2004 Mathematics. Higher. Finalised Marking Instructions

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1 Mathematics Higher Finalised Marking Instructions

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5 Higher Maths Paper Marking Scheme Final The point A has coordinates (, ). The straight lines with equations x + y + = and x + y = intersect at B. (a) Find the gradient of AB. (b) Hence show that AB is perpendicular to only one of these two lines. a C.. CN / b C..9,.. ss : pd : process strategy for solving sim. equations pd : calculate gradient ss : ss : ic : ic : 8 ic : Notes For use m. m = arrange in standard form state gradient state gradient complete proof Elimination may be used instead of substitution Evidence of a start to elimination would be the appearance of equal coefficients of x or y. For (a) equating the zeros, neither of the first two marks are available. (, ) may be obtained by inspection or trial and improvement. If it is justified by checking in both equations, and may be awarded. If is not justified in both equations, award neither of the first two marks. A general statement about perpendicular lines must have m. m = earns no marks Candidates who make a mistake in (a) may have to show in (b) that neither line is perpendicular to AB. All five marks are available. page Primary Method : Give mark for each x = y and attempt to substitute eg.. ( y )+... = B(, ) m = m = m = AB perp y = x... stated / impliedby m = l m = l 8 so only the st line is perpendicular to AB marks Alternative Method for to 8 y = x... m = l m l = l : = so AB l 8 and AB is not l Alternative Method for to 8 Continued on page may be implied by m = m = AB perp y = x stated / implied by m = l m = l AB 8 so only the ndline isperpendicular to AB marks marks marks

6 Higher Maths Paper Marking Scheme Final The point A has coordinates (, ). The straight lines with equations x + y + = and x + y = intersect at B. (a) Find the gradient of AB. (b) Hence show that AB is perpendicular to only one of these two lines. a C.. CN / b C..9,.. continued from page Alternative Method for to 8 m perp = strat: find equ. thr' B with gradient y ( ) = ( x ) leading to y + x + = 8 the first line is the ONLY line perp. to AB marks Alternative Method for to 8 y = x... m = l m l = may be implied by l : = so AB is the ONLY line l 8 implied by the" ONLY " at. marks A Poor illustration Further illustrations y = x... y = x st equ is perp. to AB nd equ is not perp to AB mark mark AB is perp. to because [ end!] x + y + = = only AB is perp. to x + y + = as its gradient is - (& ) and AB is = all of the above writing AB is perp. to the line y =... because = all of the above writing page

7 Higher Maths Paper Marking Scheme Final fx ()= x x x. (a) (i) Show that (x + ) is a factor of fx (). (ii) Hence or otherwise factorise fx ()fully. (b) One of the turning points of the graph of y = f() x lies on the x-axis. Write down the coordinates of this turning point. a C.. NC /8 b C.. ss : know to find f( ) ss : start eg synthetic division pd : complete to zero remainder ic : extract quadratic ic : fully factorise ic : state coordinates Primary Method : Give mark for each know to find f ( ).. x x ( x + )( x + )( x ) (, ) marks mark Alternative Method for, and know to find f ( ) f( ) = ( ) ( ) ( ) = a strategy for finding the quadratic factor eg inspection, long division, synthetic division Notes Treat fx () = ( x + ),( x + ),( x ) as bad form is not available for (,) or (, ) x = an unsupported (, ) Treat x = y =... = so point = (, ) as bad form page

8 Higher Maths Paper Marking Scheme Final Find all the values of x in the interval x π for which tan ( x)=. C..9,.. NC /8 ss : know to get the square root pd : solve trig equation pd : solve trig equation ic : know there is + and Primary Method : Give mark for each tanx = π x = π x = tanx = stated explicitly and x = π π, marks Alternative Method for and tanx = π x = π tanx = and x = π and π marks Notes Candidates must produce final answers in radians. If their final answer(s) are in degrees then deduct one mark. Cave Candidates who produce the four correct answers from tan( x ) = can only be awarded and. Do not penalise correct answers outside the range x π π π Do NOT accept π + for. page 8

