FP PAST EXAM QUESTIONS ON NUMERICAL METHODS: NEWTON-RAPHSON ONLY A number of questions demand that you know derivatives of functions now not included in FP. Just look up the derivatives in the mark scheme, and then you can use those questions for practice.. f() = 3 3 + 5 4 (a) Use differentiation to find f ' (). The equation f() = 0 has a root α in the interval.4 < <.5 Taking.4 as a first approimation to α, use the Newton-Raphson procedure once to obtain a second approimation to α. Give your answer to 3 decimal places. () 3. f () = 3 7, > 0 Taking.45 as a first approimation to α, apply the Newton-Raphson procedure once to 3 7 f() = to obtain a second approimation to α, giving your answer to 3 decimal places. 4. f() = 3 Taking.4 as a first approimation to α, apply the Newton-Raphson procedure once to f() to obtain a second approimation to α, giving your answer to 3 decimal places. (5) (Total 9 marks) (5) 5. Given that α is the only real root of the equation 3 6 = 0 Taking. as a first approimation to α, apply the Newton-Raphson procedure once to f () = 3 6 to obtain a second approimation to α, giving your answer to 3 decimal places. (5) 6. Find f (). 8 f ( ) 3 0 Using 0 =. as a first approimation to α, apply the Newton-Raphson procedure once to f() to find a second approimation to α, giving your answer to 3 significant figures. (3) City of London Academy
7. f() = 4 cos + e. (a) Show that the equation f() = 0 has a root α between.6 and.7 () Taking.6 as your first approimation to α, apply the Newton-Raphson procedure once to f() to obtain a second approimation to α. Give your answer to 3 significant figures. (Total 6 marks) 8. f ( ) 3 tan, Taking 0.75 as a first approimation to α, apply the Newton Raphson procedure once to f() to obtain a second approimation to α. Give your answer to 3 decimal places. (Total 8 marks) 9. f() = 3 + 8 9. (d) Obtain an approimation to the real root of f() = 0 by performing two application of the Newton-Raphson procedure to f(), using = as the first approimation. Give your answer to 3 decimal places. By considering the change of sign of f() over an appropriate interval, show that your answer to part is accurate to 3 decimal places. () 0. f() = ln + 3, > 0. (d) Taking.5 as a first approimation to α, apply the Newton-Raphson process once to f() to obtain a second approimation to α, giving your answer to 3 decimal places. Show that your answer in part gives α correct to 3 decimal places. (5) () (Total marks). f() = 0.5 + 4sin. Taking as a first approimation to a root β, use the Newton-Raphson process on f() once to obtain a second approimation to β. Give your answer to decimal places. (6) (Total marks) City of London Academy
. The temperature θ C of a room t hours after a heating system has been turned on is given by = t + 6 0e 0.5t, t 0. Taking.9 as a first approimation to α, use the Newton-Raphson procedure once to obtain a second approimation to α. Give your answer to 3 decimal places. (6) 3. f() = e + 3 sin 4. The equation f() = 0 has a root in the interval.0 < <.4. Taking. as a first approimation to, apply the Newton-Raphson procedure once to f() to obtain a second approimation to. By considering the change of sign of f() over an appropriate interval, show that your answer to part is accurate to decimal places. () y 7 6 5 4 3 O 3 4 5 6 3 The diagram above shows part of the graph of y = f(), where The equation f() = 0 has a single root. (a) f() = sin + 3. Taking = as a first approimation to, apply the Newton-Raphson procedure once to f() to find a second approimation to, to 3 significant figures. Given instead that = 5 is taken as a first approimation to in the Newton-Raphson procedure, (5) City of London Academy 3
