H SPECIMEN S87/76/ National Quali cations ONLY Mathematics Paper Date Not applicable Duration hour 5 minutes Total marks 80 Attempt ALL questions. You may use a calculator. To earn full marks you must show your working in your answers. State the units for your answer where appropriate. You will not earn marks for answers obtained by readings from scale drawings. Write your answers clearly in the spaces provided in the answer booklet. The size of the space provided for an answer is not an indication of how much to write. You do not need to use all the space. Additional space for answers is provided at the end of the answer booklet. If you use this space you must clearly identify the question number you are attempting. Use blue or black ink. Before leaving the eamination room you must give your answer booklet to the Invigilator; if you do not, you may lose all the marks for this paper. *S8776*
FORMULAE LIST Circle: The equation + y + g + fy + c = 0 represents a circle centre ( g, f ) and radius g + f c. The equation ( a) + ( y b) = r represents a circle centre (a, b) and radius r. Scalar product: a.b = a b cos θ, where θ is the angle between a and b or a b a.b = ab + ab + ab where a = a and b = b. a b Trigonometric formulae: sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B sin A sin B sin A = sin A cos A cos A = cos A sin A = cos A = sin A Table of standard derivatives: f( ) f ( ) sin a cos a a cos a asin a Table of standard integrals: f( ) f ( ) d sin a cos a cos a + c a sin a a + c page 0
Attempt ALL questions Total marks 80 MARKS. The vertices of triangle ABC are A( 5, 7), B(, 5) and C(, ) as shown in the diagram. The broken line represents the altitude from C. y A C B (a) Find the equation of the altitude from C. (b) Find the equation of the median from B. (c) Find the coordinates of the point of intersection of the altitude from C and the median from B.. Find + d, 0. page 0
. The diagram shows the curve with equation y f ( ) f ( ) = k( + a)( + b). =, where The curve passes through (, 0), (0, 0), (, ) and (, 0). MARKS f ( ) y (, ) O Find the values of a, b and k.. D,OABC is a square-based pyramid as shown. z D y C B O M A O is the origin and OA = units. M is the mid-point of OA. OD = i + j + 6k (a) Epress DB and DM in component form. (b) Find the size of angle BDM. 5 page 0
MARKS 5. The line with equation y = + is a tangent to the curve with equation y = + + + at A(0, ), as shown. y y = + + + A(0, ) O B(, ) y = + The line meets the curve again at B(, ). Find the area enclosed by the line and the curve. 5 6. (a) Epress 50 + + in the form ( ) a + b + c.. (b) Given that f ( ) = + + 50, find f ( ) (c) Hence, or otherwise, eplain why the curve with equation y f ( ) increasing for all values of. = is strictly page 05
7. The diagram below shows the graph of a quartic y h( ) (0, 5) and (, ). =, with stationary points at MARKS y 5 O (, ) y = h( ) On separate diagrams sketch the graphs of: (a) y h( ) =. (b) y h ( ) =. 8. (a) Epress 5cos sin in the form kcos ( a) +, where k > 0 and 0< a < π. (b) The diagram shows a sketch of part of the graph of y = 0 + 5cos sin and the line with equation y =. The line cuts the curve at the points P and Q. y P Q y = 0 + 5cos sin y = O Find the -coordinates of P and Q. page 06
9. A design for a new grain container is in the shape of a cylinder with a hemispherical roof and a flat circular base. The radius of the cylinder is r metres, and the height is h metres. The volume of the cylindrical part of the container needs to be 00 cubic metres. MARKS r m h m (a) Given that the curved surface area of a hemisphere of radius r is πr show that the surface area of metal needed to build the grain container is given by: A 00 r = + πr square metres (b) Determine the value of r which minimises the amount of metal needed to build the container. 6 0. Given that calculate the value of a. a π π sin d =, 0 a <, π 8 6 page 07
. Show that sin sin cos sin cos =, where 0 π < <. MARKS. (a) Show that the points A( 7, ), B(, ) and C(7, 6) are collinear. Three circles with centres A, B and C are drawn inside a circle with centre D as shown. A B D C The circles with centres A, B and C have radii r A, r B and r C respectively. r A = 0 r = r B A rc = ra + rb (b) Determine the equation of the circle with centre D. page 08
. The concentration of a pesticide in soil can be modelled by the equation MARKS P t = P 0 e kt where: P 0 is the initial concentration; P t is the concentration at time t ; t is the time, in days, after the application of the pesticide. (a) It takes 5 days for the concentration of the pesticide to be reduced to one half of its initial concentration. Calculate the value of k. (b) Eighty days after the initial application, what is the percentage decrease in concentration of the pesticide? [END OF SPECIMEN QUESTION PAPER] page 09
H SPECIMEN S87/76/ National Quali cations ONLY Mathematics Paper Marking Instructions These marking instructions have been provided to show how SQA would mark this specimen question paper. The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is reproduced, SQA should be clearly acknowledged as the source. If it is to be used for any other purpose, written permission must be obtained from permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (ie secondary copyright), this material should only be reproduced for the purposes of eamination or assessment. If it needs to be reproduced for any other purpose it is the user s responsibility to obtain the necessary copyright clearance.
