Edexcel IGCSE Mathematics - Higher (4400)igmath.com/ms/3H/2006NOV.pdf · 2016. 4. 3. · Edexcel IGCSE Mathematics - Higher (4400) November 2006 Mark Scheme and Examiner’s Report

IGCSE Edexcel IGCSE

Mathematics - Higher (4400)

November 2006 Mark Scheme and Examiner’s Report

Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers.

Through a network of UK and overseas offices, Edexcel’s centres receive the support they need to help them deliver their education and training programmes to learners.

For further information please call our Customer Services on 0870 240 9800, or visit our website at www.edexcel.org.uk.

January 2007

Publications Code UG018569

All the material in this publication is copyright © Edexcel Ltd 2006

4400 Higher Mark scheme and Examiner Report November 2006

1

IGCSE Maths November 2006 – Paper 3H Final Mark Scheme

Question No. Working Answer Mark Notes

1 a 290 ± 2 2 B2 B1 for 290 ± 5 or 360 − 70

b 226 − 180 2 M1

046 A1 Condone omission of 0

Total 4 marks

2 a x + x + x + x + x + x or 6x 2 B1

x + 7 + x + 7 + x + 7 + x + 7 or 4(x + 7) or 4x + 28 B1

bi “6x” = “4(x + 7)” 4 M1

ii 6x = 4x + 28 M1

6x − 4x = 28 oe M1

14 A1 cao

Total 6 marks

3 100 × 1.80 or 180 6 M1

60 × 4.00 or 240 M1

4.00 ÷ 5 or 0.8(0) or 3.2(0) M1 may be part of an expression

35 × 3.20 or 112 M1

“240” + “112” − “180” M1 dep on at least 2 of previous 4 M marks

172 A1 cao

Total 6 marks

4 a 3602150 ±

oe inc 125 , 0.42, , 0.417 641.0 & 2 B1

B1 numerator = 2150 ± denominator = 360

b 10×30+12×12+14×18+17×60 or 300+144+252+1020 or 1716

4 M1 finds products f×x consistently within intervals (inc end points) & sums them

use of at least 3 midpoints M1

120

"1716" M1 (dep on 1st M1) for division by Σf

14.3 A1 Accept 14 if all M marks scored

Total 6 marks


2

5 6048

or 60 − 48 3 M1

80 or 60

"12" M1

20 A1 cao

Total 3 marks

6 25

240 × 2 M1

600 A1 cao SC B1 for 52240 × or 96

Total 2 marks

7 64 <x or x46 −<− 3 M1 correctly collects x terms

M1 correctly collects constants

x < 1.5 oe A1

Total 3 marks

8 0.5 + 0.1 or 0.5 + 0.1 + 0.3 or table completed with 0.1 3 M1

1−(0.5+0.1) or1−(0.5+0.1+0.3) + 0.3 M1

0.4 A1

Total 3 marks


3

9 a BM = 5 seen or implied 4 B1

22 513 − or 144 M1 for squaring and subtracting

22Accept or 69 1013 −

22 513 − M1 for 22 513 − only

12 A1 cao

b "12"1021 ×× 4 M1 for (a) their 102

1 ××

× 4 M1 dep on first M1

10 × 10 or 100 M1 indep

340 A1 ft from "12"

Total 8 marks

10 Q correct 4 B1

R correct B1 ft from Q

Reflection B1

y = x B1 Accept eg in dotted line but, if stated, equation must be correct

ft from R if at least one transformation correct

Total 4 marks


4

11 a 1 2 2 2 5 5 5 5 5 6 6 6 6 7 9 3 M1

Attempt to find 4th (or 3¾th)

& 12th (or 11¼th) values M1

4 A1 cao

bi eg B had higher marks than A 2 B1 B0 if median for A seen and ≠ 5

ii eg B less spread or more consistent B1

Total 5 marks

12 a Attempt to find horizvert for line PQ 4 M1

(gradient =) 2 A1 (y =) 2x M1A1 ⇒

y = 2x − 4 B2 ft from “2” B1 for 2x − 4 B1 for y = mx − 4 where m ≠ 2

b Line through (0, 1) 3 M1

Attempts grad −½ or correctly finds

coordinates of another point M1

Correct line A1 Passes within 1mm of (−2, 2) and (2, 0)

