fp1 numerical methods past exam … numerical methods past exam questions questions 1-6 and q.8 are...

City of London Academy 1

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 x0 as an approximation to the curve y=f(x) for the purposes of finding x1). 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) = x3 – 3x

2 + 5x – 4

(a) Use differentiation to find f (x). (2)

The equation f(x) = 0 has a root α in the interval 1.4 < x < 1.5

(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) = 2x – 6x

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 1.5

(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)

273 x

x

273 x

x


4. f(x) = 3x2

(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 1.4

(b) Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width 0.025

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

x3 – x

2 – 6 = 0

(a) show that 2.2 < α < 2.3 (2)

(b) Taking 2.2 as a first approximation to α, apply the Newton-Raphson procedure once to

f (x) = x3 – 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)

6.

(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 x0 = 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

11

x

2018

3)( x

xxf


(4)

(Total 9 marks)

7. f(x) = 4 cos x + e– x

.

(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 to

f(x) to obtain a second approximation to α. Give your answer to 3 significant figures. (4)

(Total 6 marks)

8.

The equation f(x) = 0 has a root α in the interval [0.7, 0.8].

(a) Use linear interpolation, on the values at the end points of this interval, to obtain an

approximation to α. Give your answer to 3 decimal places. (4)

(b) Taking 0.75 as a first approximation to α, apply the Newton–Raphson procedure once to

f(x) to obtain a second approximation to α. Give your answer to 3 decimal places. (4)

(Total 8 marks)

9. f(x) = x3 + 8x – 19.

(a) Show that the equation f(x) = 0 has only one real root. (3)

(b) Show that the real root of f(x) = 0 lies between 1 and 2. (2)

(c) Obtain an approximation to the real root of f(x) = 0 by performing two application of the

Newton-Raphson procedure to f(x), using x = 2 as the first approximation. Give your

answer to 3 decimal places. (4)

(d) By considering the change of sign of f(x) over an appropriate interval, show that your

answer to part (c) is accurate to 3 decimal places. (2)

(Total 11 marks)

10. f(x) = ln x +x – 3, x > 0.

(a) Find f(2.0) and f(2.5), each to 4 decimal places, and show that the root α of the equation f(x)

= 0 satisfies 2.0 < α < 2.5. (3)

x

xxxx ,2

2tan3)(f 2


(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)

11.

f(x) = 0.25x – 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 – ex + 3 sin 2x

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)

14.

The diagram above shows part of the graph of y = f(x), where

y

x

7

6

5

4

3

2

1

–1

–2

–3

O

1 2 3 4 5 6


f(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 to

f(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-Raphson

procedure,

(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].

(a) Use linear interpolation on the values at the end points of this interval to find an

approximation to . (2)

(b) Taking x = 1 as a first approximation to , apply the Newton-Raphson procedure once to

f(x) to obtain a second approximation to . (4)

(Total 6 marks)

16. f(x) = sin 3x – 2x + 1.

(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 to

f(x) to find a second and a third approximation to . Give your answers to 4 significant

figures. (5)

(Total 8 marks)


17. f(x) = 3x – 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 2x + 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)

19.

f(x) = ex

.

(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) = 3x2

– 6x + 5 M1A1 2

(b) f(1.4) = –0.136 B1

f'(1.4) = 2.48 B1ft

1

342

x

x


x0 = 1.4, x 1 = 1.4 M1

= 1.455 (3 dpl) A1 4 [6]

2. End points: (4, –8) and (5, 2) B1

(or equiv.) M1

α = 4.8 A1 3 [3]

3. (a) and Evaluate both M1

(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] ] M1

or –0.019 or –0.02 M1

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 A0

Answer with no working is also M0M0A0

(c) M1 A1

(Special case: +2

then ) A1 ft

48.2

136.0––

2

–5

8

4–

...)4.1(f ...)5.1(f

256.0)4.1(f 32

125 ...708.0)5.1(f

17(or )

24

...221.0)45.1(f

...018.0)425.1(f

22 73)(f xxx

...636.9)45.1(f 22 73)(f xxx

f (1.45) 11.636...


M1 A1 cao 5

Note

M1: for attempt at differentiation (decrease in power) A1 is cao

Second 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 3dp

Newton Raphson used more than once – isw.

Special case: f′(x) = 3x2 + 7x

–2 + 2 then f′(1.45) = 11.636...) is M1 A0

A1ft M1 A0 This mark can also be given by implication from

final answer of 1.43 [10]

4. (a) f(1.3) = –1.439 and f(1.4) = 0.268 (allow awrt) B1 1

Note

Both answers required for B1. Accept anything that rounds to 3dp

values above.

