2009 GCE ‘A’ Level Solution Paper 1 (Contributed by Hwa Chong Institution) 1i) Let un = an2 + bn + c. u1 = a + b + c = 10 u2 = 4a + 2b + c = 6 u3 = 9a + 3b + c = 5 Using GC, a = 1.5, b = –8.5, c = 17.  un = 1.5n2 – 8.5n + 17. (ii) Let y = 1.5n2 – 8.5n + 17 – 100 Teaching Point: The set of simultaneous equations can either be solved by using the application PlySmlt2 or finding rref of the augmented matrix formed by the 3 equations. Teaching Point: Please use ZoomFit on your GC to see the entire graph clearly. Besides finding the zero of the function y = 1.5n2 – 8.5n + 17 – 100, another method is to find the point of intersection of the curve y = 1.5n2 – 8.5n + 17 and the line y = 100. Hint to students:

From GC, n > 10.79. Hence solution set = {n  ℤ+ : n  11}. 1 1 0 4 – x2 dx 1 2+x 1 = [4 ln 2 – x ]0 1 = 4 ( ln 3 – ln 1 ) 1 = 4 ln 3 1 1/2p dx 0 1 – p2x2 1 1 = p 1/2p dx 0 1 2 p2 – x 1 1/2p = p [sin–1 px]0 1 1 = p (sin–1 2 ) 1 1 = p 6 = 4 ln 3 2 p = 3 ln 3 2) 3i) 1 2 1 –n+n+1 n–1 n(n + 1) – 2(n2 – 1) + n(n – 1) = n(n2 – 1)

It is helpful to use the GC to do a quick check that the integration has been done correctly. Teaching point: Students sometimes leave out 1 the term p when integrating 1 . To safeguard this, 1 – p2x2 it is advisable to change the integrand into the form 1 before integrating. 2 a – x2

1

2 = n3 – n A = 2.

Shown

(ii)

r=2

 r3 – r

n

1

1 n 2 = 2  r3 – r r=2 1 1 n 2 1 =2  (r–1–r+r+1) r=2 1 1 2 1 =2 [1–2+3 1 2 1 +2–3+4 1 2 1 +3–4+5 : : : : 2 1 1 +n–2–n–1+n 1 2 1 +n–1–n+n+1] 1 1 1 1 =2[2–n+n+1] (iii) 1 1 As n  , n  0 and n + 1  0.

1 the series converges to the value 4 . 4i) f(27) + f(45) = f(23) + f(41) = f(19) + f(37) : : = f(3) + f(1) =5+6 = 11

2

(ii) y 7

Teaching Point: Students should be advised to sketch a clear and properly– labelled graph.

3

–7 –6 –4 –2 (iii)

2

4

6

8

10

x Hint to students: The area under the curve from x = –4 to x = –2 is exactly the same as the area from x = 0 to x = 2. Hence the factor 2 in front of 2 7 0 2 – x dx. The area from x = –2 to x = 0 is the same as the area from x = 2 to x = 4. It would be wrong to find0 2x – 1 dx. –2

3 f(x) dx –4 = 22 7 – x2 dx + 4 2x – 1 dx + 3 2x – 1 dx 0 2 2 3 4 3 x 2 = 2[7x – 3 ]0 + [x2 – x]2 + [x2 – x]2 8 = 2[14 – 3 ] + [12 – 2] + [6 – 2] 2 = 363

5) n

Let Pn be the statement: 1

r=1 When n = 1: LHS = 12 = 1 1 RHS = 6 1(2)(3) = 1 = LHS P1 is true. Assume that Pk is true for some k  ℤ + k 1 i.e.  r2 = 6 k(k + 1)(2k + 1). r=1 Prove that Pk+1 is also true k+1 1 i.e.  r2 = 6 (k + 1)(k + 2)(2k + 3). r=1 LHS k =  r2 r=1 Teaching Point: Some students make the mistake of writing “Assume that Pk is true for all k  ℤ +”. It should be either “some k” or “a k”. Clearly, if you assume that it is true for all k, then there is nothing to prove.

 r2 = 6 n(n + 1)(2n + 1).

