Notes

Mark (i) and (ii) together

(a)(i) B1: Centre \((-3,\ 4)\) Accept without brackets. May be written e.g. \(x = -3,\ y = 4\)

(a)(ii) M1: States or implies that \(r^2 = 24\) or \(r = \sqrt{24}\). A final answer of \(\sqrt{24}\) or \(2\sqrt{6}\) implies the radius

May multiply out the brackets, collect terms (\(x^2 + y^2 + 6x - 8y + 1 = 0\)) and states the radius is \(r^2 = \dfrac{6^2}{4} + \dfrac{(-8)^2}{4} - 1\) o.e. Do not condone slips for this mark.

A1: \(2\sqrt{6}\) isw if they proceed to write as a decimal

Notes

Note that if their radius is incorrect in (a) then maximum score is M1A0 unless they restart in (b)

M1: Attempts a valid method. For example

• Finds the distance (or distance \(^2\)) of the centre from the origin.
  They must be attempting \((d =)\sqrt{(\pm\text{``}3\text{''})^2 + (\pm\text{``}4\text{''})^2} = \ldots\) or \(\left(d^2 =\right)(\pm\text{``}3\text{''} - 0)^2 + (\pm\text{``}4\text{''} - 0)^2 = \ldots\) and proceed to a value.
  May be seen as substituting the coordinates of the origin into the equation for \(C\) proceeding to a value for the left hand side e.g. 25 (to be able to compare with 24)
• Sets \(y = 0\) and attempts to solve \((x+3)^2 + (-4)^2 = 24 \Rightarrow x = \ldots\) (at least one value)
• Sets \(x = 0\) and attempts to solve \((3)^2 + (y-4)^2 = 24 \Rightarrow y = \ldots\) (at least one value)

In each method the starting expression or equation must be correct but do not be concerned by slips when evaluating or processing in finding the distance, the \(x\) coordinate or \(y\) coordinate.

A1: Correct calculation(s), reason and conclusion examples:

Note if their reasoning is incorrect e.g. referring to the radius as 24 instead of \(\sqrt{24}\) then A0 but allow referencing to “the radius of C” provided their radius in (a) was correct.

Note if they give a reason that the origin does not lie inside the circle \(C\) because e.g. \(25 \neq 24\) this scores M1A0 (M1 for 25 but A0 incorrect reasoning)

Notes

(a)(i)

B1: Correct centre. Allow as a coordinate pair or written separately e.g. \(x = 3,\ y = -7\) or as a column vector \(\begin{pmatrix}3\\-7\end{pmatrix}\)
Condone missing brackets e.g. 3, \(-7\) but do not allow coordinates the wrong way round.

(ii)

M1: Uses a correct strategy to find the radius or radius\(^2\)
Requires an attempt at: \((x \pm 3)^2 + (y \pm 7)^2 - 3^2 - 7^2 \pm 33 = 0 \Rightarrow (x \pm 3)^2 + (y \pm 7)^2 = \alpha,\ \alpha \gt 0\)
Award for \((x \pm 3)^2 + (y \pm 7)^2 - a^2 - b^2 \pm 33 = 0 \Rightarrow (x \pm 3)^2 + (y \pm 7)^2 = \alpha,\ \alpha \gt 0\) with at least one of \(a = 3\) or \(b = 7\) (or 9 or 49)
You may see an attempt at "\(f^2 + g^2 - c\)" or \(\sqrt{\text{``}f^2 + g^2 - c\text{''}}\) e.g. "\(3^2 + 7^2 \pm 33\)" or \(\sqrt{\text{``}3^2 + 7^2 \pm 33\text{''}}\)

A1: Correct radius of 5. Do not allow \(\pm 5\) or \(\sqrt{25}\).
May be scored following \((x \pm 3)^2 + (y \pm 7)^2 = 25\)
Correct answers only in (a) scores B1M1A1

Notes

M1: Uses Pythagoras correctly on their centre from part (a) and the given centre to find the distance between the centres.
Look for \(\sqrt{\left(-6 - (\text{their } x)\right)^2 + \left(-8 - (\text{their } y)\right)^2}\) or e.g. \(\sqrt{\left((\text{their } x) - (-6)\right)^2 + \left((\text{their } y) - (-8)\right)^2}\) but condone one sign slip with their coordinates if the intention is clear.

A1ft: Correct distance or follow through their centre from part (a).
This may be implied by their value. Condone the use of decimals so allow 3sf accuracy e.g. awrt 9.06 for \(\sqrt{82}\) or you may need to check their value following an incorrect centre in (a)(i). Not e.g. \(\pm\sqrt{82}\) unless the positive root is subsequently used.

dM1: Correct strategy for one of the limits. E.g. adds or subtracts their 5 to their distance between centres.