Total 7 marks


5

13 a 81 1 B1

b73 1 B1

c649 1 B1

fractions

Accept equivalent

Total 3 marks

14 a 5000 −1250x 2 B2 B1 for 5000 B1 for −1250x

b 5000 − 1250x = 0 3 M1

x = 4 M1

4 10 000 A1 and a is linear

ft from a if at least B1 scored

ci max 2 B1 independent

ii

coeff of x2< 0 or 0ddy

>x

for x value < 4 and

0ddy

>x

for x value > 4 or y < 10 000 for x value

< 4 and for x value > 4 or 2

2

d

d

x

y = −1250 < 0

B1

di 4 2 B1 ft from b if at least 1 scored

ii max profit oe B1 Accept eg largest profit

Total 9 marks


6

15 23334 ÷×π + 1032

31 ××π 4 M1 for 233

34 ÷×π

or value rounding to 56.5 or 56.6

M1 for 103231 ××π

or value rounding to 94.2 or 94.3

M1 for sum (dep on first two M marks)

151 A1 for 151 or better (150.796…) (3.14 → 56.52 + 94.2 = 150.72)

Total 4 marks

16 i B ⊂ A 2 B1 cao

ii A ∩ B = Ø B1 cao

Total 2 marks


7

17 ai 211 oe 2 B1

ii 43 oe B1 Don’t accept

43

−−

b 1 1 B1 cao

ci1

1

1

−−

−

xxx

x

4 M1

1)1(

1

−−−

−

xxx

xx

or )1( −− xx

x oe M1

x A1 cao

SC B1 for ff(x) evaluated correctly for two values of x and an answer of x

ii eg f is its own inverse, ff-1 = B1 dep on correct ci

Total 7 marks

18 x2 = 2x + 15 5 M1 yy

=⎟⎠⎞

⎜⎝⎛ − 2

215

x2 − 2x − 15 = 0 M1 y2− 34y + 225 = 0

(x + 3)(x − 5) = 0 2

82 ±=x M1 (y − 25)(y − 9) = 0

x = −3 or x = 5 A1 y = 9 or y = 25

−3, 9 and 5,25 A1

Total 5 marks


8

19 a 7 − x B1 1

b 8 − x seen or 9, 13, 6 marked correctly on diagram or 50 − (10 + 9 + 13 + 6)= 50 − 38 = 12 and 8 + 7 = 15

3 M1

10+13+9+6+(7 − x)+(8 − x)+x=50 oe inc 7 − x + 8 − x + x = 12 or 15 − 12

M1 equation must be correct

3 A1

Total 3 marks

20 a k:1 1 B1 Accept k

b 2 or 2

1 seen 2 M1

7.1 A1 for 7.1 or better (7.071…) Accept 50

Total 3 marks


9

21 a 3n oe 1 B1 Accept eg n + 2n

b n − 1, 3n − 1 seen 5 B2 B1 for each

101

131

31

=−−

×nn

oe inc 101

131

3=

−−

×nn

nn

M1 for correct equation

10(n − 1) = 3(3n − 1) oe

inc 10n(n−1) = 3n(3n − 1) M1 for correctly removing fractions

(n = 7) 21 A1 cao

Total 6 marks

Total 100 marks


10

IGCSE November 2006 – Paper 4H Final Mark Scheme

Question No. Working Answer Mark Notes

1 4.346.6

1.9

2

M1 A1

for 3.4 cao

Total 2 marks

2 a 156 +t 1 B1 cao

b 23 3yy − 2 B2 B1 for , B1 for 3y 23y−

c 21372 +++ xxx

21102 ++ xx 2

M1 A1 Condone 1 error

d 87qp 2 B2 B1 for , B1 for . Allow p7p 8q 7 x q8

Total 7 marks

3 4145+

9

2

M1 A1

36 or 9:36 M1A0 cao

Total 2 marks


11

4 a for B3 )1(2 += nP oe (a&b) Ignore units

B2 for )1(2 +n oe or n = P/2 – 1 oe B2 for 12 += nP oe or 21×+= nP oe

)1(2 += nP

3

B1 for P = any f(n) (not P = n) B1 for 2n + 1 oe or 21×+n oe B0 for muddle eg n+1 = x 2 = P

