Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C Advanced Subsidiary Thursday June 0 Morning Time: hour 0 minutes Materials required for eamination Mathematical Formulae (Pink) Items included with question papers Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation or symbolic differentiation/integration, or have retrievable mathematical formulae stored in them. Instructions to Candidates In the boes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. You must write your answer for each question in the space following the question. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. The marks for the parts of questions are shown in round brackets, e.g. (). There are 8 questions in this question paper. The total mark for this paper is 75. There are pages in this question paper. Any blank pages are indicated. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Eaminer. Answers without working may not gain full credit. P406A This publication may only be reproduced in accordance with Edecel Limited copyright policy. 0 Edecel Limited.
. Given that 4 a b c 5 4 e 4 4, find the values of the constants a, b, c, d and e. (4). Given that f() = ln, > 0 sketch on separate aes the graphs of (i) y = f(), (ii) y = f(), (iii) y = f( 4). Show, on each diagram, the point where the graph meets or crosses the -ais. In each case, state the equation of the asymptote. (7) P406A
. Given that (a) Show, without using a calculator, that cos ( + 50) = sin ( + 40). tan = tan 40. (4) (b) Hence solve, for 0 θ < 60, cos (θ + 50) = sin (θ + 40), giving your answers to decimal place. (4) 4. f() = 5 e 6, R. (a) Using calculus, find the eact coordinates of the turning points on the curve with equation y = f(). (5) (b) Show that the equation f() = 0 can be written as = 4 e. 5 () The equation f() = 0 has a root α, where α = 0.5 to decimal place. (c) Starting with 0 = 0.5, use the iteration formula n+ = 4 e n 5 to calculate the values of, and, giving your answers to decimal places. (d) Give an accurate estimate for α to decimal places, and justify your answer. () () P406A
5. Given that = sec y, 0 < y < 6 (a) find d in terms of y. () (b) Hence show that 6 ( ) (4) (c) Find an epression for d y in terms of. Give your answer in its simplest form. (4) 6. Find algebraically the eact solutions to the equations (a) ln (4 ) + ln (9 ) = ln ( + ), < <, (b) e + = 0. (5) Give your answer to (b) in the form a ln b c ln d where a, b, c and d are integers. (5) P406A 4
7. The function f has domain 6 and is linear from (, 0) to (, 0) and from (, 0) to (6, 4). A sketch of the graph of y = f() is shown in Figure. Figure (a) Write down the range of f. (b) Find ff(0). () () The function g is defined by g : 4, R, 5. 5 (c) Find g (). (d) Solve the equation gf() = 6. () (5) P406A 5
8. Figure Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon runner, John, approaching at m s. Kate is 4 m ahead of John when she starts to cross the road from the fied point A. John passes her as she reaches the other side of the road at a variable point B, as shown in Figure. Kate s speed is V m s and she moves in a straight line, which makes an angle θ, 0 < θ < 50, with the edge of the road, as shown in Figure. You may assume that V is given by the formula V, 0 < θ < 50 4sin 7cos (a) Epress 4sin θ + 7cos θ in the form R cos (θ α), where R and α are constants and where R > 0 and 0 < α < 90, giving the value of α to decimal places. () Given that θ varies, (b) find the minimum value of V. () Given that Kate s speed has the value found in part (b), (c) find the distance AB. () Given instead that Kate s speed is.68 m s, (d) find the two possible values of the angle θ, given that 0 < θ < 50. (6) END TOTAL FOR PAPER: 75 MARKS P406A 6
