2007 Paper 1 Solutions

Transcription

1 27 Paper 1 Solutions 2x 2 x 19 x 2 + x = 2x2 x 19 (x 2 + x + 2) x 2 + x + 2 2x 2 x 19 x 2 + x + 2 > 1 2x 2 x 19 x 2 + x > x 2 4x 21 x 2 + x + 2 > (x + )(x 7) (x + 2)(x + 1) > = x2 4x 21 x 2 + x (Shown) x < or 2 < x < 1 or x > 7 2(i) R g = [, ) D f Hence fg does not exist. R f = IR \{} D g Hence gf exists. 1 gf(x) = g( x ) = 1 (x ) 2 gf : x 1 (x ) 2, x IR, x

2 1 (ii) Let y = x, x x = 1 y x = 1 y + f 1 : x 1 x +, x IR, x (a) 2 1 (1) Check that the circle passes through origin. When z =, 2 i = sq rt 1. Drawing accurately using the same scale in both axes also help realize this. (2) Label centre and radius. (b) ww* + 2w = + 4i (a + ib)(a ib) + 2(a+ib) = +4i a 2 + b 2 + 2a + 2ib = + 4i Comparing imag. parts: 2b = 4 b = 2 Comparing real parts: a a = a 2 + 2a + 1 = (a + 1) 2 = a = 1 w = 1 + 2i

3 4. 4 di dt = 2 I I di = dt 4 2 I di = dt 4 ln 2 I = t + c ln 2 I = 4 (t + c) 2 I = e (t+c)/4 2 I = ± e t/4 e c/4 = Ae t/4 I = 1 (2 Ae t/4 ) I = 2 when t = 2 = 1 (2 A) A = 4 I = 2 (1 + 2e t/4 ) Alternatively, you can solve for 4 c = ln 4 and then obtain t e 2 I = 4 4 which can be simplified to either 2 I = 4 t e 4 ( 2 I ) = or t 4 4e. The latter is correct as t =, I = 2 satisfy the equation. For large values of t, I 2. The current tends towards 2/.

4 5. y = 2x + 7 2(x + 2) + x + 2 = x + 2 = 2 + x + 2 y = 1 x translate 2 units in the negative x-direction 1 y = x + 2 scale // y axis by factor of y = x translate 2 units in the positive y-direction 5. y = 2 + x + 2 = 2x + 7 x + 2 A = 2, B = Your description of the transformations should be formal rather than left right up down. y = 2 (, 7 2 ) ( 7 2, ) x = 2

5 6(i) (ii) (iii) OA OB 1 2 = 1 4 = = 2 1 Hence OA OB. Check your cross product by OM = 2 OA + OB = ( ) = 1 4 performing a dot product 2 with the OA and OC vectors 1 5 and obtaining zero will Area of triangle OAC ensure that the answer is = 1 2 OA OC right. = = (2 + 8) 2 4 = = 5 = = 5 1

6 7(i) (ii) (iii) A second root is z = re iθ A quadratic factor of P(z) is (z re iθ )(z re iθ ) = z 2 z(re iθ + re iθ ) + re iθ re iθ = z 2 rz(cos θ + i sin θ + cos θ i sin θ) + r 2 e iθ iθ = z 2 2rz cos θ + r 2 (Shown) z 6 = 64 = 64e iπ = 2 6 e i(π+2kπ) z = 2e i(π 6 + kπ ), k =, 2, 1,, 1, 2 z = 2e 5iπ/6, 2e iπ/2, 2e iπ/6, 2e iπ/6, 2e iπ/2, 2e 5iπ/6 Argument of 64 is π and modulus is 1. k can also be from to 5. Both acceptable though here it is better to have the angles in + and relationship. z = (z 2 4z cos π )(z 2 4z cos π )(z 2 4z cos 5π ) = (z 2 4z 2 + 4)(z2 + 4)(z 2 + 4z 2 + 4) = (z 2 2 z + 4)(z 2 + 4)(z 2 +2 z + 4) Question says non-trigo from means evaluate cos π 6.

