FP1 PAST EXAM QUESTIONS ON NUMERICAL METHODS:
FP1 PAST EXAM QUESTIONS ON NUMERICAL METHODS: NEWTON-RAPHSON ONLYA number of questions demand that you know derivatives of functions now not included inFP1. Just look up the derivatives in the mark scheme, and then you can use those questionsfor practice.f(x) x3 – 3x2 5x – 41.(a)Use differentiation to find f ' (x).(2)The equation f(x) 0 has a root α in the interval 1.4 x 1.5(b)Taking 1.4 as a first approximation to α, use the Newton-Raphson procedure once to obtaina second approximation to α. Give your answer to 3 decimal places.(4)3f (x) x 3.(c)7 2,xx 0Taking 1.45 as a first approximation to α, apply the Newton-Raphson procedure once to73f(x) x 2 to obtain a second approximation to α, giving your answer to 3 decimalxplaces.(5)f(x) 3x2 4.(c)11x2Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to f(x)to obtain a second approximation to α, giving your answer to 3 decimal places.(5)(Total 9 marks)5.Given that α is the only real root of the equationx3 – x2 – 6 0(b)Taking 2.2 as a first approximation to α, apply the Newton-Raphson procedure once tof (x) x3 – x2 – 6 to obtain a second approximation to α, giving your answer to 3 decimalplaces.(5)f ( x) 3 x 6.18 20x(b)Find f ′(x).(3)(c)Using x0 1.1 as a first approximation to α , apply the Newton-Raphson procedure once tof(x) to find a second approximation to α , giving your answer to 3 significant figures.(4)City of London Academy1
f(x) 4 cos x e– x .7.(a)Show that the equation f(x) 0 has a root α between 1.6 and 1.7(2)(b)Taking 1.6 as your first approximation to α, apply the Newton-Raphson procedure once tof(x) to obtain a second approximation to α. Give your answer to 3 significant figures.(4)(Total 6 marks) x f ( x) 3x 2 x tan 2, x 2 8.(4)(b)Taking 0.75 as a first approximation to α, apply the Newton–Raphson procedure once tof(x) to obtain a second approximation to α. Give your answer to 3 decimal places.(4)(Total 8 marks)f(x) x3 8x – 19.9.(c)Obtain an approximation to the real root of f(x) 0 by performing two application of theNewton-Raphson procedure to f(x), using x 2 as the first approximation. Give youranswer to 3 decimal places.(4)(d)By considering the change of sign of f(x) over an appropriate interval, show that youranswer to part (c) is accurate to 3 decimal places.(2)f(x) ln x x – 3,10.(c)x 0.Taking 2.25 as a first approximation to α, apply the Newton-Raphson process once to f(x)to obtain a second approximation to α, giving your answer to 3 decimal places.(5)(d)Show that your answer in part (c) gives α correct to 3 decimal places.(2)(Total 12 marks)11.f(x) 0.25x – 2 4sin x.(c)Taking 11 as a first approximation to a root β, use the Newton-Raphson process on f(x)once to obtain a second approximation to β. Give your answer to 2 decimal places.(6)(Total 11 marks)City of London Academy2