Question Number Scheme Marks By Division + 7 4 ( + 0) 4 5 + ( 0) 4 + 0 4 + 7 + 0 + 0 + 8 7 84 7 + 08 8 + 4 Long division as far as... 4 ( + 0) 4 5 + ( 0) 4 + 0 4 +... +... a = B M Two of b= c= 7 d = 8 e= 4 A All four of b= c= 7 d = 8 e= 4 A Notes for Question B Stating a =. This can also be scored by the coefficient of in + 7 (4 marks) M Using long division by 4 and getting as far as the term. The coefficients need not be correct. Award if you see the whole number part as... +... following some working. You may also see this in a table/ grid. Long division by ( + ) will not score anything until ( ) has been divided into the new quotient. It is very unlikely to score full marks and the mark scheme can be applied. A Achieving two of b= c= 7 d = 8 e= 4. The answers may be embedded within the division sum and can be implied. A Achieving all of b= c= 7 d = 8and e= 4 Accept a correct long division for out of the 4 marks scoring BMAA0 Need to see a=, b=, or the values embedded in the rhs for all 4 marks
Question Number Scheme Marks Alt By Multiplicat ion * 4 5 4 ( a + b + c)( 4 ) + + e Compares the 4 terms a = B Compares coefficients to obtain a numerical value of one further constant = b, 5= 4 a+ c c=.., M Two of b= c= 7 d = 8 e= 4 A All four of b= c= 7 d = 8 e= 4 A Notes for Question B Stating a =. This can also be scored for writing = a 4 4 (4 marks) M Multiply out epression given to get *. Condone slips only on signs of either epression. 4 Then compare the coefficient of any term (other than ) to obtain a numerical value of one further constant. In reality this means a valid attempt at either b or c The method may be implied by a correct additional constant to a. A Achieving two of b= c= 7 d = 8 e= 4 A Achieving all of b= c= 7 d = 8and e= 4
Question Number Scheme Marks (i) ln graph crossing ais at (,0) and asymptote at =0 B Shape including cusp Bft (ii) Touches or crosses the ais at (,0) Asymptote given as =0 Bft B (iii) Shape Crosses at (5, 0) Asymptote given as =4 B Bft B (7 marks)
Notes for Question (i) B Correct shape, correct position and passing through (, 0). Graph must start to the rhs of the y - ais in quadrant 4 with a gradient that is large. The gradient then decreases as it moves through (, 0) into quadrant. There must not be an obvious maimum point but condone slips. Condone the point marked (0,) on the correct ais. See practice and qualification for clarification. Do not with hold this mark if (=0) the asymptote is incorrect or not given. (ii) Bft Correct shape including the cusp wholly contained in quadrant. The shape to the rhs of the cusp should have a decreasing gradient and must not have an obvious maimum.. The shape to the lhs of the cusp should not bend backwards past (,0) Tolerate a linear looking section here but not one with incorrect curvature (See eamples sheet (ii) number. For further clarification see practice and qualification items. Follow through on an incorrect sketch in part (i) as long as it was above and below the ais. Bft The curve touches or crosses the ais at (, 0). Allow for the curve passing through a point marked on the ais. Condone the point marked on the correct ais as (0, ) Follow through on an incorrect intersection in part (i). B Award for the asymptote to the curve given/ marked as = 0. Do not allow for it given/ marked as the y ais. There must be a graph for this mark to be awarded, and there must be an asymptote on the graph at = 0. Accept if =0 is drawn separately to the y ais. (iii) B Correct shape. The gradient should always be negative and becoming less steep. It must be approimately infinite at the lh end and not have an obvious minimum. The lh end must not bend forwards to make a C shape. The position is not important for this mark. See practice and qualification for clarification. Bft The graph crosses (or touches) the ais at (5, 0). Allow for the curve passing through a point marked 5 on the ais. Condone the point marked on the correct ais as (0, 5) Follow through on an incorrect intersection in part (i). Allow for (() i + 4,0) B The asymptote is given/ marked as = 4. There must be a graph for this to be awarded and there must be an asymptote on the graph (in the correct place to the rhs of the y ais). If the graphs are not labelled as (i), (ii) and (iii) mark them in the order that they are given.