A1: Correct limits. There is no follow through but allow decimals to 3sf e.g. awrt 4.06 and awrt 14.1

A1: Correct answer with exact values using set notation.
Allow as shown in the main scheme but also allow equivalent set notation e.g.
\(\left\{k : k \in \mathbb{R}, \sqrt{82} - 5 \lt k \lt \sqrt{82} + 5\right\},\ \left\{k :\ \sqrt{82} - 5 \lt k \lt \sqrt{82} + 5\right\},\ k \in \left(\sqrt{82} - 5,\ \sqrt{82} + 5\right)\)
and allow “|” for “:” and allow the “\(k\):” or “\(k \in\)” to be missing
e.g. \(\left\{\sqrt{82} - 5 \lt k \lt \sqrt{82} + 5\right\}\) and \(\left(\sqrt{82} - 5,\ \sqrt{82} + 5\right)\) are both acceptable.
But \(\left\{k : k \lt \sqrt{82} + 5\right\} \cup \left\{k : k \gt \sqrt{82} - 5\right\}\) or \(\left\{k :\ k \lt \sqrt{82} - 5,\ k \lt \sqrt{82} + 5\right\}\) score A0
Do not allow solutions not in set notation such as \(\sqrt{82} - 5 \lt k \lt \sqrt{82} + 5\)

Correct answers with no working should be sent to review.

Scenario for part (b) for reference:

Algebraic approach for part (b):

We will mark this as follows:

M1: This requires a valid strategy that:

• solves the 2 circle equations simultaneously to find \(y\) in terms of \(x\) and \(k\), or \(x\) in terms of \(y\) and \(k\)
• substitutes for \(y\) or \(x\) into one of the circle equations to obtain an equation in \(x\) and \(k\) only, or \(y\) and \(k\) only,
• attempts \(b^2 - 4ac = 0\) or e.g. \(b^2 - 4ac \gt 0\) or equivalent to obtain an equation in \(k\) only. You do not need to look at the details of their algebra.

A1: Correct simplified 3TQ in \(k^2\)

dM1: Solves their 3TQ in \(k^2\) by any correct method including a calculator to find \(k\).

A1: Both correct values for \(k\) (exact or decimals as in the main scheme) (they may have extras which can be ignored)

A1: As main scheme (exact and in set notation)

Note that work such as

is not a valid strategy as it greatly simplifies the problem and would generally score no marks.

Implicit differentiation approach for part (b):

We will mark this as follows:

M1: This requires a valid strategy that:

• differentiates the equations of both circles implicitly and equates the derivatives to obtain an equation connecting \(y\) and \(x\). (Note that the equation connecting \(y\) and \(x\) is the common equation through the centres which can also be found from using the coordinates of the centres)
• substitutes for \(x\) or \(y\) into the equation for \(C_1\) to obtain an equation in one variable

A1: Correct 3TQ in \(y\) or \(x\)

dM1: This requires:

• solves their 3TQ in \(y\) or \(x\) by any correct means including a calculator and finds at least one point of intersection
• substitutes this point into \(C_2\) and proceeds to a value for \(k\)

A1: Correct values for \(k\) (exact or decimals as in the main scheme)

A1: As main scheme (exact and in set notation)

Typical solution

(12, 2)

Typical solution

\((12, -8)\)

Typical solution

Typical solution

(d)(i)

(d)(ii)

Typical solution

Typical solution

(−4, 6)

Typical solution

Typical solution

Typical solution

Typical solution

So \(A\left(\dfrac{2 - \sqrt{6}}{2}, \dfrac{\sqrt{6}}{2}\right)\)

Typical solution

Typical solution

(i)

(ii)

Typical solution

(i)

Centre \((5, 8)\)

(ii)

4

Typical solution

4

Typical solution

(i)

At \(P\)

Line \(PQ\)

(10, 11)

(ii)

Typical solution

(i)

(ii)

Typical solution

Notes

B1: State two correct equations soi

M1: Attempt to solve their two linear simultaneous equations. As far as eliminating one variable. M1 implied by correct final answer, if no method shown

A1: Obtain correct value for \(a\)

A1: Obtain correct value for \(b\)

Alternative method (graphical approach)

M1: Attempt gradient of line. M1 implied by correct \(a\)

A1: Obtain \(a = 7\)

M1: Attempt to find value for \(b\). Intercept of line or algebraic method. M1 implied by correct \(b\)

A1: Obtain \(b = 73\)

Notes

(b)(i)

M1: Substitute \(t = 4\) into their model

A1FT: Condone no units, but A0 if incorrect eg 101m. FT their model. Or 1.01 (m)

(b)(ii)

B1: Any sensible reason as to why the model for ‘average height’ may, or may not, be accurate in predicting Sam’s height
eg height varies a lot between children of the same age
eg could depend on different diets and lifestyle
eg might be accurate as long as Sam is ‘average’

B0 for interpolation, including comments about Sam’s height being between 87cm and 108cm

Exemplar responses for Q5(b)(ii)

Notes

B1: Any sensible comment
eg growth spurt at 12 due to puberty
eg people grow quicker when younger

Extrapolation not reliable
B0 for comments about individual variations as question refers to ‘average height’ model

Exemplar responses for Q5(c)

Notes

B1: Both soi (i.e. may be seen embedded in equation)

M1: FT their centre and radius. OR allow 5 instead of 25

M1: Substitute from line into their circle equation in any form to reach an equation in \(x\) or \(y\) only (substitution must be correct for this mark).