General marking principles for Higher Mathematics Always apply these general principles. Use them in conjunction with the detailed marking instructions, which identify the key features required in candidates responses. For each question, the marking instructions are generally in two sections: generic scheme this indicates why each mark is awarded illustrative scheme this covers methods which are commonly seen throughout the marking In general, you should use the illustrative scheme. Only use the generic scheme where a candidate has used a method not covered in the illustrative scheme. (a) (b) (c) (d) (e) (f) (g) Always use positive marking. This means candidates accumulate marks for the demonstration of relevant skills, knowledge and understanding; marks are not deducted for errors or omissions. If a candidate response does not seem to be covered by either the principles or detailed marking instructions, and you are uncertain how to assess it, you must seek guidance from your team leader. One mark is available for each. There are no half marks. If a candidate s response contains an error, all working subsequent to this error must still be marked. Only award marks if the level of difficulty in their working is similar to the level of difficulty in the illustrative scheme. Only award full marks where the solution contains appropriate working. A correct answer with no working receives no mark, unless specifically mentioned in the marking instructions. Candidates may use any mathematically correct method to answer questions, ecept in cases where a particular method is specified or ecluded. If an error is trivial, casual or insignificant, for eample 6 6 =, candidates lose the opportunity to gain a mark, ecept for instances such as the second eample in point (h) below. page 0
(h) If a candidate makes a transcription error (question paper to script or within script), they lose the opportunity to gain the net process mark, for eample This is a transcription error and so the mark is not awarded. This is no longer a solution of a quadratic equation, so the mark is not awarded. + 5 + 7= 9 + + = 0 = The following eample is an eception to the above This error is not treated as a transcription error, as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt and all marks awarded. + 5 + 7 = 9 + + = 0 ( )( ) = 0 = or (i) Horizontal/vertical marking If a question results in two pairs of solutions, apply the following technique, but only if indicated in the detailed marking instructions for the question. Eample: 5 6 5 6 = = y = 5 y = 7 Horizontal: 5 = and = Vertical: 5 = and y = 5 6 y = 5 and y = 7 6 = and y = 7 You must choose whichever method benefits the candidate, not a combination of both. (j) In final answers, candidates should simplify numerical values as far as possible unless specifically mentioned in the detailed marking instruction. For eample 5 must be simplified to 5 or must be simplified to 5 0 must be simplified to 50 5 6 must be simplified to 8* must be simplified to 5 *The square root of perfect squares up to and including 00 must be known. page 0
(k) Do not penalise candidates for any of the following, unless specifically mentioned in the detailed marking instructions: working subsequent to a correct answer correct working in the wrong part of a question legitimate variations in numerical answers/algebraic epressions, for eample angles in degrees rounded to nearest degree omission of units bad form (bad form only becomes bad form if subsequent working is correct), for eample ( + + + )( + ) written as ( + + + ) + = + 5 + 8 + 7+ gains full credit repeated error within a question, but not between questions or papers (l) In any Show that question, where candidates have to arrive at a required result, the last mark is not awarded as a follow-through from a previous error, unless specified in the detailed marking instructions. (m) You must check all working carefully, even where a fundamental misunderstanding is apparent early in a candidate s response. You may still be able to award marks later in the question so you must refer continually to the marking instructions. The appearance of the correct answer does not necessarily indicate that you can award all the available marks to a candidate. (n) (o) You should mark legible scored-out working that has not been replaced. However, if the scored-out working has been replaced, you must only mark the replacement working. If candidates make multiple attempts using the same strategy and do not identify their final answer, mark all attempts and award the lowest mark. If candidates try different valid strategies, apply the above rule to attempts within each strategy and then award the highest mark. For eample: Strategy attempt is worth marks. Strategy attempt is worth mark. Strategy attempt is worth marks. Strategy attempt is worth 5 marks. From the attempts using strategy, the resultant mark would be. From the attempts using strategy, the resultant mark would be. In this case, award marks. page 0