Eamples of graphs in number Part (i) Condoned Not condoned Condone Y ais (0,) Part (ii),, 0,,, 0,,0, =0 =0 =0 Eample of follow through in part (ii) and (iii) (i) B0 (ii) BftBftB0 (iii) B0BftB0 0 0 4
Question Number Scheme Marks (a) cos cos50 sin sin 50 = sin cos 40 + cos sin 40 M sin (cos 40 + sin 50) = cos (cos50 sin 40) cos tan (cos 40 + sin 50) = cos50 sin 40 M cos50 sin 40 tan =, cos40+ sin50 (or numerical answer awrt 0.8) A States or uses cos50 = sin40 and cos40 =sin50 and so o o tan tan 40 = * cao A * (4) (b) Deduces tanθ = tan40 M θ = 5.6 so θ = awrt 7.8() One answer A Also θ = 95.6, 75.6, 555.6 θ =.. θ = awrt 7.8, 97.8, 87.8, 77.8 All 4 answers M A (4) [8 marks ] Alt (a) cos cos 50 sin sin 50 = sin cos 40 + cos sin 40 M cos sin 40 sin cos 40 = sin cos 40 + cos sin 40 cos sin 40 tan cos 40 = tan cos 40 + sin 40 M tan = sin 40 ( or numerical answer awrt 0.8), tan = tan 40 cos 40 A,A Alt (a) cos( + 50) = sin( + 40) sin(40 ) = sin( + 40) cos sin 40 sin cos 40 = sin cos 40 + cos sin 40 M cos sin 40 tan cos 40 = tan cos 40 + sin 40 M sin 40 tan = ( or numerical answer awrt 0.8), tan = tan 40 cos 40 A,A
(a) Notes for Question M Epand both epressions using cos( + 50) = cos cos50sin sin 50 and sin( + 40) = sin cos 40 + cos sin 40. Condone a missing bracket on the lhs. The terms of the epansions must be correct as these are given identities. You may condone a sign error on one of the epressions. Allow if written separately and not in a connected equation. M Divide by cos to reach an equation in tan. Below is an eample of MM with incorrect sign on left hand side cos cos50 + sin sin 50 = sin cos 40 + cos sin 40 cos50+ tansin50= tancos40+ sin40 This is independent of the first mark. cos50 sin 40 A tan = cos 40 + sin 50 Accept for this mark tan = awrt 0.8... as long as MM has been achieved. A* States or uses cos50=sin40 and cos40=sin50 leading to showing cos50 sin 40 sin 40 tan = = = tan 40 cos 40 + sin 50 cos 40 (b) This is a given answer and all steps above must be shown. The line above is acceptable. Do not allow from tan = awrt 0.8... M For linking part (a) with (b). Award for writing tanθ = tan40 A Solves to find one solution of θ which is usually (awrt) 7.8 M Uses the correct method to find at least another value of θ. It must be a full method but can be implied by any correct answer. Accept 80 + theirα 60 + 540,( or) theirα + θ =,( or) theirα A Obtains all four answers awrt dp. θ = 7.8, 97.8, 87.8, 77.8. Ignore any etra solutions outside the range. Withhold this mark for etras inside the range. Condone a different variable. Accept = 7.8, 97.8, 87.8, 77.8 Answers fully given in radians, loses the first A mark. Acceptable answers in rads are awrt 0.6,.7,.8, 4.85 Mied units can only score the first M
Question Number 4(a) f() 50 e = + 50e oe. Scheme MA Puts f() = 0to give = - and = 0 or one coordinate dma - Obtains ( 0,-6) and (-, 5e -6) CSO A Marks (5) (b) Puts 6 4 = = =± B* 5 5 5 e 6 0 e e () 4 (c) Subs 0 = 0.5 into e = = awrt 0.485 MA 5 = awrt 0.49, = awrt 0.489 A () (d) α = 0.49 f (0.485) = 0.487, f (0.495) = ( + )0.485, sign change and deduction Notes for Question 4 No marks can be scored in part (a) unless you see differentiation as required by the question. (a) M Uses vu ' + uv '. If the rule is quoted it must be correct. It can be implied by their u=.., v=..., u' =..., v' =... followed by their vu ' + uv ' If the rule is not quoted nor implied only accept answers of the form A e + Be A f() 50 e = + 50e. Allow un simplified forms such as f() 5 e = + 50 e dm Sets f() = 0, factorises out/ or cancels the This is dependent upon the first M being scored. e leading to at least one solution of - A Both = - and = 0 or one complete coordinate. Accept ( 0,-6) and (-, 5e -6) or (, awrt.6) B B () ( marks) A CSO. Obtains both solutions from differentiation. Coordinates can be given in any way. 5 - = -,0 y= 6, 6 or linked together by coordinate pairs ( 0,-6) and (-, 5e -6) but e the pairs must be correct and eact.
