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FP1 NUMERICAL METHODS PAST EXAM QUESTIONS
Questions 1-6 and Q.8 are standard plain-vanilla questions.
Questions 9 (part d, anyway), 14, and 16 are ones where many students failed.
They do not demand more or harder working. They demand that you think a bit, and
understand what e.g. Newton-Raphson means (taking the tangent at x 0 as an approximation
to the curve y=f(x) for the purposes of finding x 1 ).
A number of questions from no.7 onwards demand that you know derivatives of functions
now not included in FP1. Just look up the derivatives in the mark scheme, and then you can
use those questions for practice.
1. f( x ) = x^3 – 3 x^2 + 5 x – 4
(a) Use differentiation to find f ( x ). (2)
The equation f( x ) = 0 has a root α in the interval 1.4 < x < 1. (b) Taking 1.4 as a first approximation to α , use the Newton-Raphson procedure once to obtain a second approximation to α. Give your answer to 3 decimal places. (4) (Total 6 marks)
2. f( x ) = 2 x^ – 6 x
The equation f( x ) = 0 has a root α in the interval [4, 5]. Using the end points of this interval find, by linear interpolation, an approximation to α. (Total 3 marks)
3. f ( x ) = ,^ x > 0
(a) Show that f( x ) = 0 has a root between 1.4 and 1. (2)
(b) Starting with the interval [1.4,1.5], use interval bisection twice to find an interval of width 0.025 that contains α. (3)
(c) Taking 1.45 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 10 marks)
x
x
x
x
4. f( x ) = 3 x^2
(a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4). (1)
The equation f( x ) = 0 has a root α between 1.3 and 1. (b) Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0. which contains α. (3)
(c) Taking 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 equation
x^3 – x^2 – 6 = 0 (a) show that 2.2 < α < 2. (2)
(b) Taking 2.2 as a first approximation to α , apply the Newton-Raphson procedure once to f ( x ) = x^3 – x^2 – 6 to obtain a second approximation to α , giving your answer to 3 decimal places. (5)
(c) Use linear interpolation once on the interval [2.2, 2.3] to find another approximation to α , giving your answer to 3 decimal places. (3) (Total 10 marks)
(a) Show that the equation f( x ) = 0 has a root α in the interval [1.1, 1.2]. (2)
(b) Find f′( x ). (3)
(c) Using x 0 = 1.1 as a first approximation to α , apply the Newton-Raphson procedure once to f( x ) to find a second approximation to α , giving your answer to 3 significant figures.
2
x
x
f x x
(b) Use linear interpolation with your values of f(2.0) and f(2.5) to estimate α , giving your answer to 3 decimal places. (2)
(c) 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)
f( x ) = 0.25 x – 2 + 4sin √ x. (a) Show that the equation f( x ) = 0 has a root α between x = 0.24 and x = 0.28. (2) (b) Starting with the interval [0.24, 0.28], use interval bisection three times to find an interval of width 0.005 which contains α. (3)
The equation f( x ) = 0 also has a root β between x = 10.75 and x = 11.25. (c) Taking 11 as a first approximation to 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)
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.
The heating system switches off when θ = 20. The time t = α , when the heating system switches off, is the solution of the equation θ – 20 = 0, where α lies in the interval [1.8, 2]. (a) Using the end points of the interval [1.8, 2], find, by linear interpolation, an approximation to α. Give your answer to 2 decimal places. (4)
(b) Taking 1.9 as a first approximation to α , use the Newton-Raphson procedure once to obtain a second approximation to α. Give your answer to 3 decimal places. (6)
(c) Use your answer to part (b) to find, to the nearest minute, the time for which the heating
system was on. (1) (Total 11 marks)
13. f( x ) = 1 – e x^ + 3 sin 2 x
The equation f( x ) = 0 has a root in the interval 1.0 < x < 1.4.