9 Higher Maths Paper Marking Scheme Final y The diagram shows the graph of y = g(). x (b, ) (a) Sketch the graph of y = g(). x (b) On the same diagram sketch the graph of y = g( x). (, ) O y = g(x) x (a, ) a C.. CN / b C.. ic : ic : ic : ic : sketch transformed graph show new coordinates sketch transformed graph show new coordinates Primary Method : Give mark for each reflection in x -axis and any one from (,-),(, a ),(, b ) clearly annotated the remaining two from the above list translation and any one from (,),( a, ),( b, ) clearly annotated the remaining two from the above list marks marks solution Notes For (a), reflection in the y-axis earns a maximum (a, ) y of out of with all points clearly annotated Ê ˆ For (b), a translation of Ë Á - earns a maximum (a, ) (, ) O (b, ) (, ) (b, ) x of out of with all points clearly annotated Ê ± ˆ For (b), a translation of Ë Á earns no marks. For the annotated points in (a) and (b), accept a superimposed grid. g(x) needs to retain its cubic shape for and In (b) and are only available for applying the translation to the resulting graph from (a). page 9

10 Higher Maths Paper Marking Scheme Final A,B, and C have coordinates (,, ), (, 8, ), and (,, ) respectively. (a) Show that A, B, and C are collinear. (b) Find the coordinates of D such that AD = AB. a C.. CN /n b B.. ss : ic : ic : use vector approach eg for AB compare two vectors complete proof pd : find multiple of vector ic : interpret vector Alternative Method for and eg AB = = = AC = Alternative Method for d - a = ( b - a) d = b -a or 8 d - a = ( b - a) d -a = - Primary Method : Give mark for each AB = AC = = AB AB & AC have common direction and common point Hence A,B and C collinear 8 AD = D = (,, 9) Alternatives Method for and BC = AC = = BC marks marks AB = BC = = AB Notes Treat D= -9 as bad form. For accept ONLY parallel in lieu of common direction page

11 Higher Maths Paper Marking Scheme Final dy Given that y = sin( x) + cos( x), find. dx B.. CN /n pd : process simple derivative pd : start to process compound derivative Primary Method : Give mark for each cos( x) sin( x) marks ic : complete compound derivative Alternative Methods e.g. y = sin( x) + cos ( x) cos( x) cos( x) sin( x) and no further terms marks Notes For differentiating incorrectly: For y = cos( x) + sin( x),only may be awarded. For y = cos( x) sin( x) + c, treat the +c as bad form. For clearly integrating correctly or otherwise: Award no marks. If you cannot decide whether a candidate has attempted to differentiate or integrate, assume they have attempted to differentiate. page

12 Higher Maths Paper Marking Scheme Final Find x + dx. AB.. CN / ic : express in integrable form pd : integrate a composite fractional power ic : interpret the ic : substitute limts pd : evaluate Primary Method : Give mark for each x + x + ( ) ( ) ( ) ( + ) + or equivalent fraction or mixed number marks Common wrong solution x + x + x + x ( ) marks awarded ( ) dx dx Notes is available for substituting the limits correctly into any function except the original one. ( ) eg x + dx = ( x + ) = ( + ) = = ( ) + may be awarded, not (no integration) not (not dealing with fgx (())) not (original function) not (working eased) For, DO NOT accept answers like 9. page

13 Higher Maths Paper Marking Scheme Final 8 ( ) +. (a) Write x x + in the form x + b c (b) Hence show that the function gx ()= x x + x is always increasing. 8 a C..8 NC / b B.. pd : deal with the b pd : deal with the c ss : use differentiation pd : differentiate ss : ic : use previous working complete proof Primary Method : Give mark for each ( ) ( ) + x x g ( x) = STATED EXPLICITLY x x + ( ) + x g ( x) > for all x and so gx () increasing marks marks Alternative Method for to g ( x) = STATED EXPLICITLY x x + b ac = 8 = 8 no roots, concave up, g ( x) > and thus gx () increasing marks Notes For, accept g () x > in lieu of g () x > Evaluating g(), g() etc or g (), g () etc gains no credit. page

14 Higher Maths Paper Marking Scheme Final 9 Solve the equation log ( x + ) log ( ) =. 9 AB.. NC / ic : ic : ic : use log laws use log laws pd : process express in exponential form Primary Method : Give mark for each log log x + = x + = x = marks Alternative Method log log log ( x + ) = log x + log ( )= + log log ( x + )= log x = ( ) marks page