M1: Rearrange their quadratic to a 2- or 3-term quadratic \(= 0\) (this mark may be implied by correct values for \(x\) or \(y\) i.e. either of the next two A marks). The \(= 0\) may be implied by correct solutions to their quadratic. Condone one error in the terms.

A1: Both seen

A1: Both seen

M1: FT their coordinates (must see this step oe)

A1: From fully correct working (dep on all previous marks)

Alternative method

B1: Both soi

M1: Attempt find equation of perpendicular

A1: Correct equation

A1: From fully correct working (dep on all previous marks)

Notes

M1: Equate attempt at gradient of line to 2
Must be attempting to use \(\dfrac{y_2 - y_1}{x_2 - x_1}\), with consistent order in numerator and denominator; allow one sign slip
Could also use informal methods

A1: Obtain \(h = \tfrac{5}{2}\) oe

Notes

B1FT: Obtain correct \(x\) coordinate for \(C\), following their \(h\) ie \(2h - 1\)
Must be clear that their \(2h - 1\) is the \(x\) coordinate
BOD if brackets omitted

B1: Must be clear that 13 is the \(y\) coordinate
BOD if brackets omitted

SC B1 if both values correct but given as a vector not a coordinate

Notes

M1: Attempt equation of line through \(A\)
Allow one sign slip, but M0 if \(x\) and \(y\) coordinates transposed
Could use their \(B\) or \(C\) instead, but must still be using gradient of 2

M1: Equate line and curve

A1: Obtain correct quadratic, with like terms collected
Condone no ‘= 0’

M1*: Attempt discriminant of 3 term quadratic, resulting from equating line and curve
Correct discriminant for their quadratic
Condone any inequality or equality, or no, sign for this mark
M0 if using just \(x^2 - 4x + k\)
If the discriminant is initially embedded in the quadratic formula, then M1 is only awarded when it is considered in isolation

M1d*: Use \(b^2 - 4ac \gt 0\)
Inequality sign could be implied by final answer
M0 if incorrect inequality sign, including \(b^2 - 4ac \geqslant 0\)

A1: Obtain \(k \lt 14\)

Alt method for final 3 marks

M1*: Attempt completed square form
Correct expression for their quadratic
Allow unsimplified

M1d*: Set constant term \(\lt 0\)
M0 if constant term \(\leqslant 0\)
May see more informal method to determine inequality sign

A1: Obtain \(k \lt 14\)

Alt method (using differentiation to find point of intersection)

M1: Differentiate equation of curve, equate to 2 and attempt \(x\)

M1: Attempt equation of line through \(A\)
Could use their \(B\) or \(C\) instead

M1: Attempt \(y\) value from line, and use their (3, 11) in equation of curve to attempt \(k\)

A1: Obtain \(k = 14\)

M1: Clear method to determine inequality sign
Could be algebraic or a sketch
Could be implied by final answer

A1: Obtain \(k \lt 14\)

Notes

M1: Re-arranges to obtain both \(\sin t = \dfrac{x \pm 4k}{k}\) and \(\cos t = \dfrac{y \pm 2}{4}\) and use \(\sin^2 t + \cos^2 t = 1\) to eliminate \(t\)

A1: Allow any correct un-simplified cartesian form not involving trigonometric terms
ISW once a correct answer seen

SC for part (a): If M0 awarded then SC B1 for \(y = 2 + 4\cos\left(\arcsin\left(\dfrac{x - 4k}{k}\right)\right)\) or \(x = 4k + k\sin\left(\arccos\left(\dfrac{y - 2}{4}\right)\right)\) or \(\arcsin\left(\dfrac{x - 4k}{k}\right) = \arccos\left(\dfrac{y - 2}{4}\right)\) (or any correct form in terms of trig. functions – allow \(\cos^{-1}\) for \(\arccos\) etc.)