(a) Starting with the interval (1.0, 1.4), use interval bisection three times to find the value of
to one decimal place. (3)
(b) Taking your answer to part (a) 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 your answer to part ( b ) is accurate to 2 decimal places. (2) (Total 9 marks)
The diagram above shows part of the graph of y = f( x ), where
y
x
O
17. f( x ) = 3 x^ – x – 6.
(a) Show that f( x ) = 0 has a root between x = 1 and x = 2.
(2) (b) Starting with the interval (1, 2), use interval bisection three times to find an interval of
width 0.125 which contains .
(2)
(c) Taking 2 as a first approximation to , apply the Newton-Raphson procedure once to f( x ) to
obtain a second approximation to . Give your answer to 3 decimal places.
(4) (Total 8 marks)
18. f( x ) = 2 sin 2 x + x – 2.
The root of the equation f( x ) = 0 lies in the interval [2, ].
(a) Using the end points of this interval find, by linear interpolation, an approximation to .
(4)
(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)
f( x ) = e x^.
(a) Evaluate f(0.9) and f(1.0). (2)
(b) Deduce that the equation f( x ) = 0 has a root in the interval [0.9, 1.0].
(1)
(c) Taking 0.9 as a first approximation to , use the Newton-Raphson procedure once on 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) (Total 12 marks)
MARK SCHEME
1. (a) f'( x ) = 3 x^2 – 6 x + 5 M1A1 2
(b) f(1.4) = – 0.136 B f'(1.4) = 2.48 B1ft
x
x
x 0 = 1.4, x (^) 1 = 1.4 M
= 1.455 (3 dpl) A1 4 [6]
2. End points: (4, – 8) and (5, 2) B
(or equiv.) M
α = 4.8 A1 3 [3]
3. (a) and Evaluate both M
(or ),
Change of sign, root A1 2 Alternative method: Graphical method could earn M1 if 1.4 and 1.5 are both indicated A1 then needs correct graph and conclusion, i.e. change of sign root Note M1: Some attempt at two evaluations A1: needs accuracy to 1 figure truncated or rounded and conclusion including sign change indicated (One figure accuracy sufficient)
(b) or 0.2 [root is in [1.4, 1.45] ] M or – 0.019 or – 0.02 M
root is in [1.425, 1.45] A1 cso 3 Note M1: See f(1.45) attempted and positive M1: See f(1.425) attempted and negative A1: is cso – any slips in numerical work are penalised here even if correct region found. Answer may be written as 1.425 ≤ α ≤ 1.45 or 1.425 < α < 1.45 or (1.425, 1.45) must be correct way round. Between is sufficient. There is no credit for linear interpolation. This is M0 M0 A Answer with no working is also M0M0A
(c) M1 A (Special case: + then ) A1 ft
f ( 1. 4 )... f ( 1. 5 )...
f ( 1. 4 ) 0. 256
f^ (^1.^5 )^0.^708 ...
(or )
f ( 1. 45 ) 0. 221 ...
f ( 1. 425 ) 0. 018 ...
f (^ x ) 3 x^2 7 x ^2 f ( 1. 45 ) 9. 636 ... f ( x ) 3 x^2 7 x ^2
f (1.45)^ 11.636...
