1 Triangle ABC has vertices A( 1,12), B( 2, 5)

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1 Higher Mathematics Paper : Marking Scheme Version Triangle ABC has vertices A(,), B(, ) A(, ) y and C(, ). (a) (b) (c) Find the equation of the median BD. Find the equation of the altitude AE. Find the coordinates of the point of intersection of BD and AE. O D C(, ) a,b,c,, C G, G CN / B(, ) E ic interpret median ss find gradient ic state equation ss find gradient ss find perpendicular gradient ic state equation ss start to solve simultaneous equations pr solve for one variable pr process D = (, ) m = BD y = ( ) or y + = ( ) etc m = BC m = alt y = ( ( )) y = ( ) and y = ( ( )) = y = stated eplicitly or equivalent marks marks marks For candidates who find two medians,, and,, are available. For candidates who find two altitudes,, and,, are available. For candidates who find (a) altitude and (b) median see common error bo number. In (a) note that (, ) happens to lie on the median but does not qualify as a point to be used in. cont In (b) is only available as a consequence of attempting to find a perpendicular gradient. In (b) candidates who guess the coordinates for E and use these to find the equation AE, can earn no marks in this part. In (c) note that equating zeros is only a valid strategy when either the coefficients of or the coefficients of y are equal. is a strategy mark for jutaposing the two required equations. See general note at the foot of page. Common Error Finding two medians D = (, ) m = BD y = ( ) y = & + y = = y = maimum of marks Common Error Finding two altitudes m = BC m = alt y = ( ( )) y = & y = + = y = maimum of marks Common Error Finding (a) altitude and (b) median m = AC m = BD y = ( ) midpt of BC =, m = AC y = ( ( )) y = & + y = = y = maimum of marks

2 Higher Mathematics Paper : Marking Scheme Version A circle has centre C(, ) and passes through P(, ). y (a) (b) Find the equation of the circle. PQ is a diameter of the circle. Find the equation of the tangent to this circle at Q. C (, ) P (, ) a C G CN / b C G CN Q O ic enter coord. of centre in general equation ss find (radius) ss e.g. use PC = CQ to find Q pr find gradient of diameter ss know and use tangent perp. to diameter ic state equation ( a) + ( y b) = r ( ) + ( y ) r = Q = (, ) m = diameter m = tangent y = ( ) stated or implied by marks marks In (a) ( ) is not acceptable for. In (b) if the coordinates of Q are estimated (i.e. guessed) then can only be awarded if the coordinates are of the form (a, ) where a <. Alternative Method for (a) For answers of the form + y + g + fy + c = + y + y + c = c = In (b) is only available if an attempt has been made to find a perpendicular gradient. General applicable throughout the marking scheme There are many instances when follow throughs come into play and these will not always be highlighted for you. The following eample is a reminder of what you have to look out for when you are marking. eample At the stage a candidate start with the wrong coordinates for Q. Then Q = (, ) m = diameter m = tangent y = ( ) so the candidate loses but gains, and as a consequence of following through. Any error can be followed through and the subsequent marks awarded provided the working has not been eased. Any deviation from this will be noted in the marking scheme.

3 Higher Mathematics Paper : Marking Scheme Version Two functions f and g are defined on the set of real numbers by f () = + and g () =. (a) Find an epressions for (i) f g() ( ()). (ii) g f. (b) Determine the least possible value of f g() g f() a C A CN / b C A CN ic int. composition ic int. composition ic int. composition pr simplify all functions ic int. result f( g )= f( ) ( ) + g( f )= ( +) min.value = stated or implied by stated eplicitly marks marks In (a) marks are available for finding one of and the third mark is for f g() or g f() the other one. In (a) the finding of f f() and g g() earns no marks. is only available if has been awarded. In (b) for, no justification is necessary. Ignore any comments, rational or irrational. Alternative Marking [Marks -] g( f )= g( + ) ( + ) f( g )= ( )+ Common Error No. for (a) g and f transposed. f( g )= f( + ) ( + ) g( f )= ( ) + Award out of Common Error No. for (a) f( g )= f( + ) ( + ) g( f )= ( + ) Award out of Common Error No. for (a) Repeated error f( g )= f( ) ( + ) g( f( ))= ( ) + Award out of