15 Higher Maths Paper Marking Scheme Final C B A In the diagram, angle DEC = angle CEB = x and x x angle CDE = angle BEA = 9. CD = unit; DE = units. D E By writing angle DEA in terms of x, find the exact value of cos( DEA ˆ ). B..,.. CN / ic : interpret diagram pd : expand trig expression pd : simplify ss : pd : process ic : use appropriate formula interpret pd : simplify Primary Method : Give mark for each DEA ˆ = ( x + 9 ) cos( x )cos( 9 ) sin( x ) sin( 9 ) sin( x ) sin( x )cos( x ) CE = + = stated / implied by sin( x ) =( ) and cos( x ) = cosdea ˆ = ( )( )= marks Note Although unusual, it would be perfectly acceptable for a candidate to go from to without expanding (via knowledge of transformations). In this case would awarded by default. another common wrong solution common wrong solution DEA ˆ = ( x + 9 ) cos( x + 9 ) cos( x ) + cos( 9 ) cos( x ) [ working eased] eg cos x CE = + = stated / implied by cos( x ) = cosdea ˆ = ( )( ) = 8 marks awarded DEA ˆ = ( x + 9 ) cos( x )cos( 9 ) sin( x )sin( 9 ) cos( x ) sin( x ) cos( x ) eg cos x CE = + = stated / implied by cos( x ) = cosdea ˆ = ( )( ) = 8 marks awarded page

16 Higher Maths Paper Marking Scheme Final The diagram shows a parabola passing through the points (, ), (, ) and (, ). y (a) The equation of the parabola is of the form y = ax( x b). Find the values of a and b. (b) This parabola is the graph of y = f ( x). O (, ) x Given that f() =, find the formula for f(x). (, ) a B.. CN / b A..8 ss : use parabolic form pd : substitute pd : process ss : know to integrate pd : express in integrable form pd : integrate ss : 8 pd : process introduce constant and substitute Primary Method : Give mark for each b = or y = ax( x ) substitute (, ) a = ( ) ( ) fx ( ) = xx ( ) dx x x dx x x = + c 8 c = 8 marks marks Notes In the primary method, must be justified. A guess and check would be acceptable ie guess a = then check that (, ) fits the equation. In the primary method, is only available if an intention to integrate has been indicated. For candidates who fail to complete (a) but produce values for a and b ex nihilo, marks are available in (b). A deduction of mark may be made if their choice eases the working. For candidates who retain a and b in part (b), marks to are available. CAVE xx ( ) dx= x x = 8 may be awarded, and. page Alternative Method for to two simultaneous equations a( b) = and a( b) = b = a = Alternative Method for to y = k( x ) = k( ) k = y = ( x ) y = x( x ) marks marks

17 Higher Maths Paper Marking Scheme Final S The IQs of a group of students were measured and the scores recorded in the stem-and-leaf diagram as shown. Identify any outliers. replacing qu. (in position ) IQs of a group of students n= means S C..,.. CN / pd : calculate quartiles ss : know how to calculate fences pd : calculate fence/interpret outlier pd : calculate fence/interpret outlier Primary Method : Give mark for each Q =, Q = 8 eg lower fence = Q ( Q Q ) fence = 9. fence. & = is outlier marks page

18 Higher Maths Paper Marking Scheme Final S Calculate the mean and variance of the discrete random variable X whose probability distribution is as follows: x P(X = x).... replacing qu. S C.. NC / ss : know and state rule for mean pd : calculate mean ss : ss : know/state rule for variance know how to find EX ( ) pd : calculate EX ( ) pd : calculate variance Primary Method : Give mark for each EX ( ) = Σxpx ( ) Σxp( x) = VX ( ) EX ( ) EX ( ) = ( ) EX ( ) = Σxpx ( ) Σxpx ( ) = VX ( ) = marks page 8

19 Higher Maths Paper Marking Scheme Final S The committee of New Tron Golf Club consists of men and women which reflects the proportions of men and women who are members of the club. It is agreed to send a delegation of committee members to a local planning meeting. The members of the delegation are to be chosen at random and will consist of men and women. What is the probability that both committee members Mr Hook and Miss Green will be selected? replacing qu. S C..,.. NC / ic : ic : ss : pd : process interpret probability interpret probability know to multiply for independent events Primary Method : Give mark for each Pman ( )= Plady ( ) = multiply = marks Alternative Method eg C C C 9 C C! 9!.! 9!!! =!!.! 9!!! marks page 9