If considering one branch only e.g. \(y\left(= 2 + 4\sqrt{1 - \sin^2 t}\right) = 2 + 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) this scores M1 only (or equivalent expression for \(x\))
If considering both branches e.g. \(y\left(= 2 \pm 4\sqrt{1 - \sin^2 t}\right) = 2 \pm 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) then this scores M1 A1 (or equivalent expression for \(x\))
Note a (common) correct answer for M1 A1 is: \(x^2 + y^2 = 16k^2 + 8k(x - 4k) + (x - 4k)^2 + 4 + 4(y - 2) + (y - 2)^2\)

Notes

(b)(i)

M1: Setting \(c\) and \(d\) equal in their \(\dfrac{(x \pm a)^2}{c} + \dfrac{(y \pm b)^2}{d} = 1\) or stating that \(y_{\min} = -2\) and \(y_{\max} = 6\)
Possibly implied by correct value for \(r\) www

A1: \(r = \pm 4\) is A0 unless replaced with positive \(r\) only

(b)(ii) B2: B1 for either one correct centre or both \(x\)-coordinates correct
These marks are dependent on \(r = 4\) from correct working or www

Notes

B1: Substitute (4, 8) into the equation of the curve.
Seen anywhere in solution
Allow for unsimplified equation, even if error then occurs

M1: Attempt differentiation.
To obtain derivative of the form \(px + qx^{-0.5}\)
Can still be awarded if \(p\) and \(q\) now incorrect numerical values

M1: Use \(x = 4\) correctly in their derivative.
Must be an attempt at differentiation, but could still follow M0
Their derivative could now be part of an equation or an attempt at a perpendicular gradient

M1: Attempt to use the relationship between the gradients of perpendicular lines.
Attempt the gradient of the tangent, using attempt at gradient of given normal (condone \(-\tfrac{1}{13}x\) if recovered)
OR
Attempt gradient of the normal using their derivative (either algebraic or in terms of \(a\) and \(b\))
Condone slips with fractions within fractions as long as intent is clear
NB \(\left(8a + \tfrac{1}{4}b\right) \times -\dfrac{1}{13} = -1\) would imply next M1 as well

M1: Equate expressions / values for normals or for tangents, with \(x\) now substituted. Could be using gradients or equations.
Must be comparing like with like ie tangent with tangent, or normal with normal
If using equations then would need to equate expressions for either the gradients or the intercepts

A1: Obtain correct linear equation.
If using eg \(-\dfrac{1}{8a + \frac{1}{4}b} = -\dfrac{1}{13}\) then terms in the denominators must have been dealt with correctly

A1: Obtain \(a = 2,\ b = -12\)
BC so no method needed

Notes

In this question, if candidates attempt more than one method, review all their working to identify which is their ‘final’ or ‘substantially most complete’ answer, then apply one scheme only.
Candidates cannot gain marks from more than one scheme.

M1: Substitute \(y = x\) into equation of circle (need not be simplified – may see this quadratic in \(y\))

A1: A1 for either correct (condone if not specifically identified as \(A\) or \(B\) – may see \(y =\) these values)

M1: Attempt \(CM\) using their values (method must be correct)
May see \(CM = \frac{3\sqrt{2}}{2}\)

M1: Attempt \(BA\) using their values (method must be correct)
May see just \(\sqrt{2(9 - 2a)}\) oe, e.g. \(\sqrt{18 - 4a}\)
Alternative: \(AM = \sqrt{\frac{9}{2} - a}\) (and then \(\text{Area} = 2 \times \frac{1}{2} \times CM \times AM\))

M1: Attempt at area, in terms of \(a\)

A1: Must see correct expression before answer

Alternative Method

M1: Attempt complete the square for \(x\)

A1: Both soi

B1: Possibly coming from \(\dfrac{|3 - 0|}{\sqrt{1^2 + 1^2}}\)

M1: Their radius² (in terms of \(a\)) – their \(CM^2\)

M1: Attempted in terms of \(a\)

A1FT: FT their \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\)

A1: Must see one correct intermediate step

Alternative method for last three marks

M1: Attempted in terms of \(a\)

A1FT: FT their \(r\) and \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\)

A1: Must see one correct intermediate step

Alternative method for last three marks

M1: Attempted in terms of \(a\)

A1FT: FT their \(r\) and \(AM\) or \(AB\) (in terms of \(a\)), and their \(CM\) (using double angle formula for \(\sin BCA\))

A1: Must see one correct intermediate step

Notes

(b)(i)

B1: oe

(b)(ii)

B1: Must see this geometrical answer (accept e.g. the line touches the circle)
Accept a diagram that clearly shows the line is a tangent

Notes

B1: Must see this geometrical answer (condone BOD for ‘the line does not touch the circle’)
Accept a diagram that clearly shows the line does not meet the circle

Notes

M1: Complete the square (for \(x\) and \(y\)) to obtain \((x \pm 3)^2 + (y \pm 5)^2 + \ldots\)
or for \(\ldots \pm k \pm 3^2 \pm 5^2\)

