Notes

B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)

M1: Attempts to find the critical values by solving a quadratic equation for \(x \gt 5\) or for \(x \lt 5\)

M1: A complete method to find all the critical values by solving two quadratic equations. E.g. considers both \(x \gt 5\) and \(x \lt 5\)

A1: Obtains any 2 correct critical values as long as one of the previous method marks has been scored

A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.

A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct. If uses set notation it must be correct, or, \(\cup\) not \(\cap\) isw

Alternative 1 by squaring:

B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)

M1: Attempts to find square both sides and multiples up or vice versa in order to obtain a quartic equation that will give the critical values

M1: A complete method to find all four of the critical values by solving a quartic equation may use a calculator

A1: Obtains any 2 correct critical values, as long as they have scored at least one of the previous method marks. May be seen as an inequality.

A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.

A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct. If uses set notation it must be correct, or, \(\cup\) not \(\cap\) isw

Alternative 2

B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)

M1: Either attempts to multiply both sides by \((x - 5)^2\) or \((5 - x)^2\) and solves to find the critical values. Or collects onto one side with a denominator of \(x - 5\) or \(5 - x\), uses a common denominator to combine and then finds the critical values may use a calculator

M1: Attempts to multiply both sides by \((x - 5)^2\) and \((5 - x)^2\) and solves to find the critical values. may use a calculator Or collects onto one side with a denominator of \(x - 5\) and \(5 - x\), uses a common denominator to combine and then finds the critical values

A1: Obtains any 2 correct critical values, as long as they have scored one of the previous method marks and using a correct equation

A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.

A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct

Notes

M1: A complete method to form equations/an equation that will give all the critical values. For method 1, both equations must be attempted. Allow with inequalities in place of equals throughout and do not be concerned if the inequalities are correct or not. This is for the method to find an equation or equations for the CVs.

dM1: Dependent on the previous method mark. Solve their equation(s) to find at least all the non-zero critical values. If using method 1 both equations must be solved, in method 2 the quadratic after factoring out or dividing by \(x^2\) must be solved. Allow for attempts by dividing through by \(x\) or \(x^2\).

A1: Correct critical values, including 0 (although this may be later rejected), seen stated or used in the solution. Must have come from correct work.

A1: Deduces the correct range following correct work. Both parts must clearly be identified for the answer (e.g. underlined during working if not on the final line).
Accept alternative notation, e.g. set or interval notation. In words “and” or “or” may be used, but if using mathematical notation, must use union and not intersection.
E.g. \(\{0\} \cup \{x : 1 \leqslant x \leqslant 3\}\) or \([0] \cup [1, 3]\) (if intervals given, must be square brackets not round) are acceptable answers.

Notes

M1: Multiplies through by \(|x + 6|\) or \((x + 6)\) and considers either \((x + 6)\) or \(-(x + 6)\) and attempts to solve the resulting 3TQ (do not be concerned about the method of solving), or squares both sides and simplifies to a quartic and attempts to solve. May use “=” or any inequality for this and the next 3 marks. Allow if an extra factor \((x + 6)\) is included (i.e. multiplies through by \((x + 6)^2\)).

A1: Correct roots for either equation, or any two correct roots from the (correct) quartic. Allow if \(-6\) is also included but do not count this as one of the roots.

M1: Recognises the requirement to consider both \((x + 6)\) and \(-(x + 6)\) and attempts to solve the resulting 3TQ’s (allow for any two values following the quadratic), or implied by an attempt to solve their quartic equation to find for 3 or 4 answers for their quartic, and allow for decimals, 0.2749…, −7.2749… for this mark. Again allow if extra factor \((x + 6)\) included. Allow if this is carried out but later rejected (e.g. crossed out) as they think the answers are inadmissible.

A1: All correct and exact critical values (and may include \(-6\) and \(\dfrac{-7 + \sqrt{57}}{2}\) at this stage) and no other incorrect values.

M1: Produces an answer with the correct form of the solution set from the graph, allowing for non-strict inequalities for this mark, using three distinct answer from their critical values, \(\alpha, \beta, \gamma\), with \(\alpha \leqslant -6 \leqslant \beta \lt 0 \lt \gamma\) and allow with either \(\alpha = -6\) or \(\beta = -6\) (but not both) for this mark. SC allow M1 if they list as 3 separate inequalities \(x \lt \dfrac{-7 - \sqrt{57}}{2},\ x \gt -\dfrac{27}{5},\ x \lt 2\) for this mark.

A1ft: Deduces one of the correct ranges, following through on appropriate critical values. If the two quadratics are solved independently then allow for either their \(\dfrac{-27}{5} \lt x \lt\) their 2 from the “\(x \gt -6\)” equation (one positive, one negative) or for \(x \lt\) their \(\dfrac{-7 - \sqrt{57}}{2}\) from the “\(x \lt -6\)” equation choosing a root less than \(-6\). If a quartic is solved then allow with any three of their CVs satisfying the M used to get one interval of correct form. (Answer may be inexact here - follow through decimal roots.)