20 Higher Maths Paper Marking Scheme Final S The cumulative distribution function for a random variable X is given by Fx ()= ( ) x x x otherwise Show that the median is. replacing qu.9 S AB..,..,.. NC / ss : know where median is pd : substitute Primary Method : Give mark for each Fmedian ( ) = F( ) = F( ) =, hence median = ( ) marks ic : interpret result page

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24 Higher Maths Paper Marking Scheme Final (a) The diagram shows line OA with equation x y =. The angle between OA and the x-axis is a. y O a A x Find the value of a. (b) The second diagram shows lines OA and OB. The angle between these y B two lines is. A Calculate the gradient of line OB correct to decimal place. O a x a C.. CR /8 b C.. ic : ss : pd : process find gradient of a line know gradient =tan(angle) and apply pd : process angle=tan - (angle) Primary Method : Give mark for each gradient = tana = gradient stated or implied by tan. m l = tan( +. ) = () = marks mark Common Error no. m = tan a = m a = tan ( ) =. Notes Accept any answer in (a) rounded correctly, so that e.g. if a = (OK) m OB =tan(+) =. Common Error no. m = tan a = m a = tan ( ) = A candiate who states m = tan θ, and does not go on to use it, cannot be awarded. Treat tan = as very bad form. In (b) do not penalise not rounding to d.p. but accept any correct answer which rounds to. Common Error no. m = tan a = m a = tan ( ) =. or. page

25 Higher Maths Paper Marking Scheme Final P,Q and R have coordinates (,, ), (,, ) and (,, ) respectively. (a) Express the vectors QP and QR in component form. (b) Hence or otherwise find the size of angle PQR. a C..8 CR / b C..9,.. ic : ic : ss : interpret coordinates to vectors interpret coordinates to vectors know to use eg scalar product pd : process scalar product pd : process length pd : process length pd : process angle Primary Method : Give mark for each QP = QR = cospqr = QP. QR = QP = QP.QR QP QR stated or implied by marks QR = Note in (a) For calculating PQ and RQ, award mark (out of ) in (a) Treat e.g. (,, ) as bad form For candidates who do not attempt : the formula quoted at in both methods must relate to the labelling in the question to earn PQˆ R =. Alternative Method for and cospqr ˆ = p r pr q = r = p = 9 PQR ˆ =. marks stated or implied by marks CONTINUED page

26 Higher Maths Paper Marking Scheme Final P,Q and R have coordinates (,, ), (,, ) and (,, ) respectively. (a) Express the vectors QP and QR in component form. (b) Hence or otherwise find the size of angle PQR. a C..8 CR / b C..9,.. Common errors no. Common errors no. OP. OR cospor = OP OR stated or implied by cosqor = OQ. OR OQ OR stated or implied by OP. OR = OQ. OR = OP = OQ = OR = OR = POR ˆ = 99. or. c marks awarded : deduct per error QOR ˆ =. or. 8 9 c marks awarded : deduct per error Common errors no. Common errors no. cosqop = OQ. OP OQ OP stated or implied by cosprq ˆ = RP. RQ RP RQ stated or implied by OQ. OP = OQ = RP. RQ = RP = 9 OP = RQ = QOR ˆ =. or. 8 c marks awarded : deduct per error PRQ ˆ =. or. c marks awarded : deduct per error Common errors no. cosrpq ˆ = PR. PQ = 8 PR = PR. PQ PR PQ stated or implied by PQ = 9 page RPQ ˆ =. or. c marks awarded : deduct per error

27 Higher Maths Paper Marking Scheme Final Prove that the roots of the equation x + px = are real for all values of p. C,B..,.. CN /8 ss : ic : pd : simplify ic : know/use discriminant identify discriminant complete proof Primary Method : Give mark for each know to show b ac p ( ) p + p is positive so and roots real Note Evidence for will more than likely appear at the stage. Treat b ac > as bad form marks Alternative Method p ( p) ( ) x = ± p p x = ± + we need p + p is positive and so roots real marks page 8