A1: cao – www
allow equivalent in either set notation e.g. \(\{k : k \lt 34\}\) or interval notation e.g. \((-\infty, 34)\) or \((-\infty, 34]\) but not \([-\infty, 34)\) or \([-\infty, 34]\) unless \(k \lt 34\) already seen
If implying a lower limit then A0
Allow \(k \leqslant 34\)

Notes

M1*: Attempt to differentiate the equation for \(C\) implicitly – must be four terms including a \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) term and two other terms correct (condone either \(-46\) or \(k\) appearing in their deriv. as a 5^th term) but if the derivative of \(x^2 + y^2 - 6x + 10y - 46 = 0\) is put equal to \(\frac{1}{2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (and used subsequently) then M0
or applying the chain rule to an expression of the form \(\pm 5 \pm \sqrt{\lambda \pm (x \pm 3)^2}\) for some non-zero \(\lambda\), so must be of the form \(\frac{1}{2}(\mathrm{f}(x))^{-\frac{1}{2}}\mathrm{g}(x)\) where \(\mathrm{f}(x)\) is quadratic and \(\mathrm{g}(x)\) is linear

M1dep*: Sets derivative equal to \(\frac{1}{2}\)
or substitutes \(\frac{1}{2}\) for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1: Substitutes their linear expression into the given equation of \(C\) with \(k = -46\) (must be five terms with two quadratic terms and two linear terms in \(x\) oe if using completing the square form from (a)) to obtain an expression/equation in \(x\) (or \(y\)) only
Dependent on first two M marks – if chain rule used this mark is implied by setting \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to \(\frac{1}{2}\)

M1: Simplify to a 3TQ in \(x\) (or \(y\)) (allow sign errors only when simplifying from their five term equation/expression or completing the square equation/expression)
Dependent on all M marks
For \(y\) if correct: \(y^2 + 10y - 39\ (= 0)\)

A1: BC - do not need to be stated as coordinates
Two values of \(x\) and \(y\) only

Alternative method for first two marks

M1*: Finding an expression for the gradient of the line segment that passes through the centre and any point \((x, y)\) on the circumference of \(C\)
Follow through their centre from (a) – gradient expression must be correct for their centre

M1dep*: Equating gradient with \(-2\) (oe)

Notes

M1: Attempt correct process to find gradient in terms of \(t\) or \(p\)
Correctly combine attempts at two derivatives
Need \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = kt^{-2}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\)
SC B1 for gradient of \(-2x^{-2}\) if it is never seen in terms of \(t\) or \(p\)

A1: Obtain correct gradient
In terms of \(t\) or \(p\)

M1: Attempt equation of tangent
Condone still working in terms of \(t\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Using their gradient from a differentiation attempt, but not dependent on first M1
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)

A1: Obtain given answer
Must now be in terms of \(p\)
Expand brackets and simplify to given answer, or find \(c\) and substitute back into equation

Notes

B1FT: Correct (unsimplified) gradient of normal, following their derivative
Gradient in terms of \(t\) or \(p\), but not \(x\)
Could either FT on their incorrect derivative or deduce the gradient from the equation given in (a)

M1: Attempt equation of normal
Attempt to use their gradient and \(P\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)

M1: Using their attempt at normal equation
As far as finding an expression for \(x\)

A1: Correct \(x\)-coordinate for \(B\)
Any equivalent form

B1: Correct \(x\)-coordinate for \(A\)
Any equivalent form

M1: Attempt length of \(PA\) or \(PB\)
Or M1 for attempting one of \((PA)^2\) or \((PB)^2\)
Must correct distance formula
Using the given \(P\), and their coordinates for \(A\) and/or \(B\), which must involve a function of \(p\)

A1: Or correct \((PA)^2\) and \((PB)^2\)

A1: Simplify ratio to obtain given answer
Must show clear method, such as same expression in each square root before cancelling
Could also consider fraction and then cancel to deduce given ratio
Could simplify \((PA)^2 : (PB)^2\), and then square root to obtain ratio

Notes

M1*: Attempt at the midpoint (3 out of 4 values used correctly or one correct coordinate)
If correct \(\left(\dfrac{5}{2}, -2\right)\)

M1*: Attempt to calculate diameter \(d\) or radius \(r\) (or the square of either of these two)
e.g. \(r^2 = \left(2 - \frac{5}{2}\right)^2 + (-5 - (-2))^2\)
M0 if implying that \(r^2 = (2 - 3)^2 + (-5 - 1)^2\). If the value is not labelled (as either \(r\) or \(d\)) then consider how this value is used in their circle equation
3 out of 4 values used correctly – if correct \(d^2 = 37\) or \(r^2 = \frac{37}{4}\)

M1dep*: Correct form for the equation of a circle with their values for the centre and radius
Dependent on both previous M marks