(i) use the diagram to produce a rough sketch of y = f() for 3 6, and by drawing suitable tangents, and without further calculation, (ii) show the approimate positions of and 3, the second and third approimations to. () (Total 7 marks) 5. f() = + 4. The equation f() = 0 has a root in the interval [, ]. Taking = as a first approimation to, apply the Newton-Raphson procedure once to f() to obtain a second approimation to. (Total 6 marks) 6. f() = sin 3 +. (a) Show by drawing a sketch that there is just one solution of f() = 0. Taking 0.8 as a first approimation to, apply the Newton-Raphson procedure twice to f() to find a second and a third approimation to. Give your answers to 4 significant figures. (5) (Total 8 marks) (3) 7. f() = 3 6. Taking as a first approimation to, apply the Newton-Raphson procedure once to f() to obtain a second approimation to. Give your answer to 3 decimal places. (Total 8 marks) 8. f() = sin +. The root of the equation f() = 0 lies in the interval [, ]. Taking.8 as a first approimation to, apply the Newton-Raphson procedure once to f() to find a second approimation to, giving your answer to 3 significant figures. (5) (Total 9 marks) 9. (d) 4 3 f() = e. Taking 0.9 as a first approimation to the root, use the Newton-Raphson procedure once on f() to find a second approimation, giving your answer to decimal places. Investigate whether or not your answer in part gives correct to decimal places. (6) (3) City of London Academy 4
(Total marks) MARK SCHEME. (a) f'() = 3 6 + 5 A f(.4) = 0.36 B f'(.4) =.48 0 =.4, =.4 0.36.48 Bft =.455 (3 dpl) A 4 [6]. End points: (4, 8) and (5, ) B 4 5 (or equiv.) 8 α = 4.8 A 3 [3] 3. (a) f(.4)... and f(.5)... Evaluate both 3 7 f(.4) 0.56 (or ), f(.5) 0.708... (or ) 5 4 Change of sign, root A Alternative method: Graphical method could earn if.4 and.5 are both indicated A then needs correct graph and conclusion, i.e. change of sign root : Some attempt at two evaluations A: needs accuracy to figure truncated or rounded and conclusion including sign change indicated (One figure accuracy sufficient) f (.45) 0... or 0. [root is in [.4,.45] ] f (.45) 0.08... or 0.09 or 0.0 root is in [.45,.45] A cso 3 : See f(.45) attempted and positive : See f(.45) attempted and negative A: is cso any slips in numerical work are penalised here even if correct region found. Answer may be written as.45 α.45 or.45 < α <.45 or City of London Academy 5
(.45,.45) must be correct way round. Between is sufficient. There is no credit for linear interpolation. This is M0 M0 A0 Answer with no working is also M0M0A0 f ( ) 3 7 A f (.45) 9.636... (Special case: f ( ) 3 7 + then f (.45).636... ) A ft f (.45) 0....45.45 f (.45) 9.636....47 A cao 5 : for attempt at differentiation (decrease in power) A is cao Second Amay be implied by correct answer (do not need to see it) ft is limited to special case given. nd : for attempt at Newton Raphson with their values for f(.45) and f (.45). A: is cao and needs to be correct to 3dp Newton Raphson used more than once isw. Special case: f () = 3 + 7 + then f (.45) =.636...) is A0 Aft A0 This mark can also be given by implication from final answer of.43 [0] 4. (a) f(.3) =.439 and f(.4) = 0.68 (allow awrt) B Both answers required for B. Accept anything that rounds to 3dp values above. f(.35) < 0 ( 0.568...).35 < α <.4 A f(.375) < 0 ( 0.46...).375 < α <.4 A 3 f(.35) or awrt 0.6 (f(.35) and awrt 0.6) AND (f(.375) and awrt 0.) for first A.375 < α <.4 or epression using brackets or equivalent in words for second A f () 6 + 3 A f ( 0 ) 0.68 = 0.4, =.384 A,A 5 f ( ) 6.47 s 0 One term correct for, both correct for A Correct formula seen or implied and attempt to substitute for City of London Academy 6
awrt 6.4 for second A which can be implied by correct final answer awrt.384 correct answer only A [9] 5. (a) f (.) =. 3. 6 (= 0.9) f (.3) =.3 3.3 6 (= 0.877) Change of sign Root need numerical values correct (to s.f.). A for attempt at f(.) and f(.3) A need indication that there is a change of sign (could be 0.9<0, 0.88>0) and need conclusion. (These marks may be awarded in other parts of the question if not done in part (a)) f () = 3 B f (.) = 0. B f( 0 ) f' ( ) 0 0. 0.9 0. Aft =.9 Acao 5 B for seeing correct derivative (but may be implied by later correct work) B for seeing 0. or this may be implied by later work Attempt Newton-Raphson with their values Aft may be implied by the following answer (but does not require an evaluation) Final A must.9 eactly as shown. So answer of.897 would get 4/5 If done twice ignore second attempt..3 k 0. k (or equivalent such as. ) '0.9' '0.877' ' 0.9' '0.877' α (0.877 + 0.9) =.3 0.9 +. 0.877 A or k(0.877 + 0.9) = 0. 0.9, where α =. + k so α.8 (.796 ) (Allow awrt) A 3 Alternative Uses equation of line joining (., 0.9) to (.3, 0.877) and substitutes y = 0 0.9 0.877 y 0.9 (.) 0. and y = 0, so α.8 or A, A awrt as before (NB Gradient = 0.69) City of London Academy 7
