2009 A Level H2 Maths Solution - Texas Instruments



2009 GCE ‘A’ Level Solution Paper 1 (Contributed by Hwa Chong Institution)

|1i) Let un = an2 + bn + c. |Teaching Point: |

|u1 = a + b + c = 10 |The set of simultaneous equations can either |

|u2 = 4a + 2b + c = 6 |be solved by using the application PlySmlt2 |

|u3 = 9a + 3b + c = 5 |or finding rref of the augmented matrix |

|Using GC, a = 1.5, b = –8.5, c = 17. |formed by the 3 equations. |

|( un = 1.5n2 – 8.5n + 17. | |

|(ii) Let y = 1.5n2 – 8.5n + 17 – 100 |Teaching Point: |

|[pic] |Please use ZoomFit on your GC to see the |

|From GC, n > 10.79. |entire graph clearly. |

|Hence solution set = {n ( ℤ+ : n ( 11}. |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. |

|2) dx |Hint to students: |

|= [ln ] |[pic] |

|= ( ln 3 – ln 1 ) |It is helpful to use the GC to do a quick |

|= ln 3 |check that the integration has been done |

|dx |correctly. |

|= dx |Teaching point: |

|= [sin–1 px] |Students sometimes leave out the term when |

|= (sin–1 ) |integrating . To safeguard this, it is |

|= = ln 3 |advisable to change the integrand into the |

|p = |form before integrating. |

|3i) – + | |

|= | |

|= Shown | |

|(A = 2. | |

|(ii) | |

|= | |

|= ( – + ) | |

|= [ – + | |

|+ – + | |

|+ – + | |

|: | |

|: | |

|: | |

|: | |

|+ – + | |

|+ – + ] | |

|= [ – + ] | |

|(iii) As n ( (, ( 0 and ( 0. | |

|(the series converges to the value . | |

|4i) f(27) + f(45) | |

|= f(23) + f(41) | |

|= f(19) + f(37) | |

|: | |

|: | |

|= f(3) + f(1) | |

|= 5 + 6 | |

|= 11 | |

|(ii) |Teaching Point: |

|[pic] |Students should be advised to sketch a clear |

| |and properly–labelled graph. |

|(iii) f(x) dx |Hint to students: |

|= 27 – x2 dx + 2x – 1 dx + 2x – 1 dx |The area under the curve from x = –4 to x = |

|= 2[7x – ]+ [x2 – x]+ [x2 – x] |–2 is exactly the same as the area from x = 0|

|= 2[14 – ] + [12 – 2] + [6 – 2] |to x = 2. Hence the factor 2 in front of 7 – |

|= 36 |x2 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 find2x – 1 dx. |

|5) Let Pn be the statement: | |

|r2 = n(n + 1)(2n + 1). | |

|When n = 1: | |

|LHS = 12 = 1 | |

|RHS = 1(2)(3) = 1 = LHS | |

|(P1 is true. | |

| | |

|Assume that Pk is true for some k ( ℤ + | |

|i.e. r2 = k(k + 1)(2k + 1). | |

| | |

|Prove that Pk+1 is also true | |

|i.e. r2 = (k + 1)(k + 2)(2k + 3). |Teaching Point: |

| |Some students make the mistake of writing |

|LHS |“Assume that Pk is true for all k ( ℤ +”. |

|= r2 + (k + 1)2 |It should be either “some k” or “a k”. |

|= k(k + 1)(2k + 1) + (k + 1)2 |Clearly, if you assume that it is true for |

|= (k + 1)[ k(2k + 1) + 6(k + 1) ] |all k, then there is nothing to prove. |

|= (k + 1)(2k2 + 7k + 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 +. | |

| | |

| |Teaching Point: |

| |Students are advised to take out the common |

| |factor (k + 1) instead of expanding |

| |everything and then factorising later. |

| |Clearly a waste of effort. |

|6i) |Hint to students: |

|[pic] |Students must realise that x = –2 is a |

| |vertical asymptote and draw it even though it|

| |does not appear on the Graphing Calculator. |

|(ii) Substitute y = into + = 1: |Hint to students: |

|+ = 1: |The question requires students to “show |

|x2(x + 2)2 + 2(x – 2)2 = 6(x + 2)2 |algebraically”. It would be wrong if students|

|2(x – 2)2 = (x + 2)2 (6 – x2) Shown |show that the numerical values of the points |

| |of intersection obtained from the GC satisfy |

| |the given equation. |

|(iii) Method 1: |Teaching Point: |

|Find the points of intersection from the original graphs. |Students have a choice of using Method 1 |