28 Higher Maths Paper Marking Scheme Final A sequence is defined by the recurrence relation u = ku +. n+ n (a) Write down the condition on k for this sequence to have a limit. (b) The sequence tends to a limit of as n. Determine the value of k. a C.. CN / b B.. ic : ss : ic : state condition for limit know how to find limit substitute pd : process Primary Method : Give mark for each < k < b l = " " a = k k = stated or implied by mark marks Alternative Method : no. Notes k does not gain but accept between and for accept k < for < a < does not gain unless it has been replaced by k in subsequent working in (b) < k < L = kl+ stated or implied by = k + k = mark marks Guess and check : Guessing k =. and checking algebraically that this does yield a limit of may be awarded marks Guess and check : Guessing k =. and checking iteratively that this does yield a limit of may be awarded mark No working : Simply stating that k =. earns no marks Wrong formula : Work using an incorrect formula leading to a valid value of k may be awarded mark. page 9

29 Higher Maths Paper Marking Scheme Final The point P(x, y) lies on the curve with equation y = x x. (a) Find the value of x for which the gradient of the tangent at P is. (b) Hence find the equation of the tangent at P. a C..,..9 CN /9 b C.. ss : know to differentiate pd : differentiate ss : set derivative = gradient pd : start to solve pd : process pd : process ic : state equation of tangent Primary Method : Give mark for each dy = stated or implied by dx x x x x = ( x ) = x = y = y = ( x ) marks marks Common error no. dy = stated or implied by dx x x x x = x( x) x = and x = marks awarded x = y = y = ( x ) marks awarded Notes For dy = x x dx x x = followed by a guess of x = and no check, only, and can be awarded. For dy = x x dx x x = followed by a guess of x = and a check that does in fact yield,,, and can be awarded. page

30 Higher Maths Paper Marking Scheme Final (a) Express cos( x )+ sin x ( ) in the form kcos ( x a ) where k > and a (b) Hence solve the equation cos( x )+ sin ( x )= for x a C.. CR / b B.. CR ss : ic : expand equate coefficients pd : solve for k pd : solve for a ss : use transformed function pd : solve trig equation for x a pd : solve for x Primary Method : Give mark for each kcosxcosa + ksin xsina STATED EXPLICITLY k cos a =, ksina = STATED EXPLICITLY k = a = 9 cos( x 9) = x 9 = any one of,,. x = marks marks Note Using kcos( x + a ) etc: candidates may use any form of wave equation to start with, as long as their answer is in the form kcos( x a). If it is not, then is not available. ( ) is OK for k cosxcosa + sin xsina cos x cos a + sin x sin a is OK for Treat kcosxcosa + sin xsina as bad form provided is gained Accept answers which round to.8 for k at For, accept any answer which rounds to 9 Using kcos a =, ksin a =, leads to a =. Only marks, and are available Alternative Method strategy : r / a triangle,, cos x. + sin x. ( cos x.cosa + sin x.sina) and tana = a = 9 marks CONTINUED page

31 Higher Maths Paper Marking Scheme Final (a) Express cos( x )+ sin x ( ) in the form kcos ( x a ) where k > and a (b) Hence solve the equation cos( x )+ sin ( x )= for x a C.. CR / b B.. CR Common wrong solution Early rounding kcosxcosa + ksin xsina STATED EXPLICITLY kcos a =, ksina = STATED EXPLICITLY k = a = kcosxcosa + ksinxsin a STATED EX. kcosa =, ksin a = STATED EX. k = 8. a = 9 cos( x ) = x = anyoneof,. x =. ( this mark not awarded as working eased) so award marks(( ticks) cos( x 9) = x 9 = anyoneof 8., 8,. 8 x =. 8 so award marks(( ticks) (.,) y O (9.,.8) y = cos(x )+sin(x ) (9.,) P (9.,.8) x Alternative Method annotated on diagram 9.,. 8 annotated on diagram via a Graphics Calculator max at ( ) and min at ( 9.,. 8) (., ) or ( 9., ) or ( 9., ) " from the amplitude k = 8. " " from the shift a = 9. " communication : eg solution will be where y = meets the graph annotated on diagram the line withequation y = intersection gives x =. marks marks page

32 Higher Maths Paper Marking Scheme Final The graph of the cubic function y = f() x is shown in the diagram. There are turning points at (, ) and (, ). (, ) Sketch the graph of y = f ( x). y O (, ) y =f (x) x B.. CN /8 ic : ic : interpret stationary points interpret between roots ic : know f ( cubic) = parabola Primary Method : Give mark for each asketchwith the following details only two intercepts on the x - axis at and function is + ve between the roots and ve outwith a parabola (symmetrical about midpoint of x-intercepts), stated or implied by the accuracy of the diagram marks y Note The evidence for may be on a diagram or in a y =f (x) table or in words For, with the intercepts unknown, they must lie on the positive branch of the x-axis O (, ) (, ) x For a parabola passing through (, ) and (, ) award ONLY MARK. page