7 AB = 2 = Equation of l is r = 2 + λ λ Substitute eqn of l into eqn of p: 2 + λ 1 = 17 4 λ 2 9λ 2 λ + 8 6λ = 17 16λ = 8 λ = 1 2 point of intersection = (1 + 2, 2 1 2, ) = (2.5, 1.5, 5.5) = = θ = Answer = = 78.8 (to 1 decimal place) (ii) cos θ = 2 (iii) One point, C on plane p is (1,, 7). 1 1 CA = 2 = Distance = 1 CA n 8 i = i + + = or 2.14 (to s.f.) 2 14

8 9(i) From GC α =.619, β = (to decimal places) (ii) Let the sequence converge to a value x. i.e. x n x and x n+1 x as n Then x satisfies x = 1 ex x = e x e x x =. Since α and β are the roots to e x x =, x is either α or β.

9 (iii) Key into GC and change u(nmin) =, 1, 2 to see the respective behaviours in the table. If x 1 = or 1, the sequence converges to the value α =.619 If x 1 = 2, the sequence diverges. In GC, when x 1 = 2, the terms gets very large by x 6 and error after that because it is too big for the GC to display. (iv) x n+1 x n = 1 exn x n = 1 (exn x n ) From the graph, this is 1 (exn x n ) < if α < x n < β Treat x n as x they are just dummy variables. But 1 (exn x n ) > if x n < α or x n > β Hence x n+1 < x n if α < x n < β, x n+1 > x n if x n < α or x n > β (v) Since < α =.619, if x 1 =, the sequence increases in value and approaches the value α =.619. Since α < 1 < β, if x 1 = 1, the sequence decreases in value and approaches the value α =.619. Since 2 > β = 1.512, if x 1 = 2, the sequence increases in value and approaches. This part is quite tough.

10 1. a + d = ar d = 1 (ar a) a + 5d = ar 2 d = 1 5 (ar2 a) Make d the subject so that we can eliminate it to get an equation in r. 1 5 (ar2 a) = 1 (ar a) [a is non zero, so can cancel away on both sides] (r 2 1) = 5(r 1) r 2 5r + 2 = (Shown) (ii) (r 2)(r 1) = r = 2, 1 (r = 1 is rejected since d = 1 (ar a) = but Q states d is non zero) Since r = 2 < 1, the geometric series is convergent. Here you have to give a reason why r = 1 is rejected as the question ask a you to show convergent so it is S = 1 2 = a obvious r =1 has to be rejected. (iii) d = 1 ( 2 a a) = a 9 S = n 2 [2a (n 1) a 9 ] > 4a n 2 ( 19 9 n 9 ) > 4 [a can be cancelled away without affecting the inequality sign as a>] 19n n 2 > 72 n 2 19n + 72 <. Show graph below. From GC, 5.2 < n < 1.8 Hence set of possible values of n = {6, 7, 8,..., 1}.

11 11i) 1 Sketched for the required domain for t. Go windows to change. 1 It is a curve, not a straight line. Label intercepts. (ii) dx dt = 2 cos t ( sin t) dy dt = sin2 t cos t dy dx = sin2 t cos t 2 cos t sin t = 2 sin t 1 1 Gradient at the point is 2 sin θ. Equation of tangent is y sin θ = 2 sin θ (x cos2 θ) When y =, 2 sin θ (x cos2 θ) = sin θ x cos 2 θ = 2 sin2 θ x = cos 2 θ + 2 sin2 θ When x = : y sin θ = 2 sin θ cos2 θ y = sin θ + 2 sin θ cos2 θ

12 Area of triangle OQR = 1 2 OQ OR = 1 2 (cos2 θ + 2 sin2 θ)(sin θ + 2 sin θ cos2 θ) = 1 12 sin θ ( cos2 θ + 2 sin 2 θ)(2 sin 2 θ + cos 2 θ) = 1 12 sin θ ( cos2 θ + 2 sin 2 θ) 2 (iii) Area = 1 y dx When x =, t = π 2 When x = 1, t = = π/2 sin t ( 2 cos t sin t dt) = 2 π/2 cos t sin 4 t dt (Shown) Let u = sin t du dt = cos t When t =, u = When t = π 2, u = 1 π When x =, cost = so t corresponds to. 2 When x = 1, cost = 1so t corresponds to. = 2 1 u4 du = 2 [ u5 5 ]1 = 2 5