12.The temperature θ C of a room t hours after a heating system has been turned on is given by t 26 – 20e –0.5t, t 0.(b)Taking 1.9 as a first approximation to α, use the Newton-Raphson procedure once to obtaina second approximation to α. Give your answer to 3 decimal places.(6)f(x) 1 – ex 3 sin 2x13.The equation f(x) 0 has a root in the interval 1.0 x 1.4.(b)Taking 1.2 as a first approximation to , apply the Newton-Raphson procedure once to f(x)to obtain a second approximation to .(4)(c)By considering the change of sign of f(x) over an appropriate interval, show that youranswer to part (b) is accurate to 2 decimal places.(2)14.y7654321O123456x–1–2–3The diagram above shows part of the graph of y f(x), wheref(x) x sin x 2x – 3.The equation f(x) 0 has a single root .(a)Taking x1 1 as a first approximation to , apply the Newton-Raphson procedure once tof(x) to find a second approximation to , to 3 significant figures.(5)(b)Given instead that x1 5 is taken as a first approximation to in the Newton-Raphsonprocedure,City of London Academy3
(i)use the diagram to produce a rough sketch of y f(x) for 3 x 6,and by drawing suitable tangents, and without further calculation,(ii)show the approximate positions of x2 and x3, the second and third approximations to .(2)(Total 7 marks)15.f(x) 2x x – 4.The equation f(x) 0 has a root in the interval [1, 2].(b)Taking x 1 as a first approximation to , apply the Newton-Raphson procedure once tof(x) to obtain a second approximation to .(4)(Total 6 marks)f(x) sin 3x – 2x 1.16.(a)Show by drawing a sketch that there is just one solution of f(x) 0.(3)(b)Taking 0.8 as a first approximation to , apply the Newton-Raphson procedure twice tof(x) to find a second and a third approximation to . Give your answers to 4 significantfigures.(5)(Total 8 marks)f(x) 3x – x – 6.17.(c)Taking 2 as a first approximation to , apply the Newton-Raphson procedure once to f(x) toobtain a second approximation to . Give your answer to 3 decimal places.(4)(Total 8 marks)f(x) 2 sin 2x x – 2.18.The root of the equation f(x) 0 lies in the interval [2, ].(b)Taking 2.8 as a first approximation to , apply the Newton-Raphson procedure once to f(x)to find a second approximation to , giving your answer to 3 significant figures.(5)(Total 9 marks)19.f(x) (c)4x 3x 12 e x .Taking 0.9 as a first approximation to the root , use the Newton-Raphson procedure onceon f(x) to find a second approximation, giving your answer to 2 decimal places.(6)(d)Investigate whether or not your answer in part (c) gives correct to 2 decimal places.(3)City of London Academy4
(Total 12 marks)MARK SCHEME1.(a)f'(x) 3x2 – 6x 5(b)f(1.4) –0.136M1A1B1f'(1.4) 2.48x0 1.4, x 1 1.4 –2B1ft– 0.1362.48M1 1.455 (3 dpl)A14[6]2.End points: (4, –8) and (5, 2) –48 B15– (or equiv.)2M1α 4.8A13[3]3.(a)f (1.4) . and f (1.5) .f (1.4) 0.256 (or Evaluate bothM11732), f (1.5) 0.708 . (or)24125Change of sign, rootA12Alternative method:Graphical method could earn M1 if 1.4 and 1.5 are both indicatedA1 then needs correct graph and conclusion, i.e. change of sign rootNoteM1: Some attempt at two evaluationsA1: needs accuracy to 1 figure truncated or rounded and conclusionincluding sign change indicated (One figure accuracy sufficient)(b)f (1.45) 0.221. or 0.2 [ root is in [1.4, 1.45] ]f (1.425) 0.018. or –0.019 or –0.02 root is in [1.425, 1.45]M1M1A1 cso3NoteM1: See f(1.45) attempted and positiveM1: See f(1.425) attempted and negativeA1: is cso – any slips in numerical work are penalised here evenif correct region found.Answer may be written as 1.425 α 1.45 or 1.425 α 1.45 orCity of London Academy5
