Note: Students must indicate that the centre of the circle is at (7, 3) and that the radius is 4.

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1 0 GCE A Level H Maths Solution Paper (a) dy C d 3 y 8 C D (b) du dt 3u 3u ln 3 8 3u t C Substitute u, t 0 Note: Students must add in the arbitrary constant each time they work out an indefinite integral. Note: The modulus sign has to be placed for ln. C ln 7 3u ln t ln 7 3u 3u T ln ln 7 3u Note: To simplify the answer, combine the two logarithms into a single term. 3u ln 7 3u (i) Im O P (7, 3) Re Note: Students must indicate that the centre of the circle is at (7, 3) and that the radius is. (ii)(a) The smallest value of z is the length OP (ii)(b) 3 3 z 7 cos tan i3 sin tan i Note: Make use of the centre of the circle and its radius. Note: Here is the part which tests on the conversion of one trigonometric ratio to another using the method of right angle triangle. (iii) 3 arg z tan sin correct to s.f.

2 3(i) y y f Note: This is a one-mark question and it is not epected to have lots of details input into the graph. O (ii) By inspection,. 3 0 Discriminant of Hence 3 0 has no real solutions. Hence f ( ) has no other real solutions ecept. 3(iii) Let 3 3(iv) y Note: We can also use completing the square method to argue there are not real zeros from the quadratic 3 7 factor: 3 0 Note: This new equation makes use of the replacement of by in the original cubic equation f 0. Note: The curve has a sharp point (cusp) at.66, 0. O.66 3(v) The equations are 3 and 3 From (ii), 3 3 From GC,,0, Hence roots of are, 0,,. Note: f f or f.

3 (i) n 00 0n 5000 From GC, minimum n. Mrs A's account became > $5000 on December 00. (ii) Amount at end of st month Amount at end of nd month Amount at end of 3rd month Amount at end of nth month n n n Mr B's account became > $5000 at the end of the 5th month i.e. September 00. (iii) From GC,.0796 The interest rate is.80% per month. Note: We can also solve this inequality algebraically: 000(.005 n ) 5000 n 000(.005 ) 5000 n n ln ln (i) 0.00 disease p p positive negative p no disease p positive negative 3

4 5(i)(a) 5(i)(b) 5(ii) P(test positive) P(disease test positive) P(disease test positive) P(test positive) correct to 3 s.f. P(disease test positive) P(disease test positive) P(test positive) 0.00p 0.00p0.999( p) 0.00p p p p From GC, p correct to 6 d.p. 6(i) Let be the mean tail length (in cm) of the squirrels. 6(ii) Thus the hypotheses are H : and H : 0 Under H 0, 3.8 X ~ N,. 0 For H 0 not to be rejected, Note: InvNor(0.975) but we do not write the GC key strokes in the working. z correct to 3 s.f. Thus the set of values is :.3 5.7

5 6(iii) If 5.8, H 0 will be rejected. There is sufficient evidence at the 5% level to say that the squirrels on the island do not have the same tail length as the species known to her. 7(i) Required probability 7(ii)!! 5! 5 Required probability P all brothers are together 3! 3! 5! Note: We have to give the conclusion in the contet of the question. Note: Treat the two sisters as one item and also note the internal permutation of the sisters within the item. Note: The method of complement is used here which calculates cases which include the case of two of the three brothers as one item, and the remaining brother separate from this item. 7(iii) 7(iv) 7(v) 8(i) Required probability!! 3! 5! 55 Required probability Required probability ( )!! (5 )! 7 8 y Note: Treat the two sisters as one item, and the three brothers as one item. Note: We may use the result PA B PA PB PA B Note: Alternatively, we can also compute this probability by finding the total number of possible ways in the numerator of the probability as follows: Arrange the remaining 3 people plus one of the sisters, and then multiply by the number of ways of arranging the second sister among the people

6 8(ii) The irregularity occurred in Week. That practice paper may be more difficult than the other papers. 8(iii) The marks cannot eceed 00, and so a linear model, which models an indefinite upward/downward trend of data, is not appropriate. The marks are likely to plateau off or stay constant as the weeks go by, rather than in the case of a quadratic model which is epected to fit data with an increase and then a decrease (or the other way round) trend. Thus a quadratic model is also not appropriate. 8(iv) For L 9, r correct to 6 d.p. Note: Another possible reason could be that Amy was not prepared academically for the first practice paper in Week. 8(v) Since r is closest to for L 9, this is the most appropriate value for L. 8(vi) From GC, a.0, b 0.80 Thus ln correct to 3 s.f. Amy will get at least 90% in Week 3. 6

7 8(vii) 9(i) L is the percentage mark she will get if she continues practising indefinitely. The probability that a voter supports the Alliance Party is constant, and each voter s decision to vote for Alliance Party A is independent of the other voters. 9(ii) P A 3 P A correct to 3 s.f. 9(iii)(a) Since n 30 is large, np and nq , it is possible to approimate A with a normal distribution. 9(iii)(b) Since np 6.5 5, we cannot approimate A with a Poisson distribution. 9(iv) Given P A 5 30 p ( p) p p p p p p p p 0.39 or 0.6 Since p0.5, therefore p

8 0(i) 0(ii) 0(iii) 0(iv) 0(v) Gold coins are scattered randomly in a randomly chosen region of area square metre. Secondly, the mean number of gold coins found in any randomly chosen region of area square metre is equal to the mean number of the same type of coins found in another randomly chosen region of area square metre. Let X = number of gold coins in a region of square metre. Thus, P( X 3) P( X ) 0.07 correct to 3 s.f. Let Y = number of gold coins found in a region of square metres. Thus Y ~ Po(0.8 ) P( Y ) e From GC, 0.3 correct to 3 s.f. Let W = number of gold coins found in a region of 00 square metres. W ~ Po(80) Since 80 0, W ~ N(80,80) appro. P( W 90) P( W 89.5) 0. Let C and S denote the number of gold coins and pottery shards in 50 square metres respectively. Since 0 0, C ~ N(0,0) appro. Since 50 0, S ~ N(50,50) appro. C S ~ N(90,90) appro. P( CS 00) P( C S 99.5) 0.5 0(vi) From (v), C and S follow approimately the respective normal distributions. S 3 C~ N(50 3 0,50 3 0) S-3 C~ N(30,50) appro. P( S 3 C) P( S3C 0) P( S3C 0.5) 0.9 Note: Remember to perform continuity correction. Note: Without the use of suitable approimations, C S ~ Po(90). Note: Can we approimate S 3C directly to a normal distribution without mentioning that S and C are following approimate normal distributions? 8