H2 Math Paper 2 Solution

1 Section A: Pure Mathematics [40 marks] 1 Let x be the distance between A and P. 22 286 4PW x x=+ =+ Cost of laying the pipe, () 260000 64 45000 12Cxx= ++ − d 0d C x = 2 60000 45000 0 64 x x −= + () 2 2 22 2 60000 45000 64 43 6 4 16 9 64 576 7 576 576 (reject as 0)77 xx xx xx x xx =+ =+ =+ = =− > To show that C is minimum when 576 7x= : Method 1: First Derivative Test x 576 7x − ⎛⎞=⎜⎟⎜⎟⎝⎠ 576 7x= 576 7x + ⎛⎞=⎜⎟⎜⎟⎝⎠ 2 d 60000 45000d 64 Cx x x =− + < 0 0 > 0 ∴C is a minimum when 576 7x= . Land W P B A 8 Ocean x 12 – x

2 Method 2: Second Derivative Test () () () 2 2 2 2 22 22 33 22 22 6000060000 64 d 64 d6 4 60000 64 60000 3840000 0 for all 64 64 xx C x xx xx x xx +− += + +− == > ++ ∴C is a minimum when 576 7x= . 2 (i) t y d d = k(10 – y) 1 d d10 yk ty =−∫∫ – ln |10 – y| = kt + c |10 – y| = e kt c−− 10 – y = e kt c−−± 10 – y = ee eck t k t A−− −±= , where A = ± e c− y = 10 – ktA −e When t = 0, y = 0 ⇒ A = 10 When t = 2, y = 5 ⇒ 225 10 10e 10e 5kk−−=− ⇒ = k = 1 ln 22 ∴ y = 10 – 1 ln 2210e t⎛⎞−⎜⎟⎝⎠ (ii) The amount of material me morised tends to 10 units. (iii) t y d d = k(10 – y) – αy, α > 0 y t 10 0 y = 10 – 1 ln 2210e t⎛⎞−⎜⎟⎝⎠ y = 10

3 3 (i) ()02 i 4z−+ < and () ( )46 i 02 izz−−+ ≤ − + (ii) 1 4tan 1.10712AOC − ⎛⎞∠= = ⎜⎟⎝⎠ rad Least possible arg z = π 2 Largest possible arg z = π 2+1.1071 = 2.68 π arg 2.682 z∴ << (iii) ()() () 3arg 2 4i arg 1 i π4w−+ = − −= − ()04 26,2 , 422MM −+⎛⎞ ⇒−⎜⎟⎝⎠ Least value of zw− = CM () ( ) 22 20 42=− − +− = 22 4 (i) 2 14 1 11 4 10 5 n −−⎛⎞ ⎛ ⎞⎛⎞ ⎜⎟ ⎜ ⎟⎜⎟=× =⎜⎟ ⎜ ⎟⎜⎟⎜⎟ ⎜ ⎟⎜⎟ −⎝⎠ ⎝ ⎠⎝⎠ Acute angle between 1Π and 2Π 1 11 24 15cos 11 2 4 15 − −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−•⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠= −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠ 1 14cos 64 2 − ⎛⎞= ⎜⎟⎝⎠ 28.1= D (to 1 dp) D C(0,2) y 3 4 π− 6 -2 (4 , 6 )− 4 O A (2,4) M x