Change of sign Root need numerical values correct (to 1 s.f.). A1 2 Note M1 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 in other parts of the question if not done in part (a))
(b) f′( x ) = 3 x^2 – 2 x B f′(2.2) = 10.12 B
M1 A1ft
= 2.219 A1cao 5 Note B1 for seeing correct derivative (but may be implied by later correct work) B1 for seeing 10.12 or this may be implied by later work M1 Attempt Newton-Raphson with their values A1ft may be implied by the following answer (but does not require an evaluation) Final A1 must 2.219 exactly as shown. So answer of 2. would get 4/ If done twice ignore second attempt
(c) (or equivalent such as ) M
α (0.877 + 0.192) = 2.3 × 0.192 + 2.2 × 0.877 A or k (0.877 + 0.192) = 0.1 × 0.192, where α = 2.2 + k so α ≈ 2.218 (2.21796…) (Allow awrt) A1 3 Alternative Uses equation of line joining (2.2, – 0.192) to (2.3, 0.877) and M substitutes y = 0
and y = 0, so α ≈ 2.218 or A1, A
awrt as before (NB Gradient = 10.69) Note M1 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 M A1 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. If done twice ignore second attempt [10]
f'( )
f( )
0 1 ^0 x
x x x
k k
y 0. 192 x
6. (a) attempt evaluation of f(1.1) and f(1.2) (– looking for sign change) M
f(1.1) = 0.30875, f(1.2) = – 0.28199 Change of sign in f( x ) root in the interval A1 2 Note awrt 0.3 and – 0.3 and indication of sign change for first A
(b) f’( x ) = M1 A1 A1 3
Note Multiply by power and subtract 1 from power for evidence of differentiation and award of first M
(c) f (1.1) = 0.30875.. f′ (1.1) = – 6.37086... B1 B
M
= 1.15(to 3 sig.figs.) A1 4 Note awrt 0.309 B1and awrt – 6.37 B1 if answer incorrect Evidence of Newton-Raphson for M Evidence of Newton-Raphson and awrt 1.15 award 4/ [9]
7. (a) f(1.6)= ... f(1.7) = ... (Evaluate both) M 0.08… (or 0.09), –0.3… One +ve, one –ve or Sign change, root A1 2 Any errors seen in evaluation of f(1.6) or f(1.7) lose A mark so – 0.32 is A Values are 0.0851 and – 0.3327 Need concluding statement also.
(b) f( x ) = – 4 sin x – e– x^ B
1.6 – M
= 1.6 – A
= 1.62 A1 4
B1 may be awarded if seen in N–R as – 4sin 1.6 – e–1.6^ or as – 4. M1 for statement of Newton Raphson (sign error in rule results in M0) First A1 may be implied by correct work previously followed
2
x x
x 1 1. 1 –
f(1.6)
f(1.6)
- 2 ...
( 4 sin 1. 6 e )
4 cos 1. 6 e
6
6
(c) x 1 = 2 – , = 2 – (= 1.75) M1, A
x 2 = x 1 – = 1.729 (ONLY)( ) M1, A1 4
1 st^ 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) M
(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. A1 2 For 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……. = – 0.3069 AWRT 3 d.p. M f(2.5) = 0.41629……… = 0.4163 both correct 4 d.p. A States change of sign, so root (between 2 and 2.5) B1 3 Note: B1 gained if candidate’s 2 values do show a change of sign and statement made
(b) or equivalent M
Or and x found
= 2.212 AWRT A1 2
(c) f(2.25) = 0.06093……. ( 3 d.p.) [Allow ln.2.25 + 2.25 – 3] B
f( x ) = (allow 1.444) M1,A
= 2.20781.... = 2.208 AWRT M1A1 5
First M in (c) is just for + 1
If no intermediate values seen B1M1A1M1A0 is possible for 2.209 or 2.21, otherwise as scheme (B1 eased to award this if not evaluated) MR 2.5 instead of 2.25 (Answer 2.203) award on ePen B0M1A0M1A
f(2)
f(2) 20
f( )
f( ) 1
1 x
x
- 5 or f(2) f(2.5)
f(2) ( 2 ) f
f
f(2.5)
f(2)
x x
or 4
1 ,f( 2. 25 ) 1. 4 or 1
x
, f(2.25)
f(2.25)
- 25
x
(d) f(2.2075) =, {–6.3…. × 10–^4 } M f(2.2085) = , {8.1…. × 10–^4 } Correct values ( > 1 s.f.), (root in interval) so root is 2.208 to 3 d.p. A1 2 A1 requires values correct ( > 1 s.f.) and statement (need not say change 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. or better and statement [12]
11. (a) f(0.24) 0.058, f (0.28) = 0.089 accept 1sf M