(1.425, 1.45) must be correct way round. Between is sufficient.There is no credit for linear interpolation. This is M0 M0 A0Answer with no working is also M0M0A0(c)f ( x) 3x 2 7 x 2M1 A1f (1.45) 9.636 . (Special case: f ( x) 3x 7 xthen f (1.45) 11.636. )2x1 1.45 2 2f (1.45)0.221 . 1.45 1.427f (1.45)9.636 .A1 ftM1 A1 cao5NoteM1: for attempt at differentiation (decrease in power) A1 is caoSecond A1may be implied by correct answer (do not need to see it)ft is limited to special case given.2nd M1: for attempt at Newton Raphson with their values for f(1.45)and f ′(1.45).A1: is cao and needs to be correct to 3dpNewton Raphson used more than once – isw.Special case: f′(x) 3x2 7x–2 2 then f′(1.45) 11.636.) is M1 A0A1ft M1 A0 This mark can also be given by implication fromfinal answer of 1.43[10]4.(a)f(1.3) –1.439 and f(1.4) 0.268 (allow awrt)B11M1A1A13NoteBoth answers required for B1. Accept anything that rounds to 3dpvalues above.(b)f(1.35) 0 (–0.568.) 1.35 α 1.4f(1.375) 0 (–0.146.) 1.375 α 1.4Notef(1.35) or awrt –0.6 M1(f(1.35) and awrt –0.6) AND (f(1.375) and awrt –0.1) for first A11.375 α 1.4 or expression using brackets or equivalent in wordsfor second A1(c)f′(x) 6x 22x–3f ( x0 )0.268x1 x0 –, 1.384 1.4 f ( x 0 )16.417M1A1M1A1,A15NotesOne term correct for M1, both correct for A1Correct formula seen or implied and attempt to substitute for M1City of London Academy6
awrt 16.4 for second A1 which can be implied by correct final answerawrt 1.384 correct answer only A1[9]5.(a)f (2.2) 2.23 – 2.22 – 6( –0.192)f (2.3) 2.33 – 2.32 – 6( 0.877)M1Change of sign Rootneed numerical values correct (to 1 s.f.).A12NoteM1 for attempt at f(2.2) and f(2.3)A1 need indication that there is a change of sign – (could be– 0.19 0, 0.88 0) and need conclusion. (These marks may be awarded inother parts of the question if not done in part (a))(b)f′(x) 3x2 – 2xf′(2.2) 10.12B1B1f( x 0 )– 0.192 2.2 –f' ( x 0 )10 .12x1 x 0 –M1 A1ft 2.219A1cao5NoteB1 for seeing correct derivative (but may be implied by later correct work)B1 for seeing 10.12 or this may be implied by later workM1 Attempt Newton-Raphson with their valuesA1ft may be implied by the following answer (but does not requirean evaluation)Final A1 must 2.219 exactly as shown.would get 4/5So answer of 2.21897If done twice ignore second attempt(c) – 2.2 '0.192 ' 2.3 – '0.877 '(or equivalent such as0.1 – kk .) '0.192 ' '0.877 'α (0.877 0.192) 2.3 0.192 2.2 0.877M1A1or k(0.877 0.192) 0.1 0.192, where α 2.2 kso α 2.218 (2.21796 )(Allow awrt)A13AlternativeUses equation of line joining (2.2, –0.192) to (2.3, 0.877) andsubstitutes y 0M10.192 0.877( x – 2.2) and y 0, so α 2.218 or0.1awrt as before (NB Gradient 10.69)A1, A1y 0.192 City of London Academy7
NoteM1 Attempt at ratio with their values of f(2.2) and f(2.3).N.B. If you see 0.192 – α or 0.877 – α in the fraction then this is M0A1 correct linear expression and definition of variable if not α(may be implied by final correct answer– does not need 3 dp accuracy)A1 for awrt 2.218If done twice ignore second attempt[10]6.(a)attempt evaluation of f(1.1) and f(1.2) (– looking for sign change)M1f(1.1) 0.30875, f(1.2) –0.28199 Change of sign inf(x) root in the intervalA12M1 A1 A13Noteawrt 0.3 and –0.3 and indication of sign change for first A11(b)1–13 –f’(x) x 2 – 9 x 22NoteMultiply by power and subtract 1 from power for evidenceof differentiation and award of first M1(c)f (1.1) 0.30875. f′ (1.1) –6.37086.x1 1.1 –0.30875.