6665/01 Edexcel GCE Core Mathematics C3 Silver Level S4

Size: px

Start display at page:

Download "6665/01 Edexcel GCE Core Mathematics C3 Silver Level S4"

Samantha Parrish
6 years ago
Views:

1 Paper Referene(s) 6665/0 Edexel GCE Core Mathematis C Silver Level S4 Time: hour 0 minutes Materials required for examination papers Mathematial Formulae (Green) Items inluded with question Nil Candidates may use any alulator allowed by the regulations of the Joint Counil for Qualifiations. Calulators must not have the faility for symboli algebra manipulation, differentiation and integration, or have retrievable mathematial formulas stored in them. Instrutions to Candidates Write the name of the examining body (Edexel), your entre number, andidate number, the unit title (Core Mathematis C), the paper referene (6665), your surname, initials and signature. Information for Candidates A booklet Mathematial Formulae and Statistial Tables is provided. Full marks may be obtained for answers to ALL questions. There are 8 questions in this question paper. The total mark for this paper is 75. Advie to Candidates You must ensure that your answers to parts of questions are learly labelled. You must show suffiient working to make your methods lear to the Examiner. Answers without working may gain no redit. Suggested grade boundaries for this paper: A* A B C D E Silver 4 This publiation may only be reprodued in aordane with Edexel Limited opyright poliy Edexel Limited.

2 . (a) Express 7 os x 4 sin x in the form R os (x + α) where R > 0 and 0 < α < π. Give the value of α to deimal plaes. (b) Hene write down the minimum value of 7 os x 4 sin x. () Solve, for 0 x < π, the equation () () 7 os x 4 sin x = 0, giving your answers to deimal plaes. (5) January 0. (a) Use the identity os θ + sin θ = to prove that tan θ = se θ. (b) Solve, for 0 θ < 60, the equation tan θ + 4 se θ + se θ =. () (6) June 009. (a) Express 5 os x sin x in the form R os(x + α), where R > 0 and 0 < α < π. (b) Hene, or otherwise, solve the equation 5 os x sin x = 4 for 0 x < π, giving your answers to deimal plaes. (5) January f(x) = x + x x x +. x (a) Express f(x) as a single fration in its simplest form. (b) Hene show that f (x) =. ( x ) () January 009 Silver 4: 8/

3 5. The mass, m grams, of a leaf t days after it has been piked from a tree is given by where k and p are positive onstants. m = pe kt, When the leaf is piked from the tree, its mass is 7.5 grams and 4 days later its mass is.5 grams. (a) Write down the value of p. () (b) Show that k = 4 ln. () Find the value of t when dm dt = 0.6 ln. (6) June 0 6. (a) Prove that (b) Hene, or otherwise, (i) show that tan 5 =, (ii) solve, for 0 < x < 60, os θ = tan θ, θ 90n, n Z. sin θ sin θ ose 4x ot 4x =. () (5) June 0 Silver 4: 8/

4 7. h(x) = x x + 5 ( x 8, x )( x + ) (a) Show that h(x) = x. x + 5 (b) Hene, or otherwise, find h (x) in its simplest form. () Figure Figure shows a graph of the urve with equation y = h(x). () Calulate the range of h(x). (5) January 0 Silver 4: 8/ 4

5 8. The amount of a ertain type of drug in the bloodstream t hours after it has been taken is given by the formula x = - t D 8 where x is the amount of the drug in the bloodstream in milligrams and D is the dose given in milligrams. A dose of 0 mg of the drug is given. e, (a) Find the amount of the drug in the bloodstream 5 hours after the dose is given. Give your answer in mg to deimal plaes. A seond dose of 0 mg is given after 5 hours. () (b) Show that the amount of the drug in the bloodstream hour after the seond dose is.549 mg to deimal plaes. () No more doses of the drug are given. At time T hours after the seond dose is given, the amount of the drug in the bloodstream is mg. () Find the value of T. () June 007 END TOTAL FOR PAPER: 75 MARKS Silver 4: 8/ 5