(b) f(1.35) < 0 (–0.568...) 1.35 < α < 1.4 M1A1

f(1.375) < 0 (–0.146...) 1.375 < α < 1.4 A1 3

Note

f(1.35) or awrt –0.6 M1

(f(1.35) and awrt –0.6) AND (f(1.375) and awrt –0.1) for first A1

1.375 < α < 1.4 or expression using brackets or equivalent in words

for second A1

(c) f′(x) 6x + 22x–3

M1A1

x1 = x0 – , = 1.384 M1A1,A1 5

Notes

One term correct for M1, both correct for A1

Correct formula seen or implied and attempt to substitute for M1

awrt 16.4 for second A1 which can be implied by correct final answer

awrt 1.384 correct answer only A1 [9]

5. (a) f (2.2) = 2.23 – 2.2

2 – 6 (= –0.192)

f (2.3) = 2.33 – 2.3

2 – 6 (= 0.877) M1

427.1...636.9

...221.045.1

)45.1(f

)45.1(f45.11

x

417.16

268.04.1

)(f

)(f

0

0 x

x


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) = 3x2 – 2x B1

f′(2.2) = 10.12 B1

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 2.21897

would get 4/5

If done twice ignore second attempt

(c) (or equivalent such as ) M1

α (0.877 + 0.192) = 2.3 × 0.192 + 2.2 × 0.877 A1

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 M1

substitutes y = 0

and y = 0, so α ≈ 2.218 or A1, A1

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 M0

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 2.218

If done twice ignore second attempt [10]

12.10

192.0––2.2

)(f'

)f(–

0

001

x

xxx

'877.0'

–3.2

'192.0'

2.2–

.

'877.0'

–1.0

'192.0'

kk

)2.2–(1.0

877.0192.0192.0 xy


6. (a) attempt evaluation of f(1.1) and f(1.2) (– looking for sign change) M1

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 A1

(b) f’(x) = M1 A1 A1 3

Note

Multiply by power and subtract 1 from power for evidence

of differentiation and award of first M1

(c) f (1.1) = 0.30875.. f′ (1.1) = –6.37086... B1 B1

M1

= 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 M1

Evidence of Newton-Raphson and awrt 1.15 award 4/4 [9]

7. (a) f(1.6)= ... f(1.7) = ... (Evaluate both) M1

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 A0

Values are 0.0851 and –0.3327 Need concluding statement also.

(b) f(x) = –4 sin x–e–x

B1

1.6 – M1

= 1.6 – A1

= 1.62 A1 4

B1 may be awarded if seen in N–R as –4sin 1.6 – e–1.6

or as –4.2

M1 for statement of Newton Raphson (sign error in rule results in M0)

First A1 may be implied by correct work previously followed

2

11–

2

1–

9–2

3xx

6.37086..–

0.30875...–1.11 x

(1.6)f

f(1.6)

...2.4

...085.06.1

)e6.1sin4(

e6.1cos46.1

6.1


by correct answer

Do not accept 1.620 for final A1. It must be given and correct to 3sf.

1.62 may follow incorrect work and is A0

No working at all in part (b) is zero marks. [6]

8. (a) f (0.7) = –0.195028497 and f (0.8) = 0.297206781 M1

Use to obtain = A1 4

(= 0.739620991) = 0.740 Answer required to 3 dp or better

Bs for 3dp or better First M for reasonable attempt using

fractions and differences.

(b) f(x) = 6x + M1A1

Use x2 = 0.75 (= 0.741087218) = 0.741 M1A1 4

Answer required to 3 dp or better

First 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 + 8 3x

2 + 8 = 0........ or 3x

2 + 8 > 0....... M1

Correct derivative and, e.g., ‘no turning points’ or ‘increasing

function’.

Simple sketch, (increasing, crossing positive x-axis)

(Or, if the M1A1 has been scored, a reason such as ‘crosses

x-axis only once’). B1 3

M: Differentiate and consider sign of f(x), or equate

f(x) to zero.

Alternative

M1: Attempt to rearrange as x3 – 19 = –8x or x

3 = 19 – 8x

(condone sign slips), and to sketch a cubic graph and a

straight line graph.

A1: Correct graphs (shape correct and intercepts ‘in the

right place’).

B1: Comment such as “one intersection, therefore one root”).

