MATHEMATICS Higher Grade - Paper I

Transcription

1 Higher Mathematics - Practice Eaminatin B Please nte the frmat f this practice eaminatin is different frm the current frmat. The paper timings are different and calculatrs can be used thrughut. MATHEMATICS Higher Grade - Paper I Time allwed - hurs Read Carefully. Full credit will be given nly where the slutin cntains apprpriate wrking.. Calculatrs may be used.. Answers btained by readings frm scale drawings will nt receive any credit. Pegasys 005

2 FORMULAE LIST The equatin ( g f c) +. + y + g + fy + c = 0 represents a circle centre ( g, f ) and radius The equatin ( a) + ( y b) = r represents a circle centre ( a, b ) and radius r. Scalar Prduct: a. b = a b csθ, whereθ is theanglebetweenaandb. r a. b = ab+ ab + ab where a= a a a and b = b b b Trignmetric frmulae: ( ) ( ) sin A± B = sin AcsB ± cs AsinB cs A± B = cs AcsB m sin AsinB csa = cs A sin A = cs A = sin A sina = sin Acs A Table f standard derivatives: f ( ) f ( ) sin a a cs a cs a a sin a Table f standard integrals: f ( ) f ( ) d sin a cs a a cs a + a sin a + C C Pegasys 005

3 All questins shuld be answered. Differentiate +, with respect t, epressing yur answer with psitive indices. (4). Tw functins f and g are defined n the set f real numbers as fllws : + 9 f ( ) =, g( ) =. 4 (a) Evaluate f ( g( ) ). () (b) Find an epressin, in its simplest frm, fr ( f ()) g. (). Given that = and = are tw rts f the equatin + a + + b= 0, establish the values f a and b and hence find the third rt f the equatin. (5) 4. A sequence is defined by the recurrence relatin Un+ = 0 8 Un +. (a) Eplain why this sequence has a limit as n. () (b) Find the limit f this sequence. () (c) Taking U 0 = 0 and L as the limit f the sequence, find n such that L U n = 56 () 5. Find the equatin f the line which passes thrugh the pint P(,-5) and is parallel t the line passing thrugh the pints (-,4) and (7,-). (4) Pegasys 005

4 6. The diagram belw shws the circle with equatin + y + 4 y = 0. The line L is a tangent t the circle at the pint T and has as its equatin + y=5. T y L C O (a) Find the crdinates f pint T. () (b) Given that the centre f the circle is C (-,), shw that angle TCO is a right angle. () 7. The graph f y = f ( ) is shwn belw. Sketch the graph f y = f (). () y Tw circles, which d nt tuch r verlap, have as their equatins ( 5) + ( y 6) = 40 and + y 6 4y+ = 0. (a) Shw that the eact distance between the centres f the tw circles is 4 0 units. () (b) Hence shw that the shrtest distance between the tw circles is equal t the radius f the smaller circle. (4) Pegasys 005

5 9. Evaluate ( 4 ) d 0 (4) 0. The picture belw shws a small sectin f a larger circuit bard. P Q R Relative t rectangular aes the pints P, Q and R have as their crdinates (-8,,), (-,-6,4) and (,-,6) respectively. Prve that the pints P, Q and R are cllinear, and find the rati PQ : QR. (4). A functin is defined as f ( ) = p, fr R and p is a cnstant p (a) Epress the functin in the frm f ( ) =, and hence ( a) + b state the maimum value f f in terms f p. (4) (b) Given nw that p = f f is. + shw that the eact maimum value (). The diagram belw shws the parabla with equatin y = 8 and the line which is a tangent t the curve at the pint T(,5). y T(,5) θ y = 8 Find the size f the angle marked θ, t the nearest degree. (4) Pegasys 005

