2009 GCE A Level Solution Paper 1

Size: px

Start display at page:

Download "2009 GCE A Level Solution Paper 1"

Claribel Stephens
5 years ago
Views:

1 2009 GCE A Level Solution Paper i) Let u n = an 2 + bn + c. u = a + b + c = 0 u 2 = 4a + 2b + c = 6 u 3 = 9a + 3b + c = 5 Using GC, a =.5, b = 8.5, c = 7. u n =.5n 2 8.5n + 7. (ii) Let y =.5n 2 8.5n From GC, n > Hence solution set = {n Z + : n }. 2) /2p x 2 dx = [ 4 ln 2 + x 2 x ] 0 = 4 ( ln 3 ln ) = 4 ln 3 p 2 x 2 dx = p 0 /2p dx p 2 x2 = p [sin px] /2p 0 = p (sin 2 ) = π p 6 = 4 ln 3 2π p = 3 ln 3 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 =.5n 2 8.5n , another method is to find the point of intersection of the curve y =.5n 2 8.5n + 7 and the line y = 00. 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 the term p when integrating p 2 x 2. To safeguard this, it is advisable to change the integrand into the form a 2 x 2 before integrating. 3i) n 2 n + n + = n(n + ) 2(n2 ) + n(n ) n(n 2 )

2 A = 2. = 2 n 3 n Shown n (ii) r 3 r r=2 = n 2 2 r 3 r r=2 = n 2 ( r 2 r + r + ) r=2 = 2 [ : : : : + n 2 2 n + n + n 2 n + n + ] = 2 [ 2 n + n + ] (iii) As n, n 0 and n + 0. the series converges to the value 4. 4i) f(27) + f(45) = f(23) + f(4) = f(9) + f(37) : : = f(3) + f() = = 2

3 (ii) y 7 Teaching Point: Students should be advised to sketch a clear and properly labelled graph x (iii) 3 f(x) dx 4 = x 2 dx x dx x dx 2 = 2[7x x3 3 ]2 0 + [x2 x] [x2 x] 3 2 = 2[4 8 3 ] + [2 2] + [6 2] = ) Let P n be the statement: n r 2 = 6 n(n + )(2n + ). r= When n = : LHS = 2 = RHS = 6 (2)(3) = = LHS P is true. Assume that P k is true for some k Z + k i.e. r 2 = 6 k(k + )(2k + ). r= Prove that P k+ is also true k+ i.e. r 2 = 6 (k + )(k + 2)(2k + 3). r= LHS k = r 2 + (k + ) 2 r= 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 x 2 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 dx. 2 Teaching Point: Some students make the mistake of writing Assume that P k 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. 3

4 = 6 k(k + )(2k + ) + (k + )2 = 6 (k + )[ k(2k + ) + 6(k + ) ] = 6 (k + )(2k2 + 7k + 6) = 6 (k + )(k + 2)(2k + 3) = RHS Since P is true and P k true P k+ true, hence by Math Induction, P n is true for all n Z +. Teaching Point: Students are advised to take out the common factor 6 (k + ) instead of expanding everything and then factorising later. Clearly a waste of effort. 6i) Students must realise that x = 3 2 is a vertical asymptote and y = draw it even though it does not appear on the Graphing Calculator. 3 x = 2 (ii) Substitute y = x 2 x2 x + 2 into 6 + y2 3 = : x x 2 x = : x 2 (x + 2) 2 + 2(x 2) 2 = 6(x + 2) 2 2(x 2) 2 = (x + 2) 2 (6 x 2 ) Shown (iii) Method : Find the points of intersection from the original graphs. 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 (since the graphs have already been plotted on the GC) or Method 2 which is also relatively straight forward.. From GC, x = 0.55 or