b 22 += nP M1 22 +n seen or M2 for

22 −= Pn

22−P

or 2P

- 1

3

M1

A1 1

2+= nP

or P – 2 ÷ 2

1

SC ft from 2 += nP or 2n + 1 only M1 for 12 −= Pn or P - 1 ÷ 2

A1 ft for 2

1−P oe

Total 6 marks

5 75.7

5456

M1for

time5456

or 732 seen

B1 for 7.75 or 465 if ...x 60 or “km/m”

704 3 A1 cao

Total 3 marks

6 ai eg “9 is not a member of ℰ”, “It is not an even number” “ℰ is only even nos”, “9 is odd”

1

B1 for either interpreting statement or for giving a reason

ii 6, 12, 18 1 B1 Condone omission of brackets

b 6, 12 2 B2 B1 for 6 or 3, 6, 12

Total 4 marks


12

7 3.84.72 ××π M2 for 3.84.72 ××π

M1 for or 2303 - 2305 3.89.42 ××π

576 3 A1 for 575.7-576.1

Total 3 marks

8 147 −=− x M1 for substituting correctly

− 1½ oe 2 A1

Total 2 marks

9 a 8348× M1

18 2 A1 cao ans 18/48: M1A0

b eg 48 −18 − 18, )18(248 +=+ xx M1

12 2 A1f ft from “18”

Total 4 marks


13

10 eg 3 225 M2 for full systematic method of at least 3

3 75 divisions by prime numbers oe

5 25 (factor trees)

5 Condone 1 error

Or for 3 × 3 × 5 × 5 or 3, 3, 5, 5

M1 for 225 written as correct product with only one non-prime

22 53 × 3 A1

Total 3 marks

11 a eg enlargement, (scale factor) 3, (centre) (1,2) B3 B1 for enlargement

Not single trans: B0B0B0

3 B1 for 3, B1 for (1,2)

b Correct triangle B22 B1 for 1 to the left B1 for 3 up

Total 5 marks

12 101012 =+ yx 30206 =− yx M1 Correctly equating coefficients of x or y or rearranging to x = … or y = …

( 2515 =x ) ( 2525 −=y )

321=x

(or 1.7 or better), 1 −=y

3

A1 A1

Condone 1.66 cao

Total 3 marks


14

13 a 7108.7 × 1 B1 cao

b 0.004 oe 1 B1 cao

c 121075.3 −× 1 B1

Total 3 marks

14 a 4.53.9tan =∠LMN M1 sinLMN =

)4.53.9(3.9

22 +or cos etc M1A1

9.3/5.4 or 1.722… A1

59.9 3 for 59.85-59.9A1

bi 5.45 1 Accept , 5.4499… B1 944.5 &

ii 5.35 1 caoB1

c "35.5"35.9

M1

1.74766… 2 A1 for 1.74 or 1.75 or better

Total 7 marks

15 10)210(180 −×

or 180 – 360/10 360/10 M1

144 36 A1

180 − [360 − (60 + 144)] or 24 60 – 36 (= 24) M1 360 – 204 = 156

"24"

360

M1 180 x (n-2)/ n = 156 or 180 – 360/n = 156 or 2340/15 = 156

15 5 caoA1

Total 5 marks


15

16 a 28, 50, 64, 74, 80 1 B1 cao

b

Points B1

In (b) incr’ing y’s nec’y. Not blocks end pts + ½ square ft from sensible table condone one error