A1: Must have \(= 0\)
Order of terms on the lhs irrelevant
\(a\), \(b\) and \(c\) need not be explicitly stated

Notes

M1: \(\mathbf{b} - \mathbf{a}\) or \(\mathbf{a} - \mathbf{b}\) attempted, using \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) or column vectors
May be implied by calculation seen

A1: www. Correct answer, no working: M1A1
Mark(s) cannot be gained retrospectively in (b)

Notes

M1: Attempt \(AB^2 = BP^2 + PA^2\) (involving \(p\)) ft their \(AB\)

A1f: Correct simplified equation, ft their (a), ie:
or \(p^2 - 7p + \dfrac{46 - \text{their } a^2}{2} = 0\) or \(\left(p - \frac{7}{2}\right)^2 = \dfrac{\text{their } a^2 - 17}{4}\)

A1f: ft only their (a)

Alternative methods for M1

Notes

B1: Both. Allow midpoint \(= \left(\frac{0 + 7}{2}, \frac{6 + 5}{2}\right)\) ISW

M1: \((= 7)\)

A1: cao. Correct answer, no working or inadequate working: SC B2

Alternative

B1: Both
M1: ft their midpt and gradient, NOT \(-\frac{1}{7}\)
A1: cao. Any correct form

Alternative

M1: Attempt expansion
A1: cao. Any correct form eg \(y = 7x - 19\)

Notes

B1: Any correct form for another perp bisector

M1: Attempt solve simultaneously equations of two perpendicular bisectors. Can be implied

B1: cao. NB, if centre \(= (3, 2)\) without clear working, B0M0B1

M1: Correct method for \(r^2\) or \(r\) using their centre & \(A\) or \(B\) or \(C\)

A1ft: ISW. ft their centre & radius, dep both M1 marks

Alternative method for 1^st two marks

Notes

M1: Attempts to solve simultaneously

M1: Collects terms. Must see \(=0\).
May be implied by correct discriminant

M1: Forms discriminant of their equation and equates to zero

A1: cao

Alternative method

M1: Uses gradient of 3 to form an equation linking \(k\) and \(x\)

M1: Forms second equation for the point of intersection oe

M1: (Either) Uses above quadratic equation and an expression for \(x\) obtained from gradient to form a quadratic in \(k\) only

A1: cao

M1: (Or) Uses above quadratic equation and an expression for \(k\) to form a quadratic in \(x\) only, leading to values for \(k\)

A1: cao

Notes

M1: Uses their values for \(k\) and attempts to solve simultaneously

A1: Note – using any values of \(k\) will give the same point of intersection for full credit

Notes

B1: Line through (3, 0) and (0, 6), mark intent
Condone good freehand line
Condone line stopping at axes
Ignore additional lines, perhaps as part of solution to (b) or (c), as long as it is not several attempts at (a)

Notes

B1: Coordinates soi by correct midpoint
Condone labels A and B reversed or omitted
Condone missing brackets for coordinates throughout if intent is clear

M1: Correct midpoint or correct method to find the midpoint. e.g. \(\left(\frac{3+0}{2}, \frac{0+6}{2}\right)\)
Allow embedded in the equation for perp. bisector

M1: Correct gradient or correct method to find gradient e.g. \(y = -2x + c\) or \(\frac{6-0}{0-3}\) or \(\frac{6-3}{0-1.5}\) or \(\frac{0-3}{3-1.5}\) oe
FT their midpoint if used
Implied by correct gradient of perpendicular bisector

M1: FT \(\dfrac{-1}{\textit{their}\text{ gradient}}\)
May be seen embedded in the equation.

M1: Using their midpoint and their gradient of perpendicular bisector
Allow \(y = \frac{1}{2}x + c\) and \(3 = \frac{1}{2} \times \frac{3}{2} + c\) oe

A1: OR \(2x - 4y = -9\)
Allow any correct equivalent simplified form e.g. \(4y = 2x + 9\) or \(y - \frac{1}{2}x = \frac{9}{4}\) or \(\frac{1}{2}x - y + \frac{9}{4} = 0\) or any multiple of these.
Allow decimals e.g. \(y = 0.5x + 2.25\)
isw after correct equation seen

Notes

B1: Use of Pythagoras which will help to calculate side e.g. AC², BC², AB² or AC, BC, AB soi
FT their midpoint from (a) if used

M1: Further working may not be seen as likely to be done by calculator

A1: Accept any exact correct answer e.g. \(\frac{3\sqrt{10}}{2}\), \(\frac{3}{4}\sqrt{40}\), \(1.5\sqrt{10}\)

Alternative method 1 for M1

M1: Using \(\mathrm{AC} = \sqrt{1.5^2 + 3^2}\) where C is the midpoint of the square