+ (k + 1)2

3

1 = 6 k(k + 1)(2k + 1) + (k + 1)2 1 = 6 (k + 1)[ k(2k + 1) + 6(k + 1) ] 1 = 6 (k + 1)(2k2 + 7k + 6) 1 = 6 (k + 1)(k + 2)(2k + 3) = RHS Since P1 is true and Pk true  Pk+1 true, hence by Math Induction, Pn is true for all n  ℤ +. 6i)
3 – 6 x = –2 –1 – 3 2 y=1 6

Teaching Point: Students are advised to take 1 out the common factor 6 (k + 1) instead of expanding everything and then factorising later. Clearly a waste of effort.

Hint to students: Students must realise that x = –2 is a vertical asymptote and draw it even though it does not appear on the Graphing Calculator.

(ii)

x–2 x2 y2 Substitute y = x + 2 into 6 + 3 = 1:

x2 1  x – 2  2 6 + 3 x + 2 = 1:   x2(x + 2)2 + 2(x – 2)2 = 6(x + 2)2 2(x – 2)2 = (x + 2)2 (6 – x2) Shown

(iii) Method 1: Find the points of intersection from the original graphs.

Hint to students: The question requires students to “show algebraically”. It would be wrong if students show that the numerical values of the points of intersection obtained from the GC satisfy the given equation. Teaching Point: Students have a choice of using Method 1 (since the graphs have already been plotted on the GC) or Method 2 which is also relatively straight forward..

From GC, x = –0.515 or 2.45

4

Method 2: Sketch y = 2(x – 2)2 – (x + 2)2 (6 – x2)

From GC, x = –0.515 or 2.45 7i) f (x) = –sin x ecos x = – sin x f(x) f (x) = – sin x f (x) – cos x f(x) f(0) = e f (0) = 0 f (0) = –e e f(x) = e + 0x – 2 x2 + ... e = e – 2 x2 + ... 1 a + bx2 1 1 =a bx2 1+ a 1 b = a (1 + a x2)–1 1 b = a (1 – a x2 + ...) 1 b e = a – a2 x2 + ... = e – 2 x2 + ... 1 1 a=ea=e 1 e e b – a2 = – 2  be2 = 2  b = 2e (ii) 8i) ar24 = 20r24 = 5 1 r24 = 4 Hint to students: It is helpful to use the GC to check the value of the derivative at x = 0.

Hint to students: It is easier to use the expansion (1 + x)–1 = 1 – x + x2 –... instead of the general Binomial Expansion formula in this case.

5

12 1 2 Total length of all bars < S 20 = 12 1 1– 2 = 356.343 < 357 r=  12 25 1 1– 2  12 1 1– 2

Hint to students: Since all the lengths are positive, students need to realise that the total length of all the bars is less than the sum to infinity. Shown

(ii)

L = 20

= 272.2573

 12  12 1 Length of 13th bar = 20  2  1 = 20  2 = 10 Let b = length of first bar of instrument B. 25 2 [ 2b + 24d ] = L L b + 12d = 25 Also b + 24d = 5 L So 12d = 5 – 25 272.2573 =5– 25  d = –0.49086 Length of longest bar =b = 5 – 24(–0.49086) = 16.8 cm 9i) =1+i = 2 ei/4 = 2 ei(2k+/4) 2k  i( 7 + 28) 1/14 z=2 e , k = 0, 1, 2, 3 z7

Hint to students: Since the arguments must lie in the principal range, k must take the values 0, 1, 2, 3.

6

(ii)

2/7 2/7 2/7 2/7 2/7 2/7 /28

Hint to students: Students must realise that all the roots have the same modulus 21/14 and are all 2 spaced 7 radians apart on the Argand Diagram.

/4

Substituting z = 0 into | z – z1 | = | z – z2 |: | 0 – z1 | = | z1 | = 21/14 | 0 – z2 | = | z2 | = 21/14 Since z = 0 satisfies the equation | z – z1 | = | z – z2 |, the locus passes through the origin. z2 (iii)

/7 /7 /28

z1

  Equation of locus is y = x tan ( 28 + 7 ) 5 = x tan 28 2 –1 1  2      3  1  cos  = 4+1+9 1+4+1 3 = 14 6  = 70.9

10i)

Hint to students: The question specified “acute angle”. Hence the modulus sign. In this case, it makes no difference to the answer though.

7

(ii)

Solving 2x + y + 3z = 1 –x + 2y + z = 2

Teaching Point: Students can either use the application PlySmlt2 or find the rref of the augmented matrix formed by the 2 equations.