– 6.37086. 1.15(to 3 sig.figs.)B1 B1M1A14Noteawrt 0.309 B1and awrt –6.37 B1 if answer incorrectEvidence of Newton-Raphson for M1Evidence of Newton-Raphson and awrt 1.15 award 4/4[9]7.(a)f(1.6) .f(1.7) .(Evaluate both)0.08 (or 0.09), –0.3 One ve, one –ve or Sign change, rootM1A12Any errors seen in evaluation of f(1.6) or f(1.7) lose A markso –0.32 is A0Values are 0.0851 and –0.3327 Need concluding statement also.(b)f (x) –4 sin x–e–xCity of London AcademyB18
f(1.6)f (1.6)1.6 – 1.6 –M14 cos 1.6 e 1.6 0.085 . 1.6 1.6 4.2. ( 4 sin 1.6 e ) 1.62A1A14B1 may be awarded if seen in N–R as –4sin 1.6 – e–1.6 or as –4.2M1 for statement of Newton Raphson (sign error in rule results in M0)First A1 may be implied by correct work previously followedby correct answerDo not accept 1.620 for final A1. It must be given and correct to 3sf.1.62 may follow incorrect work and is A0No working at all in part (b) is zero marks.[6]8.(a)f (0.7) –0.195028497 and f (0.8) 0.297206781f (0.8) 0.8 f (0.7) 0.7 f (0.8)0.8 Useto obtain 0.7 f (0.7)f (0.8) f (0.7)( 0.739620991) 0.740 Answer required to 3 dp or betterM1A14Bs for 3dp or better First M for reasonable attempt usingfractions and differences.(b)1 x f (x) 6x 1 sec 2 2 2 f (0.75 )Use x2 0.75 ( 0.741087218) 0.741f (0.75 )Answer required to 3 dp or betterM1A1M1A14First M attempt to differentiate f(x), term in x is enough.Lose last A if either or both not to 3 dp[8]9.(a)f (x) 3x2 83x2 8 0. or 3x2 8 0.Correct derivative and, e.g., ‘no turning points’ or Simple sketch, (increasing, crossing positive x-axis)(Or, if the M1A1 has been scored, a reason such as ‘crossesx-axis only once’).M:B13Differentiate and consider sign of f (x), or equatef (x) to zero.City of London Academy9
AlternativeM1: Attempt to rearrange as x3 – 19 –8x or x3 19 – 8x(condone sign slips), and to sketch a cubic graph and astraight line graph.(b)(c)A1:Correct graphs (shape correct and intercepts ‘in theright place’).B1:Comment such as “one intersection, therefore one root”).Calculate f(1) and f(2) (Values must be seen)F(1) –10, f(2) 5, Sign change, Rootx1 2 –f(2)5, 2–( 1.75)20f (2)x2 x1 –f( x1 ) 0.359375 , 1.75 1.729 (ONLY)( )f ( x1 ) 17.1875 M1A12M1, A1M1, A141st A1 can be implied by an answer of 1.729, provided N.R.has been used.Answer only: No marks. The Newton-Raphson method must be seen.(d)Calculate f( – 0.0005) and f( 0.0005)(Or a ‘tighter’ interval that gives a sign change).f(1.7285) –0.0077. and f(1.7295) 0.0092., Accurate to 3 d.p.M1A12For A1, correct values of f(1.7285) and f(1.7295) must be seen,together with a conclusion. If only 1 s.f. is given in the values,allow rounded (e.g. – 0.008) or truncated (e.g. – 0.007) values.[11]10.(a)f(2.0) – 0.30685 . – 0.3069AWRT 3 d.p.f(2.5) 0.41629 0.4163both correct 4 d.p.States change of sign, so root (between 2 and 2.5)M1A1B13Note:B1 gained if candidate’s 2 values do show a change of signand statement made(b) (2) f(2)f(2) f(2.5) 0.5 orf (2.0) 2 or equivalent2.5 f (2.5)M1x0.5 xand x found f(2)f(2.5) 2.212 AWRTA1f(2.25) 0.06093 . ( 3 d.p.) [Allow ln.2.25 2.25 – 3]B1Or(c)City of London Academy210