5 Method 2: Sketch y = 2(x 2) 2 (x + 2) 2 (6 x 2 ) From GC, x = 0.55 or i) f (x) = sin x e cos x = sin x f(x) f (x) = sin x f (x) cos x f(x) f(0) = e f (0) = 0 f (0) = e f(x) = e + 0x e 2 x It is helpful to use the GC to check the value of the derivative at x = 0. = e e 2 x (ii) = a a + bx 2 + bx2 a = a ( + b a x2 ) = a ( b a x2 +...) = a b a 2 x = e e 2 x a = e a = e It is easier to use the expansion ( + x) = x + x 2... instead of the general Binomial Expansion formula in this case. b a 2 = e 2 be2 = e 2 b = 2e 8i) ar 24 = 20r 24 = 5 r 24 = 4 5

6 r = 2 2 Total length of all bars < S 20 = 2 2 = < 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 2 = Length of 3th bar = = 20 2 = 0 Let b = length of first bar of instrument B [ 2b + 24d ] = L b + 2d = L 25 Also b + 24d = 5 So 2d = 5 L 25 = d = Length of longest bar = b = 5 24( ) = 6.8 cm 9i) z 7 = + i = 2 e iπ/4 = 2 e i(2kπ+π/4) z = 2 /4 e i(2kπ 7 + π 28 ), k = 0, ±, ±2, ±3 Since the arguments must lie in the principal range, k must take the values 0, ±, ±2, ±3. 6

7 (ii) 2π/7 2π/7 2π/7 π/28 2π/7 π/4 2π/7 2π/7 Students must realise that all the roots have the same modulus 2 /4 and are all spaced 2π radians apart on 7 the Argand Diagram. (iii) Substituting z = 0 into z z = z z 2 : 0 z = z = 2 /4 0 z 2 = z 2 = 2 /4 Since z = 0 satisfies the equation z z = z z 2, the locus passes through the origin. z 2 π/7 π/7 π/28 z Equation of locus is y = x tan ( π 28 + π 7 ) = x tan 5π 28 0i) cos θ = = θ = 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 = 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.

Substitute x = 2, y = 3, z = 4 into the equation of p 3 : LHS = 4 + 3 + 2 + k( 2 + 6 + 4 2) = 8 + 6k = 0 k = 3 equation of plane is 2x + y + 3z 3( x + 2y + z 2) = 0 i.e. 5x 5y + 5 = 0 i.e. x y + = 0 i) Teaching Point: Students must show clearly the asymptotic nature of the graph as it approaches the x- axis.

8 (ii) Solving 2x + y + 3z = 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 From GC, the equation of l is r = + λ. 0 (iii) Substitute x = λ, y = λ, z = λ into the equation of p 3 : LHS = 2λ + λ + 3λ + k(λ + 2( λ) + λ 2) = 0 for any λ Hence l lies in p 3 for all k. Substitute x = 2, y = 3, z = 4 into the equation of p 3 : LHS = k( ) = 8 + 6k = 0 k = 3 equation of plane is 2x + y + 3z 3( x + 2y + z 2) = 0 i.e. 5x 5y + 5 = 0 i.e. x y + = 0 i) Teaching Point: Students must show clearly the asymptotic nature of the graph as it approaches the x- axis. (ii) dy dx = e x2 + x e x2 ( 2x) Students can use the GC to check that these turning 8

9 = e x2 ( 2x 2 ) = 0 x 2 = 2 points are numerically correct. x = 2, 2 turning points = ( 2, 2 e /2 ) y = 2 e /2, 2 e /2 and ( 2, 2 e /2 ) (iii) n x e x2 dx 0 u = x 2 du dx = 2x When x = 0, u = 0 When x = n, u = n 2 = 0 n 2 2 e u du Students can use the GC to check that the area is numerically correct. = 2 [ e u ] n2 0 = 2 [ e n2 ] As n, e n2 0, so area = 2 (iv) 2 f(x) dx 2 = 2 2 f(x) dx 0 = e 4 The GC can be used to do a quick check that the answer is numerically correct. (v) Volume = π 0 x 2 e 2x2 dx =