0 –1 1 + –1 . From GC, the equation of l is r =      0 1 (iii) Substitute x = –, y = 1 – , z =  into the equation of p3 : LHS = –2 + 1 –  + 3 – 1 + k( + 2(1 – ) +  – 2) for any  =0 Hence l lies in p3 for all k. Substitute x = 2, y = 3, z = 4 into the equation of p3 : LHS = 4 + 3 + 12 – 1 + k(–2 + 6 + 4 – 2) = 18 + 6k = 0  k = –3 equation of plane is 2x + y + 3z – 1 –3(–x + 2y + z – 2) = 0 i.e. 5x – 5y + 5 = 0 i.e. x – y + 1 = 0 11i) Teaching Point: Students must show clearly the asymptotic nature of the graph as it approaches the xaxis.

(ii)

dy dx

2 2 = e–x + x e–x (–2x)

Hint to students: Students can use the GC to 8

= e–x (1 – 2x2) = 0 1 x2 = 2 1 1 x= ,– 2 2 1 –1/2 1 –1/2 y= e ,– e 2 2 1 1 turning points = ( , e–1/2 ) 2 2 1 –1/2 1 ,– e ) and (– 2 2 (iii) du u = x2  dx = 2x

2

check that these turning points are numerically correct.

When x = 0, u = 0 When x = n, u = n2 2 n x e–x dx 0 2 1 = n 2 e–u du 0 1 n2 = 2 [–e–u ]0 2 1 = 2 [1 – e–n ] 2 1 As n  , e–n  0, so area = 2 (iv) 2 | f(x) | dx –2 = 2 2 f(x) dx 0 = 1 – e–4

Hint to students: Students can use the GC to check that the area is numerically correct.

Hint to students: The GC can be used to do a quick check that the answer is numerically correct.

(v)

2 Volume = 1 x2 e–2x dx 0 = 0.363

9

2009 GCE ‘A’ Level Solution Paper 2 1i)

(ii) Method 1: dx dt = 2t + 4 dy 2 dt = 3t + 2t dy 3t2 + 2t dx = 2t + 4 dy 16 When t = 2: x = 12, y = 12, dx = 8 = 2 Equation of l is y – 12 = 2( x – 12) i.e. y = 2x – 12 Method 2:

Hint to students: The gradient of the curve at t = 2 can either be obtained algebraically (Method 1) or from the GC (Method 2).

dy When t = 2: x = 12, y = 12, dx = 2 Equation of l is y – 12 = 2( x – 12) i.e. y = 2x – 12 (iii) Substitute x = t2 + 4t, y = t3 + t2 into equation of l : t3 + t2 = 2(t2 + 4t) – 12 t3 – t2 – 8t + 12 = 0 10

(t – 2)(t2 + t – 6) = 0 (t – 2)(t – 2)(t + 3) = 0 t = 2 (reject since this gives us the point P), –3 x = 9 – 12 = –3 y = –27 + 9 = –18 coordinates of Q is (–3, –18). By Ratio Theorem,    OA + 2OB OP = 3 14  11  1   14 + 2–13 = 3     14  2  36  1 –12 =3   18  12 = –4   6 coordinates of P is (12, –4, 6). 11  14  –3  1   –13 – 14 = –27 = –39 (ii) AB =          2  14 –12  4 1  6    AB  OP = –39  2–2      4  3  = –6(6 – 18 + 12) =0 Hence AB & OP are perpendicular. 6 1 –2 c= 36 + 4 + 9  3    2i) Hint to students: Some students have the mistaken notion  2  that OP = 3 AB .

A

P

B

 They should realise that OP may not be  parallel to AB at all, and that the Ratio  Theorem is the quickest way to find OP . Hint to students: This is a good place to check that the answer obtained in part (i) is correct. If students cannot show that AB and OP are perpendicular, then they should suspect that they have made mistakes in their working in part (i).

(iii)

6 1  –2 =7  3 | a  c | is the length of the projection of the vector a on OP. 14 12 a  p = 14  –4     14  6  1 6 = 14 1  2–2     1 3  3+2  = 28 –(3 – 6)    –2 – 6  Hint to students: It is advisable to remove common factors before finding cross product to simplify the multiplication and reduce chances of computational mistakes.