Curve

or line segments

2

B1 dep end pts or midpts thro’ pts + ½ square; ignore x < 5 dep on 4 pts correct or ft

c cf for time of 17h found from graph

M1 In (c) incr’ing cf graph essential eg line, mark on graph

~12 2 A1f 12 or consistent with curve

Total 5 marks

17 (36067 or 0.186…) x… M1 or … ÷ (360/67 or 5.37…)

22.836067

×× π M1 or π x 8.22 ÷ 360/67

39.3 3 A1 for 39.2 - 39.32

Total 3 marks

18 a 0.25, 2.5, 8, 15.25 2 B2 Accept rounding or truncating

B1 for 2 or 3 correct

b

Points B1f

Allow + ½ square Condone 1 error or omission ft if at least B1 in (a)

Curve 2 B1f ft if at least B1 in (a)

c 1.4 – 1.47 1 B1

d xx

x 232 =− or indication of xy 2= M1 indication may be mark or line on graph Must see 2x or indic’n of line y = 2x

~2.5 2 A1 ft if at least B1 in (b)

Total 7 marks


16

19 100x = 23.2323… M1

9923 2 A1

Total 2 marks

20 ai 61 1 caoB1

ii opp angles of a cyclic quad (add to 180° or are suppl) 1 B1

b 90 – “61” 29

2

M1 A1f

°=∠ 90ACB stated or indicated on diagram

Total 4 marks

21 a 128, 72 2 B1 for 128 cao B2 B1 for 72 cao

b bar correct 1 34 little squares high B1

Total 3 marks

22 36.0 or 0.6 M1

(1 − “0.6”) × (1 − “0.6”) or 0.4 × 0.4 M1 dep

0.16 3 for 0.16 oe A1

Total 3 marks

23 )32)(32()4)(32(

−+−+

xxxx M1

M1 for )4)(32( −+ xx

)for 32)(32( −+ xx

32

4−

−x

x 3 A1

Total 3 marks


17

24 eg °

=° 48sin

""75sin6.8 a or

°57sin""b M1

°°

75sin48sin6.8 or 6.61(.…) or

°°

75sin57sin6.8 or 7.46(….) A1

°××× 57sin"616.6"6.821 or °××× 48sin"467.7"6.82

1 M1 dep M1 or ½ x “6.616” x “7.467” x sin75o

23.9 4 A1

Total 4 marks

25 a

222 )4()6()5( +=−+− xxx 22 12361025 xxxx +−++− 1682 ++= xx

4

B2 B1 B1

two of )5( x− , )6( x− , )4( +x seen or equiv, eg (10 –x – 4) B1 for one of these correct equn not expanded correct equn expanded

b 2

4543030 2 ×−± 3 M1 Allow –302

272030 ±

or 28.4 & 1.584 A1

1.58 A1

Total 4 marks


18


19

IGCSE Mathematics 4400 Paper 3H The candidates (almost 1100, 75% of them Higher tier) found the demands of all four papers reasonable and took the opportunity to show what they knew. Many questions had a high success rate and very few candidates were entered for Higher tier when Foundation would have been more appropriate. Working was generally well presented and methods clearly shown. Introduction This paper gave candidates the opportunity to demonstrate positive achievement and many gained high marks. Most of the questions proved accessible and had pleasing success rates although only a minority of candidates scored full marks on Q17(c) (Functions) and Q21(b) (Conditional probability). Methods were generally clearly explained and sufficient working shown. Report on Individual Questions Question 1 Although well answered, this question caused more problems than most others in the first half of the paper and it was not unusual for candidates to lose marks on this question, usually on part (b), and then go on to do very well on the rest of the paper. In part (a), 250° (180° + 70°) appeared occasionally. In part (b), 134° (360° − 226°) was the most popular wrong answer and 44° (270° − 226°) also appeared regularly. No credit was given for answers such as 45° and 47° obtained by drawing as the instructions were to work out the bearing. Omission of the leading zero in the bearing was not penalised. Question 2 Many candidates scored full marks, the only error which appeared with any regularity being the omission of the brackets in the expression for the height in the first part i.e. 4x + 7 instead of 4(x + 7). Question 3 The only consistent error was with the selling price of each cake on Tuesday, 0.8 or 3.8 (4 − 0.2) sometimes being used instead of 3.2 (4 − 0.8). A minority had problems dealing with the unsold cakes but full marks were common. Question 4

