + x^2}\) and \(n\) integer soi e.g. by equating to an integer

M1: Rearranging to get their 17 on one side and factorising

A1: OR \(n - x = 17\) and \(n + x = 1\) AND \(n - x = -17\) and \(n + x = -1\)
Implies M1. If the final B2 is earned following only one pair of correct equations then award A1

B2: Both values and no others
B1 if only one value given (ignore extras)

Alternative method 1 for M1 A1

M1: At least two different square numbers added to their 17 e.g. two from: \(17 + 1\), \(17 + 4\), \(17 + 9\), \(17 + 16\), \(17 + 25\), \(17 + 49\), \(17 + 64\), \(17 + 81\) and so on
Or tested two using \(17 + x^2 = 25\), \(17 + x^2 = 36\), \(17 + x^2 = 49\), \(17 + x^2 = 64\), \(17 + x^2 = 81\) and so on

A1: e.g. Since 64 and 17 + 64 are adjacent square numbers there will be no larger values of \(x^2\)

Alternative method 2 for M1 A1

M1: Uses adjacent square numbers
Implied by \(2x + 1 = 17\)
Condone not testing earlier squares for this M1

A1: If the final B2 is earned following M1 then award A1

Notes

M1: or identify AP with \(n\) terms and \(a = 1\), \(d = 2\); may be implied by award of first A1

A1: use of formula for sum of arithmetic progression with \(n\) terms and \(a = 1\) and \(l = 2n - 1\); or \(a = 1\) and \(d = 2\); allow \(\frac{n}{2}(2 + 2n - 2)\) but no further contraction

A1: correct intermediate step needed for A1

Alternative method (Further Maths 1)

M1: from \(\sum_{r=1}^{n}(2r-1)\); allow for \(\sum_{r=1}^{n}(2r+1)\); may be implied by award of first A1

A1: all correct;

A1: correct intermediate step needed for A1

Alternative method (Further Maths 2)

A1: correct intermediate step and statement needed

Notes

M1: Dealing with the \(a\) terms convincingly by e.g. collecting terms with factor \(a\), expanding the brackets and collecting etc.

A1: Squaring brackets or difference of two squares demonstrated to give a convincing completion to given result

Additional guidance

To get the M1 they must work with the two terms with factor a. They may be moved next to each other, but they could also work with them in their original positions.

The A1 is for manipulating the factor a terms correctly. This may be squaring of brackets or difference of two squares. But it must be convincing to get the required result. Give A0 for any error seen in the working.

Notes

M1: Attempts to complete the square for either \(x\) or \(y\) terms Soi

A1: Correct coordinates of the centre seen or used. Condone incorrect or missing radius

M1: Attempt to solve the equations simultaneously
\(\left[x^2 + \left(\dfrac{x+25}{7}\right)^2 + 2x - 14\left(\dfrac{x+25}{7}\right) + 25 = 0\right]\) or oe

M1: Simplifies equation leading to two roots.
Allow arithmetic errors \([50x^2 + 50x - 600 = 0]\)

A1: Could be solved by calculator \([x = -4, 3]\)

A1: FT their \(y\) values

M1: Attempt to find gradient(s) of at least one of these lines

A1: Both correct gradients (not OA or OB)

A1: Clear argument based on perpendicular lines www

Alternative for last three marks

M1: Attempt to find the length of one of the sides of the triangle
FT their coordinates

A1: All three correct lengths found

A1: Clear argument based on Pythagoras’ theorem or the cosine rule leading to a value for angle ACB www

Notes

M1: e.g. Use of \(\sin 2A = 2\sin A\cos A\)
Use of means more than just quoting the formula - usually it will be part of an argument
Small angle approximations score M0

M1: e.g. Use of \(\tan A = \dfrac{\sin A}{\cos A}\)
The 2 double angle formulas along with the tan identity may be applied in any order. If proof is sound award the 3 marks for equivalent steps.

A1: e.g. Use of any correct identity for \(\cos 2A\) and convincing completion to given result
Working it through as an equation with the ‘ = 1 ‘ is not ‘convincing’
Alternative incorrect approach.
If \(\sin 8\theta = 4\sin 2\theta\), and \(\sin 2\theta\) is then expanded then SC1 can be awarded. A similar approach applies to \(\cos 8\theta\). However the SC1 can only be awarded once.