4 (ii) 31 14 1 2 15 −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−= −⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠ i 2Π∴ : 1 41 2 4 51 2 5 xyz −⎛⎞ ⎜⎟ =− ⇒− + − =−⎜⎟⎜⎟−⎝⎠ ri 1Π : 2 4xy z−+ = 2Π : 45 1 2xyz−+ − = − 1 43 Using GC, eqn of is 4 2 , where 01 l αα −⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟=− + ∈⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ \r . (iii) Given 6 3 5 OA ⎛⎞ ⎜⎟=⎜⎟⎜⎟−⎝⎠ JJJG and 2 3 1 OB ⎛⎞ ⎜⎟=⎜⎟⎜⎟⎝⎠ JJJG , 26 4 33 0 15 6 AB −⎛⎞⎛ ⎞⎛ ⎞ ⎜⎟⎜ ⎟⎜ ⎟=− =⎜⎟⎜ ⎟⎜ ⎟⎜⎟⎜ ⎟⎜ ⎟−⎝⎠⎝ ⎠⎝ ⎠ JJJG Length of projection of AB JJJG onto the line 1l 22 2 43 02 61 321 −⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟•⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠= ++ 6 14 −= 31 4 7= 1Π : 1 24 1 ⎛⎞ ⎜⎟−=⎜⎟⎜⎟⎝⎠ ri l2 : 3 21 3 3 p q pt q ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟=+ + −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟−⎝⎠ ⎝ ⎠ r Let D be a point on the plane 1Π . Since 41 02 4 01 ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟−=⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ i , D is (4, 0, 0). ∴ 44 21 0 21 30 3 pp DC p p −⎛⎞ ⎛ ⎞ ⎛⎞ ⎜⎟ ⎜ ⎟ ⎜⎟=+ −=+⎜⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜ ⎟ ⎜⎟−−⎝⎠ ⎝ ⎠ ⎝⎠ JJJG

5 Perpendicular distance from C to 1Π = Length of projection of DC JJJG onto the normal of 1Π 14 1 22 1 2 13 1 15 1 66 2 1 p DC p −⎛⎞ ⎛ ⎞ ⎛⎞ ⎜⎟ ⎜ ⎟ ⎜⎟−+ −⎜⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜ ⎟ ⎜⎟ −⎝⎠ ⎝ ⎠ ⎝⎠== = ⎛⎞ ⎜⎟−⎜⎟⎜⎟⎝⎠ JJJG . . (4 ) 2 ( 21 ) 3 1 5 39 1 5 39 1 5 o r 391 5 8 (rej 0) 2 pp p pp pp p −− +−= −− = −− = −− = − =− > = ∵ Acute angle between l2 and 1Π 11 31 32 1 2sin sin 31 6 32 1 q q q q −− ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−−⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎛⎞⎝⎠ ⎝⎠== ⎜⎟⎛⎞⎛⎞ ⎝⎠ ⎜⎟⎜⎟−−⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠ . ()22 36 2 36 36 s i n c e 0 699 6 qq qq qq q qq ++ =+ + = + + > ++ () () 2 2 2 2 46 2 1 0 9 46 4 1 0 9 20 2 (since 0) qq qq qq qq += + += + −= =>

6 Section B: Statistics [60 marks] 5 Systematic sampling of every 30th employee arranged in alphabetical order by name may result in too many employees chosen from a particular age group. The method is stratified sampling. To carry out a stratified sample, stratify the 1500 employees into the 3 age groups. Take a random sample from each stratum with sample size proportional to the relative size of the stratum: Age group 21-40 41-60 60 and above No. of employees 35% × 50 = 17.5 ≈ 18 50% × 50 = 25 15% × 50 = 7.5 ≈ 7 6 (i) ENDANGERED: 3E, 2N, 2D, 1A, 1G, 1R No of 4-letter code-words from E, N, D, A, G, R 6 4 4! 360 C=× = (ii) Select 1 letter from N, D, A, G, R. No of 4-letter code-words with the chosen letter and 3 “E”s 5 1 4! 3!C=× 20= (iii) Case I: 3 same letters No. of such code-words = 20 Case II: 1 pair of same letters No. of such code-words 35 12 4! 3602!CC=×× = Case III: 2 pairs of same letters No. of such code-words 3 2 4! 182!2!C=× = No. of 4-letter code-words that contain at least 1 repeated letter 20 360 18 398=+ +=

7 7 (i) For a 3 set match, probability of A winning = probability of B winning () () ( ) () ( )( ) ( ) ( )( )( )0.3 0.4 1 0.4 0.7 0.3 0.6 1 0.4 0.3pp pp +− = +− 0.12 0.28 0.28 0.18 0.12 0.12p pp p+ −=+ − 0.22 0.16p= 8 11p= (ii) () wins the first set | wins the matchPB A () () wins the first set wins the match wins the match PB A PA ∩= () () () () () () () 3 0.4 0.711 88 30.7 0.3 0.4 0.4 0.711 11 11 ⎛⎞⎜⎟⎝⎠= ⎛⎞ ⎛⎞ ⎛⎞ ++⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ 0.11351 0.114=≈ 0.3 A 0.7 A A A A B B B B B 0.4 0.6 0.6 0.3 0.4 0.7 p 1 p−