A1: Fully correct solution set, in any suitable form. May give the two inequality statements, in which case accept with “or” or “and” between. Accept in interval notation. In formal set notation accept with union (\(\cup\)) but intersection (\(\cap\)) is A0. Do not accept \(x \gt -\dfrac{27}{5},\ x \lt 2\) for the second interval for this mark. Values must be exact.

(Corrected from the printed mark scheme: the second M1 note prints the decimal root as 0.2729…; the root \(\dfrac{-7 + \sqrt{57}}{2} = 0.2749\ldots\))

Notes

If both case are attempted mark for the correct one.

M1: Considers \(x^2 - 8 = -(mx + c)\) collects terms, finds the discriminant and sets \(= 0\). Must see a correct equation (without modulus) initially, though allow if subsequent slips rearranging occur.

A1*: Correct result with no incorrect working seen.

Notes

M1: Substitutes \(c = 3m\) into the equation (or their equation as long as it came from an attempt at using the correct equation in (a)) and solves the resulting 3TQ to find a value for \(m\) or \(c\).

M1: Finds the corresponding value of \(c\) (or \(m\)) or solved the other 3TQ to get values for \(c\) (or \(m\))

A1: Deduces the correct values for \(m\) and \(c\). If two sets of values are stated this mark is not achieved until the extra set \(m = 8\) and \(c = 24\) are rejected (correct reason for rejection is not needed).

Notes

M1: Correct method to find all the critical values (so solves both equations). Allow if both cases are included and more than three critical values are found. Allow if relevant work was seen in (b).

A1ft: Correct three critical values only. May be implied by their final answer. Follow through on \(m = 8\) and \(c = 24\) only. Allow if the \(-2\) was seen in (b).

A1cao: Deduces the correct region. Accept any correct notation. Accept with “and” or “or”, but not with \(\wedge\)

Notes

M1: Considers the two cases of \(x \lt 0\) and \(x \geqslant 0\) to find critical values. Don’t be concerned which side the \(x = 0\) case is considered part of. Allow if “=” used when considering C.V.s. This mark is for the overall strategy, so both cases must be considered, or equivalent complete longer methods.

M1: Correct method for intersection of line and curve for \(x\) positive.

A1: Line and curve intersect at \(x = \dfrac{5 \pm \sqrt{5}}{2}\)

M1: Correct method for intersection of line and curve for \(x\) negative.

A1: Line and curve intersect at \(x = -\sqrt{5}\) Must have rejected the positive value for this mark (though may be done later)

M1: Uses the graph (or other method) to identify at least one correct region, which must include consideration of the vertical asymptotes. Implied by two correct intervals being given for their critical values. Allow if \(y = 2x - 5\) is added to the sketch and at least two (not necessarily correct) intervals produced as long as the points \(x = \pm 2\) are excluded.

A1ft: At least one correct interval identified following through their solutions (as long as it is sensible).

A1: Fully correct solution, all three intervals given – accept alternative notations, may be just listed (no need for unions shown).

Multiplying both sides by \((x - 2)^2\) or \((|x| - 2)^2\) can score a maximum of M0 M1 A1 M0 A1 M1 A1ft A0

M0 M1: for multiplying through by \((x - 2)^2\)
\((2x - 5)(x - 2)^2 \gt x(x - 2)\)
\((x - 2)\left[(2x - 5)(x - 2) - x\right] \gt 0\) leading to a value for \(x\)
\((x - 2)\left(x^2 - 5x + 5\right) \gt 0\)

A1: Line and curve intersect at \(x = \dfrac{5 \pm \sqrt{5}}{2}\)

M0A0: Not finding the point of intersection for negative \(x\)

M1 A1ft: for either “\(x \gt \dfrac{5 + \sqrt{5}}{2}\)” or “\(\dfrac{5 - \sqrt{5}}{2} \lt x \lt 2\)”

A0:

If they multiply through by \((-x - 2)^2\) the other marks can be scored

Notes

M1: Sets problem up as an inequalities problem, and forms complete strategy to solve – must see attempt at all critical values and some attempt to form at least one range from them. May be scored if algebra not used. This mark is for showing an overall awareness of the problem.

M1: Attempts to find C.V.’s for the “positives”. Any valid method using algebra. Must see an attempt to find a 3TQ (oe), but allow answers from calculator once a 3TQ \(= 0\) is seen.

A1: Correct C.V.’s, both required.