13 27 Paper 2 Solutions Let x, y, z = price of 1 kg of pineapples, mangoes, lychees respectively 1.15x +.6y +.55z = x +.45y +.z = x +.9y +.65z = Lee Lian paid = = $7.65 By GC, x=.5, y = 2.6, z = 4.9

14 2(i) Let P n be the statement: u n = 1 n 2, n 1 When n = 1: LHS = u 1 = 1 RHS = = 1 = LHS P 1 is true. Assume that P k is true for some k 1 i.e. u k = 1 k 2 Prove that P k+1 is also true i.e. u k+1 = LHS = u k+1 = u k = 1 k 2 2k + 1 k 2 (k + 1) 2 1 (k + 1) 2 2k + 1 k 2 2 (k + 1) [Recurrence relation given by Q] = (k + 1)2 2 (2k + 1) k 2 (k + 1) 2 = k + 2k+ 1 2k 1 2 ( k + 1) k 2 = k 2 (k + 1) 2 = 1 (k + 1) 2 = RHS P k true P k+1 true also. [By assumption] Since P 1 is true, and P k true P k+1 true, by Mathematical Induction, P n is true for all n 1.

15 (ii) N 2n + 1 n 2 (n + 1) 2 = N (u n u n+1 ) n=1 n=1 = u 1 u 2 + u 2 u + u u 4 : + u N u N+1 = u 1 u N+1 = 1 1 (N + 1) 2 Remember it is N for the last term not n. 1 (iii) As N, (N + 1) 2 N 2n + 1 Hence n 2 (n + 1) 2 = (N + 1) is convergent. n=1 Sum to infinity = 1 = 1 For sum to be convergent, it means when N goes to infinity, the sum must tends towards a finite number. The sum to infinity is equal to 1, NOT approximately 1. r < 1 is only a condition for sum to be convergent when it is a GP SUM. Not applicable in (iii) as it is not a GP sum! (iv) N 2n 1 n 2 (n 1) 2 = N 1 N 2 (N 1) 2 n=2 Compare to N 2n + 1 = 2 2 n ( n+ 1) n= N 1 N 2 (N 1) 2 + 2N N ( N + 1) Therefore, N 2n 1 N 1 n 2 (n 1) 2 = 2n + 1 n 2 (n + 1) 2 = 1 1 = n=2 n=1 ( N 1+ 1) N 2 [Replace N by N 1 in the sum in (ii)]

16 (i) Let y = (1 + x) n dy dx = n(1 + x)n 1 d 2 y 2 dx = n(n 1)(1 + x)n 2 d y dx = n(n 1)(n 2)(1 + x)n When x =, y = 1, dy dx = n, d2 y dx 2 = n(n 1), d y dx y = 1 + nx + n(n 1) 2! x 2 + n(n 1)(n 2)! = n(n 1)(n 2). x +... You should also know how to derive the expansions of those in the formula list by 1 st principle like e x, sin x, ln(1+x), cos x. You can double check your answer with the expansion in the formula list! (ii) (4 x) /2 (1 + 2x 2 ) /2 = 4 /2 (1 x 4 )/2 ( x2 +...) 1 1 = 8(1 + 2 ( x 4 ) ! ( x )2 +! ( x 4 ) +...)(1 + x ) = 8(1 x 8 + x x ) (1 + x2 +...) = 8(1 x x x ) = 8 x + 87x x (iii) The expansion is valid for x 4 < 1 and 2x2 < x < 4 4< x < 4 and 2x < 1 x < < x < Therefore, 1 2 < x < 1 2 [ you can also give answer as 1 x <.] 2

17 4(i) 5π/ sin 2 x dx = 5π/ 1 cos 2x 2 dx = 1 sin 2x 2 [ x 2 = 5π π/ = [ x ] 5π/ (ii)(a) Area = π/2 ] 5π/ cos 2 x dx = 5π/ = [ x 2 cos x ] π/2 = + 2 π/2 = 2[ x sin x] π/2 = 2[ π 2 = 1 2 [ 5π + 4 ] 1 sin 2 x dx [ 5π ] = 5π 5π 6 8 = 5π 6 8 x 2 sin x dx u = x 2 v = sinx u' = 2x, v = cosx π/2 2x cos x dx x cos x dx 2 π/2 ] 2 [ cos x ]π/2 sin x dx u = x v = cos x u' = 1, v = sin x = π + 2 [ 1] = π 2 Exact value means you have to perform integration and also evaluate sin π as 2, etc. You should know you can check your answer using GC! (b) Volume = π π/2 ( x 2 sin x ) 2 dx = 5.91 ( decimal places) Give your answer to decimal places hints that GC can be used to evaluate the answer!