4 Higher Mathematics Paper : Marking Scheme Version A sequence is defined by the recurrence relation u. u, u. n+ n (a) State why the recurrence relation has a limit. (b) Find this limit. a C A NC / b C A NC sequence has limit since <. < mark ic state limit condition ss know how to find L pr process limt L =. L+ limit = marks For (a) Accept. < Alternative Method for (b) L =. limit = <. <. lies between and. is a proper fraction Bad Form L =. limit = Do NOT accept award marks. < a <. < unless a is clearly identifed/replaced by. anywhere in the answer. Common Error L =. limit = In (b) L b = and nothing else gains no marks. a L = or or. etc does NOT gain. An answer of without any working gains NO marks. Any calculations based on wrong formulae gain NO marks.

5 Higher Mathematics Paper : Marking Scheme Version A function f is defined by f ()=.Find the coordinates of the stationary point on the graph with equation y = f() and determine its nature. C C, C NC / B ss know to start to differentiate pr differentiate ss set derivative = pr solve pr evaluate ic justification ic state conclusion f =... ( ) f = = f( ) = nature table pt of infleion at (,) marks The = shown at must appear at least once somewhere in the working between and (but not necessarily at ). is only available as a consequence of solving f () =. A wrong derivative which eases the working will preclude at least from being awarded. Common Error No. f =... ( ) f = =, and are still available For marks and, a nature table is mandatory. The minimum amount of detail that is required is shown here: < > f () + + Candidates who use only f () = and try to draw conclusions from this cannot gain or. [ f () = is a necessary but not sufficient condition for identifying points of infleion]. is ONLY available subsequent to a correct nature table for the candidate s own derivative. Common Error No. f =... ( ) f = =, and are still available is lost in each of the following cases for the candidate s solution to the equation at. (i) (ii) (iii) = and = something else two wrong values for guess a value for Only one value for needs to be followed through for, and.

6 Higher Mathematics Paper : Marking Scheme Version The graph shown has equation y = + +. y The shaded area is bounded by the curve, the -ais, the y-ais and the line =. S O (,) (,) (a) Calculate the shaded area labelled S. (b) Hence find the total shaded area. a C C NC / b B C NC ss know to integrate pr integrate ic substitute limits pr evaluate ic use result from with new limits pr evaluate ss deal with the ve sign and evaluate total area for (a) Only a limited number of marks are available to candidates who differentiate see Common Error No.. In (a) candidates who transpose the limits can still earn if the deal with the ve sign appropriately. In (b) is lost for such statements as =. + + d + + ( ) or equivalent... d Alternative Method for (b) stated or implied by.. ( +. + ) ( )= or equivalent... d... ( ) +. + In (b) using...d earns no marks. Common Error No. + + d + (.. + )... d or equivalent (.. + ) (.. + )= Alternative Method for (b) ( ) Alternative Method for (b)... d... ( ) +. +

7 Higher Mathematics Paper : Marking Scheme Version Solve the equation sin sin = in the interval. C T NC / ss know to use double angle formula pr factorise pr solve ic know eact values An = must appear somewhere between the start and evidence. The inclusion of etra answers which would have been correct with a larger interval should be treated as bad form and NOT penalised. The omission of a correct answer (e.g. ) means the candidates loses a mark ( in the Primary Method). Candidates may embark on a journey with the wrong formula for sin( ). With an equivalent level of difficulty it may still be worth a maimum of marks. See Common Error No.. Candidates who draw a sketch of y = sin( ) and y = sin( ) giving,, may be awarded and. sin( ) sin( )cos( = ) sin( )( cos( ))= sin( ) = or cos( ) =. =,,,, Alternative Marking Method (Cross marking for and ) sin( ) sin( )cos( = ) sin( )( cos( ))= sin( ) = and =,, cos( ) =. and =, Alternative Method Division by sin() sin( ) sin( )cos( = ) either sin( ) = or sin( ) sin( ) = =,, cos( ) =. =, Common Error No. sin( ) ( sin ( ))= sin ( + ) sin( ) = ( sin( ) ) ( sin( + ) )= sin( ) = or sin( ) = =,, = award marks Common Error No. sin( ) sin ( = ) sin( ) sin( ) = sin( ) = or sin( ) = =,,, award marks Common Error No. sin sin = sin =, sin = etc gainsnomarks