Change of sign (and continuity) α (0.24, 0.28) Al 2
(b) f (0.26) 0.017 ( a (0.24,0.26)) accept lsf M
f (0.25) 0.020 ( a (0.25, 0.26))
f (0.255) 0.001 a (0.255, 0.26) M1 A1 3
(c) f (11) 0.0534 at least 3sf B
M1 A
f(11) 0.3438 at least 2sf A
M1 A1 6
If f (11) – 0.3438 is produced without working, this is to be
accepted for three marks M1 A1 A1. [11]
12. (a) f(1.8) = 19.6686... – 20 = – 0.3313... Allow awrt ± 0.33 B
f(2) = 20.6424... – 20 = 0.6424... Allow awrt ± 0.64 B
1.87M1, A1 4
(b) f(1.9) 0.1651795..., or just 1.9 + 6 – 20e–05×1.9^ Allow awrt 0.165 B
f( t ) = 1 + 10e–0.5×1.9^ M1 A
f(1.9) = 4.8674..., or just 1 + 10e–05×1.9^ Allow awrt 4.87 A
2 = 1.9 – M1 A1 6
2 cos 1 f '( )
x
x x
, 1. 8 0.^33
[Notes: Answer 1.047, 1.05 implies second A mark]
(b) Two tangents drawn, one at {5, f(5)}, the other at { x 2 , f( x 2 )} M x 2 , x 3 marked in appropriate positions A1 2 [7]
15. (a) f(1) = -1 and f(2) = 2 B
= 1 B1 2
(b) f ( x ) = 2 x^ ln 2 + 1 M
Attempts f (x)
f ( x ) = 2 ln 2 + 1 A
= 1 – dM
Uses Newton Raphson
= = 1.419 A1 4
any correct answer [6]
16. (a)
y = sin 3 x B y = 2 x – 1 B 1 point where they meet B1 3
2 ln 2 1
2 ln 2 1
2 ln 2 2
f( ) x
x
x (^) 3
or
Shape B Asymptotic behaviour to y = – 2 x + 1 B Cross x -axis once + comment B
(b) f´( x ) = 3cos 3 x – 2 Attempt to diff. cos3 x + two terms for M1 M1, A
u 1 = 0.8 – M
= 0.8179 A u 2 = 0.8177 A1 5 [8]
17. (a) f(1) = – 4, f(2) = 1
both M
change of sign (and continuity) implies 1 < < 2 A1 2
(b) f(1.5) = –2.3… 1.5 < < 2 B
f(1.75) = –0.9… 1.75 < < 2
f(1.875) = –0.03… 1.875 < < 2 B1 2
(c) f´( x ) = 3 x^ ln 3 – 1 M attempt at differentiation using lns can be implied by next line f´(2) = 8.8875… A accept 3^2 ln 3 – 1, awrt 8. use of numerical differentiation button is acceptable
2 = 2 = 1.887 M1, A1 4
use of N-R, cao [8] Notes: Incorrect method of differentiation is M0, A0, M1, A For example, f( x ) = x 3 x –^1 – 1 x = 1.8 is ¼. The exact answer is 1.8789…
Alternative to 4 (a): Use of a diagram
f( ) x
x
= 0.9 + M
= 0.93 (2 decimal places) A1 6
(d) f(0.925) = e0.925^ 0.019 M
(0.935) = e0.935^ 0.0022 A
Since root lies in the interval (0.925, 0.935) due to sign change
= 0.93 is correct to 2 decimal places A1 3
[12]
1. No Report available for this question. 2. No Report available for this question. 3. Part (a) was straightforward and generally well done. Very few errors were made in the numerical evaluations (we do need to see these). There were, however, a minority of candidates who did not give the required conclusion, which ideally requires the sign change to be noted and a statement made of the interval for the root. In part (b) a few candidates attempted linear interpolation, maybe indicating a lack of practice with interval bisection. The numerical evaluations of f(1.45) and f(1.425), which are required, were well done by the majority of candidates. A noticeable number, however, did not produce a correctly stated conclusion – commonly no statement at all or just a single x value. Candidates should be made aware that, if the interval notation is used, the smaller number should be first – in this case [1.425, 1.45]. Part (c) was generally very well done with the vast majority knowing the Newton-Raphson iteration. The most common
error was not differentiating the constant (+2). A few had problems differentiating and a small number continued beyond one iteration. Candidates should be encouraged to evaluate and write down intermediate values in their working, so that if a slip is made the examiner can see where. Failure to do so will have lost a mark here for some candidates. Candidates should also be encouraged to check that their answers throughout a question are consistent.