8 8 (i) Let A = weight of an apple in kg. () 2N 0.2,0.05A ∼ Let O = weight of an orange in kg. () 2N 0.3,0.08O ∼ () ( )() 22 123 N 3 0.2 0.3,3 0.05 0.08AAAO++− − + ∼ ()N 0.3,0.0139= () 123P0 . 2AAAO++−< = () 123P0 . 2 0 . 2AAAO−<++− < 0.19816 0.198=≈ (ii) T = () ()1 2345 1234523 AAAAA OOOOO++++ + ++++ () ( ) () ( ) () () () ( )() 22 22N2 , 25 0.2 3 5 0.3 5 0.05 3 5 0.08T ++∼ ()N 6.5, 0.338= ()P 5 0.99506 0.995T >= ≈ (iii) 20.05N0 . 2 ,A n ⎛⎞ ⎜⎟⎝⎠ ∼ () 2 P0 . 2 1 0 . 3 0.21 0.2P0 . 3 0.05 A Z n >< ⎛⎞ −⎜⎟ ><⎜⎟ ⎜⎟⎝⎠ 1P 0 . 3 5 P0 . 7 5 0.52440 6.87495 Least 7. nZ nZ n n n ⎛⎞−≤<⎜⎟⎜⎟⎝⎠ ⎛⎞ ≤>⎜⎟⎜⎟⎝⎠ >⇒ > ∴ =

9 9 (i) 1. The defects in a reference book are independent of each other. 2. The average number of defects in a reference book is constant. (ii) ()~P o 3R () () () P7 P8 1 P 7 0.011905 0.0119 RR R >= ≥ =− ≤ = ≈ (iii) Let X = no. of reference books donated away out of 1000. ()B 1000,0.011905X ∼ Since n is large, np = 11.905 > 5 and nq = 988.095 > 5, ( )N 11.905,11.763X ∼ approximately. () ( ) ..P 10 15 P 9.5 15.5 0.61114 0.611 ccXX≤≤ ⎯ ⎯ → ≤≤ =≈ (iv) Let C = no. of defects in a children’s book. ()~P o 5C Hence, ( )( )~P o 3 5 P o 8RC++ = () ( )10 9 0.71662 0.717PR C PR C+< = +≤= ≈ (v) Now, ()~P oC λ () 01 P1 0 . 1 P( 0) P( 1) 0.1 ee 0.10! 1! C CC λλλλ−− ≤= =+ == += ee 0 . 1λλ λ−− += Using GC, 3.8897 3.89λ =≈

10 10 (i) Let X = payment made on a motor insurance claim. Assumption: Payments made on motor insurance claims are normally distributed. ( 10000) 10000 9737.625 xx −=+ =∑ [ ] 2 22 ( 10000)1 ( 10000) 240506.5 24 25 x sx − =− − = ⎧⎫⎪⎪⎨⎬ ⎪⎪⎩⎭ ∑∑ H 0 : μ = 10000 H 1 : μ < 10000 Under H 0, 2 ~ N 10000, 25 X σ⎛⎞ ⎜⎟ ⎝⎠ . Hence, test statistic 2 10000 ~( 2 4 ) 25 XTt S −= . α = 0.01 From GC, t = 9737.6 10000 2.6753 240506.5 25 − =− . p -value = 0.00662 Since p = 0.00662 < α = 0.01, we reject H0 at 1% level of significance and conclude there is sufficient evidence that the mean payment made is less than $10000. (ii) It means that there is a probability of 0.01 of concluding that the mean payment made is less than $10000, given that the mean payment made is $10000. (iii) p-value for the two-tailed test = 2 × 0.0066172 = 0.013234 Since p-value = 0.013234 > α = 0.01, we do not reject H 0 at 1% level of significance, i.e. the conclusion in part (i) would not be the same. (iv) Let μ0 be the mean payment made that the insurance company should declare. H 0 : μ = μ0 H 1 : μ < μ0 Under H 0, 2 0 500~N , 25 X μ⎛⎞ ⎜⎟ ⎝⎠ . Hence, test statistic 0 2 ~N ( 0 , 1 ) 500 25 XZ μ−= .