8 Higher Mathematics Paper : Marking Scheme Version (a) Epress + in the form a( + b) + c. (b) Write down the coordinates of the turning point on the parabola with equation. y = + a B A NC / b C A NC ss know how to complete (deal with the a ) pr process the value of b pr process the value of c ic interpret equation of parabola a = b = c = (, ) Note Alternative Method should be used for assessing part marks/follow throughs. For, no justification is required. Candidates may choose to differentiate etc. but may still earn only one mark for the correct answer. For, accept ( b, c). Alternative Method for (a) ( + ) ( + ) ( + ) (, ) Alternative Method for (a) : Comparing coefficients + = a + ab + ab + c a = ab = ab + c = b = c = (, )

9 Higher Mathematics Paper : Marking Scheme Version k u and v are vectors given by u = and v = k, where k >. k + (a) If u.v = show that k + k k =. (b) Show that (k + ) is a factor of k + k k and hence factorise k + k k fully. (c) Deduce the only possible value of k. (d) The angle between u and v is θ. Find the eact value of cos θ. θ u v marks marks mark marks a C G CN / b C A NC c C A CN d C G NC pr find scalar product ic complete proof ss know to use k = pr complete evaluation and conclusion ic start to find quadratic factor ic complete quadratic factor pr factorise completely ic interpret k ic interpret vectors pr find magnitudes ss use formula No eplanation is required for k but the chosen value must follow from the working for or. Do not accept. In primary method and alternative ( ) candidates must show some acknowledgement of the resulting zero. Although a statement w.r.t. the zero is preferable, accept something as simple as underlining the zero. Only numerical values are acceptable for, and ; answers stated or implied by u.v = k. +. ( k )+ k +. before completion k + k k = and complete marks know to use k = + ( ) = + is a factor ( k + )( k...) ( k + ) k ( k + )( k + )( k ) k = u =,v = u = and v = cosθ = N.B. and may be cross-marked. stated eplicitly stated or implied by marks mark marks are acceptable in unsimplified form eg cosθ = Alternative method (marks ) Long Division k + k + k k k + k k k k remainder iszeroso( k+ ) isa factor k ( k + )( k + )( k ) stated eplicitly Alternative method (marks ) Synthetic Division " f( )" = so ( k + ) is a factor k ( k + )( k + )( k ) stated eplicitly

10 Two variables, and y, are connected by the law y = a. A graph of Higher Mathematics Paper : Marking Scheme Version log against is a straight line passing through the origin and the point A(,). Find the value of a. A(, ) log y A A NC / O ss know to take logarithms ic substitute known point pr solve pr solve log ( y) = log ( a ) = log ( a ) a = a = marks Note m = and nothing else gains no marks. Alternative Method For, a correct answer without any legitimate evidence gains NO marks. For, ignore the inclusion of a negative answer. log ( y) = log ( a ) = log ( a) log ( a) = a = Alternative Method log ( y) = m + c m =, c = y = y = = a = Common Error log ( y) = log ( a ) log = log ( a ) = a a = Alternative Method At A log ( y) = y = a = a = Common Error log ( y) = y = a = Alternative Method log ( y) = log ( a ) log ( y) = log ( a) log ( a) = a = =

11 Higher Mathematics Paper : Marking Scheme Version PQRS is a parallelogram. P is the point (, ), S is (, ) y and Q lies on the -ais, as shown. The diagonal QS is perpendicular to the side PS. (a) Show that the equation of QS is + y =. (b) Hence find the coordinates of Q and R. O P(, ) S(, ) Q R a,b, C G CN / pr find gradient from two points ss use m m = ic state equation of the line ic completes proof ic interpret diagram ic interpret diagram m PS = m = QS y = ( ) completes proof Q = (, ) R = (, ) marks marks In (a) In the Primary method, is only available if an attempt has been made to find and use a perpendicular gradient. In the Primary method and the Alt. method, is only available for reaching the required equation. To gain, some evidence of completion needs to be shown e.g. y = ( ) ( y ) = ( ) + y = Alternative Method m = m = = + c completes proof PS QS y = + c Q = (, ) R = (, ) Sometimes candidates manage to find R first. Provided the coordinates of R are of the form (?, ), only then is available as a follow through. Alternative Method and are available to candidates who use their own erroneous equation for QS. General applicable throughout the marking scheme There are many instances when follow throughs come into play and these will not always be highlighted for you. The following eample is a reminder of what you have to look out for when you are marking. eample At the stage a candidate may switch the coordinates round so we have Q(, ) R(, ) repeated error so the candidate loses for switching the coordinates but gains as a consequence of following through. Any error can be followed through and the subsequent marks awarded provided the working has not been eased. Any deviation from this will be noted in the marking scheme. Let Q = (, q ) ( q ) = + + ( q ) + q = Q = (, ) and R = (, ) = QS m y = ( ) leading to y + = N.B. The coordinates of Q can also be arrived at by right-angled trig. Use the alt. method marking scheme with replaced by appropriate trig. work. The only acceptable value for q is.