(iv)

11

5 = 28  3    –8 | a  p | is the area of the parallelogram formed by the vectors a and p. 1 Area of triangle OAP = 2 | a  p |  5  1 = 2  28  3    –8 = 14 25 + 9 + 64 = 14 98 = 98 2 3i) ax Let y = bx – a Hint to students: Students should be able to spot that f(x) = f –1(x) immediately. This enables them to work out f 2(x) effortlessly without any computation at all.

bxy – ay = ax bxy – ax = ay x(by – a) = ay ay x = by – a ax f –1(x) = bx – a f(x) = f –1(x)  f 2(x) = x a Range of f 2 = ℝ\{ b } (ii) a Rg = ℝ\{ 0 } ⊈ Df = ℝ\{ b } since a  0.

Hint to students: f 2(x) is the identity function. Hence range of f 2 = domain of f. Teaching Point: Students are reminded to give Rg and Df instead of just quoting “Rg ⊈ Df ”.

Hence fg does not exist. f –1(x) = x ax bx – a = x ax = bx2 – ax bx2 – 2ax = 0 x(bx – 2a) = 0 2a x = 0 or b (iii) 4) d2n dt2 = 10 – 6t where c = constant where d = constant

dn 2 dt = 10t – 3t + c, n = 5t2 – t3 + ct + d,

12

When t = 0, n = 100  100 = d  n = 5t2 – t3 + ct + 100 c = 10 c=0 100 c = –10

Hint to students: If students choose to plot using c = –1, 0, 1, the three curves tend to be very close together. Students can try using c = –10, 0, 10 instead.

(ii)

dn dt = 3 – 0.02n

1  3 – 0.02n dn =  dt ln | 3 – 0.02n | =t+c –0.02 ln | 3 – 0.02n | = –0.02t – 0.02c | 3 – 0.02n | = e–0.02t e–0.02c 3 – 0.02n =  e–0.02t e–0.02c = Ae–0.02t where A =  e–0.02c When t = 0, n = 100: 3–2=A 3 – 0.02n = e–0.02t 0.02n = 3 – e–0.02t n = 150 – 50e–0.02t As t  , n  150 the population will stabilise at 150 000. 5) The manager can choose to survey, say, 50 male cinema–goers and 50 female cinema– goers. He is free to choose anyone convenient to meet his quota. One disadvantage of quota sampling is that the sample obtained is likely to be biased. 6i) t 40

Hint to students: Students must include modulus sign when 1 integrating 3 – 0.02n since 3 – 0.02n may be negative. However they are reminded to introduce A =  e–0.02c before substituting t = 0, n = 100.

Hint to students: There are many ways to answer these two parts of the question. All reasonable answers will be accepted.

20 x 1930 2000

Hint to students: Students are reminded to label and indicate the scale on the axes. It is advisable to draw the scatter diagram to scale and to copy what appears on the screen of the GC as closely as possible.

13

(ii) A linear model may not be appropriate since there is a certain limit to how fast a person can complete the distance. This is also evident in the scatter diagram, which shows a slight reduction in the rate of decrease of the record time. (iii)

A quadratic model would show the record time increasing again in the future, which is impossible. Hence a quadratic model would not be appropriate. (iv)

By GC, the regression line is ln t = 34.853 – 0.016128x. The predicted world record time as at 1st January 2010 is 3 min 30 + e34.853 – 0.016128(2010) = 3 min 30 s + 11.424 s = 3 min 41.4 s.

14

Since x = 2010 is outside the range of the data values, the prediction is not reliable. 7i) P(faulty) = 0.25  0.05 + 0.75  0.03 = 0.035 (ii) f(p) = P(supplied by A | faulty) P(supplied by A  faulty) = P(faulty) p 100  0.05 = p 100 – p  0.05 + 100  0.03 100 0.05p = 0.05p + 3 – 0.03p 0.05p = 0.02p + 3 Shown (0.02p + 3)0.05 – 0.05p(0.02) f (p) = (0.02p + 3)2 0.15 = (0.02p + 3)2 > 0 for all 0  p  100. Hence f is an increasing function for 0  p  100. This means that as we buy more and more components from A, it is more likely that a randomly chosen component that is faulty was supplied by A. 8i) Since there are 8 letters, including 3 E’s, 8! no. of ways = 3! = 6720 (ii) Method 1: No. of ways where T & D are together 7! = 3!  2! = 1680 Hence no. of ways where T & D are not next to one another = 6720 – 1680 = 5040 Method 2: E L E V A E        6! The 6 letters besides T & D can be in 3! ways. The letters T & D can be arranged in the 7 spaces available in 7P2 ways. Hint to students: Using the Complementary Method (Method 1) depends on the assumption that the answer to part (i) is correct, which should be the case. If there are any doubts, students can use Method 2.