18 5(i) To survey, say a sample of 5 shoppers who visit a particular shopping mall to find out whether how many times they visit the mall in a month. 5 shoppers can be 25 females and 25 males. State the population which is the shopper who visits the mall State the purpose of survey which is how many times they visit the mall in a month State the mutually exclusive subgroups females and males. Quota sampling is appropriate in this situation because the sampling frame is not available (do not have information on the total population of shoppers visiting the mall) so random sampling methods cannot apply. Your answer should show appreciation that sampling frame is not available so you cannot do simple random and systematic because you need the list of all the people and number them. You cannot do stratified because you do not have proportion of each subgroup in the population! A disadvantage of quota sampling is that the sample is biased as it is up to the surveyor to choose who he prefers to survey to meet the quota. (ii) Impossible since we do not have a list of all the shoppers. Do state possible or not. Possible as answers is not acceptable.

19 6 Let X = no. of people out of 1 with gene A. X ~ B(1,.24) P(X 4) =.9 (i) Let A = no. of people out of 1 with gene A. A~B(1,.24) A~N(24, 182.4) since n large, np=24>5, n(1 p)= 76>5 P(2 A 26) = P(229.5 A 26.5) =.717 (ii) Let B = no. of people out of 1 with gene B. B ~ B(1,.) B ~ Po() since n is large, np = < 5 Q says using approximation, so approximation must be used. Remember to state the conditions including n is large which is often left out. Remember to do continuity correction for (i)!! P(2 B < 5) = P(2 B 4) = PB ( 4) PB ( 1) =.616

20 7) x = =.84 (exact) = unbiased estimate for population mean s 2 = ( ) =.7259 =.7 ( s.f.) = unbiased estimate for population variance H : μ = H 1 : μ > X μ.84 Z = = = s n p-value =.8 Since p value =.8 <.5, we reject H. There is sufficient evidence at the 5% level to conclude that the population mean time for a student to complete the project exceeds hours. Since n = 15 is large, there is no need to assume that the population is normal because by Central Limit Theorem, X is normally distributed approximately.

21 8) Let C, T = mass of a chicken, turkey respectively. C ~ N(2.2,.5 2 ) T ~ N(1.5, ) (i) P(C > 7) = P(C > 7 ) =.9486 =.95 (ii) P(C > 7) P(5T > 55) =.9486 P(T > 11) =.16 (iii) C + 5T ~ N( , ) = N(59.1, 112.5) P(C + 5T > 62) =.92 (iv) The event in (iii) includes the case in (ii). So the probability in (iii) is higher.

22 9(i)(a) 12! = (b) 6! (2!) 6 = 46 8 (ii)(a) (12 1)! = (b) (6 1)! 6! = 86 4 (c) (6 1)! 2 = 24 It is only 2, not 2 6. It is because once the wife of the first man choose to his left, the rest of the wives has no choices. So either left or right 2 choices.

23 1) 1/8 S 1/4 S /4 F 1/2 S 1/2 F 1/4 S /4 F (i) P(SSS) = = /8 F 1/8 S 7/8 F 1/4 S /4 F 1/8 S 7/8 F (ii) P(SSF) + P(SFS) + P(FSS) = = (iii) P(A B) = PA ( B) PB ( ) = P(SFS) + P(FSS) 17/256 = /256 = 1 17

24 11. The regression line is x =.2597t x =.26 t (ans to s.f.) Here x is dependent and t is independent. Read the Q. So x on t. When t =, x = 11.7 which is negative. But the concentration cannot be negative. Hence the linear model is not suitable.

25 (i) Correlation coefficient r =.994 which is closer to 1. This indicates that t and y = ln x have a stronger linear correlation than t and x. (ii) Regression line is ln x =.124t ln 15 =.12t t = 155 min. Here x is dependent and t is independent. Read the Q. So line of ln x on t is to be used even if you are estimating t given x = 15.