(b) B (c) Notes for Question 4 Continued This is a show that question and all elements must be seen Candidates must ) State that f()=0 or writes 5 e 6 = 0 or 5 e = 6 ) Show at least one intermediate (correct) line with either 6 6 or the subject. Eg = e, = e oe 5 5 or square rooting 5 e = 6 5e = ± 4 or factorising by DOTS to give (5e + 4)(5e 4) = 0 4 ) Show the given answer e =±. 5 Condone the minus sign just appearing on the final line. A reverse proof is acceptable as long as there is a statement that f()=0 4 M Substitutes 0 = 0.5 into e = =... 5 4 0.5 This can be implied by e =, or awrt 0.49 5 A = awrt 0.485 dp. Mark as the first value given. Don t be concerned by the subscript. A = awrt 0.49, = awrt 0.489 dp. Mark as the second and third values given. (d) B States α = 0.49 B Justifies by either calculating correctly f (0.485) and f (0.495) to awrt sf or dp, f (0.485) = 0.5, f (0.495) = ( + )0.5 rounded f (0.485) = 0.4, f (0.495) = ( + )0.4 truncated giving a reason accept change of sign, >0 <0 or f (0.485) f (0.495) <0 and giving a minimal conclusion. Eg. Accept hence root or α = 0.49 A smaller interval containing the root may be used, eg f (0.49) and f (0.495). Root = 0.49007 or by stating that the iteration is oscillating or by calculating by continued iteration to at least the value of 4 = awrt 0.49 and stating (or seeing each value round to) 0.49
Question Number Scheme = = 6siny oe cos y 5(a) sec y sec y tan y ( 6sec y tan y) Marks MA () (b) Uses d = 6sec ytany = to obtain d M tan y sec y = = B Uses sec y = and = = to get d y or d in just. M tan y sec y = 6 ( ) CSO A* (4) (c) d y 0 [6( ) + ( ) ] = 6 ( ) MA d 69 y = = 6 ( ) ( ) dma (4) (0 marks) Alt to 5(a) = (cos y) = (cos y) sin y MA Alt to 5 (a) = secy secy = secy secytany+ secy secytany MA Alt To 5 (c) d y 6 = [ ( )( ) + ( ) ( ) ] MA 6 = ( ) [ ( ) + ( )( )] dm ( ) [ ] = oe A (4)
(a) Notes for Question 5 M Uses the chain rule to get Asecysecytan y ( Asec ytan y) A (b) M =. There is no need to get the lhs of the epression. Alternatively could use the chain rule on (cos y) A(cos y) sin y ± Acosysiny or the quotient rule on 4 (cos y) (cos y) = secysecytan yor equivalent. There is no need to simplify the rhs but both sides must be correct. Uses = to get an epression for d y. Follow through on their d d Allow slips on the coefficient but not trig epression. y B Writes tan y sec y = or an equivalent such as tany sec y = and M uses = sec y to obtain either All elements must be present. tan y tany= = or ( ) Accept y cosy= tan y = If the differential was in terms of sin y,cosy it is awarded for sin y = Uses sec y = and tan y = sec y = or equivalent to get d y just. Allow slips on the signs in tan y = sec y. It may be implied- see below A* CSO. This is a given solution and you must be convinced that all steps are shown. Note that the two method marks may occur the other way around Eg. = 6sec y tan y = 6 ( ) = 6 ( ) Scores the nd method Scores the st method The above solution will score M, B0, M, A0
Notes for Question 5 Continued Eample - Scores 0 marks in part (b) = y y = = = 6 ( ) 6sec tan 6sec tan 6sec sec Eample - Scores MBMA0 = y y = = = ( ) sec tan sec y tan y sec y sec y (c) Using Quotient and Product Rules vu ' uv' M Uses the quotient rule with u = and v= 6 ( ) and achieving v u' = 0and v' = A( ) + B( ). If the formulae are quoted, both must be correct. If they are not quoted nor implied by their working allow epressions of the form d y 0 [ A ( ) + B ( ) ] = 6 ( ) or d y 0 A ( ) ± B ( ) = C ( ) A Correct un simplified epression d y 0 [6( ) + ( ) ] = 6 ( ) oe dm Multiply numerator and denominator by ( ) producing a linear numerator which is then simplified by collecting like terms. Alternatively take out a common factor of ( ) from the numerator and collect like terms from the linear epression This is dependent upon the st M being scored. d y A Correct simplified epression = oe ( )