6 Question Number. (a) 7 os x 4sin x = R os( x+ α ) Sheme Marks 7 os x 4 sin x = R os x osα R sin x sinα Equate os x: 7 = Rosα Equate sin x: 4 = Rsinα R = ; = 5 R = 5 B tanα = α = tanα = or tanα = 7 4 M awrt.87 A Hene, 7 os x 4sin x = 5os( x +.87) () (b) Minimum value = 5 5 or R Bft () () 7 os x 4sin x = 0 5os( x +.87) = ( x + ) = os ( x their α ) os.87 0 ± = M ( their R) PV = or For applying 0 os their R M So, x +.87 = , , { } gives, x = { , } either or their PV π + or 60 + or their PV M awrt.84 OR 6.6 A awrt.84 AND 6.6 A (5) [9] Silver 4: 8/ 6

7 Question Number. (a) os θ + sin θ = ( os θ) Sheme Marks os θ sin θ + = M os θ os θ os θ + tan θ = se θ tan θ se θ = (as required) (b) tan θ 4seθ se θ, ( eqn *) + + = 0 θ < 60 A so () θ + θ + θ = M (se ) 4se se se θ + 4seθ + se θ = se θ 4seθ = M ( θ )( θ ) se + se = 0 M seθ = or seθ = = or = osθ osθ os θ ; or os = θ = A; α = 0 or α = no solutions θ = 0 θ = 40 A B ft (6) [8] Silver 4: 8/ 7

8 Question Number Sheme. (a) 5os sin = os ( + α ), > 0, 0 < < x x R x R x π Marks 5os x sin x = Ros xosα Rsin xsinα Equate os x: 5 = Rosα Equate sin x: = Rsinα R = + = = { } 5 ; tanα = α = Hene, 5os x sin x = 4 os( x ) (b) 5os x sin x = 4 M; A M A 4 os( x ) = 4 4 os( x ) = { = } M 4 ( x ) = M x = A ( ) x π { } + = = M x = A Hene, x = { 0.7, 4.9} (5) [9] Silver 4: 8/ 8

9 Question Number 4. (a) Sheme x+ x+ x+ x+ = + ( )( ) x+ ( x+ )( x+ ) ( x )( x+ ) ( x+ )( x) ( x )( x+ ) x x x x x x = = x = Aept x x, x x x Marks M A M A (b) ( x )( ) ( x) ( ) d x = dx x x x+ + x = = ( x ) ( x ) M A so A () 5. (a) p = 7.5 B 4k (b).5 = 7.5e M () e = M 4k = ln dm 4k = ln 4k ( ) k = ln ( ) 4 A dm kt = kpe dt ft on their p and k MAft 4 (ln ) t ln 7.5 e = 0.6ln 4 4 (ln ) t.4 e = = ( 0.) 7.5 MA ( ) t = ln ( ) 4 dm t = or awrt 4. A (6) [] [7] () Silver 4: 8/ 9

10 Question Number 6. (a) (b)(i) Sheme Marks os θ os θ = sin θ sin θ sin θ M sin θ sinθosθ MA sinθ = = tanθ osθ so A* os0 tan5 = sin0 sin0 M tan5 = = so dm A () (b)(ii) tan x = M x = 45 A x = M x =.5,.5, 0.5, 9.5 A(any two) A (5) [] Silver 4: 8/ 0

11 Question Number 7. (a) Sheme 4 8 ( x + 5) + 4( x+ ) 8 + = 5 ( )( 5) ( )( 5) x+ x + x+ x + x+ x + Marks MA = xx ( + ) + + ( x )( x 5) M (b) h '( x) = = x ( x + 5) x + x x ( x + 5) ( 5) A* MA 0 x h '( x) = ( x + 5) so A () ( ) Maximum ours when x = x = x= M h '( ) x = 5 A When 5 x= 5 h( x) = M,A 5 Range of h(x) is 0 hx ( ) 5 5 Aft (5) [] 8. (a) 8 D = 0, t = 5, x = 0e 5 M = 5.5 awrt A () 5 (b) D = 0 0e 8 +, t =, x = e M x =.549 ( ) A so () T () e = M T 8 e = = ln T = M T =.06 or. or A () [7] Silver 4: 8/