M1: Attempts to find the other C.V.’s (same conditions as above)

A1: Correct C.V.’s. Need not see the negative value stated as \(t \gt 0\) is required. (If both given, they must be correct)

M1: Selects correct critical regions, shows the idea the “insides” are needed. \((-1 \lt)\,\alpha \lt t \lt \beta,\ \gamma \lt t \lt \delta\) where \(\alpha \lt \beta \lt \gamma \lt \delta\) are their four critical values, possibly truncated at 0 as long as no more than 1 is negative.

A1: One correct interval. Allow with loose or strict inequalities. Allow this mark if \(-1 \lt t \lt 5 - 2\sqrt{3}\) is given. Allow any variable for this mark.

A1: Fully correct solution. Must start at zero for the leftmost interval but accept \(\lt\) or \(\leqslant\) here. Must be using \(t\)

Alternative

M1: Sets problem up as an inequalities problem, and forms complete strategy to solve – must see attempt at all critical values and some attempt to form at least one ranges from them. May be scored if algebra not used. This mark is for showing an overall awareness of the problem.

M1: Attempts to find C.V.’s by squaring both sides and forming a quartic equation.

A1: Correct quartic equation

M1: Attempts to solve their quartic equation

A1: All 4 correct exact C.V.’s. Need not see the negative value stated as \(t \gt 0\) is required.

M1: Selects correct critical regions, shows the idea the “insides” are needed. \((-1 \lt)\,\alpha \lt t \lt \beta,\ \gamma \lt t \lt \delta\) where \(\alpha \lt \beta \lt \gamma \lt \delta\) are their four critical values, possibly truncated at 0 as long as no more than 1 is negative.

A1: One correct interval. Allow with loose or strict inequalities. Allow this mark if \(-1 \lt t \lt 5 - 2\sqrt{3}\) is given. Allow any variable for this mark.

A1: Fully correct solution. Must start at zero for the leftmost interval but accept \(\lt\) or \(\leqslant\) here. Must be using \(t\)

Notes

M1: Uses their result from (a) to determine how long particle \(B\) is closer to \(O\) than particle \(A\) is.

A1: Draws a suitable conclusion for their answer to (a) – if correct in (a) it is that the model is not very suitable. Must not have a negative time.

Notes

ALT

Notes

Alternative 1: Multiplies both sides by \(x^2(x + 2)^2\)

Alternative 2: using a sketch graph

(probably from calculator)

NB  As above, but with no sketch graph shown: CVs \(x = 0, -2\) must be stated somewhere. (B1)
Otherwise no marks available.

Notes

M1: Mult through by \((x + 3)^2\) and collect on one side or use any other valid method (NOT calculator)
Eg work from \(\dfrac{(x + 3)(x + 2) - 12}{(x + 3)} \gt 0\)
NB: Multiplying by \((x + 3)\) is not a valid method unless the two cases \(x \gt -3\) and \(x \lt -3\) are considered separately or \(-3\) stated to be a cv

B1: for -3 seen anywhere

A1A1: other cvs  (A1A0 if only one correct)

dM1: obtaining inequalities using their critical values and no other numbers. Award if one correct inequality seen or any valid method eg sketch graph or number line seen

A1: correct inequalities and no extras. Use of \(\leqslant\) or \(\geqslant\)  scores A0. May be written in set notation.

No marks for candidates who draw a sketch graph and follow with the cvs without any algebra shown. Those who use some algebra after their graph may gain marks as earned (possibly all)

(Corrected from the printed mark scheme: the two cases are printed as \(x \gt 3\) and \(x \lt 3\); they are \(x \gt -3\) and \(x \lt -3\).)

Notes

B1: correct answer only shown. Allow \(x \geqslant 1\) if already penalised in (a)

Notes

M1  obtaining two non-zero cvs by any valid method (not calculator)
A1  non-zero cvs correct
B1  \(x = 0\)
M1  deducing one appropriate range from their cvs
A1  both ranges correct
First 3 marks – award with inequalities or =
M1A0 if strict inequality not used

ALT: If multiplied through by \(x\):

Notes

If no algebra seen (implies a calculator solution) no marks.

With algebra:
M1 Squaring and reaching a quartic = 0
M1 Attempt to factorise and obtain at least one solution for \(x\). Coefficient of \(x^4\) and constant term correct for their quartic.
A1 Any 2 correct values
A1 All 4 correct values
Final 2 A marks as above

Accept set notation for the final 2 A marks. \(x \in (1, 2)\), \(x \in \left(\tfrac{11}{3}, 6\right)\) not \([1, 2]\)

Notes

M1 for multiplying through by \((x + 1)^2(3 - x)^2\)
OR: for collecting one side of the inequality and attempting to form a single fraction (see alternative in mark scheme)

M1dep for collecting on one side of the inequality and factorising the result of the above (usual rules for factorising the quadratic)
OR: for factorising the numerator – must be a three term quadratic – usual rules for factorising a quadratic (see alternative in mark scheme)
Dependent on the first M mark

B1 for the critical values \(3,\ -1\)

A1 for either \(-\dfrac{3}{2}\) or \(\dfrac{1}{3}\)

A1 for the second of these

NB: the critical values need not be shown explicitly – they may be shown on a sketch or just appear in the ranges or in the working for the ranges.