10 2009 GCE A Level Solution Paper 2 i) (ii) Method : dx dt = 2t + 4 dy dt = 3t2 + 2t dy dx = 3t2 + 2t 2t + 4 When t = 2: x = 2, y = 2, dy dx = 6 8 = 2 Equation of l is y 2 = 2( x 2) i.e. y = 2x 2 The gradient of the curve at t = 2 can either be obtained algebraically (Method ) or from the GC (Method 2). Method 2: When t = 2: x = 2, y = 2, dy dx = 2 Equation of l is y 2 = 2( x 2) i.e. y = 2x 2 (iii) Substitute x = t 2 + 4t, y = t 3 + t 2 into equation of l : t 3 + t 2 = 2(t 2 + 4t) 2 t 3 t 2 8t + 2 = 0 0

11 (t 2)(t 2 + t 6) = 0 (t 2)(t 2)(t + 3) = 0 t = 2 (reject since this gives us the point P), 3 x = 9 2 = 3 y = = 8 coordinates of Q is ( 3, 8). 2i) By Ratio Theorem, OA + 2OB OP = 3 = = = 4 6 coordinates of P is (2, 4, 6). 4 3 (ii) AB = 3 4 = 27 = AB OP 6 = = 6( ) = 0 Hence AB & OP are perpendicular. Some students have the mistaken notion that OP = 2 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. 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) c = = a c is the length of the projection of the vector a on OP. 4 2 (iv) a p = = = 28 (3 6) 2 6 It is advisable to remove common factors before finding cross product to simplify the multiplication and reduce chances of computational mistakes.

12 5 = a p is the area of the parallelogram formed by the vectors a and p. Area of triangle OAP = 2 a p = = = 4 98 = 98 2 ax 3i) Let y = bx a bxy ay = ax bxy ax = ay x(by a) = ay ay x = by a f ax (x) = bx a f(x) = f (x) f 2 (x) = x Range of f 2 = R\{ a b } (ii) R g = R\{ 0 } D f = R\{ a b } since a 0. Hence fg does not exist. Students should be able to spot that f(x) = f (x) immediately. This enables them to work out f 2 (x) effortlessly without any computation at all. f 2 (x) is the identity function. Hence range of f 2 = domain of f. Teaching Point: Students are reminded to give R g and D f instead of just quoting R g D f. (iii) f (x) = x ax bx a = x ax = bx 2 ax bx 2 2ax = 0 x(bx 2a) = 0 x = 0 or 2a b d 2 n 4) 2 dt = 0 6t dn dt = 0t 3t2 + c, n = 5t 2 t 3 + ct + d, where c = constant where d = constant 2

13 When t = 0, n = = d n = 5t 2 t 3 + ct c = 0 c = 0 If students choose to plot using c =, 0,, the three curves tend to be very close together. Students can try using c = 0, 0, 0 instead. c = 0 dn (ii) dt = n n dn = dt ln n 0.02 = t + c ln n = 0.02t 0.02c n = e 0.02t e 0.02c n = ± e 0.02t e 0.02c = Ae 0.02t where A = ± e 0.02c Students must include modulus sign when integrating since n may n be negative. However they are reminded to introduce A = ± e 0.02c before substituting t = 0, n = 00. When t = 0, n = 00: 3 2 = A n = e 0.02t 0.02n = 3 e 0.02t n = 50 50e 0.02t As t, n 50 the population will stabilise at ) 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. There are many ways to answer these two parts of the question. All reasonable answers will be accepted. One disadvantage of quota sampling is that the sample obtained is likely to be biased. 6i) t 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. 930 x

(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.