14 13(allow 1.444) 1, f (2.25) 1.4 or 1 orx49f(2.25) 2.25 , 2.20781. 2.208f (2.25)f (x) First M in (c) is just forM1,A1AWRTM1A151 1xIf no intermediate values seen B1M1A1M1A0 is possible for2.209 or 2.21, otherwise as scheme (B1 eased to award thisif not evaluated)MR 2.5 instead of 2.25 (Answer 2.203) award on ePen B0M1A0M1A1(d)f(2.2075) , {–6.3 . 10–4}f(2.2085) , {8.1 . 10–4} Correct values ( 1 s.f.), (root in interval) so root is 2.208 to 3 d.p.M1A12A1 requires values correct ( 1 s.f.) and statement (need not saychange of sign)M can be given for candidate’s f(2.2075) and f(2.2085)Allow N–R applied at least twice more, but A1 requires 2.20794or better and statement[12]11.(a)f(0.24) 0.058, f (0.28) 0.089accept 1sfM1Change of sign (and continuity) α (0.24, 0.28)(b)f (0.26) 0.017( a (0.24,0.26))Alaccept lsf2M1f (0.25) 0.020 ( a (0.25, 0.26))f (0.255) 0.001(c)f (11) 0.0534f ' ( x) a (0.255, 0.26)M1 A1at least 3sf2 cos x 1 4 xf (11) 0.3438 11 3B1M1 A1at least 2sf0.0534 11 .160.3438If f (11) –0.3438 is produced without working, this is to beaccepted for three marks M1 A1 A1.A1M1 A16[11]12.(a)f(1.8) 19.6686. – 20 – 0.3313.City of London AcademyAllow awrt 0.33B111
f(2) 20.6424. – 20 0.6424. 1.8 2 , 1.8 "0.33"(b)"0.64" Allow awrt 0.640.33 0.2 0.33 0.64 B11.87M1, A1f(1.9) 0.1651795., or just 1.9 6 – 20e–05 1.9 Allow awrt 0.165f (t) 1 10e–0.5 1.9(c)Allow awrt 4.870.16518 1.8664.867410112 (min)B1M1 A1f (1.9) 4.8674., or just 1 10e–05 1.9 2 1.9 –4A1M1 A16B11(1 h 52 m)[11](a)“Answer only” does not score any marks.Do not allow (α 1.8) or (2 α) for the M mark, but allow ‘minus slips’.‘0.33’ and ‘0.64’ used the ‘wrong way round’ scores the M mark.Further applications of linear interpolation: isw, and accept 1.87 if seen at end.(b)“Answer only” does not score any marks.M for differentiation: evidence from one non–constant term is sufficient.The B1 for f(1.9) and the A1 for f (1.9) can be implied by the answer 1.866.For failure to round answers to the required accuracy, penalise a maximum of one mark inthe question (at the first occurrence).Special case in part (a):‘19.67’ and ‘20.64’ used instead of 0.33 and 0.64, to give 1.90 scores B0B0M1A1.13.N.B. f(1) 1.0 ., f(1.1) 0.42 , f(1.2) – 0.2937 .f(1.15) 0.078 ., f(1.4) – 2.05.(a)f(1.2) – 0.2937 .f(1.2) to 1sf or betterB1f(1.1) 0.42 . and f(1.15) 0.078 .Attempt f(1.1), f(1.15)M1 1.2(b)f (x) 6cos2x – ex 0.2937 .x2 1.2 –f (1.2) 1.16(2 .)A.W.R.T. 1.16City of London AcademyA1 c.a.o.3M1 A1M1A1412
(c)f(1.155) 0.04. . (change of sign) 1.16f(1.165) –0.029., M1, A12N.B. f (1.2) –7.744 [9]14.(a)Correct method for f (x); xcosx sinx 2f(1) –0.1585, f (1) 3.382 or better seenUsing N-R correctly: u1 1 –" 0.1585 "; 1.05 (3 s.f.)"3.382 "M1A1A1M1A15M1A12[Notes: Answer 1.047, 1.05 implies second A mark](b)Two tangents drawn, one at {5, f(5)}, the other at {x2, f(x2)}x2, x3 marked in appropriate positions[7]15.(a)(b)f(1) -1 and f(2) 22 2 1 1 1 13B1f (x) 2x ln 2 1Attempts f (x)M1f (x) 2 ln 2 1( 1) 1–2 ln 2 1Uses Newton RaphsonA1 2 ln 2 2 1.4192 ln 2 1any correct answerB12dM1A14[6]City of London Academy13
f(x)x3x16.(a)–1y sin 3xy 2x – 11 point where they meetB1B1B13f(x)xorShapeAsymptotic behaviour to y – 2x 1Cross x-axis once comment(b)f (x) 3cos 3x – 20.075u1 0.8 – 4.212 0.8179u2 0.8177B1B1B1Attempt to diff. cos3x two terms for M1M1, A1M1A1A15[8]17.(a)(b)(c)f(1) –4,f(2) 1bothM1change of sign (and continuity) implies 1 2A1f(1.5) –2.3 1.5 2f(1.75) –0.9 1.75 2f(1.875) –0.03 1.875 2B1f (x) 3x ln 3 – 1attempt at differentiation using lns can be implied by next lineM1f (2) 8.8875 A1B1222accept 3 ln 3 – 1, awrt 8.888City of London Academy14