M1 using their 4 critical values to obtain appropriate ranges e.g. use a sketch graph of a quartic, (which must be the correct shape and cross the \(x\)-axis at the cvs) or a table or number line

A1cso for both of \(-\dfrac{3}{2} < x < -1,\quad \dfrac{1}{3} < x < 3\)

Set notation acceptable i.e. \(\left(-\dfrac{3}{2}, -1\right) \cup \left(\dfrac{1}{3}, 3\right)\) All brackets must be round; if square brackets appear anywhere then A0.

If both ranges correct, no working is needed for the last 2 marks, but any working shown must be correct.

Purely graphical methods are unacceptable as the question specifies “Use algebra…”.

Q2 – Alternative 1

Notes

NB: Marks for (a) can only be awarded for work shown in (a):

M1 for \(2x^2 + 6x - 5 = 5 - 2x\)

M1 for obtaining a 3 term quadratic and attempting to solve by factorising, formula or completing the square

A1 for \(x = -5,\ x = 1\)

M1 for considering the part of the quadratic that needs to be reflected ie for \(-2x^2 - 6x + 5 = 5 - 2x\) oe

A1 for a correct 2 term quadratic, terms in any order \(2x^2 + 4x = 0\) oe

A1 for \(x = 0\quad x = -2\)

NB: The question demands that algebra is used, so solutions which do not show how the roots have been obtained will score very few if any marks, depending on what is written on the page.

Alternative: Squaring both sides:
M1 Square both sides and simplify to a quartic expression
M1 Take out the common factor \(x\)
A1 \(x\), a correct linear factor and a correct quadratic factor
M1 \(x\) and 3 linear factors
A1 any two of the required values
A1 all 4 values correct

Notes

B1 for a line drawn, with negative gradient, crossing the positive \(y\)-axis

B1 for the quadratic curve, with part reflected and the correct shape. It should cross the \(y\)-axis at the same point as the line and be pointed where it meets the \(x\)-axis (ie not U-shaped like a turning point)

B1ft for showing the \(x\) coordinates of the points where the line crosses the curve. They can be shown on the \(x\)-axis as in the MS (accept \(O\) for 0) or written alongside the points as long as it is clear the numbers are the \(x\) coordinates
The line should cross the curve at all the crossing points found and no others for this mark to be given.

Notes

NB: No follow through for these marks

B1 for any one of \(x < -5,\quad -2 < x < 0,\quad x > 1\) correct

B1 for a second one of these correct

B1 for the third one correct

Special case: if \(\leqslant\) or \(\geqslant\) is used, deduct the last B mark earned.

Notes

1^st M1 accept \(\pm(x^2 - 4) > 3x\) or \(\pm(x^2 - 4) = 3x\). Require modulus of parabola and straight line with positive gradient through origin for graphical method.

1^st B1 for \(x = 1\), 2^nd B1 for \(x = 4\)

2^nd M1 dependent upon first M1

A0 for error in solution of quadratic leading to correct answer.

Notes

1^st M1 for \(\pm(x^2 - 4x - 12)\) – ‘=0’ not required.

B marks can be awarded for values appearing in solution e.g. on sketch of graph or in final answer.

2^nd M1 for attempt at method using graph sketch or +/-

If cvs correct but correct inequalities are not strict award A1A0.

Alternative using squaring method

Notes

Use of \(\leqslant\) instead of \(<\) lose last B1 Extra inequalities lose last B1

Notes

Second M attempt to factorise quadratic expression with 3 terms (usual rules).

Second A don’t require −2 outside but can be part of factors.   B1, B1

Notes

First M mark can be implied by the two correct values, but otherwise a method must be seen. (The method may be graphical, but either \((x =)\) 4 or \((x =)\) 0 needs to be clearly written or used in this case).
Ignore ‘extra values’ which might arise through ‘squaring both sides’ methods.

\(\leqslant\) appearing: maximum one A mark penalty (final mark).

Notes

“Squaring”

\(x = 1\) allowed anywhere, no penalty in (b)

Notes

Correct answers seen with no working is independent of (a) (graphical calculator) mark as scheme.
Only allow the accuracy mark if no other interval, in both parts
\(\leqslant\) used penalise first time used

Notes

First M mark can be implied by the two correct values, but otherwise a Method must be seen.

\(\leqslant\) appearing: maximum 1 mark penalty (at first occurrence).

Notes

(corrected from the printed mark scheme: the third B1 line reads “\(-z, z\)”; the graph meets the \(x\)-axis at \(-2\) and \(2\))