(c) Using Product and Chain Rules Notes for Question 5 Continued M Writes = = A ( ) 6 ( ) and uses the product rule with vor u = ( ). If any rule is quoted it must be correct. uor v A = and If the rules are not quoted nor implied then award if you see an epression of the form ( ) B ± C( ) A d y 6 = [ ( )( ) + ( ) ( ) ] dm Factorises out / uses a common denominator of ( ) must be simplified by collecting like terms. Need a single fraction. producing a linear factor/numerator which A Correct simplified epression d y == ( ) [ ] oe (c) Using Quotient and Chain rules Rules vu ' uv' M Uses the quotient rule with u = ( ) andv= 6and achieving v u' = A( ) and v' = B. If the formulae is quoted, it must be correct. If it is not quoted nor implied by their working allow an epression of the form A d y C( ) D( ) = E Correct un simplified epression d y d 6 ( ) ( ) 6 = ( 6) dm Multiply numerator and denominator by ( ) producing a linear numerator which is then simplified by collecting like terms. Alternatively take out a common factor of ( ) linear epression This is dependent upon the st M being scored. A Correct simplified epression from the numerator and collect like terms from the ( ) d y d y ( ) = oe = ( )
Notes for Question 5 Continued (c) M M Using just the chain rule Writes = = = (6 6 ) 6 ( ) (6 6 ) A(6 6 ) ( B C). Would automatically follow under this method if the first M has been scored and proceeds by the chain rule to
Question Number 6(a) ln(4 )(9 ), ln( ) Scheme Marks = + M, M So 6 0 6 + = + + and 5 5 0 + = A Solve 5 5 0 + = to give 7 = oe ( Ignore the solution = 5 ) MA 5 (b) Take log e s to give ln +lne + = ln0 M ln +( +)lne = ln0 M (ln +lne) = ln0 ln e =.. dm (5) and uses lne = + ln0 = + ln M A (5) (a) Note that the 4 th M mark may occur on line Notes for Question 6 (0 marks) M Uses addition law on lhs of equation. Accept slips on the signs. If one of the terms is taken over to the rhs it would be for the subtraction law. M Uses power rule for logs write the ln( + ) term as ln( + ). Condone invisible brackets A Undoes the logs to obtain the TQ =0. 5 5 0 + =. Accept equivalences. The equals zero may be implied by a subsequent solution of the equation. M Solves a quadratic by any allowable method. The quadratic cannot be a version of (4 )(9 ) = 0 however. A Deduces =.4 or equivalent. Accept both =.4 and =5. Candidates do not have to eliminate = 5. You may ignore any other solution as long as it is not in the range < <. Etra solutions in the range scores A0.
(b) Notes for Question 6 Continued M Takes logs of both sides and splits LHS using addition law. If one of the terms is taken to the other side it can be awarded for taking logs of both sides and using the subtraction law. M Taking both powers down using power rule. It is not wholly dependent upon the first M but logs of both sides must have been taken. Below is an eample of M0M + ln lne = ln0 ln (+ ) ln e= ln0 dm This is dependent upon both previous two M s being scored. It can be awarded for a full method to solve their linear equation in. The terms in must be collected on one side of the equation and factorised. You may condone slips in signs for this mark but the process must be correct and leading to = M Uses ln e =. This could appear in line, but it must be part of their equation and not just a statement. Another eample where it could be awarded is e + = + =... 0 + ln0 ln0 loge0 A Obtains answer = = = oe. DO NOT ISW HERE + ln + ln + loge Note : If the candidate takes log 0 s of both sides can score MMdMM0A0 for out of 5. Answer = log e + log0 log e + = = log e + log log e + log + log0 Note : If the candidate writes = + log the last A mark, scoring 4 out of 5. without reference to natural logs then award M4 but with hold
Question Number Alt to 6(b) Scheme Marks Writes lhs in e s + ln + e = 0 e e = 0 st M ln + + e = 0, ln+ + = ln0 nd M, 4 th M (ln +) = ln0 =.. dm M M + ln0 = + ln Notes for Question 6 Alt ln Writes the lhs of the epression in e s. Seeing = e in their equation is sufficient Uses the addition law on the lhs to produce a single eponential A (5) dm Takes ln s of both sides to produce and attempt to solve a linear equation in You may condone slips in signs for this mark but the process must be correct leading to =.. M Uses ln e =. This could appear in line