|[pic][pic] |(since the graphs have already been plotted |

|From GC, x = –0.515 or 2.45 |on the GC) or Method 2 which is also |

| |relatively straight forward.. |

|Method 2: | |

|Sketch y = 2(x – 2)2 – (x + 2)2 (6 – x2) | |

|[pic][pic] | |

|From GC, x = –0.515 or 2.45 | |

|7i) f ((x) = –sin x ecos x = – sin x f(x) |Hint to students: |

|f ((x) = – sin x f ((x) – cos x f(x) |It is helpful to use the GC to check the |

|f(0) = e |value of the derivative at x = 0. |

|f ((0) = 0 |[pic] |

|f ((0) = –e | |

|(f(x) = e + 0x – x2 + ... | |

|= e – x2 + ... | |

|(ii) | |

|= | |

|= (1 + x2)–1 | |

|= (1 – x2 + ...) | |

|= – x2 + ... = e – x2 + ... | |

|( = e ( a = | |

|– = – ( be2 = ( b = |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. |

|8i) ar24 = 20r24 = 5 | |

|r24 = | |

|r = | |

|Total length of all bars | |

|< S( | |

|= | |

|= 356.343 |Hint to students: |

|< 357 Shown |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. |

|(ii) L = 20 = 272.2573 | |

|Length of 13th bar = 20 | |

|= 20 ( | |

|= 10 | |

| | |

|Let b = length of first bar of instrument B. | |

|[ 2b + 24d ] = L | |

|b + 12d = | |

|Also b + 24d = 5 | |

|So 12d = 5 – | |

|= 5 – | |

|( d = –0.49086 | |

| | |

|Length of longest bar | |

|= b | |

|= 5 – 24(–0.49086) | |

|= 16.8 cm | |

|9i) z7 = 1 + i | |

|= ei(/4 |Hint to students: |

|= ei(2k(+(/4) |Since the arguments must lie in the principal|

|z = 21/14 e, k = 0, (1, (2, (3 |range, k must take the values 0, (1, (2, (3. |

|(ii) |Hint to students: |

|[pic] |Students must realise that all the roots have|

| |the same modulus 21/14 and are all spaced |

| |radians apart on the Argand Diagram. |

|(iii) 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. | |

| | |

|[pic] | |

|Equation of locus is y = x tan ( + ) | |

|= x tan | |

|10i) cos ( = | |

|= |Hint to students: |

|( = 70.9( |The question specified “acute angle”. Hence |

| |the modulus sign. In this case, it makes no |

| |difference to the answer though. |

|(ii) Solving 2x + y + 3z = 1 |Teaching Point: |

|–x + 2y + z = 2 |Students can either use the application |

|[pic] |PlySmlt2 or find the rref of the augmented |

|[pic] |matrix formed by the 2 equations. |

|From GC, the equation of l is r = + (. | |

|(iii) Substitute x = –(, y = 1 – (, z = ( into the equation of p3 : | |

|LHS | |

|= –2( + 1 – ( + 3( – 1 + k(( + 2(1 – () + ( – 2) | |

|= 0 for any ( | |

|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: |

|[pic] |Students must show clearly the asymptotic |

| |nature of the graph as it approaches the |

| |x-axis. |

|(ii) = e+ x e(–2x) |Hint to students: |

|= e(1 – 2x2) = 0 |Students can use the GC to check that these |

|x2 = |turning points are numerically correct. |

|x = , – |[pic] |

|y = e–1/2 , – e–1/2 | |

|(turning points = (,e–1/2 ) | |

|and (– ,– e–1/2 ) | |

|(iii) u = x2 ( = 2x | |

|When x = 0, u = 0 | |

|When x = n, u = n2 | |

|x edx |Hint to students: |

|= e–u du |Students can use the GC to check that the |

|= [–e–u ] |area is numerically correct. |

|= [1 – e ] |[pic] |

|As n ( (, e( 0, so area = | |

|(iv) | f(x) | dx |Hint to students: |

|= 2 f(x) dx |The GC can be used to do a quick check that |

|= 1 – e–4 |the answer is numerically correct. |

| |[pic] |

|(v) Volume = ( x2 e dx | |

|= 0.363 | |

2009 GCE ‘A’ Level Solution Paper 2

|1i) | |

|[pic] | |

|(ii) Method 1: |Hint to students: |

|= 2t + 4 |The gradient of the curve at t = 2 can either be obtained |

|= 3t2 + 2t |algebraically (Method 1) or from the GC (Method 2). |

|= | |

|When t = 2: x = 12, y = 12, = = 2 | |

|Equation of l is | |

|y – 12 = 2( x – 12) | |

|i.e. y = 2x – 12 | |

|Method 2: | |

|[pic] | |

|[pic] | |

|When t = 2: x = 12, y = 12, = 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 | |