12 Higher Mathematics Paper : Marking Scheme Version Find the value of k such that the equation k + k + =, k, has equal roots. C A CN /new ss know to use discriminant = ic interpret a,b,c pr substitute & factorise ic interpret solution " b ac" = a= kb, = kc, = kk ( ) k = and k = k = marks Alternative Method (completing the square) The evidence for and/or may not appear until the working immediately preceding the evidence for. i.e. a candidate may simply start, k k = kk ( ) ( ) + ( ) + = k + + equal roots + = k = k or The = has to appear at least once, at the stage or at the stage. k k, kk ( ) = In the Primary method, candidates who do not deal with the root k = cannot obtain. [see Common Errors and ] Minimum evidence for would be scoring out k = or k = underlined. Some candidates may start with the quadratic formula. Apply the marking scheme to the part underneath the square root sign. The use of any epression masquerading as the discriminant can only gain at most. Acceptable alternative for " b ac" = a= kb, = kc, = kk ( ) k = or Common Error at the stage " b ac" = a= kb, = kc, = kk ( ) k = or Common Error at the stage " b ac" = a= kb, = kc, = kk ( ) k = Common Error Division by k " b ac" = a= kb, = kc, = k k = k = k k =

13 Higher Mathematics Paper : Marking Scheme Version The parabola with equation y = + has a tangent at the point P(,). (a) Find the equation of this tangent. y O P(,) y (b) Show that the tangent found in (a) is also a tangent to the parabola with equation y = + and find the coordinates of the point of contact Q. O Q P(,) a C C CN / b C A CN ss know to differentiate pr differentiate pr evaluate gradient ic state equation of tangent ss arrange in standard form ss substitute into quadratic pr process ic factorise & interpret ic state coordinates dy = d m = y = ( ) y = = + + = stated or implied by ( ) = equal roots so tgt Q= (, ) marks marks In (a) is only available if an attempt has been made to find the gradient from differentiation. In (b) is only available for a numerical value of m. An = must occur somewhere in the working between and. is awarded for drawing a conclusion from the candidate s quadratic equation. Candidates may substitute the equation of the parabola into the equation of the line. This is a perfectly acceptable approach. Common Error dy = d = so = so m = y = ( ) y = = + = b ac = line is not tgt so award marks Alternative Marking [Marks ] b ac = = line is a tangent Alternative Method for (b) = y + y = ( y + y + ) + y + y + + = ( y + ) = equal roots so tgt Q= (, ) Alternative Method for (b) Find the equ. of the tgt to nd curve with grad. + = Q = (, ) y = ( ) y = which is the same equ. as ( a) stated or implied by stated eplicitly

14 Higher Mathematics Paper : Marking Scheme Version The circles with equations ( ) + ( y ) = and + y k y k =. have the same centre. Determine the radius of the larger circle. C G CN / ic state centre of circle ss equate -coordinates, find k. ic find radius of circle ic substitute into the radius formula ic process radius formula and compare. C, k = R R = ( ) + ( ) ( ) = = > or " nd circle " or equivalent marks requires no justification. Evidence for may appear for the first time at the stage. If R = is clearly stated at the stage, then it does not have to appear at the stage for the conclusion to be drawn. For any formula masquerading as the radius formula (e.g. see Common Error ), and are NOT available. Alternative Method + y y + = k = R R = ( ) + ( ) = > or " nd circle " or equivalent Common Error C =(, ) k = R = R = ( ) + ( ) < or " st circle " Common Error C =(, ) k = R = R = ( ) + ( ) + > or " nd circle "