Attempt at ratio with their values of ± f(.) and ± f(.3). N.B. If you see 0.9 α or 0.877 α in the fraction then this is M0 A correct linear epression and definition of variable if not α (may be implied by final correct answer does not need 3 dp accuracy) A for awrt.8 If done twice ignore second attempt [0] 6. (a) attempt evaluation of f(.) and f(.) ( looking for sign change) f(.) = 0.30875, f(.) = 0.899 Change of sign in f() root in the interval A awrt 0.3 and 0.3 and indication of sign change for first A 3 f () = 9 A A 3 Multiply by power and subtract from power for evidence of differentiation and award of first f (.) = 0.30875.. f (.) = 6.37086... B B. 0.30875... 6.37086.. =.5(to 3 sig.figs.) A 4 awrt 0.309 Band awrt 6.37 B if answer incorrect Evidence of Newton-Raphson for Evidence of Newton-Raphson and awrt.5 award 4/4 [9] 7. (a) f(.6)=... f(.7) =... (Evaluate both) 0.08 (or 0.09), 0.3 One +ve, one ve or Sign change, root A Any errors seen in evaluation of f(.6) or f(.7) lose A mark so 0.3 is A0 Values are 0.085 and 0.337 Need concluding statement also. f() = 4 sin e B City of London Academy 8
f(.6).6 f (.6).6 4 cos.6 e 0.085... =.6.6.6 ( 4 sin.6 e ) 4... A =.6 A 4 B may be awarded if seen in N R as 4sin.6 e.6 or as 4. for statement of Newton Raphson (sign error in rule results in M0) First A may be implied by correct work previously followed by correct answer Do not accept.60 for final A. It must be given and correct to 3sf..6 may follow incorrect work and is A0 No working at all in part is zero marks. [6] 8. (a) f (0.7) = 0.9508497 and f (0.8) = 0.970678 0.8 f (0.8) 0.8 f (0.7) 0.7 f (0.8) Use to obtain = 0.7 f (0.7) f (0.8) f (0.7) A 4 (= 0.7396099) = 0.740 Answer required to 3 dp or better Bs for 3dp or better First M for reasonable attempt using fractions and differences. f() = 6 + sec A f (0.75) Use = 0.75 (= 0.740878) = 0.74 A 4 f (0.75) Answer required to 3 dp or better First M attempt to differentiate f(), term in is enough. Lose last A if either or both not to 3 dp [8] 9. (a) f() = 3 + 8 3 + 8 = 0... or 3 + 8 > 0... Correct derivative and, e.g., no turning points or increasing function. y 3 Simple sketch, (increasing, crossing positive -ais) (Or, if the A has been scored, a reason such as crosses -ais only once ). B 3 M: Differentiate and consider sign of f(), or equate f() to zero. City of London Academy 9
Alternative : Attempt to rearrange as 3 9 = 8 or 3 = 9 8 (condone sign slips), and to sketch a cubic graph and a straight line graph. A: Correct graphs (shape correct and intercepts in the right place ). B: Comment such as one intersection, therefore one root ). Calculate f() and f() (Values must be seen) F() = 0, f() = 5, Sign change, Root A f() 5 =, = (=.75), A f () 0 f( ) 0.359375 =,.75 =.79 (ONLY)(), A 4 f ( ) 7. 875 st A can be implied by an answer of.79, 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(.785) = 0.0077... and f(.795) = 0.009..., Accurate to 3 d.p. A For A, correct values of f(.785) and f(.795) must be seen, together with a conclusion. If only s.f. is given in the values, allow rounded (e.g. 0.008) or truncated (e.g. 0.007) values. [] 0. (a) f(.0) = 0.30685. = 0.3069 AWRT 3 d.p. f(.5) = 0.469 = 0.463 both correct 4 d.p. A States change of sign, so root (between and.5) B 3 : B gained if candidate s values do show a change of sign and statement made f() f (.0) () 0.5 or or equivalent f() f(.5).5 f (.5) 0.5 Or and found f() f(.5) =. AWRT A f(.5) = 0.06093. ( 3 d.p.) [Allow ln..5 +.5 3] B City of London Academy 0