4. Part (a) was usually correct, though the required interval was not always stated. A small number of candidates failed to follow the requested method in part (b), in general trying to use linear interpolation. For most, however, the correct method was used, but it was surprising to see how many candidates failed to give an interval in their final answer. Newton-Raphson was usually attempted correctly in part (c), though sometimes a few candidates had problems differentiating the negative index. 5. Most candidates were clear about the steps necessary to show that the root of the given equation was between the values 2.2 and 2.3 in part (a). Almost all substituted 2.2 and 2.3 into the left hand side of the equation and gave their numeric answers. A few did not complete the solution by stating that one answer was positive and one negative and that the sign change indicated the presence of a root between 2.2 and 2.3. The Newton Raphson method in part (b) was well understood and most answered this part of the question correctly. Candidates are advised to show their expression for f′(x) and for f′(2.2). They are also advised to quote the formula and show their substitution. The final answer 2.219 was not acceptable with no working. There were many good answers to part (c), with most solutions using similar triangles. Those who had learned and quoted a formula often made sign slips. Some used the equation of the line joining (2.2, – 0.192) and (2.3, 0.877) and found where it crossed the x axis. This was an acceptable alternative method. A small number of candidates tried interval bisection however, which was not linear interpolation! 6. This was another very accessible question with many candidates gaining full marks. In part (a) most solutions stated the conclusion about change of sign implying a root. In part (b) there were few errors in finding f′ ( x ) but occasionally the second term power was incorrect or the constant term of 20 was left at the end of the answer. In part (c) there were a few candidates who did not give the answer to the required accuracy. Many candidates showed no values of f(1.1) and f ′ (1.1) in their working and a small number applied Newton-Raphson twice. 7. In part (a) it was expected that candidates would evaluate f(1.6), f(1.7) and declare a sign change, resulting in the conclusion that there was a root in the interval (1.6, 1.7). Most candidates did this and earned both marks. There were a few errors evaluating the values, but almost all candidates worked in radians rather than in degrees. A number of candidates did not draw an adequate conclusion after doing the evaluation. In solutions to part (b) there were some sign errors in the derivative f( x ). Newton Raphson was usually stated correctly
x
but there were some numerical slips in using the procedure and this frequently resulted in the wrong answer 1.58. The answer here was required to be accurate to 3 significant figures and needed to be 1.62 (not 1.620). In this question the answer 1.62, with no working, resulted in zero marks, as did the answer 1.62 following incorrect work.