(b) Calculate f(1) and f(2) (Values must be seen) M1

F(1) = –10, f(2) = 5, Sign change, Root A1 2

)7.0(

)8.0(

7.0

8.0

f

f

)7.0()8.0(

)8.0(7.0)7.0(8.0

ff

ff

2sec

2

11 2 x

)75.0(

)75.0(

f

f

y

x

2

1

–1

–2

–2 –1 1 2 3


(c) x1 = 2 – , = 2 – (= 1.75) M1, A1

x2 = x1 – = 1.729 (ONLY)() M1, A1 4

1st 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) M1

(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. M1

f(2.5) = 0.41629……… = 0.4163 both correct 4 d.p. A1

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 M1

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] B1

f(x) = (allow 1.444) M1,A1

= 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 B0M1A0M1A1

(2)f

f(2)

20

5

1875.17

359375.075.1,

)(f

)f(

1

1

x

x

)5.2(

)0.2(

5.2

2or 5.0

f(2.5)f(2)

f(2))2(

f

f

f(2.5)

5.0

f(2)

xx

9

13or

4

41or 4.1)25.2(f,1

1 x

,(2.25)f

f(2.25)25.2

x

1


(d) f(2.2075) =, {–6.3…. × 10–4

} M1

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 2.20794

or better and statement [12]

11. (a) f(0.24) 0.058, f (0.28) = 0.089 accept 1sf M1

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 M1

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 B1

M1 A1

f(11) 0.3438 at least 2sf A1

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 B1

f(2) = 20.6424... – 20 = 0.6424... Allow awrt ± 0.64 B1

1.87 M1, A1 4

(b) f(1.9) 0.1651795..., or just 1.9 + 6 – 20e–05×1.9

Allow awrt 0.165 B1

f(t) = 1 + 10e–0.5×1.9

M1 A1

f(1.9) = 4.8674..., or just 1 + 10e–05×1.9

Allow awrt 4.87 A1

2 = 1.9 – M1 A1 6

4

1cos2)('f

x

xx

16.113438.0

0534.011

2.0

64.033.033.08.1,

"64.0"2

"33.0"8.1

866.1867410.416518.0


(c) 112 (min) (1 h 52 m) B1 1 [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 in

the 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…. B1

f(1.2) to 1sf or better

f(1.1) = 0.42…. and f(1.15) = 0.078…. M1

Attempt f(1.1), f(1.15)

= 1.2 A1 c.a.o. 3

(b) f(x) = 6cos2x – ex M1 A1

x2 = 1.2 – M1

= 1.16(2…..) A1 4

A.W.R.T. 1.16

(c) = 1.16 M1, A1 2

N.B. f(1.2) = –7.744… [9]

14. (a) Correct method for f(x); xcosx + sinx + 2 M1A1

f(1) = –0.1585, f(1) = 3.382 or better seen A1

Using N-R correctly: u1 = 1 – ; = 1.05 (3 s.f.) M1A1 5

(1.2)f

...2937.0

sign) of change(–0.029..,f(1.165)

.0.04......f(1.155)

"382.3"

"1585.0"


[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)} M1

x2, x3 marked in appropriate positions A1 2

[7]

15. (a) f(1) = -1 and f(2) = 2 B1

= 1 B1 2

(b) f(x) = 2x ln 2 + 1 M1

Attempts f(x)

f(x) = 2 ln 2 + 1 A1

= 1 – dM1

Uses Newton Raphson

= = 1.419 A1 4

any correct answer [6]

16. (a)

y = sin 3x B1

y = 2x – 1 B1

1 point where they meet B1 3

1

2

1

2

3

1

12ln2

)1(

12ln2

22ln2

f( )x

x

–1

x3


or

Shape B1

Asymptotic behaviour to y = – 2x + 1 B1

Cross x-axis once + comment B1

(b) f´(x) = 3cos 3x – 2 Attempt to diff. cos3x + two terms for M1 M1, A1

u1 = 0.8 – M1

= 0.8179 A1

u2 = 0.8177 A1 5

[8]

17. (a) f(1) = –4, f(2) = 1

both M1

change of sign (and continuity) implies 1 < < 2 A1 2

(b) f(1.5) = –2.3… 1.5 < < 2 B1

f(1.75) = –0.9… 1.75 < < 2

f(1.875) = –0.03… 1.875 < < 2 B1 2

(c) f´(x) = 3x

ln 3 – 1 M1

attempt at differentiation using lns can be implied by next line

f´(2) = 8.8875… A1

accept 32 ln 3 – 1, awrt 8.888

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, A0

For example, f(x) = x3x–1

– 1 x = 1.8 is ¼.