15 Higher Mathematics Paper : Marking Scheme Version The curvey = f() is such that dy =. The curve passes through the d point (, ). Epress y in terms of. C/B C CN / ss know to integrate pr integrate ic substitute values pr process constant y =... = ( ) ( ) + c y = + stated or implied by stated eplicitly marks The equation y = must appear somewhere in the solution. Common Error Missing equation y =... d = ( ) ( ) + c c = award marks Common Error : Not using (, ) y =... d ( ) ( ) + c = y = award marks Alternative Marking y =... y = + c and = ( ) ( ) + c c = stated eplicitly

16 Higher Mathematics Paper : Marking Scheme Version P is the point (,, ) and Q is (,, ). (a) Write down PQ in component form. (b) Calculate the length of PQ. (c) Find the components of a unit vector which is parallel to PQ. a C G CN / b C G c B G ic state vector components pr find the length of a vector ic state unit vector PQ = PQ = mark mark mark Note In (a) It is perfectly acceptable to write the components as a row vector eg PQ Treat PQ = ( ). = (,, ) as bad form (i.e. not penalised). In (b) is not awarded for an unsimplified. Beware of misappropriate use of the scalar product In (c) where, by coincidence, p.q =. Accept for.

17 Higher Mathematics Paper : Marking Scheme Version The diagram shows the graph of a function y = f(). Copy the diagram and on it sketch the graphs of () a y = f( ) () b y = + f( ) Q(,) y O P(, a) y = f() a C A CN /new b C A ic know translate parallel to -ais, +ve dir. ic annotate points ic know translate parallel to y-ais, +ve dir. ic annotate points translate units right and annotate one point annotate the other point [ P'(, a) Q'(, )] translate (a) units up and annotate one point marks annotate the other point [P"(, a + ) Q"(, )] marks For (a) A translation of earns a maimum of mark with both points clearly annotated and f() retaining its shape. Any other translation gains no marks. Alternative Method translate units right and annotate one point annotate the other point [ P'(, a) Q'(, )] translate original and annotate one point annotate the other point [P"(, a + ) Q"(, )] In the Primary method For (b) and are only available for applying the translation to the resultant graph from (a). A translation of earns a maimum of mark with both points clearly annotated and the resultant graph from (a) retaining its shape. Any other translation gains no marks. In the Alternative method For (b) A translation of, or applied to the original graph earns a maimum of mark with both points clearly annotated and the resultant graph retaining its original shape. Any other translation gains no marks. In either method For (a) and (b) For the annotated points, accept a superimposed grid or clearly labelled aes. A candidate may choose to use two separate diagrams. This is acceptable.

18 Higher Mathematics Paper : Marking Scheme Version The diagram shows a right-angled triangle with height unit, base units and an angle of a at A. () a Find the eact values of () i sina A ( ii) sina. ( b) By epressing sin a as sin( a + a), find the eact value of sin a. a a C T CN / b B T CN Note ic interpret diagram for sin(a ) ss use double angle formula for sin(a) ic interpret diagram for cos(a ) pr substitute and complete ss use compound angle formula pr use double angle formula for cos(a) ic substitute pr complete Calculating approimate angles using arcsin and arccos gains no credit. sin( a ) = sin( a ) = sin( a )cos( a ) cos( a ) = sin( a ) = marks sin( a ) = sin( a )cos( a + ) cos( a )sin( a ) cos( a ) = sin( a ) =. +. sin( a ) = marks There are processing marks, and. None of these are available for an answer >. sin(a) =. and cos(a) =. are the only two decimal fractions which may receive any credit. Some candidates may double the height of the triangle and then call the base angle a. This error is equivalent to Common Error illustrated on the right. Common Error An eample based on a numerical error in Pythagoras sin( a ) = sin( a ) = sin( a )cos( a ) cos( a ) = sin( a ) = sin( a ) = sin( a )cos( a ) + cos( a )sin( a ) cos( a ) = cos ( a ) = or equivalent sin( a ) =. +. sin( a ) = Common Error An eample of Incorrect formulae sin( a ) = sin( a ) = sin( a ) sin( a ) = sin( a ) = sin( a )cos( a ) + cos( a )sin( a ) cos( a ) = cos( a ) = sin( a ) =. +. sin( a ) =

19 Higher Mathematics Paper : Marking Scheme Version dy y = cos,, find. d C/B C,C CN / ss epress in differentiable form pr differentiate a term with a negative power pr start to process a compound derivative pr complete compound derivative + sin marks For clearly integrating, correctly or otherwise, only is available. If you cannot decide whether a candidate has attempted to differentiate or integrate, assume they have attempted to differentiate.