f() = 4 3, f (.5).4 or or (allow.444),a 4 9 f(.5).5, =.078... =.08 AWRT A 5 f (.5) First M in is just for + If no intermediate values seen BAA0 is possible for.09 or., otherwise as scheme (B eased to award this if not evaluated) MR.5 instead of.5 (Answer.03) award on epen B0A0A (d) f(.075) =, { 6.3. 0 4 } f(.085) =, {8.. 0 4 } Correct values (> s.f.), (root in interval) so root is.08 to 3 d.p. A A requires values correct (> s.f.) and statement (need not say change of sign) M can be given for candidate s f(.075) and f(.085) Allow N R applied at least twice more, but A requires.0794 or better and statement []. (a) f(0.4) 0.058, f (0.8) = 0.089 accept sf Change of sign (and continuity) α (0.4, 0.8) Al f (0.6) 0.07 ( a (0.4,0.6)) accept lsf f (0.5) 0.00 ( a (0.5, 0.6)) f (0.55) 0.00 a (0.55, 0.6) A 3 f () 0.0534 at least 3sf B f '( cos ) 4 A f() 0.3438 at least sf A 0.0534.6 0.3438 If f() 0.3438 is produced without working, this is to be accepted for three marks A A. A 6 []. (a) f(.8) = 9.6686... 0 = 0.333... Allow awrt ± 0.33 B City of London Academy
f() = 0.644... 0 = 0.644... Allow awrt ± 0.64 B.8, "0.33" "0.64".8 0.33 0. 0.33 0.64.87, A 4 f(.9) 0.65795..., or just.9 + 6 0e 05.9 Allow awrt 0.65 B f(t) = + 0e 0.5.9 A f(.9) = 4.8674..., or just + 0e 05.9 Allow awrt 4.87 A 0.658 =.9.866 4.86740 A 6 (min) ( h 5 m) B [] (a) Answer only does not score any marks. Do not allow (α +.8) or ( + 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.87 if seen at end. Answer only does not score any marks. M for differentiation: evidence from one non constant term is sufficient. The B for f(.9) and the A for f(.9) can be implied by the answer.866. For failure to round answers to the required accuracy, penalise a maimum of one mark in the question (at the first occurrence). Special case in part (a): 9.67 and 0.64 used instead of 0.33 and 0.64, to give.90 scores B0B0A. 3. N.B. f() =.0., f(.) = 0.4, f(.) = 0.937. f(.5) = 0.078., f(.4) =.05. (a) f(.) = 0.937. B f(.) to sf or better f(.) = 0.4. and f(.5) = 0.078. Attempt f(.), f(.5) =. A c.a.o. 3 f() = 6cos e A 0.937... =. f (.) =.6(..) A 4 A.W.R.T..6 City of London Academy
f(.55) 0.04... (change of sign) =.6 f(.65) 0.09..,, A N.B. f(.) = 7.744 [9] 4. (a) Correct method for f(); cos + sin + A f() = 0.585, f() = 3.38 or better seen A "0.585" Using N-R correctly: u = ; =.05 (3 s.f.) "3.38" A 5 [s: Answer.047,.05 implies second A mark] Two tangents drawn, one at {5, f(5)}, the other at {, f( )}, 3 marked in appropriate positions A [7] 5. (a) f() = - and f() = B = B 3 f() = ln + Attempts f() f() = ln + A ( ) = ln d Uses Newton Raphson ln = ln =.49 A 4 any correct answer [6] City of London Academy 3
f( ) 3 6. (a) y = sin 3 B y = B point where they meet B 3 f( ) or Shape Asymptotic behaviour to y = + Cross -ais once + comment B B B f () = 3cos 3 Attempt to diff. cos3 + two terms for, A 0.075 u = 0.8 4. = 0.879 A u = 0.877 A 5 [8] 7. (a) f() = 4, f() = both change of sign (and continuity) implies < < A f(.5) =.3.5 < < B f(.75) = 0.9.75 < < f(.875) = 0.03.875 < < B f () = 3 ln 3 attempt at differentiation using lns can be implied by net line f () = 8.8875 accept 3 ln 3, awrt 8.888 A City of London Academy 4
use of numerical differentiation button is acceptable = =.887, A 4 8.8875... use of N-R, cao [8] s: Incorrect method of differentiation is M0, A0,, A0 For eample, f() = 3 =.8 is ¼. The eact answer is.8789 Alternative to 4 (a): Use of a diagram y 7 8 9 3 y = 3 y= + 6 0 Two graphs with domain at least [, ] with one intersection 3, 9 and 7, 8 A 8. (a) f() =.54 B f() =.4 B.54.4.4.54.54 +.4 = (.4 +.54) =.65 A 4 f() = 4 cos + k cos + c f(.8) = 0.465 f(.8) = 4.03 (0.465) =.8 4.03 =.9 only A 5 [9] B A 0.6 9. (a) f(0.9) = e 0.9 0.075 B.8 City of London Academy 5
f(.0) = 0.5 e 0.3 B f() is continuous and changes sign between = 0.9 and = root eists B 4( ) (4 3) f() = + e ( ) f(0.9) =.869 Newton-Raphson process gives nd approimation 0.075 = 0.9 +.869 = 0.93 ( decimal places) A 6 0.7 (d) f(0.95) = e 0.95 0.09.8556 0.74 (0.935) = e 0.935 0.00 A.874 Since root lies in the interval (0.95, 0.935) due to sign change = 0.93 is correct to decimal places A 3 [] City of London Academy 6