6 . (a) Pints E, F and G have crdinates (-,,), (,,0) and (-,-,) respectively. Given that EF = GH, find the crdinates f the pint H. () (b) Hence calculate EH. () 4. Slve algebraically the equatin cs 9cs = fr 0 < 60. (6) 5. A circle has as its centre C(-,k). y A chrd PQ is drawn with end - pints P(,-) and Q(,4) as shwn in the diagram. Q(,4) C(-,k) (a) Establish the gradient f CM. () M (b) Write dwn the crdinates f M. () (c) Find the value f k. P(,-) () 4 6. Given that g ( ) = ( ), find the value f g ( ). (4) 7. (a) Given that lg y = lgy + find a relatinship cnnecting and y. (4) (b) Hence find y when = y and y > 0. () 4 [ END OF QUESTION PAPER] Pegasys 005

7 Higher Mathematics Practice Eam B Marking Scheme - Paper. Fr ( + ) Fr + d Fr = d Then ans. = [ 4 marks ]. (a) Fr answer f ( g( )) = 0 [ mark ] ( ) + 9 (b) Fr g ( f ( ) ) = 4 fr ans. g f ( ) = ( + ) (r equiv.) [ marks ] ( ). Fr realising t use synthetic divisin Fr setting up synth. div. tables crrectly Fr a + b = and 4a + b = Fr slving system t ans. a = 5 and b = 8 Fr finding third rt i.e. = 4 [ 5 marks ] 4. (a) Eplanatin i.e. < a < etc. (r just wrds) [ mark ] b (b) Fr knwing L = (r equiv.) a Fr answer L = 5 [ marks ] (c) Fr U = 080 ( ) + = then t U = = 56 fr stating the requested ans. i.e. n = [ marks ] 5. Fr m L = 4 Fr y+ 5= ( )... () 4 i.e. crrect pint with crrect gradient Fr answer + 4y + = 0 (r equiv.) [ 4 marks ] Pegasys 005

8 6. (a) + y = 5 Fr slving a system i.e. + y + 4 y = 0 Fr first crd. i.e. y = Fr cmplete pint T(-,) [ marks ] (b) Fr strategy e.g. using grad. r pyth etc. Fr seeing thrugh t i.e. moc mot = then ans. moc mot = = right angle ( pupils may simply equate gradients i.e. L and OC... full marks ) [ marks ] 7. Fr crrect shape y Fr crrect rts Fr anntatin -4 [ marks ] 8. (a) Fr finding centres, ne mark each - (5,6) (,) Fr Pyth. r dis. frm. t answer [ marks ] (b) Fr each radius ne mark - r = 40 = 0 frm r = f + g c - r = 0 Fr gap = 4 0 (r equiv.) shrt. d = 0 = the radius f the smaller circle (nte : n marks ff fr apprimatin thrughut) [ 4 marks ] 9. Fr ( ) d 0 [ 6 + ] 0 ( ) ( 0) Fr answer 8 [ 4 marks ] Pegasys 005

9 0. Fr selecting crrect displacements i.e PQ and QR Fr bth crrect 6 4 PQ = 9 and QR = 6 Fr cllinear statement i.e. since PQ = QR then P, Q and R are cllinear (r equiv.) Fr crrect rati PQ : QR = : [ 4 marks ] +. (a) Fr [ ( 9) 8] 87 p fr f ( ) = ( 9) + 6 then stated r implied ( 9) + 6 min = 6 p f ma = = p [ 4 marks ] 6 (b) Fr f ma =. = + + then f ma =. = = ans. [ marks ] + dy. Fr = 8 6 d Fr m tan = 8-6() = Fr tanθ = m = (stated r implied) Fr ans. θ = 6 [ 4 marks ] (nte : n marks ff fr nt runding). (a) Fr EF = Fr 6 = h ~ then fr ans. 4 h = H(,, ) ~ = 4 [ marks ] (b) 5 Fr h e = ~ ~ then fr ans. mag. = 0 [ marks ] Pegasys 005