33 Higher Maths Paper Marking Scheme Final 8 The circle with centre A has equation T y x + y x y + =.The line PT is a tangent to this circle at the point P(, ). O P A x (a) Show that the equation of this tangent is x + y =. The circle with centre B has equation x + y + x + y + =. (b) Show that PT is also a tangent to this circle. (c) Q is the point of contact. Find the length of PQ. T B Q y O P A x 8 a C..,.. CN / b C..8 c C.. ic : ic : ss : interpret circle equation find gradient know/find perpendicular gradient pd : complete proof pd : start solving process ss : know/substitute pd : arrange in standard form 8 ss : 9 ic : know how to justify tangency complete proof ic : interpret solution from (b) pd : process distance formula Primary Method : Give mark for each A(, ) m = AP STATED EXPLICITLY m = PQ y + = ( x ) and complete marks x = y ( y) + y + ( y)+ y + = y y + = 8 solve and get double root tangent 9 ( y ) = marks Q = (, ) PQ = 8 marks Alternative method for to Notes is ONLY AVAILABLE if has been awarded. is only available if an attempt has been made to find a perpendicular gradient completion at : the minimum acceptable would be ( ) y + = x y + = x + y + x = ( y) + y ( y) y + = ( y + ) = double root tangent x = y = ( ) = marks page

34 Higher Maths Paper Marking Scheme Final 8 The circle with centre A has equation T y x + y x y + =.The line PT is a tangent to this circle at the point P(, ). O P A x (a) Show that the equation of this tangent is x + y =. The circle with centre B has equation x + y + x + y + =. (b) Show that PT is also a tangent to this circle. (c) Q is the point of contact. Find the length of PQ. T B Q y O P A x Alternative for 8 and 9 8 use discriminant,and get zero 9 b = ( ) = ac.. tangent Notes cont An = must appear at either the or stage. Failure to appear will forfeit one of these marks. Evidence for (b) may appear in the working for (c) Alternative for (c) ( and ) 8 BP = units, BQ = radius = units 9 by Pythagoras PQ = 8 Alternative Method for (b) ( to 9) ( ) y = x ( x) + ( x x x ( )) + ( )+ ( ( ))+ = x + x + = 8 ( x + ) = 9 double root tangency or b -ac = tangency marks Alternative Method for (b) ( to 9) centre B = (, ) diam : y + = ( x + ) x x + 9= 8 Q = (, ) 9 check : = marks Common error for (b) x = y ( y ) + y + ( y )+ y + = y + y = 8 ( y + )( y ) = 9 intersects in two pts ( y= and y=-) not a tgt marks awarded page

35 Higher Maths Paper Marking Scheme Final 9 An open cuboid measures internally x units by x uits by h units and has an inner surface area of units. (a) Show that the volume, V units, of the cuboid is given by ( ) Vx ()= x x. (b) Find the exact value of x for which this volume is a maximum. x x h 9 a AB.. CN /n b C.. ss : ss : ic : use area facts use volume facts complete proof pd : arrange in standard form pd : differentiate ss : pd : process 8 ic : set derivative to zero justification Primary Method : Give mark for each A= x + xh + xh = V = x x h x V = x = & complete V = x x dv = x dx dv dx = STATED EXPLICITLY x = 8 x < > dv dx + ve ve tgt / \ max marks marks Alternative for, and x + xh + xh = x h = x x V = x x = & complete x marks Notes Do not penalise the non-appearance of at the stage. dx dx = x < maximum may be accepted for 8. page

36 Higher Maths Paper Marking Scheme Final The amount A t micrograms of a certain radioactive substance remaining after t years decreases according to the formula A t = Ae. t, where A is the amount present initially. (a) If micrograms are left after years, how many micrograms were present initially? (b) The half-life of a substance is the time taken for the amount to decrease to half of its initial amount. What is the half-life of this substance? a C.. CR / b AB.. ss : substitute pd : change the subject pd : process exponential power ic : pd : process ss : interpret half life switch to logarithmic form pd : solve logarithmic equation Primary Method : Give mark for each. = Ae A =. e. t A = Ae. t. = e. t = ln. t = years marks marks Alternative method for (a) = Ae. lna = ln lne A =. marks Notes Accept any correct answer which rounds to. For any other answer, rounding must be indicated. A trial and improvement approach : For = Ae award For eg e = e = 9 leading to an answer which rounds to, award At, A may be replaced by any real number For (b) an answer obtained by trial and improvement which rounds to or may be awarded mark. page