8. This question was generally well done by most candidates. In part (a) a variety of methods using linear interpolation were used, with some working from first principles. The most successful used similar triangles or a standard formula. Some misinterpreted the question and continued to apply their method until successive roots agreed to 3 decimal places. The most common errors were due to rounding or to calculators being in degree mode. In part (b) the majority of candidates were able to apply Newton-Raphson successfully. Most differentiated correctly and gained full marks. Where the differentiation was incorrect the most common error was in omitting the ½ and some missed out the +1. In both parts of the question there was some evidence of incorrect calculator use. This was particularly noticeable in part (b). 9. There were very few completely correct responses to part (a). Solutions based on the function f( x ) = x^3 + 8 x – 19 needed to establish convincingly that the graph crossed the x -axis at only one point. The most efficient way to do this was to show that there were no turning points, but few candidates considered the sign of the derivative. Those who did differentiate sometimes argued that because 3 x^2 + 8 = 0 had no real roots the original equation had only one real root. Other possible methods included sketching, for example, the graphs of y = x^3 and y = 19 – 8 x , showing that these had just one point of intersection. Weaker candidates tried to solve x^3 + 8 x – 19 = 0 by using the quadratic formula. Answers to part (b) were usually correct, but occasionally lacking a reason or conclusion. The Newton-Raphson procedure was well known in part (c), where many excellent solutions were seen. Most candidates were able to choose a suitable interval to establish the accuracy of their answer to part (c), although a few thought that the interval from 1.728 to 1.730 was sufficient. Conclusions here should have referred to the accuracy of the root rather than just stating that ‘there is a root in the interval’. 10. This question was almost universally well done, with only part (b) seeming to be less familiar. Candidates are now well prepared for such questions and good complete answers to parts (a) and (c) were the norm. Although part (c), testing knowledge of the Newton-Raphson process, was well done here, it is of some concern that some candidates show no, or very little evidence of their working. It is a dangerous strategy to feed all the data into the calculator because then wrong answers clearly gain no credit; it is advisable to show f ( x ) and relevant numerical intermediate results as often marks may then be gained if the final answer is incorrect. 11. The majority of candidates gained both marks in part (a). It is essential to realise that, when answering parts (a) and (b), statements like f (0.4 )<0 are inadequate. They have a fifty per cent chance of being accurate and offer no way that an examiner can evaluate a candidate’s response. Here one significant figure is enough - f (0.28 ) ≈ 0.09 >0 is a sufficient statement – but anywhere in a question on numerical analysis, intermediate results should be given which show that working is being done with sufficient accuracy to obtain the result required in the question. Interval bisection is not always well understood and, again, questions need to be read carefully. Some candidates having given completely correct working gave their answer as an approximation to the root, α ≈ 0.255, instead of giving, as asked, an interval of width 0.005 which contained α, (0.255, 0.26 ). Some candidates produced three linear interpolations instead of three interval bisections and the amount of time this took must have seriously affected their ability to complete the paper. In part (c), nearly all candidates knew how to use the Newton-Raphson method but the majority of candidates were unable to differentiate 4sin x correctly.
This question was almost universally well done, with only part (b) seeming to be less familiar. Candidates are now well prepared for such questions and good complete answers to parts (a) and (c) were the norm. Although part (c), testing knowledge of the Newton-Raphson process, was well done here, it is of some concern that some candidates show no, or very little evidence of their working. It is a dangerous strategy to feed all the data into the calculator because then wrong answers clearly gain no credit; it is advisable to show f ( x ) and relevant numerical intermediate results as often marks may then be gained if the final answer is incorrect.
12. Some candidates had difficulty in interpreting the context of this question and worked with θ rather than α in part (a) and/or part (b). In part (a), a few were unfamiliar with linear interpolation or confused it with interval bisection, but most were able to use a correct formula to find an approximation to α. Some candidates persisted with further iterations, wasting time but still reaching the required answer 1.87. Numerical slips were not uncommon.
The Newton-Raphson procedure was well known in part (b) and many excellent solutions were seen. The required differentiation was usually correct, but a few candidates showed insufficient detail in their numerical working, penalising themselves if their final answer was wrong. Occasionally part (c) was omitted, but most candidates were able to refer back to the context and calculate the required time.
13. It was disappointing that in a question about numerical methods a number of candidates failed to give values to justify their statements. In part (a), for example, simply stating f(1.2)<0, f(1.1)>0 and f(1.15)>0 is not sufficient. However, apart from the small minority who worked in degrees, most candidates did evaluate the function at the appropriate points and were able to score well on this question.