use of numerical differentiation button is acceptable1 1.8878.8875 .use of N-R, cao 2 2 M1, A14[8]Notes: Incorrect method of differentiation is M0, A0, M1, A0For example, f (x) x3x–1 – 1 x 1.8 is ¼.The exact answer is 1.8789 Alternative to 4 (a): Use of a diagramy987 3 01 y x 6y 3x2xTwo graphs with domain at least [1, 2] with one intersection3, 9 and 7, 818.(a)f(2) 1.514f( ) 1.142M1A12B1B11.1422–1.514 1.142 2 1.514 1.514 2 1.142 (1.142 1.514) 2.65(b)f (x) 4 cos 2x 1k cos 2x cM1A1M1f(2.8) 0.4625B1f (2.8) 4.1023A1x2 2.8 ( 0.4625 )4.1023 2.91 only4M1A15[9]19.(a)f(0.9) 0 .6 e 0.9 0.07511.81City of London AcademyB115
(b)(c)f(1.0) 0.5 – e 1 0.1321B12f(x) is continuous and changes sign between x 0.9 and x 1 root existsB11f (x) 4( x 2 1) 2 x(4 x 3) e x22( x 1)f (0.9) 2.2869Newton-Raphson process gives 2nd approximation0.0751 0.9 2.2869 0.93 (2 decimal places)(d)f(0.925) (0.935) 0 .7 e 0.925 0.0191.85560.74 e 0.935 0.00221.8742Since root lies in the interval (0.925, 0.935) due to sign change 0.93 is correct to 2 decimal placesM1A16M1A1A13[12]City of London Academy16
City of London Academy 3 12. The temperature θ C of a room t hours after a heating system has been turned on is given by t 26 – 20e –0.5t t 0. (b) Taking 1.9 as a first approximation to α, use the
Past exam papers from June 2019 GRADE 8 1. Afrikaans P2 Exam and Memo 2. Afrikaans P3 Exam 3. Creative Arts - Drama Exam 4. Creative Arts - Visual Arts Exam 5. English P1 Exam 6. English P3 Exam 7. EMS P1 Exam and Memo 8. EMS P2 Exam and Memo 9. Life Orientation Exam 10. Math P1 Exam 11. Social Science P1 Exam and Memo 12.
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Certification Exam Structure? Exam Name CompTIA A Exam Code 220-1101 Exam Price 239 (USD) Duration 90 mins Number of Questions 90 Passing Score 675 / 900 Books / Training CertMaster Learn for A Schedule Exam Pearson VUE Sample Questions CompTIA A Core 1 Sample Questions Practice Exam CompTIA 220-1101 Certification Practice Exam
Questions 1- 307 have been taken from the previous set of Exam C sample questions . Questions no longer relevant to the syllabus have been deleted. Some of the questions in this study note are taken from past examinations. The weight of topics in these sample questions is not representative of the weight of topics on the exam.
Past Exam Questions on Sections 6b and 6c 360 §6d Outstanding Balance, Interest Paid, and Principal Repaid under the Sinking Fund Method 369 Summary of Concepts and Formulas in Section 6d 372 Past Exam Questions on Section 6d 373 §6e Varying Series of Payments 375 Summary of Concepts and Formulas in Section 6e 378 Past Exam Questions on .
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FINAL EXAM: The final exam will cover chapter 11, 13 and 15. There will be no make-up exam for the final exam. The final exam will count 100 points. The final exam will be 40 questions. The format will be multiple-choice. Only the materials covered in the lectures will be on the exam and you will have designated class time to finish the exam.