Errors were rare in part (a), the answer usually being given as a fraction, often 125

195

, but decimals

and percentages were, of course, accepted. In part (b), many candidates realised that the 1 <≤ x interval was wider than the others and took 17 as the halfway value. Those who took 16 and used it correctly could still score 3 marks out of 4. Those using halfway values of 9.5, 11.5, 13.5 and 16.5 correctly could score 2 marks.

Question 5 The majority of candidates obtained the correct answer and many of the rest scored 2 marks out

of 3 for an answer of 80% ⎟⎠⎞

⎜⎝⎛ ×100

6048

. Occasionally, 1251004860

=× and 4810060

× resulted

in answers of 25% and 28.8% respectively. Question 6 The vast majority of candidates scored full marks. Those who did not usually shared £240 in the ratio 2 : 5 and gave an answer £171.43. Question 7 Most candidates were able to simplify the inequality to 4x < 6 but some lost a mark by then

writing 211 or x =

211 as the answer instead of x <

211 . A few solved 4x < 6 as

x < 32 .

Question 8 The majority found the probability (0.1) that Danielle will win the race and, although a few gave this as their final answer, most went on to use it correctly and gain full marks. A small minority multiplied 0.3 and 0.1, instead of adding them. Question 9 Most used Pythagoras’ theorem successfully in the first part. If a mistake were made in the second part, it was usually in calculating the area of the triangle, not failing to divide by 2, but

using 13 cm as the “height”. A less common, but still noticeable, error was the use of 12521 ××

to find the area of a triangular face but there were many completely correct solutions. Question 10 This was very well answered, although occasionally the reflection was in the x-axis or the rotation anticlockwise. When a single transformation is asked for, no marks are awarded if more than one transformation appears in the answer, even if the correct one is included. Question 11 Although this question was quite well answered, familiarity with interquartile range varied widely. In part (a), the marks were sometimes not put in order and their total (72) used in a variety of ways. Unfortunately, it was possible to obtain the correct answer using the range, 8,

and finding 8841

43 ×−× . Answers clearly obtained in this way received no credit.

In part (b), although comparisons such as “Class B’s marks were less spread.” were hoped for, quantitative comparisons like “B’s median was 2 more than A’s median.” were accepted. Some candidates used words like “they” and “it” without making clear which class they were referring to.


20

Question 12 Knowledge of varied but it was well understood by the many candidates, full marks frequently being awarded. Very occasionally, the line

cmxy +=12 +−= xy was drawn in the second

part. Question 13 A small minority gave their answers as decimals and received no credit. Most candidates scored

the mark in the first part with an answer of 81 but any equivalent fraction was accepted. 32

1,

however, was not. In the second part, 3439

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ×

343

2731

sometimes appeared but the final part

was usually correct. Question 14 Few candidates were unable to score at least one mark on part (a) and many gained full marks

on the whole question. In part (b), y, instead ofxy

dd

, was sometimes equated to 0 and answers of

(4, 0) were not unusual, the y-coordinate resulting from the substitution of x = 4 into 5000 − 1250x. In part (c)(ii), any reasonable explanation was accepted e.g. “a negative parabola” and

the evaluation of xy

dd

for values of x on each side of x = 4. Finding 1250dd

2

2−=

xy

and the

comment that this is negative was, of course, accepted, even though knowledge of this approach is not included in the specifications. In part (d)(ii), candidates were expected to relate their answer for part (i) to the context with comments such as “This price gives the greatest profit.” Question 15 Many candidates gained full marks but the others made a wide variety of errors. The most frequent one was failing to halve the volume of a sphere to find the volume of the hemisphere. The information on the formula sheet was not always used correctly or, perhaps, it was ignored.