Hint to students: The question requires the candidate to “prove by differentiation”. It would be wrong to sketch a graph to show that it is increasing.

15

6! Hence total no. of ways = 3! 7P2 = 5040. (iii) The consonants and vowels must be arranged like this: CVCVCVCV or VCVCVCVC The 4 consonants can be arranged in 4! ways. 4! The 4 vowels can be arranged in 3! ways. 4! Hence total no. of ways = 4! 3!  2 = 192. (iv) The letters can be arranged as follows: _E__E__E E___E__E E__E___E E__E__E_ Hint to students: First make sure there are two letters between any two E’s. This takes up 4 letters. The remaining letter can be placed before the first E, between the first 2 E’s, between the last 2 E’s, or after the last E.

or or or

Total no. of ways = 5!  4 = 480. 9i) Method 1: 0.12 M ~ N(2.5, n ) Teaching Point: Students have a choice of either solving algebraically (Method 1) or listing out the probabilities and searching for the answer (Method 2). It is individual preference.

P( M > 2.53) = 0.0668 P( M < 2.53) = 0.9332 2.53 – 2.5 P( Z < ) = 0.9332 0.1 n 0.3 n = 1.500056 n = 25 Method 2:

16

(ii) Let M, S = thickness of a mechanics and statistics textbook respectively. M1 +... + M21 + S1 +...+ S24 ~ N(212.5 + 242, 210.12 + 240.082) = N(100.5, 0.3636) P(M1 +... + M21 + S1 +...+ S24  100) = 0.203 (iii) S1 +...+ S4 – 3M ~ N(42 – 32.5, 40.082 + 320.12) = N(0.5, 0.1156) P(S1 +...+ S4 < 3M) = P(S1 +...+ S4 – 3M < 0) = 0.0707 (iv) We assume that the thicknesses of mechanics and statistics textbooks are independent. 10i) Unbiased estimate of the mean 86.4 = x = 9 = 9.6 Unbiased estimate of the variance 1 86.42 = s2 = 8 (835.92 – 9 ) = 0.81 (ii) We assume that the distribution of the mass of sugar in a packet is normal. H0 :  = 10 H1 :   10 X – 10 Under H0 , T = 0.81 9 ~ t(8) Teaching Point: The formula for the unbiased estimate of the variance is found in MF15. There is no need for students to memorise it. Hint to students: Students need to square 3 when computing variance since we are dealing with the multiple of a normal variable. Hint to students: There is no need to square 21 and 24 when computing variance since we are dealing with sums of normal variables.

17

Since p–value = 0.219 > 0.05, we do not reject H0 . There is insufficient evidence at the 5% level to say that the mean mass of sugar in a packet is not 10 grams. The Central Limit Theorem does not apply in this case since the sample size 9 is small. (iii) We would carry out a Z–test instead.

11i) Two assumptions are:  the colours of the n cars are independent of one another.  the probability of a car being red is constant. (ii) R ~ B(20, 0.15) P(4  R < 8) = P(R  7) – P(R  3) = 0.346 (iii) Since n = 240 is large, np = 72 > 5, nq = 168 > 5, R ~ N(72, 50.4) approximately. P(R < 60) = P(R < 59.5) (continuity correction)  0.0391 (iv) Since n = 240 > 50, np = 4.8 < 5, R ~ Po(4.8) approximately. The Poisson approximation is appropriate in this case since mean = 4.8  4.704 = variance. P(R = 3)  0.1517 to 4 decimal places. (v) P(R = 0 or 1) = P(R = 0) + P(R = 1) = (1 – p)20 + 20p(1 – p)19 = 0.2 18 Hint to students: Students have to be careful here since P(R < 8) = P(R  7) and they have to subtract P(R  3) instead of P(R  4). Hint to students: Some students mix up the use of nq and npq. Also it is common for students to forget to do continuity correction.

 (1 – p)20 + 20p(1 – p)19 – 0.2 = 0 Method 1:

From GC, p = 0.142. Method 2:

From GC, p = 0.142. Method 3:

19

From GC, p = 0.142.

20