12 Question Examiner reports Question was a familiar one to most andidates. It was generally well done by the majority of andidates although part (b) and finding answers in the range 0 to π in part () did disriminate. In part (a) most andidates were able to find R and to make a worthwhile attempt at α usually via the tangent ratio. Degrees were oasionally used despite the range being given in radians. Some andidates were undeided and gave both degrees and radians, sometimes ontinuing with this throughout the question. Part (b) was frequently inorret with +5 as ommon as the orret answer of 5. Another ommon answer was and more surprisingly 0. A less ommon error was to identify the value of x for whih the maximum/minimum would our. The majority of andidates attempted part () and realised the need to use the form found in part (a). There were therefore some very good solutions, in many of whih the only error was to omit the seond orret answer. Candidates should remember to derive additional values from their prinipal value before rearranging their equation. Not many gave all three values of.6, 5. and 7.44 for (x +.87). Rounding errors were ommon with.8 and 6.5 popular answers. Question In part (a), the majority of andidates started with os θ sin θ + = and divided all terms by os θ and rearranged the resulting equation to give the orret result. A signifiant minority of andidates started with the RHS of se θ to prove the LHS of tan θ by using both se θ = and sin θ = os θ. There were a few andidates, however, who used os θ more elaborate and less effiient methods to give the orret proof. In part (b), most andidates used the result in part (a) to form and solve a quadrati equation in seθ and then proeeded to find 0 or both orret angles. Some andidates in addition to orretly solving seθ = found extra solutions by attempting to solve se θ =, usually by proeeding to write osθ =, leading to one or two additional inorret solutions. A signifiant minority of andidates, however, struggled or did not attempt to solve seθ =. sinθ A signifiant minority of andidates used tanθ = and sin θ = os θ to ahieve osθ both answers by a longer method but some of these andidates made errors in multiplying both sides of their equation by os θ. Question This question was takled with onfidene by most andidates, many of whom gained at least 8 out of the 9 marks available. In part (a), almost all andidates were able to obtain the orret value of R although + 5 = 6 was a ommon error for a few andidates, as was the use of the subtration form of Pythagoras. A minority of andidates used their value of α to find R. Some andidates inorretly wrote tanα as either 5 5, or 5 In all of these ases, suh Silver 4: 8/.

13 andidates lost the final auray mark for this part. A signifiant number of andidates found α in degrees, although many of them onverted their answer into radians. Many andidates who were suessful in part (a) were usually able to make progress with part (b) and used a orret method to find the first angle. A signifiant minority of andidates struggled to apply a orret method in order to find their seond angle. These andidates usually applied an inorret method of (π their 0.7) or (π their α their 0.7), rather than applying the orret method of (π their prinipal angle their α). Premature rounding aused a signifiant number of andidates to lose at least auray mark, notably with a solution of 0.8 instead of 0.7. Question 4 This type of question has been set quite frequently and the majority of andidates knew the method well. Most approahed the question in the onventional way by expressing the x x+. This question an, however, be made frations with the ommon denominator ( )( ) simpler by anelling down the first fration by ( x + ), obtaining x + ( x + ) = =. Those who used the ommoner method often had + ( )( ) x x x x x diffiulties with the numerator of the ombined fration, not reognising that x + = x = x + x an be used to simplify this fration. If part (a) was ompleted ( )( ) orretly, part (b) was almost invariably orret. It was possible to gain full marks in part (b) from unsimplified frations in part (a), but this was rarely ahieved. Question 5 This question tested andidates on a real life example of exponentials. Part (a) was very rarely inorret. A few andidates did write p =.5 and then in (b) substituted 7.5 on the lefthand side. This seemed to be in an attempt to get straight to 4 ln. A similar question to part (b) was asked in January and andidates had seemed to learn from that experiene. Most seemed to sore the first three marks. Some andidates still need to learn however that when an answer is given, it must be shown without doubt. For example justifying ln = ln proved diffiult for many. It was not unommon to see k = 4 ln going straight on to k = 4 ln without any explanation. Of those who did provide an adequate proof (a large minority), it was ommon to see e 4k = e 4k = used, and also k = 4 ln k = 4 ln ( ) k = 4 ln. Less ommon was k = 4 ln ( ) k = k = 4 (ln ln ). Part () was one of the more demanding parts of the paper; the derivative was not diffiult but the numbers used made the question triky. There were a pleasing number of ompletely orret solutions by and large using the method shown on the mark sheme. A small number were able to proeed suessfully with a hange to powers of. The latter method did 4 ause a lot of problems for most who attempted it; many had A ommon diffiulty was proessing the Silver 4: 8/ t rather than the orret. (ln )t and this was often aused by the lak of a 4 t 4