Thus, sometimes, volume of cone = hrπ 221 was used or the formula for the surface area of a

sphere used, instead of that for the volume. 234 rπV = was seen regularly, either for the volume

of a sphere or as the “simplified” form of 2

.3

4 3rπV = . Even when the correct formula for the

volume of a sphere was used, 33 was occasionally evaluated as 9. Question 16 Few candidates failed to score at least one mark. If an error were made, it was more likely to be on the first part, with a wrong answer of ABA =∩ or BA ⊂ , both of which suggested some knowledge of subsets. =∪ BA E was the most likely wrong answer to the second part.


21

Question 17 Parts (a) and (b) were well answered. In part (a)(ii),

43− and

43

−− appeared sometimes; both

scored no marks. In part (b), 0 was the most common incorrect answer. Those giving both 0 and 1 as their answer also scored no marks. The algebraic manipulation in part (c)(i) was beyond all but the strongest candidates but there

were a small number of elegant solutions. Even those starting with the expression 1

1

1

−−

−

xxx

x

frequently “simplified” it to 0, 1 or −1. In part (c)(ii), candidates were expected to state, either in words or symbolically, that f is its own inverse. Question 18 Although, like Q17, the algebra is demanding, it was probably more familiar to candidates and the solutions of the simultaneous equations were found competently and accurately by many of them. Question 19 7 − x was the simplest answer to the first part but any unsimplified expression, if correct, was accepted, as was a formula such as n = 7 − x but not x = 7 − x. In the second part, many scored one mark for marking 13, 6 and 9 correctly on the Venn Diagram but finding the value of x proved much more demanding. Arithmetical or trial methods were as likely to be successful as algebra. Some candidates trying to construct an equation used 6 − x instead of 6, which led to 1 as the value of x. Those giving an answer of x = −3 would have been well advised to reflect on their answer. Question 20 Areas of similar shapes is a topic that many students find difficult and part (a) was intended to help with part (b). In fact, it was not unusual to see part (b) answered correctly after an incorrect

part (a). The most common wrong answer to part (a) was k21:1 while, in part (b), 10

appeared regularly. Question 21 As in Question 20, the aim of the first part was to help candidates with the second part. In this case, the first part was quite well answered, any expression, such as n + 2n, which is equivalent

to 3n being accepted. 32+n was a regular wrong answer.

Only a minority answered the second part successfully and, although the first part directed candidates towards an algebraic method, the correct answer was obtained by trial methods as often as by building up an equation and solving it. Some candidates did not appreciate that this was a “without replacement” situation.


22

IGCSE Mathematics 4400 Paper 4H Introduction The majority of candidates showed a very good understanding of most of the mathematics tested in this paper and marks were generally high. For the ablest candidates, only Q15 and Q25(a) gave serious pause for thought. There were a few cases of candidates failing to show working and thereby losing method marks. Question 1 Just a few candidates calculated 6.46 ÷ 1.8 + 1.6 = 5.18. Question 2 (a) 6t + 5 was occasionally seen but most candidates obtained 6t + 15. (b) Errors in the powers were sometimes seen. (c) A few candidates clearly did not know how to remove brackets of this type. Some gave

2x instead of x2. Others expanded correctly but then “collected up” incorrectly. (d) This part was usually correctly answered. Question 3 A few found 45 ÷ 4 = 11.25 Question 4 (a) P = n + 1 x 2 was probably the most common mistake. Those whose notation was slack

in this way also tended to struggle with the rearrangement in part (b). (b) Those who scored 3 marks for (a) tended to succeed in (b), although n = (P – 1)/2 was

common. Question 5 This question was answered correctly by most candidates. Division by 7.45 (or 465 without considering units) accounted for almost all mistakes. Very few candidates failed to gain a mark for dividing by time. Question 6 Some centres seemed more familiar with set notation than others. A few candidates insisted that 9 was a member of M (perhaps misunderstanding the ‘∈’) and some invented more obscure reasons for (a)(i). Despite this hint that the universal set was involved, it was not uncommon to see odd numbers listed for (a)(ii) and (b). {3, 6, 9, 12, 15, 18} was probably the most common wrong answer for (a)(ii), often accompanied by {3, 6, 12} for (b). A few scored 1 mark for {6} in (b).


