6665/01 Edexcel GCE Core Mathematics C3 Bronze Level B3

Transcription

1 Paper Reference(s) 6665/0 Edecel GCE Core Mathematics C3 Bronze Level B3 Time: hour 30 minutes Materials required for eamination papers Mathematical Formulae (Green) Items included with question 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, differentiation and integration, or have retrievable mathematical formulas stored in them. Instructions to Candidates Write the name of the eamining body (Edecel), your centre number, candidate number, the unit title (Core Mathematics C3), the paper reference (6665), your surname, initials and signature. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. There are 7 questions in this question paper. The total mark for this paper is 75. 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 gain no credit. Suggested grade boundaries for this paper: A* A B C D E Bronze 3 This publication may only be reproduced in accordance with Edecel Limited copyright policy Edecel Limited.

2 . The curve C has equation y = ( 3) 5 The point P lies on C and has coordinates (w, 3). Find (a) the value of w, (b) the equation of the tangent to C at the point P in the form y = m + c, where m and c are constants. (5) January 03. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f() = 0 in the interval [.,.3]. (b) Show that the equation f() = 0 can be written in the form = sin + 4 (c) Use the iterative formula n+ = sin n + 4, 0 =.5, to calculate the values of, and 3, giving your answers to 4 decimal places. (d) By considering the change of sign of f() in a suitable interval, verify that α =.9 correct to 3 decimal places. June 00 Bronze 3: 3/

3 3. The functions f and g are defined by f : + 3, g : 3 4, (a) State the range of f. (b) Find fg(). (c) Find g, the inverse function of g. (d) Solve the equation gg() + [g()] = 0 (5) June 03 (R) 4. (i) Differentiate with respect to (a) y = 3 ln, (b) y = ( + sin ) 3. (6) Given that = cot y, (ii) show that d y = d +. (5) January 03 Bronze 3: 3/ 3

4 5. The functions f and g are defined by (a) State the range of f. f: e +, R, g : ln, > 0. (b) Find fg(), giving your answer in its simplest form. (c) Find the eact value of for which f( + 3) = 6. (d) Find f, the inverse function of f, stating its domain. (e) On the same aes sketch the curves with equation y = f() and y = f (), giving the coordinates of all the points where the curves cross the aes. (4) () (4) June 0 6. f() = (a) Show that f () = 0 has a root, α, between =.4 and =.45. (b) Show that the equation f () = 0 can be written as = (c) Starting with 0 =.43, use the iteration +, 0. 3 n + = n + 3 to calculate the values of, and 3, giving your answers to 4 decimal places. (d) By choosing a suitable interval, show that α =.435 is correct to 3 decimal places. June 008 Bronze 3: 3/ 4

5 7. (a) Epress cos 3 3 sin 3 in the form R cos (3 + α), where R and α are constants, R > 0 and 0 < α < π. Give your answers to 3 significant figures. (4) f() = e cos 3. (b) Show that f () can be written in the form f () = Re cos (3 + α ), where R and α are the constants found in part (a). (c) Hence, or otherwise, find the smallest positive value of for which the curve with equation y = f() has a turning point. (5) June 0 END TOTAL FOR PAPER: 75 MARKS Bronze 3: 3/ 5

6 Question Number Scheme Marks. (a) (b) 3 = (w 3) w= oe MA dy d ( 3) or 0( 3) = MA When =, Gradient = 60 M Equation of tangent is '60' = y ( 3) ' ' oe dm y = 60 cso A. (a) f (.) = , f (.3) = Sign change (and as f( ) is continuous) therefore a root α is such that α [.,.3] MA (b) 4cosec 4 + = 0 4 = 4cosec + M = cosec + = + A 4 sin 4 (c) = + M sin(.5) 4 = , = A 3 = A (d) f (.905) = , f (.95) = Sign change (and as f( ) is continuous) therefore a root α is such that (.905,.95) α α =.9 (3 dp) M A (5) [7] [9] Bronze 3: 3/ 6

7 Question Number Scheme Marks 3. (a) f( ) 3 MA (b) An attempt to find when = M (c ) Correct answer fg() = 5 A 3 y y = 3 4 4= 3 y = 4 M 3 g ( ) = 4 A (d) [ ] g ( ) = (3 4 ) B gg( ) = 3 4(3 4 ) M [ ] gg ( ) + g ( ) = = 0 = A (i)(a) (i)(b) (ii) Uses in 8 ( ) = 0 = 0, 0.5 oe MA (5 ) [] dy 3 = 3 ln + MAA d = 3 ln + dy 3 ( cos ) d = ( + sin ) + BMA d cosec y dy = MA dy d = cosec y M cosec y = + cot yand = cot y in d y d or d to get an epression dy dy = = = d cosec y + cot y + cso M, A* (5) [] Bronze 3: 3/ 7