37 Higher Maths Paper Marking Scheme Final An architectural feature of a building is a wall with arched windows. The curved edge of each window is parabolic. The second diagram shows one such window. The shaded part represents the glass. The top edge of the window is part of the parabola with equation y = x x. y m m Find the area in square metres of the glass in one window. m 8 O x 8 A..,..9 CN / ss : find intersections pd : process quadratic to solution ss : ss : ic : decide on appropriate areas know to integrate state limits pd : integrate pd : evaluate using limits 8 pd : evaluate area Primary Method : Give mark for each x x = x =, x = " split area up " ( ) x x dx stated or implied by... dx x x x (.. ).. 8 ( ) 8 marks Notes The first two marks may be obtained as follows: Guess x = and check that y =., award Guess x = and check that y =., award In the Primary method, is clearly not available for subtracting the wrong way round. 8 will also be lost for statements such as = so ignore the negative = squ units 8 can still be gained for statements such as and so the area = Alternative Method x x = x =, x = ( ) x x dx choose limits, a and b: a b x x b evaluate x x a for chosen values of a and b state areas to be added/subtracted 8 st / imp by 8 8 marks page 8

38 Higher Maths Paper Marking Scheme Final S A die has three red faces, two blue faces and one yellow face. An experiment consists of noting the uppermost colour after a roll of the die. Random Numbers (a) Use the given random numbers to simulate 8 trials of the experiment. Explain your strategy. (b) How closely do the results of your simulation agree with the theoretical probability of obtaining blue? replacing qu. S a C. CR / b C. ic : define simulation pd : process simulation pd : find probability Primary Method : Give mark for each define simulation results of simulation P( blue ) = theoretical cf simulation P( blue ) exp erimental with marks ic : comment page 9

39 Higher Maths Paper Marking Scheme Final S As part of a study on intensive exercise, a sports scientist recorded the peak heart rates of a random selection of sixteen volunteers of different ages who took regular exercise. The linear regression equation was calculated for the data shown in the scatter diagram and found to be y = 9 x. However after considering the scatter diagram for the data, it was realised that one piece of data had been misrecorded and this volunteer's data was ignored. (a) State the approximate age of the volunteer whose data was ignored. (b) Calculate the new regression equation using the values (c) Σx = 9, Σx = 8, Σy = 8, Σy = 9, Σxy = 9 9. Comment on the difference this makes to the prediction for the average peak heart rate of a year old volunteer. replacing qu. Peak heart rate y (beats per minute) 8 Age x (years) S a B.. CR / b B.. c A.. ic : ic : estimate from graph state n pd : process pd : process pd : determine regression coefficients pd : determine regression coefficients ic: state regression equation 8 pd : use regression equation 9 ic : interpret results Primary Method : Give mark for each 8 n = S = 9 xx S = xy a = b = y = x 8 est old =, est new = 9 removing outlier improves estimate mark marks marks page

40 Higher Maths Paper Marking Scheme Final S The selection procedure for a Police force consists of independent tests, Intelligence(I), Fitness (F) and Communication(C). The outcome of each test is an independent event and is either pass or fail. A candidate must pass all three tests to enter training. It has been established that the probability of failing each test is as follows: Test I F C P(failing) (a) Calculate the probability that a candidate will be selected for training. (b) Five candidates are being tested for selection. Find the probability that (i) all five candidates will be accepted (ii) all five candidates will be rejected. replacing qu. S a B.. CN / b B.. ss : pd : process use approp. strategy P(Pass) or P(fail) pd : process all pass pd : process one fail pd : process all fail Primary Method : Give mark for each Pselected ( ) = 8 8 or = P( not selected) =.. = 8 marks marks page

41 Higher Maths Paper Marking Scheme Final S Show that the diagram represents a probabilty density function for a continuous random variable X. replacing qu. y O x S A.. CN /9 ic : ic : state requirement for pdf state requirement for pdf pd : complete proof Primary Method : Give mark for each function above x axis total area must be + + = + + = marks page