10 4. Fr re-arranging t zer and remving the cmmn factr (whenever) Fr using crrect replacement cs Fr line - ( cs cs 5) = 0 5 then fr factrising and slving t cs = r cs = Fr rejecting cs = 5 as having n slutin Fr final answer - = 80 [ 6 marks ] 5. (a) Fr gradient f PQ m PQ = Fr gradient f perpendicular m = (b) Fr crdinates f mid-pint (,) (c) Fr selecting a strategy Fr equat. gradients r Fr finding and using the equatin f the Fr k = perpendicular y + = 5... () [ 6 marks ] 6. Fr g ( ) = 4( ).... () (split as fllws 4( ) ;... markeach) Fr ( ) = 4( ) ( ( ) ) g Fr answer g ( ) = 8 [ 4 marks ] 7. (a) Fr lg y lg y = then lg y lg y = y then lg = lg y = y Fr = y y = (r equiv.) [ 4 marks ] (b) Fr y = ( 4 y) (r equiv.) then fr y = 8 y ans. y = 8 [ marks ] Ttal 8 marks Pegasys 005

11 Higher Mathematics - Practice Eaminatin B Please nte the frmat f this practice eaminatin is different frm the current frmat. The paper timings are different and calculatrs can be used thrughut. MATHEMATICS Higher Grade - Paper II Time allwed - hurs 0 minutes Read Carefully. Full credit will be given nly where the slutin cntains apprpriate wrking.. Calculatrs may be used.. Answers btained by readings frm scale drawings will nt receive any credit. Pegasys 005

12 FORMULAE LIST The equatin ( g f c) +. + y + g + fy + c = 0 represents a circle centre ( g, f ) and radius The equatin ( a) + ( y b) = r represents a circle centre ( a, b ) and radius r. Scalar Prduct: a. b = a b cs θ, where θ is the angle between a and b. r a. b = ab+ ab + ab where a= a a a and b = b b b Trignmetric frmulae: ( ) ( ) sin A± B = sin AcsB ± cs AsinB cs A± B = cs AcsB m sin AsinB csa = cs A sin A = cs A = sin A sina = sin Acs A Table f standard derivatives: f ( ) f ( ) sin a a cs a cs a a sin a Table f standard integrals: f ( ) f ( ) d sin a cs a a cs a + a sin a + C C Pegasys 005

13 All questins shuld be answered. In the diagram belw triangle PQR has vertices as shwn. Q (, 0 ) T R ( 8, ) P ( -7,-5 ) (a) Find the equatin f the median frm P t QR. () (b) Find the equatin f the altitude frm Q t PR. (4) (c) Hence find the crdinates f the pint T where these tw lines crss. (). A circle, centre C, has as its equatin + y 4 y 5= 0. (a) Shw that the line with equatin y + = 5 is a tangent t this circle and state the crdinates f T, the pint f tangency. (5) (b) The pint P(k, -5) lies n this line f tangency. Find k. () (c) Establish the equatin f the circle which passes thrugh the pints C, T and P. (4) Pegasys 005

14 . Industrial clant is a water/il based liquid used t cl dwn metal cmpnents during their manufacture. It is cntinuusly pured ver the cmpnent and the cutting tl. The diagram belw (which is nt t scale) shws hw the clant is pumped frm a main hlding tank t each machine in the factry and is then recycled back t the tank. M M M M 4 Filters Cling Heli The clant has ne ther imprtant and epensive ingredient, namely an anti-bacterial agent. This agent, while active, hinders the grwth f bacteria which thrive in the heated clant. If this bacteria is allwed t multiply it pses a real health risk fr the machine peratrs. A cmpany wrks the fllwing system : At the beginning f each week (Mnday mrning) the liquid is drained frm the system and replaced by new clant, 40 units f the anti-bacterial agent is immediately added via the filter system. It is knwn that the anti-bacterial agent decays at the rate f % per wrking day and that this decayed amunt is nw said t be nn-active. (a) Hw many units f the anti-bacterial agent are still active at the end f a nrmal wrking week ( Friday evening)? Give yur answer t the nearest hundredth f a unit. () (b) New gvernment guidelines have just been issued which state that the minimum amunt f active anti-bacterial agent which shuld be present in industrial clant is 8 units. The cmpany decides t meet these new guidelines by tpping up the agent at the end f each wrking day. Calculate the minimum number f whle units which shuld be added at the end f each day in rder t keep the active agent abve the 8 units until the end f the wrking week. (5) (c) The cmpany had f curse lked at the alternative f simply adding mre than the 40 units f the anti-bacterial agent at the beginning f the week. Cnsider this ptin and cmment. () Pegasys 005