14 braket around the ln, so ln t was sometimes proessed in error. Most andidates went on to give a numerial answer, but it was possible to ahieve the final mark for a orret exat answer. Question 6 This question was attempted by most andidates. The big diffiulty here was in not using part (a) to help solve part (b). This is similar to what has happened in previous years and indiates that students are not aware that the question has been speifially designed to be solved in this way. Most andidates oped well with the show that in part (a). They were able to ombine two frations suessfully and onvert the expression from sin θ and os θ into single angles, although the onversion of os θ to a more appropriate form often required more than one stage of working. Those realising they should use sin θ for os θ sometimes spoiled their proof by writing sin sinθ θ on the numerator thus reahing and then osθ onveniently losing the sign. Most of those reahing the final line gave suffiient evidene of method but a few left out ruial lines. It must be stressed that show that questions require all neessary steps. Most andidates reognised the link between part (b)(i) and part (a) and used standard identities for sin 0 and os 0, subsequently providing a orret proof using or equivalents expliitly on an intermediate line, followed by simplifiation to. Note os0 that = gained only mark as the intermediate step was missing. Some sin 0 sin 0 andidates followed the alternative path offered by expanding tan( 45 0) or tan( 60 45). This group usually failed to rationalise the denominator in the surd fration of the marks. There were also some instanes of sin5 os5 +, gaining being used, with surd values used from the alulator gaining no redit at all. Similar unaeptable methods used tan (0 5) or tan (75 60). Part (b)(ii) was done very simply by good andidates who were able to write down tan x = and the 4 orret values within a ouple of lines. However, many others failed to reognise that the given equation was similar to the original expression in part (a), inevitably returning to first priniples with many attempting the Pythagorean identity ot x = ose x, often with little or no suess. Question 7 Q7(a) was usually ompleted well and most andidates were able to sore full marks. A few andidates found forming the single fration hallenging as they failed to reognise the lowest ommon denominator at the outset. In Q7(b) the majority of andidates were able to use the quotient rule orretly and a number of andidates started by quoting the rule. A number of andidates used an inorret form of the quotient rule, usually reversing the terms in the numerator. Some andidates failed to fully simplify their answer and a larger number who anelled inorretly whih resulted in the final mark being lost. The ommon error seen was to hange + 0 to ( + 5). It Silver 4: 8/ 4

15 was also ommon to see responses where andidates misunderstood the notation and tried to find the inverse funtion. Some of these did however proeed to find h (x) in Q7() and then went on to omplete Q7() suessfully. Q7() was the most demanding part of this question. Those andidates who had anelled inorretly in Q7(b) found they had an unsolvable equation and tried to rearrange their equation in an attempt to form an equation that they ould solve. Some andidates set their h (x) = 0 but then set the denominator of their derivative = 0. A number of andidates failed to reognise that the maximum value of h(x) would be at the turning point and tried evaluating h(x) = 0 or h (0). It was ommon to see andidates forming inequalities for the range using their x values instead of evaluating h(x). Of those with otherwise orret solutions, some lost the final mark by omitting the lower boundary for the range or by inorretly using a strit inequality. Question 8 Part (a) was well answered, although andidates who gave the answer to signifiant figures lost a mark. In part (b) those andidates who realised that x = but a ommon misoneption was to think that 0 (a). e 8 e 8 usually gained both marks, should be added to the answer to part Part () proved a hallenging final question, with usually only the very good andidates soring all three marks. From those who tried to solve this in one stage it was more ommon 8 to see D = or 0 or 0 or.549, instead of than , substituted into x = De t. Many andidates split up the doses but this, unfortunately, often led to a omplex expression in T, T ( T + 5) 8 8 = 0e + 0e, whih only the very best andidates were able to solve. One mark was a ommon sore for this part. Silver 4: 8/ 5

16 Statistis for C Pratie Paper Silver Level S4 Mean sore for students ahieving grade: Qu Max Modal Mean sore sore % ALL A* A B C D E U Silver 4: 8/ 6

Time: 1 hour 30 minutes

Paper Reference(s) 6665/01 Edecel GCE Core Mathematics C3 Gold Level (Harder) G3 Time: 1 hour 30 minutes Materials required for eamination Mathematical Formulae (Green) Items included with question papers