Alternative method 2 for M1

M1: Use of trigonometry to find \(a\)

Alternative method 3 for M1

M1: Use of trigonometry to find \(a\)
Using \(\mathrm{AC} = \sqrt{1.5^2 + 3^2}\) where C is the midpoint of the square

Alternative method 4 for M1

M1: Use of their AC and their equation of perpendicular bisector to form a quadratic equation to give the \(x\)-coordinates of the other two vertices of the square

Alternative method using vectors

B1: FT their midpoint from (a)
Vector from C to B or A or vice versa and rotating to get vector to one of the other two vertices

M1: Or \((-1.5, 1.5)\)

A1: Accept any exact correct answer e.g. \(\frac{3\sqrt{10}}{2}\), \(\frac{3}{4}\sqrt{40}\), \(1.5\sqrt{10}\)

Notes

B1: allow \(x = 4,\ y = -5\); condone omission of brackets

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: Allow \(2at + b\)

M1: Use of a form of equation of straight line with their gradient and \(x = t\)

A1: Convincing completion to correct result

Alternative method

M1: Getting everything to one side

A1: Convincing completion to correct result

Additional guidance

The first M1 is for differentiating. We are allowing the gradient as 2at + b as well as 2ax + b. Some are just writing 2ax+b/2at+b and that isn’t convincing that they have differentiated – rather just copying from the given answer so give M0.

The second M1 is for substituting into the equation of a straight line using their gradient and x = t. They could use \((y - y_1) = m(x - x_1)\) or y= mx + c. Condone a lack of brackets around their m if recovered.

The final A1 is for getting to the correct answer. As the answer is given, we need to see some working here.

Notes

M1: Use of tangent formula with distinct values of \(t\)
Condone alternative notation for co-ordinates.
e.g. P and Q, m and n, and other subscript versions

M1: Getting terms in \(x\) on one side

A1: Convincing completion to given result.
Accept the final result in terms of their variables.

Additional guidance

The first M1 is for equating their two tangents with distinct values of t. These could be \(x_p\) and \(x_q\) as per the scheme, but they might choose other pairs of symbols – e.g. P and Q, m and n, \(t_1\) and \(t_2\).

The second M1 is for getting the x terms on one side and the non-x terms on the other.

To get the A1 they must show the working out including the difference of two squares and cancelling to get the required result. Allow the final result in terms of their variable.

Notes

B1: Convincingly showing that the formula gives 7.5

Additional guidance

Candidates must use \(t_1\) and \(t_2\) with values 1 and -3 and show the substitution to get to 7.5 for the B1. Using -1 and 3 is not correct and scores B0.

Notes

B1: Convincingly showing that the formula gives \(-3\)

Additional guidance

Candidates must use \(t_1\) and \(t_2\) with values 1 and -3 to show \(t_1t_2 = -3\) for the B1. Using -1 and 3 is not correct and scores B0.

Notes

M1: must be sum of two squares;
may be implied by correct answer or by \((\pm 4)^2 + (\pm 2)^2\)

A1: mark the final answer; B2 for correct answer unsupported

Notes

B1: first term correct; allow \(y^{\prime}\) for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1: Product Rule; allow one coefficient error or one index error

M1*: NB \(-\frac{4}{3}\)
dependent on at least two of 3 terms correct on LHS following differentiation;
if expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) or evaluation of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) is incorrect, need to see substitution for award of M1

M1dep*: FT negative reciprocal of their \(-\frac{4}{3}\) and their 2
may see eg \(1 = \frac{3}{4} \times 2 + c\)

A1: must be in required form, but coefficients may be fractions

Notes

M1: Attempt to find equation of the line using correct gradient formula

M1: Eliminating one variable

M1: oe Three term quadratic seen or implied by correct \(x\)-values

A1: cao

A1: Both \(y\)-coordinates seen.

M1: Uses distance formula for their points (not given points)

A1: Must be exact (allow \(\sqrt{360}\) oe)

Notes

M1: allow sign errors in their expressions for \(\sin\theta\) and \(\cos\theta\); allow if just see brackets expanded, but must be 3 terms in each case

A1: allow SC2 for \((x+3)^2 + (y-1)^2 = 4\) unsupported

Alternatively

M1: allow one sign error in bracket;

Notes

(i) B1: Any midpoint correct

(i) B1: All midpoints correct
If no labels BOD for 1 or 2 marks

(ii) B1: Gradient, length or vector of any one side of PQRS
Or SC1 if KLMN used

(ii) M1: Gradient of opposite side of PQRS shown to be equal
or length or vector of opposite side shown to be equal.
ie some working must be seen