Question Number Scheme Marks 7(a) 0 f( ) 0 B (b) (c) ff(0) = f(5), = 4+ y = y(5 ) = 4 + 5 B,B () () 5y 4= y+ M 5y 4 5y 4 = ( y+ ) = y + dm 5 4 g ( ) = + A () (d) gf ( ) 6 f ( ) g (6) 4 = = = oe MA f( ) = 4 = 6 B f ( ) = 4 5.5 = 4 = 0.4 oe MA (5) ( marks) Alt to 7(d) 4+ ( a + b) gf ( ) = 6 = 6 5 ( a + b) M a + b = or 5.5 A = 6 B 4 + (5.5 ) = 6 =... 5 (5.5 ) M = 0.4 oe A (5)
Notes for Question 7 (a) B Correct range. Allow 0 f( ) 0, 0 f 0, 0 y 0, 0 range 0 Allow f ( ) 0 and f ( ) 0 but not f ( ) 0 or f ( ) 0 Do Not Allow 0 0. The inequality must include BOTH ends (b) B For correct one application of the function at =0 Possible ways to score this mark are f(0)=5, f(5) 0 5... B: ( can score both marks as long as no incorrect working is seen.), [ 0,0 ] (c) M dm A For an attempt to make or a replaced y the subject of the formula. This can be scored for putting y = g(), multiplying across, epanding and collecting terms on one side of the equation. Condone slips on the signs Take out a common factor of (or a replaced y) and divide, to make subject of formula. Only allow one sign error for this mark 5 4 5 4 Correct answer. No need to state the domain. Allow g ( ) = y = + + 4 4 5 5 Accept alternatives such as y = and y = + (d) M Stating or implying that f ( ) = g (6). For eample accept 4 + f( ) = 6 f ( ) =.. 5 f( ) A Stating f( ) = 4or implying that solutions are where f( ) = 4 B = 6 and may be given if there is no working M Full method to obtain other value from line y = 5.5 5.5 = 4 =.... Alternatively this could be done by similar triangles. Look for = ( oe ) =.. 5 4 A 0.4 or /5 Alt to (d) M Writes gf ( ) = 6 with a linear f( ). The order of gf() must be correct Condone invisible brackets. Even accept if there is a modulus sign. A Uses f( ) = or f( ) = 5.5 in the equation gf ( ) = 6 B = 6 and may be given if there is no working M Attempt at solving 4 + (5.5 ) = 6 =..... The bracketing must be correct and there must be 5 (5.5 ) no more than one error in their calculation A = 0.4, or equivalent 5
Question Number Scheme Marks 8(a) ( ) R = 7 + 4 = 5 B (b) 4 tan α =, α = awrt 7.74 MA 7 maimum value of 4sin + 7cos = 5 so V min = = (0.84) MA 5 () () (c) Distance AB = 7 sinθ, with θ = α M, B So distance = 7.9m 75 = m A 4 (d) R cos( θ α) = cos( θ α) = 0.5 M, A.68 θ α = 60 θ =.., θ α =60 θ =.. dm, dm () θ = awrt.7,.7 Notes for Question 8 (a) B 5. Accept 5.0 but not 65 or answers that are not eactly 5. Eg 5.000 M 4 7 For tan α =±, tanα =±. 7 4 4 7 If the value of R is used only accept sin α = ±, cosα =± R R A Accept answers which round to 7.74 must be in degrees for this mark (b) M Calculates V their ' R ' A, A (6) (4 marks) A Obtains correct answer. V = Accept 0.84 5 Do not accept if you see incorrect working- ie from cos( θ α) =or the minus just disappearing from a previous line. Questions involving differentiation are acceptable. To score M the candidate would have to differentiate V by the quotient rule (or similar), set V =0 to find θ and then sub this back into V to find its value.
(c) M Uses the trig equation Notes for Question 8 Continued 7 sinθ = with a numerical θ to find AB =... AB B Uses θ = their value of α in a trig calculation involving sin. (sinα = AB 7 is condoned) A Obtains answer 75 or awrt 7.9 4 (d) M Substitutes V =.68and their answer to part (a) in V = to get an equation 4sinθ + 7cosθ of the form R cos( θ ± α) = or.68r cos( θ ± α ) = or cos( θ ± α) =..68.68R Follow through on their R and α A Obtains cos( θ ± α ) = 0.5 oe. Follow through on their α. It may be implied by later working. dm Obtains one value of θ in the range 0< θ < 50from inverse cos +their α It is dependent upon the first M being scored. dm Obtains second angle of θ in the range 0< θ < 50from inverse cos +their α It is dependent upon the first M being scored. A one correct answer awrt θ =.7or.7 dp A both correct answers awrt θ =.7 and.7 dp. Etra solutions in the range loses the last A. Answers in radians, lose the first time it occurs. Answers must be to dp For your info α =.87, θ =.4, θ = 0.40