The exact answer is 1.8789…

Alternative to 4 (a): Use of a diagram

f( )x

x

212.4

075.0

...8875.8

1


Two graphs with domain at least [1, 2] with one intersection M1

3, 9 and 7, 8 A1 2

18. (a) f(2) = 1.514 B1

f() = 1.142 B1

M1

× 1.514 + 2 × 1.142 = (1.142 + 1.514)

= 2.65 A1 4

(b) f(x) = 4 cos 2x + 1 k cos 2x + c M1

f(2.8) = 0.4625 B1

f(2.8) = 4.1023 A1

x2 = 2.8 M1

= 2.91 only A1 5 [9]

19. (a) f(0.9) = e0.9

0.0751 B1

f(1.0) = 0.5 – e1

0.1321 B1 2

(b) f(x) is continuous and changes sign between x = 0.9 and x = 1

root exists B1 1

(c) f(x) = + ex

f(0.9) = 2.2869

Newton-Raphson process gives 2nd approximation

×

×

×× y x = + 6

y = 3x

x

y

3

789

0 1 2

2

–1.514

1.142

514.1

142.1

2

1023.4

)4625.0(

81.1

6.0

22

2

)1(

)34(2)1(4

x

xxx


= 0.9 + M1

= 0.93 (2 decimal places) A1 6

(d) f(0.925) = e0.925

0.019 M1

(0.935) = e0.935

0.0022 A1

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.


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

2869.2

0751.0

8556.1

7.0

8742.1

74.0

x

1


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) = x3 + 8x – 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 3x2

+ 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 = x3

and y = 19 – 8x, showing that these had

just one point of intersection. Weaker candidates tried to solve

x3

+ 8x – 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 4sinx 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.


The method of interval bisection was generally used correctly in part (a) although there were a number of candidates

who used linear interpolation instead and wasted time and marks in the process. The differentiation in part (b) was

usually correct, but sometimes the 1 did not disappear. The Newton Raphson technique was used efficiently and

accurately. Most candidates were able to give a suitable interval in part (c) and the evaluations were often accompanied by a suitable comment to secure both marks.

14. Part (a) was straightforward and, on the whole, answered well. A small minority differentiated incorrectly, but the more

common error was to work in degree mode rather than in radians, so the answer 1.33 was frequently seen; this scored 3

out of the 5 marks. Part (b) lost a little “in the translation” due to poor copying of the original curve. Marking was

generous to accommodate this fact, but many candidates did not appreciate what was being asked of them and there

seemed to be a very liberal interpretation of “tangents”. Many tangents did not intersect the x-axis, and although the first

may have been drawn at the point {5, f(5)}, the second was often drawn at a random point on the curve, sometimes at {5, f(5)} again !

15. (a) Most candidates found f(1) and f(2) correctly and a number of these went on to use linear interpolation as

required. The most common error in calculating was to deduce that 3=4 gives =3/4. A significant number,

however, seemed not to know what linear interpolation was and carried out interval bisection, in some cases, several times…

(b) The inability to differentiate was widespread, often by Centre, although candidates who went back to

consider lny = xln2 (and not lny =ln +lnx - ln4) were able to proceed. Many candidates knew the formula required for Newton Raphson but there was some confusion between f(x) and f ′(x).

16. In (a) there was a disappointing response to the drawing of the graph. Most students tried to draw y = sin3x – 2x + 1 but

the asymptotic behavior was not evident and the maxima and minima were smoothed out to appear more like points of

inflexion. Very few candidates made a comment about there being only one point of intersection indicating that there was only one root.

In part (b) the newton raphson procedure was well understood but solutions lost marks by the failure to give both

answers to 4 significant figures. A number of candidates used degrees rather than radians.

17. All candidates were familiar with the method in part (a), although not all wrote down that the change of sign implied a

root in the interval or some equivalent of this. The majority of candidates understood the procedure to be used in part (b)

and the calculations involved were generally accurately carried out. A minority of candidates did however confuse the

method with linear interpolation. Although the Newton-Raphson method is popular with candidates and well

understood, part (c) was one of the worst answered on the paper as only a minority of candidates could differentiate

. Both and x 3x1

were commoner than the correct .

18. Many candidates were familiar with the method of interpolation although several were caught out by the negative value for f (2). Many candidates wasted time by doing more than one linear interpolation when one would have sufficed.


2x

2x

3x

3x 3 ln3x

fp1 numerical methods past exam … numerical methods past exam questions questions 1-6 and q.8 are...

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