14 (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 = x. The predicted world record time as at st January (200) 200 is 3 min 30 + e = 3 min 30 s s = 3 min 4.4 s. Since x = 200 is outside the range of the data 4

15 values, the prediction is not reliable. 7i) P(faulty) = = (ii) f(p) = P(supplied by A faulty) P(supplied by A faulty) = P(faulty) p = p 00 p p = 0.05p p 0.05p = 0.02p + 3 Shown (0.02p + 3) p(0.02) f (p) = (0.02p + 3) = 2 (0.02p + 3) > 0 for all 0 p 00. Hence f is an increasing function for 0 p 00. 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. 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 = 8! 3! = 6720 (ii) Method : No. of ways where T & D are together = 7! 3! 2! = 680 Hence no. of ways where T & D are not next to one another = = 5040 Using the Complementary Method (Method ) 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. Method 2: E L E V A E The 6 letters besides T & D can be in 6! 3! ways. The letters T & D can be arranged in the 7 spaces available in 7 P 2 ways. Hence total no. of ways = 6! 3! 7 P 2 =

(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 4! 3! ways. Hence total no.

16 (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 4! 3! ways. Hence total no. of ways = 4! 4! 3! 2 = 92. (iv) or or or The letters can be arranged as follows: _ E E E E _ E E E E _ E E E E _ Total no. of ways = 5! 4 = i) Method : M ~ N(2.5, 0.2 n ) P( M > 2.53) = P( M < 2.53) = P( Z < 0. ) = n 0.3 n = n = 25 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. Teaching Point: Students have a choice of either solving algebraically (Method ) or listing out the probabilities and searching for the answer (Method 2). It is individual preference. Method 2: 6

(ii) Let M, S = thickness of a mechanics and statistics textbook respectively. M +... + M 2 + S +...+ S 24 ~ N(2 2.5 + 24 2, 2 0. 2 + 24 0.08 2 ) = N(00.5, 0.3636) P(M +... + M 2 + S +...+ S 24 00) = 0.

17 (ii) Let M, S = thickness of a mechanics and statistics textbook respectively. M M 2 + S S 24 ~ N( , ) = N(00.5, ) P(M M 2 + S S 24 00) = (iii) S S 4 3M ~ N( , ) = N(0.5, 0.56) P(S S 4 < 3M) = P(S S 4 3M < 0) = There is no need to square 2 and 24 when computing variance since we are dealing with sums of normal variables. Students need to square 3 when computing variance since we are dealing with the multiple of a normal variable. (iv) We assume that the thicknesses of mechanics and statistics textbooks are independent. 0i) Unbiased estimate of the mean = x = = 9.6 Unbiased estimate of the variance = s 2 = 8 ( ) = 0.8 Teaching Point: The formula for the unbiased estimate of the variance is found in MF5. There is no need for students to memorise it. (ii) We assume that the distribution of the mass of sugar in a packet is normal. H 0 : µ = 0 H : µ 0 Under H 0, T = X 0 ~ t(8)

18 Since p value = 0.29 > 0.05, we do not reject H 0. There is insufficient evidence at the 5% level to say that the mean mass of sugar in a packet is not 0 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. i) 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, 0.5) P(4 R < 8) = P(R 7) P(R 3) = (iii) Since n = 240 is large, np = 72 > 5, nq = 68 > 5, R ~ N(72, 50.4) approximately. P(R < 60) = P(R < 59.5) (continuity correction) 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). Some students mix up the use of nq and npq. Also it is common for students to forget to do continuity correction. (iv) Since n = 240 > 50, np = 4.8 < 5, R ~ Po(4.8) approximately. The Poisson approximation is appropriate in this case since mean = = variance. P(R = 3) 0.57 to 4 decimal places. (v) P(R = 0 or ) = P(R = 0) + P(R = ) = ( p) p( p) 9 = 0.2 ( p) p( p) = 0 Method : 8

19 From GC, p = Method 2: From GC, p = Method 3: 9

20 From GC, p =

2009 GCE A Level H1 Mathematics Solution

2009 GCE A Level H1 Mathematics Solution 1) x + 2y = 3 x = 3 2y Substitute x = 3 2y into x 2 + xy = 2: (3 2y) 2 + (3 2y)y = 2 9 12y + 4y 2 + 3y 2y 2 = 2 2y 2 9y + 7 = 0 (2y 7)(y 1) = 0 y = 7 2, 1 x = 4,