15 4 4. The curve shwn belw has as its equatin y = 8. y A B (a) Find the crdinates f the pints A and B. (7) (b) Find the equatin f the tangent t the curve at the pint where =. (5) 5. A cubid is placed relative t a set f crdinate aes as shwn in the diagram. The cubid has dimensins 8 cm by 4 cm by 6 cm. z H y G E A D 8 cm B F C 4 cm 6 cm The rigin f the aes is at the intersectin pint f the diagnals ED and HA and unit represents cm. (a) Find the crdinates f B, D and G. () (b) Hence calculate the size f angle BDG. (6) Pegasys 005

16 6. The main pwer surce used t run the lms at the New Lanark cttn mills was water. Each mill had its wn large water wheel which cnsisted f a central hub and an array f trugh like blades. The wheels were driven by water being pured ver the wheel thus filling the trughs and allwing the weight f the water t turn the wheel in a cntinuus prcess. The diagram belw shws the structure f ne f these trugh-blades. diagram The shaded end-sectin cnsists f part f a curve and a straight line. This end-sectin can be rtated and placed n a set f rectangular aes as shwn in diagram. The aes are nt drawn t scale. y A. diagram B. The curved sectin frm A t B is part f the curve y = (a) State the crdinates f pint A. () (b) The line AB is a tangent t the curve at A. Hence, r therwise, shw that this line has as its equatin y = 8 4. () (c) Calculate, in square units, the shaded area in diagram. (5) (d) (e) Given that the scale in diagram is unit = 0cm, write dwn the area f the end-sectin in square centimetres. () Hence calculate the vlume f water this trugh can hld when full, given that the trugh has a length f 6 metres. Epress yur answer in litres. () Pegasys 005

17 7. The tw diagrams shwn are design lg backgrunds. Design is a square f side Design is a rectangle measuring by c. All lengths are in centimetres. c Design Design (a) The area f Design is 4 cm mre than that f Design. By equating the areas shw that the fllwing equatin can be cnstructed ( c+ 4) + ( c+ 8) = 0 () (b) Hence find the value f c if the equatin ( c+ 4) + ( c+ 8) = has equal rts. 0 (5) (c) Using this value fr c, slve the equatin fr and hence calculate the area f each design. () 8. The luminsity, L units, emitting frm a pulsing light surce is given by the frmula L = cs6t + sin 6t +, where t is the time in secnds frm switch n. (a) Epress L in the frm Rcs( 6t α ) +, where R > 0 and 0 α 60. (4) (b) Sketch the graph f L fr 0 t 0, indicating all relevant pints. (5) (c) If L = 5 the light surce has the same luminsity as the surrunding light. When will this first ccur during this ten secnd perid? Give yur answer crrect t the nearest tenth f a secnd. () [ END OF QUESTION PAPER] Pegasys 005