23

Question 7 Nearly all candidates recognised that 9.4 had to be halved to give the radius. Just a few candidates failed to carry out the calculation correctly having gained the method marks. Question 8 The very few mistakes that were made tended to be mishandling negative signs. In these cases the method mark was usually gained. Question 9 Part (a) was usually correct. In part (b) weaker candidates tended to calculate ½ x 48 = 24, 24 – 18 = 6, failing to realise that there will be more than 48 beads after the new red beads are added. Some others gave long answers, possibly using an algebraic approach. A few candidates gave concise answers, recognising that there were 30 blue beads, so 12 more red ones were needed. Question 10 This question provided easy marks for most candidates. A few of the weaker ones failed to reduce to primes, giving products such as 152 or 9 x 5 x 5, or failed to use indices in their answers. Just a few listed the factors without expressing them as a product. Question 11 (a) A few candidates (mostly at the weaker end) lost all marks by combining a rotation with

a translation. Odd marks were lost by not including all three parts of the answer, with the centre being most likely to be omitted. Some incorrect centres were given (such as (0 , 0 ) or (4 , 5)). A few scale factors of 2 were given, even by some better candidates.

(b) Some sort of translation was nearly always shown, usually by 3 units in one direction and 1 unit in another, but not always in the correct directions.

Question 12 Nearly all candidates knew what to do. The better ones were nearly always correct. Errors tended to be with signs or failing to multiply coefficients when adjusting an equation. Question 13 Most candidates scored well on this question. In part (c) 3.75-12 was seen occasionally. Some candidates rounded to 3.8; others found 0.375 x 10-11 correctly and either stopped or gave 3.75 x 10-10. Question 14 (a) This part was usually answered correctly. A few candidates used the fraction upside

down. Fewer still got muddled with sine or cosine, sometimes having worked out the hypotenuse.

(b) The most common problem was in the upper bound, with answers such as 5.445, 5.44 and 5.5

(c) A surprising number calculated the length of the hypotenuse. Some simply used the working values of 9.3 and 5.4. A few found tan-1(gradient). Those who lost marks in (b) also tended to score zero in (c).


24

Question 15 This question was found difficult by many candidates. Many got as far as 36o, but thereafter made errors. The more perceptive candidates quickly recognised the sequence of 36o, 144o, 156o, 24o, 15 sides. Others laboured through more convoluted working to reach a correct answer, whilst the weaker candidates got lost on the way. It was one of the few questions that a few candidates failed to attempt at all. Question 16 Only the weakest candidates failed to complete the table correctly. Points were normally plotted correctly, nearly always using the interval end-points, although a significant minority plotted at the midpoints. A few drew bars. Method was usually shown in part (c). Quite a few answers were left as 68. Some others used the scale incorrectly and took readings at 16, 17.5 or 18.5 . Question 17 This question was well answered on the whole. A few candidates used 2π r. Some lost accuracy (and a mark) through premature rounding of 67/360 or 360/67. A few approximated to 1/6. Question 18 The table was usually completed correctly. The graph was often correct although careless plotting sometimes lost a mark. A surprisingly large number of candidates plotted at (1.5. - 0.25) instead of (1.5, 0.25) despite having the correct value in the table. A few plotted -5.75 at -7.5. Curves were usually reasonable (but the quality of curve drawing is rarely good). Numerous candidates joined some or all points with line segments. Very thick lines and discontinuous lines were not unusual. Candidates generally looked for the correct reading in (c), although inaccurate graph-drawing often led to the loss of a mark. Part (d) was only answered correctly by the ablest candidates. There were attempts at drawing the new curve, often resulting in a correct but unrewarded answer, and y = -2x was not uncommon. Some candidates made no attempt. The better candidates easily identified the correct line and earned both marks. Question 19 Many candidates had no difficulty with this question. Some simply wrote 23/100. More inventively, the argument 90