8 Question Number Scheme Marks 5. (a) f()> B () (b) = e + = + M, A ln fg( ), (c) e + = 6 e = 4 M A + 3 = ln 4 ln = or ln- M A (4) (d) Let y = e + y = e ln( y ) = M (e) = > A, Bft f ( ) ln( ),. Shape for f() B (0, 3) B Shape for f - () B (3, 0) B (4) [4] Bronze 3: 3/ 8

9 Question Number 6. (a) f (.4) = < 0 Scheme f (.45) = 0.45 > 0 M Marks Change of sign (and continuity) α (.4,.45) A 3 (b) 3 = = + 3 = 3 + M A = + 3 (c) =.437 B =.4347 B A cso 3 =.4355 B (d) Choosing the interval (.4345,.4355 ) or appropriate tighter interval. M f (.4345) = f (.4355) = M Change of sign (and continuity) α (.4345,.4355) α =.435, correct to 3 decimal places cso A [] Bronze 3: 3/ 9

10 Question Scheme Marks Number 7. (a) R = + 3 M R = 3 or 3.6 A tan α = 3 M α = A (4) (b) f ( ) cos3 3 sin 3 = e e MAA = e (cos3 3sin 3 ) M = e ( Rcos(3 + a)) = Re cos(3 + a) A * cso (c) f ( ) = 0 cos(3 + a) = 0 M π 3+ a = M = 0.96 awrt 0.0 A (5) [] Bronze 3: 3/ 0

11 Question Eaminer reports This was an accessible question for candidates with many gaining all 7 marks. Q(a) was very well done and usually correct. Occasionally +3 was used instead of giving w = 5. Some candidates reached w = as a result of writing ( 3) =. dy For Q(b) the correct epression for was usually achieved with nearly all candidates d recognising the need to apply the chain rule. The substitution of w = occasionally produced dy 3 60 instead of +60. Some candidates mistakenly set = 0, leading to = and d proceeded to use this as their gradient. A few used the result mm = and went on to find the gradient of the normal. The method mark for finding the equation of a line was gained by almost all candidates but a number of candidates lost the final accuracy mark as a result of errors made in arranging the answer in the form y = m + c. An eample of this was y + 3 = followed by y = Question All four parts of this question were well answered by the overwhelming majority of candidates who demonstrated their confidence with the topic of iteration with around 65% of candidates gaining at least 8 of the 9 marks available. Some candidates in parts (a), (c) and (d) worked in degrees even though it was stated in the question that was measured in radians. In part (a), the majority of candidates evaluated both f (.) and f (.3), although a very small number choose instead to evaluate both f (.5) and f (.35). A few candidates failed to conclude sign change, hence root as minimal evidence for the accuracy mark. Most candidates found the proof relatively straightforward in part (b). A small number of candidates lost the accuracy mark by failing to eplicitly write 4cosec 4 + as equal to 0 as part of their proof. Part (c) was almost universally answered correctly, although a few candidates incorrectly gave as.3037 or 3 as.98. The majority of candidates who attempted part (d) choose an appropriate interval for and evaluated f( ) at both ends of that interval. The majority of these candidates chose the interval (.905,.95 ) although incorrect intervals, such as ( ).90,.9 were seen. There were a few candidates who chose the interval(.905,.94 ). This probably reflects a misunderstanding of the nature of rounding but a change of sign over this interval does establish the correct result and this was accepted for full marks. To gain the final mark, candidates are epected to give a reason that there is a sign change, and give a suitable conclusion such as that the root is.9 to 3 decimal places or α =.9or even QED. A minority of candidates who attempted part (d) by using a repeated iteration technique received no credit because the question required the candidate to consider a change of sign of f ( ). Bronze 3: 3/