18 Higher Mathematics Practice Eam B Marking Scheme - Paper. (a) Fr mid-pt. f QR - M( 4 4, 6 4 ) Fr m PM = Fr equatin f median y = + [ marks ] (b) Fr m PR = Fr m alt. = Fr crr. pt. & grad. t y 0 = ( ) Fr answer y + = (r equiv.) [ 4 marks ] y = + (c) Fr slving a system i.e. y+ = Fr first crdinate r y, eg. = Fr cmplete pint T(,5) [ marks ]. (a) Fr knwing t slve as a system Fr + ( 5 ) 4 ( 5 ) 5 = 0 then fr 5( 6)( 6) = 0 = 6 Fr finding y - crdinate - y = then T(6,) Fr stating that ne pint represents a tangent r equivalent (i.e. b 4a c= 0, etc. ) [ 5 marks ] (b) Fr answer k = 0 [ mark ] (c) Fr realising CP is a diameter Fr establishing centre as (6,-) Fr calculating r rr i.e. r = 5 then fr ans. ( 6) + ( y+ ) = 5 [ 4 marks ] (nte: ther methds pssible i.e. system unknwns, intersectin f the perpendicular bisectrs). (a) Fr crrect multiplier a = Fr calc. U 5 = (0 87) 40 (r equiv.) Fr ans = 9 94 units [ marks ] (b) Fr setting up U n + = (0 87)40 + r (r equiv.) Fr lking at different values f b Fr setting ut calculatins t answers Fr cnsidering the lwer value befre the additin Fr discvering that + desn't wrk b = units [ 5 marks ] (c) Fr realising ttal fr (b) is 40 + = 5 units Fr calculatins as (a) fr different U 0 until U 0 = 57 Fr cnclusin i.e. (b) is better saves 5 units/week [ marks ] Pegasys 005

19 4. (a) Fr B Fr 8 4 = 0 Fr = 0 r = 4 Fr giving B(4,0) Fr A Fr dy d = 4 8 then 4 8 = 0 Fr slv. t = 0 r = Fr using = and finding y A(,54) [ 7 marks ] (b) Fr using derivative t find m f tangent. Fr sub. = int dy d giving m tan = 5 Fr sub. = 7 int " y = " fr pint (, ) 8 Fr using y - b = m( - a) (r equiv.) with pint and gradient Fr answer frm abve 8y = 40 (r equiv.) [ 5 marks ] 5. (a) Fr B(8,-,-), D(0,,-) and G(8,,) ( each) [ marks ] (b) Fr selecting suitable vectrs i.e. DB and DG 8 8 Fr DB = 4 and DG = Fr scalar prduct DB DG = 64 Fr bth magnitudes 4 5 and 0 64 Fr cs θ = (r equiv.) 40 5 Fr ans. angle BDG = 44 [ 6 marks ] 6. (a) Fr ans. A(0,8) [ mark ] dy (b) Fr m = = 4 4 d = 0, m = 4 fr using (0,8) and m = -4 t ans. [ marks ] (pupils may use ply. t est. pint B etc. - assign wn marks) (c) Fr Are a = (( 8 4 ) ( )) d... () 0 ( r tw separate integrals ) ( fr limits + fr setting up Int.) 4 fr Are a = 4 0 fr Area = ( 6 4) ( 0) then ans. Are a = units [ 5 marks ] (d) Fr 00 = cm (.k.) [ marks ] (e) Fr V = 600 fr ans.v = c m = 80 litres [ marks ] Pegasys 005

20 7. (a) Fr ( ) = c( ) 4 then c + c + 4 = 0 fr ( c + 4) + ( c+ 8) = 0 [ marks ] (b) Fr stating "fr equal rts - b 4ac = 0" Fr a =, b = ( c+ 4) and c = c+ 8 Fr sub. then t c + 4c 96 = 0 Fr slving t c = 8 r c = Fr discarding - (-ve length) c = 8 [ 5 marks ] (c) Fr (8 + 4) + (8 + 8) = 0 The slved t = 6 cm Fr areas D = 6 c m, D = 40 cm [ marks ] 8. (a) Fr R = Fr ta nα = Fr realising st Quad. then α = 60 ( ) Fr L = cs 6t 60 + [ 4 marks ] (b) Fr L ma = 4 and L min = 0 Fr drawing f graph Fr each number circled mark... () [ 5 marks ] L (units) 4 (c) Fr ( ) cs 6t 60 + = 5 6t 60 = 75 5 Fr ans t = 8 sec nds [ marks ] t (sec.) Ttal 87 marks Pegasys 005