|(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). | |

|2i) By Ratio Theorem, |Hint to students: |

|= |Some students have the mistaken notion that = . |

|= | |

|= | |

|= |A P B |

|(coordinates of P is (12, –4, 6). | |

| |They should realise that may not be parallel to at all, and |

| |that the Ratio Theorem is the quickest way to find . |

|(ii) = – = = –3 |Hint to students: |

|( = –3( 2 |This is a good place to check that the answer obtained in part |

|= –6(6 – 18 + 12) |(i) is correct. If students cannot show that AB and OP are |

|= 0 |perpendicular, then they should suspect that they have made |

|Hence AB & OP are perpendicular. |mistakes in their working in part (i). |

|(iii) c = | |

|= | |

|| a ( c | is the length of the projection of the vector a on OP. | |

|(iv) a ( p = ( |Hint to students: |

|= 14 ( 2 |It is advisable to remove common factors before finding cross |

|= 28 |product to simplify the multiplication and reduce chances of |

|= 28 |computational mistakes. |

|| a ( p | is the area of the parallelogram formed by the vectors a and p.| |

|Area of triangle OAP = | a ( p | | |

|= ( 28 | |

|= 14 | |

|= 14 | |

|= 98 | |

|3i) Let y = |Hint to students: |

|bxy – ay = ax |Students should be able to spot that f(x) = f –1(x) |

|bxy – ax = ay |immediately. This enables them to work out f 2(x) effortlessly |

|x(by – a) = ay |without any computation at all. |

|x = | |

|(f –1(x) = | |

|f(x) = f –1(x) | |

|( f 2(x) = x | |

| | |

|Range of f 2 = ℝ\{ } | |

| |Hint to students: |

| |f 2(x) is the identity function. Hence range of f 2 = domain of|

| |f. |

|(ii) Rg = ℝ\{ 0 } ⊈ Df = ℝ\{ } since a ( 0. |Teaching Point: |

|Hence fg does not exist. |Students are reminded to give Rg and Df instead of just quoting|

| |“Rg ⊈ Df ”. |

|(iii) f –1(x) = x | |

|= x | |

|ax = bx2 – ax | |

|bx2 – 2ax = 0 | |

|x(bx – 2a) = 0 | |

|x = 0 or | |

|4) = 10 – 6t | |

|= 10t – 3t2 + c, where c = constant | |

|n = 5t2 – t3 + ct + d, where d = constant | |

| | |

|When t = 0, n = 100 | |

|( 100 = d | |

|( n = 5t2 – t3 + ct + 100 | |

|[pic] | |

| | |

| |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) = 3 – 0.02n |Hint to students: |

|( dn = ( dt |Students must include modulus sign when integrating since 3 – |

|= t + c |0.02n may be negative. However they are reminded to introduce A|

|ln | 3 – 0.02n | = –0.02t – 0.02c |= ( e–0.02c before substituting t = 0, n = 100. |

|| 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 |Hint to students: |

|female cinema–goers. He is free to choose anyone convenient to meet his |There are many ways to answer these two parts of the question. |

|quota. |All reasonable answers will be accepted. |

| | |

|One disadvantage of quota sampling is that the sample obtained is likely | |

|to be biased. | |

|6i) | |

| |Hint to students: |

|[pic] |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. |

|(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) | |

|[pic] | |

|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) | |

|[pic] | |

|[pic] | |

|[pic] | |

|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. | |

|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) | |

|= | |

|= | |

|= | |

|= Shown | |

|f ((p) = | |

|= > 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.| |

| | |

| | |

| |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. |

|8i) Since there are 8 letters, including 3 E’s, no. of ways = = 6720 | |

|(ii) Method 1: |Hint to students: |

|No. of ways where T & D are together |Using the Complementary Method (Method 1) depends on the |

|= ( 2! = 1680 |assumption that the answer to part (i) is correct, which should|

|Hence no. of ways where T & D are not next to one another = 6720 – 1680 =|be the case. If there are any doubts, students can use Method |