12 Question 3 On the whole this question was well attempted. In part (a) most candidates realised that the 3 was significant, although some stated f() > 3, or occasionally f() < 3. Many successful candidates sketched a graph. Some candidates stated f() 0 or f().5. However the majority answered f() 3 correctly. A range of appropriate notation was seen. Only a few candidates gave the range in terms of. In part (b), most candidates processed the functions in the correct order and realised the significance of the modulus. Occasionally the modulus was omitted, and some found values using both = and in f() = (3 4) + 3. Candidates generally achieved the correct function in part (c), although a few left their answer in terms of y. A few mistakenly attempted, g() or g (). g( ) Part (d) was also done well. A few candidates found only one solution as a result of cancelling the instead of factorising. There were not many numerical errors. Some candidates used as an alternative method letting g() = t and obtaining 3 4t + t = 0, which they solved for t and hence found. Question 4 Q4(i)(a) was generally answered well, with the vast majority using the product rule. Many candidates started the question by quoting the rule. There were some errors in differentiating ln, obtaining either or. In Q4(i)(b) most candidates recognised the need for the chain rule. However a considerable number obtained only cos when differentiating + sin. Of those who performed this differentiation correctly, a significant number lost marks because of incorrect bracketing in their answer. Another common error was to omit the factor when differentiating sin. Common incorrect answers were 3( + sin ) cos, 3( + sin ) + cos or 3( + sin ) ( + cos ). Candidates responses in Q4(ii) showed many concise and clear solutions. A majority d epressed in terms of y, although many did not know, or use the formula sheet, to state dy that the differential of cot y is cosec y. These candidates generally differentiated either cos y or, often correctly. A number of able candidates proceeded more directly by tan y sin y using implicit differentiation. Most candidates knew that they then had to invert their result to d y find, although there were many instances of negative signs appearing or disappearing d without any justification. Candidates who used the identity cosec y = + cot y could usually reach the correct result quite efficiently, although some failed to bracket the terms and hence obtained cosec y = + cot y. However some correct use of other trig identities was also seen. At this stage some candidates confused the and y so could not reach the stated result. Bronze 3: 3/

13 Question 5 Numerous candidates could score high marks on this question, and completely correct solutions were frequently seen. The range and domain were the least well done parts of this question. In part (a) there was often poor use of notation, with many candidates still confusing the appropriate use of y or f() with that of for the range and domain of the functions ( > was unacceptable in part (a)). A surprising number of candidates gave the range of f() in part (a) as f() 3 rather than f() >. In part (b) most candidates applied the functions in the correct order and were able to simplify their epression correctly. Zero scoring attempts were very rare, but there was a small proportion of candidates who only got as far as e ln + and either did not simplify, or tried to solve e ln + = 0 instead. Part (c) tested candidate s use of ln. It was answered etremely well, although with some very poor ln work was evident amongst weaker candidates. Errors in this question were essentially of three types: incorrect epressions formed by not understanding composite functions, using (f) + 3 instead of f( + 3); missing +, giving e + 3 = 6; and incorrect ln work in solving. Many candidates appeared to forget to state the domain in part (d). Some of those who gave a domain followed through their answer to part (a), but some gave a correct answer even if they had part (a) incorrect. Missing brackets was etremely rare. In part (e) some candidates produced very careful and accurate graphs, but a large variety of shapes were produced by a minority, particularly for f () Some candidates also failed to give correct coordinates for the intersections with the aes; (0, ) and (, 0) were often seen, but other values did occur, or else no coordinates were given at all. Generally there was less success with the intercepts than the shapes. It was also not uncommon to find the two sketches intersecting. There were a few cases only of sketches having a ma or min. Some candidates also illustrated the asymptotes, which were not required, but showed a full understanding of the functions Question 6 In parts (a) and (d), candidates need to be aware that showing that something is true requires them to give reasons and conclusions. In this part (a), it is sufficient to say that a change of sign in the interval (.4,.45) implies that there is a root in the interval (.4,.45). In part(c), it would be sufficient to argue that a change of sign in the interval (.4345,.4355) implies that there is a root in the interval (.4345,.4355) and, hence, that =.435 is accurate to 3 decimal places. Part (b) was very well done but candidates must put all steps in a proof and not leave it to the eaminers to fill in important lines. Part (c) was very well done. Some candidates attempted part (d) using repeated iteration but the wording of the question precludes such a method and no marks could be gained this way. Question 7 Part (a) seemed well understood and there was a lot of correct work here. R was almost always found correctly, although there were some simple arithmetic errors where candidates had not use a calculator, for eample R = leading to R = 0 or 5. There were more errors in the tan alpha part; the most common mistake was giving tan θ as 3 or 3. A significant number used degrees here while others gave α in terms of π. Several candidates rounded their radian answer to 0.98, or Bronze 3: 3/ 3

14 In part (b) the differentiation was well done and most candidates achieved the first three d marks. There were some sign mistakes and (e ) = e also appeared. Only a few d candidates omitted the and the 3, possibly because they could see where they were heading. Most were then able to complete the proof, connecting their answer with part (a), although there was some unnecessary repetition of the working from part (a). As so often happens, with show that questions, many lost marks by failing to show or missing out the factorising stage, candidates possibly unaware of the importance of it. Quite surprisingly part (c) was found to be very demanding to all but the best of candidates. Many wrote e cos (3 + α) = 0 but did not seem to realise that e = 0 had no solution and hence went no further. Some confused smallest with finding the minimum value so wrote cos (3 + α) = ± followed by 3 + α = π or π. Some put 3 + α = 0. Bronze 3: 3/ 4

15 Statistics for C3 Practice Paper Bronze Level B3 Mean score for students achieving grade: Qu Ma Modal Mean score score % ALL A* A B C D E U Bronze 3: 3/ 5