Marking instructions for each question Question Generic scheme Illustrative scheme. (a) calculate gradient of AB m AB = Ma mark use property of perpendicular lines m alt = determine equation of altitude y=. (b) calculate midpoint of AC ( 5, ) 5 calculate gradient of median 5 m BM = 6 determine equation of median 6 y=. (c) 7 find or y coordinate 7 = or y = 8 find remaining coordinate 8 y= or =. write in integrable form + integrate one term eg... + integrate other term... complete integration and simplify + c. value of a value of b calculate k page 05
Question Generic scheme Illustrative scheme. (a) state components of DB 6 Ma mark state coordinates of M ( 00,, ) stated or implied by state components of DM 0 6. (b) evaluate DBDM. 5 evaluate DB 5 5 6 evaluate DM 6 0 7 use scalar product 7 cosbdm = 0 8 calculate angle 8 0 or 0. 70 rads page 06
5. Question Generic scheme Illustrative scheme know to integrate and interpret limits use upper lower integrate substitute limits 0... d 0 ( + + + ) ( + ) d + 0 ( ) + ( ) Ma mark 5 5 evaluate area 5 7 units page 07
Question Generic scheme Illustrative scheme 6. (a) Method Method Ma mark identify common factor ( + 8... stated or implied by complete the square ( + )... process for c and write in required form + + ( ) Method Method epand completed square form a + ab + ab + c equate coefficients a=, ab=, ab + c= 50 process for b and c and write in required form + + ( ) 6. (b) differentiate two terms +... 5 complete differentiation 5... + 50 6. (c) Method 6 link with (a) and identify sign of ( + ) 7 communicate reason Method 6 identify minimum value of f ( ) 7 communicate reason Method 6 f () ( + ) ( + ) 0 = + and + + > 0 always strictly increasing 7 ( ) Method 6 eg minimum value = or annotated sketch 7 0 ( f ( ) 0) > > always strictly increasing page 08
Question Generic scheme Illustrative scheme 7. (a) evidence of reflecting in -ais vertical translation of units identifiable from graph reflection of graph in -ais graph moves parallel to y-ais by units upwards Ma mark 7. (b) identify roots 0 and only interpret point of infleion turning point at ( 0, ) 5 complete cubic curve 5 cubic passing through origin with negative gradient page 09
Question Generic scheme Illustrative scheme 8. (a) use compound angle formula kcos cos a ksin sin a stated eplicitly Ma mark compare coefficients kcos a= 5, ksin a= stated eplicitly process for k k = 9 process for a and epress in required form 9 cos( + 0 8) 8. (b) 5 equate to and simplify constant terms 5 5cos sin= or 5cos sin = 0 6 use result of part (a) and rearrange cos + = 6 ( 0 805... ) 9 7 solve for + a 7 8 7 90, 5 098 8 solve for 8 0 8097, 7 page 0
Question Generic scheme Illustrative scheme 9. (a) equate volume to 00 V =π r h= 00 Ma mark obtain an epression for h demonstrate result h 00 = π r 00 A=π r + π r + π r leading to πr 00 A = + π r r 9. (b) start to differentiate A ( r) = 6 πr... 6 5 complete differentiation 00 5 A () r = 6 πr r 6 set derivative to zero 6 00 6πr = 0 r 7 obtain r 7 r = 00 π ( 0 ) metres 8 verify nature of stationary point 9 interpret and communicate result 8 table of signs for a derivative when 97... r = 9 minimum when r or 0 (m) 00 8 A () r = 6π+ r 9 A ( 97... ) > 0 minimum when r 0 (m) page
0. Question Generic scheme Illustrative scheme start to integrate complete integration process limits simplify numeric term and equate to cos... π cos π cos cos π π a + 8 π cos a + = Ma mark 6 5 start to solve equation 6 solve for a. Method substitute for sin simplify and factorise π 5 cos a = 6 a = π 8 Method sin cos sin cos cos stated eplicitly as above or in a simplified form of the above sin ( cos ) substitute for and simplify cos sin sin leading to sin Method substitute for sin simplify and substitute for cos epand and simplify Method sin cos sin cos cos stated eplicitly as above or in a simplified form of the above sin sin ( sin ) sin sin + sin leading to sin page
Question Generic scheme Illustrative scheme. (a) Method calculate calculate m AB m BC Method eg m AB = = 9 5 eg m BC = = 5 Ma mark interpret result and state conclusion AB and BC are parallel (common direction), B is a common point, hence A, B and C are collinear. Method Method calculate an appropriate vector, eg AB calculate a second vector, eg BC and compare eg eg 9 AB = 5 BC = 5 AB = BC 5 interpret result and state conclusion AB and BC are parallel (common direction), B is a common point, hence A, B and C are collinear. Method Method calculate m AB find equation of line and substitute point m AB = = 9 y = leading to 6 = ( 7 ) eg, ( ) communication since C lies on line A, B and C are collinear. (b) find radius 6 0 5 determine an appropriate ratio 5 eg : or 5 (using B and C) or : 5 or 8 5 (using A and C) 6 find centre 7 state equation of circle 6 ( 8, ) 7 ( ) ( y ) 8 + = 60 page
Question Generic scheme Illustrative scheme. (a) interpret half-life process equation write in logarithmic form process for k. (b) 5 interpret equation 5k P0 = Pe 0 stated or implied by e = 5k log e = 5k k 0 08 5 P = Pe t 0 80 0 08 Ma mark 6 process 7 state percentage decrease P 0 065P 6 t 0 7 89% [END OF SPECIMEN MARKING INSTRUCTIONS] page