Additional guidance

The first M1 can be awarded for using \(\sin 8\theta = 2\sin 4\theta\cos 4\theta\) or for using \(\cos 8\theta = 1 - 2\sin^2 4\theta\) (or equivalent form) or for using \(\tan 4\theta = \sin 4\theta / \cos 4\theta\). The mark is not available for just writing the formula down.

The second M1 is for using another one.

The final A1 is for using the third one and completing the solution to show it equals 1. (We saw an incorrect factorisation of \(\sin 8\theta = 4\sin 2\theta = 4 \times 2\sin\theta\cos\theta\) which qualifies for SC1. A similar approach can be used for \(\cos 8\theta\) but the SC1 can only be awarded once).

Some candidates are using small angle approximations which does lead to an answer of 1, but is the incorrect method, so scores M0M0A0.

Notes

M1: Using a negative value or zero

A1: Must see an explicit comparison between correct values (could be in words or symbols)

Notes

M1: Use of double angle formula
Allow 1 error

M1: Use of given result with sin and cos
Both sides seen but might not be equated

M1: Use of \(\sin^2\theta + \cos^2\theta = 1\)

E1: Convincing completion

Notes

A1: \(\dfrac{1}{4}\) unsupported does not score

Notes

M1*: Setting up proof by contradiction

M1*: factorising

E1: Considers the possibility that one factor might be 1

E1: Condone missing reference to \(n = 2\) (or \(p = 3\)) for this step.
Conclusion must be clear.

SC1: Allow SC1 where M1M0 or M0M0 has been awarded and \(3 = 2^2 - 1\) is established

Notes

B1: \(n \geqslant 3\)
B0 if only rounded number in standard form seen

B1: eg \(2^9 - 1 = 511\), eg \(2^{15} - 1 = 32767\) eg \(2^{21} - 1 = 2097151\)

B1: NB 32767 and 2097151 both divisible by 7;
2047 divisible by 23
correct value of \(n\) may be embedded in formula
NB B0 if answer spoiled by eg so 511 is prime

Notes

B1: Cao
Must be clear that the limit is equal to 1. Therefore tending to or approaching 1 or 0.999 or \(\to 1\) etc do not score

Notes

M1: Convincing explanation that \(\dfrac{1}{n+1}\) decreases

E1: Convincing completion (A.G.)

Special Case (Max 2 for this method)
Correct differentiation and states \(> 0\) to demonstrate increasing SC B1
Goes on to consider [positive] integers as a subset of reals to complete their argument SC B1 dep on first B1.

Alternative method 1

M1: This method is testing \(\dfrac{n}{n+1} < \dfrac{n+1}{n+2}\)
Formulating expression for \((n+1)\)th or \((n-1)\)th term

M1: For formulating inequality and cross multiplying

E1: Probably see \(\dfrac{n+1}{n+2} > \dfrac{n}{n+1}\) [i.e. increasing]

Alternative method 2

M1: Formulating expression for \((n+1)\)th term

M1: Finding difference and attempts to form a single fraction eg may be \(a_n - a_{n+1}\)

E1: Convincing completion (A.G.)

Notes

M1: eg \(\arctan 2 + \arctan\left(\dfrac{1}{2}\right) = \dfrac{\pi}{2}\)
Allow valid alternative methods

M1: Oe

M1: Oe

E1: Convincing completion possibly using the three results above (AG)

Alternative method

M1: Use of \(\arctan x + \arctan y = \arctan\left(\dfrac{x+y}{1-xy}\right) + \pi\)

Notes

M1: Stating any pair of numbers where \(p^2 > q^2\) and \(p < q\)

E1: Fully convincing argument – do not allow for only disproving the converse

Also accept general statement about a negative number [for \(p\)]

Notes

B1: Subbing values may score B1 but not E1

E1: Convincing completion; AG

Notes

B1: Possible values
• \(a = 2,\ b = 3\)
• \(a = 2,\ b = 4\)