20 Higher Mathematics Paper : Marking Scheme Version A curve has equation y = sin cos. π (a) Epress sin cos in the form ksin( a) where k > and a. (b) Hence find, in the interval π, the -coordinate of the point on the curve where the gradient is. a C T CR / b A/B T CR ss epand ic compare coefficients pr process k pr process a ic state result ss set derivative = gradient pr process from the derivative ksincos( a) kcossin( a) kcos( a) =, ksin( a) = k = a =. sin(. ) dy = cos(. ) = d =. stated eplicitly stated eplicitly marks marks In (a) is acceptable for. k sincos( a) cossin( a) Treat ksincos( a) cossin( a) as bad form if is gained. No justification is required for. is not available for an unsimplified. is acceptable evidence for sincos( a) cossin( a) and. Candidates may use any form of the wave equation to start with as long as their final answer is in the form ksin( a). If it is not, then is not available. Common Error Working in degrees ( sincos( a) cossin( a) ) kcos( a) =, ksin( a) = k = a =. sin(. ) dy = cos(. ) = d =. is only available for (i) an answer in radians which rounds to. OR (ii) an answer given as a multiple of π eg.. π. Award () a marks and () b marks kcos( a) = and ksin( a) = leading to a =. can only gain if a comment intimating that this answer is not in the given interval is given. In (b) In (b) candidates have a choice of two starting points. They can either start from y = sin(. ) as shown in the Primary method they can start from dy d OR = cos + sin. Either of these starting positions may be awarded. Candidates who work in degrees will lose for attempting to differentiate. is only available as a consequence of solving dy d =. Do not penalise etra solutions at the stage (e.g..).

21 It is claimed that a wheel is made from wood which is over years old. To test this claim, carbon dating is used. Higher Mathematics Paper : Marking Scheme Version The formula At () = Ae. t is used to determine the age of the wood, where A is the amount of carbon in any living tree, At ()is the amount of carbon in the wood being dated and t is the age of the wood in years For the wheel it was found that At () was % of the amount of carbon in a living tree. Is the claim true? A/B A CR / ic interpret information ic substitute ss take logarithms pr process ic interpret result At =. A. t e =.. t ln( e )= ln (. ). t = ln(. ) t = years so claim valid stated or implied by stated or implied by marks Candidates may choose a numerical value for A at the start of their solution. Accept this situation. is only available if has been awarded. In following through from an error, is only available for a positive value of t. Alternative Method Graph and Calculator Solution A( ) = Ae.. A and year old piece of wood try a point where t > contains.% carbon. eg.. A ( ) getting. A. sketch of y= Ae t showing. a monotonic decreasing function. points representing eg (,.%) etc observation that the point lies between the two plotted values for t and so claim valid.

22 Higher Mathematics Paper : Marking Scheme Version PQRS is a rectangle formed according to the following conditions: it is bounded by the lines = and y = y = y (, ) Q R (, ) P lies on the curve with equation y R is the point (, ). = between (, ) and (, ) (a) (i) Epress the lengths of PS and RS in terms of, the -coordinate of P. (b) (ii) Hence show that the area, A square units, of PQRS is given by A =. Find the greatest and least possible values of A and the corresponding values of for which they occur. O P, (, ) S = y = a A C CN / b A/B C ic interpret diagram to find PS ic interpret diagram to find RS ic complete proof ic epress in differentiable form ss know to set derivative to zero pr differentiate pr process equation pr evaluate area at the turning point pr evaluate area at the end point pr evaluate area at the end point ic state conclusion PS = RS = Area = da = d + = A() = A = A( ) = ma. A= at = and min. A = at = or = and complete marks marks For there needs to be clear evidence that candidates have multiplied out the brackets in order to complete the proof. An = must appear somewhere in the working between and. At the stage, ignore the omission or inclusion of =. has to be as a consequence of solving da d =. is only available if both end points have been considered.