|5040 |2. |

| | |

|Method 2: | |

| | |

|E | |

| | |

|L | |

| | |

|E | |

| | |

|V | |

| | |

|A | |

| | |

|E | |

| | |

| | |

|( | |

| | |

|( | |

| | |

|( | |

| | |

|( | |

| | |

|( | |

| | |

|( | |

| | |

|( | |

| | |

| | |

|The 6 letters besides T & D can be in ways. | |

|The letters T & D can be arranged in the 7 spaces available in 7P2 ways. | |

|Hence total no. of ways = 7P2 = 5040. | |

|(iii) The consonants and vowels must be arranged like this: | |

|C V C V C V C V | |

|or V C V C V C V C | |

|The 4 consonants can be arranged in 4! ways. | |

|The 4 vowels can be arranged in ways. | |

|Hence total no. of ways = 4! ( 2 = 192. | |

|(iv) The letters can be arranged as follows: |Hint to students: |

| |First make sure there are two letters between any two E’s. This|

|_ E _ _ E _ _ E |takes up 4 letters. |

|or E _ _ _ E _ _ E |The remaining letter can be placed before the first E, between |

|or E _ _ E _ _ _ E |the first 2 E’s, between the last 2 E’s, or after the last E. |

|or E _ _ E _ _ E _ | |

| | |

|Total no. of ways = 5! ( 4 = 480. | |

|9i) Method 1: |Teaching Point: |

|~ N(2.5, ) |Students have a choice of either solving algebraically (Method |

|P(> 2.53) = 0.0668 |1) or listing out the probabilities and searching for the |

|P(< 2.53) = 0.9332 |answer (Method 2). It is individual preference. |

|P( Z < ) = 0.9332 | |

|0.3= 1.500056 | |

|n = 25 | |

|Method 2: | |

|[pic] | |

|[pic] | |

|(ii) Let M, S = thickness of a mechanics and statistics textbook | |

|respectively. | |

| | |

|M1 +... + M21 + S1 +...+ S24 |Hint to students: |

|~ N(21(2.5 + 24(2, 21(0.12 + 24(0.082) |There is no need to square 21 and 24 when computing variance |

|= N(100.5, 0.3636) |since we are dealing with sums of normal variables. |

| | |

|P(M1 +... + M21 + S1 +...+ S24 ( 100) | |

|= 0.203 | |

|(iii) S1 +...+ S4 – 3M |Hint to students: |

|~ N(4(2 – 3(2.5, 4(0.082 + 32(0.12) |Students need to square 3 when computing variance since we are |

|= N(0.5, 0.1156) |dealing with the multiple of a normal variable. |

| | |

|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 |Teaching Point: |

|== = 9.6 |The formula for the unbiased estimate of the variance is found |

|Unbiased estimate of the variance |in MF15. There is no need for students to memorise it. |

|= s2 = (835.92 – ) = 0.81 | |

|(ii) We assume that the distribution of the mass of sugar in a packet is | |

|normal. | |

|H0 : ( = 10 | |

|H1 : ( ( 10 | |

| | |

|Under H0 , T = ~ t(8) | |

|[pic] | |

|[pic] | |

| | |

|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) |Hint to students: |

|P(4 ( R < 8) |Students have to be careful here since P(R < 8) = P(R ( 7) and |

|= P(R ( 7) – P(R ( 3) |they have to subtract P(R ( 3) instead of P(R ( 4). |

|= 0.346 | |

|(iii) Since n = 240 is large, np = 72 > 5, nq = 168 > 5, R ~ N(72, 50.4) |Hint to students: |

|approximately. |Some students mix up the use of nq and npq. |

|P(R < 60) |Also it is common for students to forget to do continuity |

|= P(R < 59.5) (continuity correction) |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 | |

|( (1 – p)20 + 20p(1 – p)19 – 0.2 = 0 | |

|Method 1: | |

|[pic] | |

|From GC, p = 0.142. | |

|Method 2: | |

|[pic] | |

|[pic] | |

|[pic] | |

|From GC, p = 0.142. | |

|Method 3: | |

|[pic] | |

|[pic] | |

|[pic] | |

|From GC, p = 0.142. | |

[pic]

-----------------------

100

–4

8

x

y

4

3

–7

10

–6

6

2

–2

7

t

x

40

1930

2000

20

z2

c = –10

c = 0

c = 10

z1

(/7

(/28

(/7

2(/7

2(/7

2(/7

2(/7

2(/7

(/4

(/28

2(/7

–

–

–1

2

x = –2

y = 1

................
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