23903

9230.02.032.0 =+=+= &&&&

& arose from time to time. This would

appear to be due to a muddling of the ‘short cut’ methods for and . 2.0 & 32.0 &

Question 20 Many fully correct answers were seen. 1190 was often given in (a) – “opposite angles equal” or even “came from the same chord”. Some lost a mark for the reason by omitting the word “opposite” or failing to describe or specify a cyclic quadrilateral, even though they were using the correct reasoning. 1190/2 = 59.5o was sometimes seen. Part (b) was usually answered correctly, although it was rarely possible to award the method mark when the answer was wrong. 1190/2 = 59.5o was common here also.


25

Question 21 Nearly all of the stronger candidates understood the concept of frequency density. Many scored full marks although there were frequent slips in calculations. Weaker candidates did not have a grasp of what was required, rarely scoring any marks. The better of these thought they could equate height to frequency, usually using the second bar to create a scale. Luckily for them, this provided the correct frequency of 72 but it gave a frequency of 90 for the first bar and a height of 222/3 small squares for the third bar. Question 22 Some candidates sensibly used a tree diagram to help. Weaker candidates tended to favour 0.642, or possibly just 0.64 . Only the many better candidates could take the step of finding 0.6, and they usually completed the question correctly. Question 23 This question produced varied responses. Some candidates did not recognise the need to factorise, attempting to cancel individual terms. Others tried to factorise and failed. Some made minor mistakes, most commonly getting the signs wrong in the numerator. Not all of those who were able to factorise the numerator could manage the difference of two squares in the denominator. Question 24 Better candidates were usually successful, though some lost a mark due to premature rounding. Others offered a range of responses, many not helped by disorganised working. The weakest stumbled because the 750 angle was not given. Some dropped a perpendicular to the base, expecting it to bisect either the base or the angle. Some curiously used the sine rule without any sines (8.6/75 = a/48). Others simply made mistakes in manipulating their equations and calculating values. Most candidates recognised the area formula using two sides and the included angle, although some wasted time (and risked errors) by finding both unknown sides and using these, rather than finding just one and using this and the given side. Question 25 The very able candidates picked up the first four marks effortlessly, and for them the second part was routine. More commonly, only x + 4 provided a mark in part (a). Many candidates spent much time trying various unproductive attempts at the proof. Few identified the crucial right-angled triangle. It was common to see the diagonal given as 2x + 8, leading to (2x + 8)2 = 92 + 102. In part (b) not all candidates who found correct roots were able to identify that only the smaller value was useful. The weaker candidates had no idea what to do with (a) and often failed to treat (b) as a quadratic equation at all, even if they realised that they were meant to solve this equation. A disappointing number who failed in part (a) seemed not to realise that part (b) was still available to them. A large minority of candidates (in either part (a) or (b)) found the areas of the circles and attempted to equate their sum with the area of the box. In part (b) incorrect use of the formula was not uncommon, usually involving an incorrect sign. A few candidates attempted trial and improvement to solve the quadratic equation - almost always unsuccessfully. This method usually gains no marks at all and should be strongly discouraged.


26


Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email [email protected] Order Code UG018569 January 2007 For more information on Edexcel qualifications, please visit www.edexcel.org.uk/qualifications Alternatively, you can contact Customer Services at www.edexcel.org.uk/ask or on 0870 240 9800 Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH


Documents

Edexcel IGCSE Mathematics - Higher (4400)igmath.com/ms/3H/2006NOV.pdf · 2016. 4. 3. · Edexcel IGCSE Mathematics - Higher (4400) November 2006 Mark Scheme and Examiner’s Report