(ii) E1: Convincing completion dep on M1

Condone confusion of labels (eg length for gradient etc) for M1/B1

Watch out for valid alternatives e.g. 2 sides equal length and equal gradient

Notes

(i) B1: Convincing completion
\(\overrightarrow{\mathrm{WX}} = \overrightarrow{\mathrm{WV}} + \overrightarrow{\mathrm{VT}} + \overrightarrow{\mathrm{TX}}\)

(ii) E1: \(\overrightarrow{\mathrm{DE}}\) from any correct route, must be shown
Convincing completion with conclusion

(iii) B1: \(\overrightarrow{\mathrm{GF}}\) from any correct route, must be shown

(iii) E1: Convincing completion with conclusion

Notes

M1: At least two terms correct

M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

A1: Need reasoning eg \(y - 0 = 6(x - 0)\) or use of \(y = mx + c\)
\(y = 6x\) implies previous M mark

Notes

M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from (a)

M1: FT negative reciprocal of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

A1: oe but constant terms should be collected

Notes

M1: Eliminate a variable
FT their equations from (a) and (b) for M1 only

Notes

M1: completing the square twice soi

A1: NB \((x-6)^2 - 36 + (y+4)^2 - 16 + 3 = 0\) oe
allow B2 for \(r = 7\) unsupported

Alternatively

A1: NB \(r^2 = 6^2 + 4^2 - 3\)

Notes

B1: FT \((x \pm 6)^2 + (y \pm 4)^2\)
or FT \(\pm 2a = -12\) oe and \(\pm 2b = 8\) oe

Notes

M1: allow sign errors

M1: or \(\left(\frac{x-2}{5}\right)^2 + \left(\frac{y-1}{5}\right)^2 = \cos^2\theta + \sin^2\theta\) oe

A1: may see eg \(x^2 - 4x + y^2 - 2y = 20\)
if M0M0 allow SC1 for \(y = 1 + 5\sin\left(\cos^{-1}\left(\frac{x-2}{5}\right)\right)\) or \(x = 2 + 5\cos\left(\sin^{-1}\left(\frac{y-1}{5}\right)\right)\)

Alternatively

M1: if only seen in expanded form, allow one coefficient error; allow sign errors

M1: must have terms in \(\cos\theta\) and \(\sin\theta\)

Alternatively

M1: allow sign error in coordinates of centre

M1: FT their centre

A1: all correct

Notes

M1: FT \(1 \div\) their \(-\frac{4}{3}\)

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)

Alternatively

B1: or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2-x}{y-1}\) oe
eg \(2(x - 2) + 2(y - 1)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3/5}{-(-4/5)}\) or \(\dfrac{2-5}{-3-1}\) oe; allow one sign error;

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)

Notes

M1: Attempt to re-arrange

A1: Ignore labelling of this expression eg \(\mathrm{fh}(x) = \sqrt[3]{(x+8)}\) scores 2

Notes

B1: Condone lack of brackets if meaning is clear

B1: C: FT their B

B1: D: FT their A

Notes

B1: FT their A and B. May be implied by later work

B1: FT -ve reciprocal of their 4

M1: Or using \(y = mx + c\) and attempting to evaluate \(c\)
Must be using their midpoint and their \(-\dfrac{1}{4}\)

A1: Final answer

Notes

M1: May solve all 4 which also earns 3^rd method mark

Or

M1: Identifying 2 perp bisectors of 2 chords eg BD and AC, or AB and CD

Notes

B1: soi

M1: Attempt to find length of AB, AC or BC

M1: Uses their midpoint and radius (do not allow for diameter used)

A1: Need not be simplified

Notes

M1: Substituting \(y = 2x + 5\) and attempting to collect terms oe
Allow for a quadratic solved BC providing it is seen in form \(ax^2 + bx = 0\) or \(ay^2 + by + c = 0\)

A1: Both values correct

A1: Correct \(y\) coordinates FT their \(x\)-coordinates

Notes

Note QAB \(= 81.9^\circ\) and QBA \(= 8.1^\circ\)

M1: Attempt to find two lengths to be used in their area calculation (excluding AB)

M1: Correct method for finding the area

A1: FT their Q

Notes

M1: allow sign error

Alternative

A1: if M0M0 allow SC1 for
\(y = 5\sin\left\{\cos^{-1}\left(\dfrac{x-7}{5}\right)\right\} - 3\)
\(x = 5\cos\left\{\sin^{-1}\left(\dfrac{y+3}{5}\right)\right\} + 7\)

Notes

B1: FT their \((x-7)^2 + (y+3)^2 = 25\)

Notes

M1: Elimination of one of \(x\), \(y\)

A1: May be as part of a pair of coordinates

A1: May be as part of a pair of coordinates
Allow correct unsimplified answers

Notes

\((0, k)\) unsupported earns B1 SC
Need not be as coordinates but must be clear which \(y\) goes with which \(x\)
Substituting for \(x\) gets M0 unless it leads to correct values of \(y\)