RI 9740/02 2015

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) 32 m Note that we need 1 216 0 h−≥ (ii) ( ) ( ) 1 2 1 2 1 2 1 21 2 1 2 16d 10 16 d dd 10 20 16 40 16 where hh hh tt h tc tb h b c −−= ⇒− = ⇒ −= +− ∴=− − =− ∫∫ 1 2 0 when 0 160 Thus, 160 40 16 h tb th = =⇒= = −− ( )1 2 Half its maximum height, that is, 16 we have 160 40 16 16 160 40 8 46.9 years h t = = − −= −≈ Raffles Institution H2 Mathematics Solution for 2015 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Remarks for Student 12 : 2 3 , 46 Lr λλ    = −+ ∈  −−   (i) 11 21 30 60 2Acute angle = cos cos 73.4721 30 60 −−           −   = =           −     (ii) 2 5 6 OP  =−   Any point, , on has position vector, 12 2 3 for some . 46 RL OR λ λλ λ + =−+ ∈−−    ( ) ( ) ( ) ( )( ) 222 22 2 2 2 12 2 33 2 3 5 = 33 46 6 2 1 3 7 2 6 33 4 4 1 9 42 49 4 24 36 33 49 70 21 0 7 10 3 0 73 10 3 or 17 PR λ λ λ λλ λ λλ λ λ λλ λλ λλ λλ λλ +− = ⇒ −+ −−− + ⇒ −+ − +− = ⇒− + +−+ + −+ = ⇒ − += ⇒ − += ⇒ − −= ⇒= =   

3 | Page 312 7 12 3 323 o r 23 17 4 6 10 346 7 13 7 5 7 46 7 OR OR  +   +      =−+ = −+ =        −− −   −−     = −  −    13 17 773 1 51'1 2 77 10 46 58 77 OR                = +− =   −      −−        (iii) ( ) ( ) ( ) 2 1 2 1 2 36 52 3 7 3 2 6 46 26 1 1 Required plane is 36 2 11 36 2 2 5 11 6 4.xy z −−             + ×=×=            −+ − − − −       − + − = − + − −=

4 | Page Question 3 No. Suggested Solution Remarks for Student 2 1f : , , 1 1x xx x ∈>− (a)(i) Method 1: ( ) 22 2f '( ) 0 fo r 1 1 xxx x = >> − Thus, f is an increasing function for x > 1. Thus f has an inverse. Method 2: Sketch graph of f and explain using horizontal line. (ii) ( ) ( )1 2 2 2 1 ff 11Let 1 1 11 111 1f1 ,0 yx xy x y xx y x x DR− − = ⇒= −− ⇒= − ⇒= − > = − = = −∞  (b) 2 2g: , , 1 1 xx xx x + ∈ ≠±− ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2Let 1 21 20 For the equation to be defined, discriminate 0, thus, 1 4 2 0 4 8 10 4 2 130 4 1 30 223 2230 23 23 or 22 xy yx x x yx x y yy yy yy y yy yy += ⇒−= +− ⇒ ++ − = ≥ − −≥ − +≥ − +−≥ − −≥ −− −+ ≥ +−≥≤

5 | Page Question 4 No. Suggested Solution Remarks for Student (a) Let nP be the statement ( )( ) ( ) ( ) 2 1 12 5 1 3 31 7412 n r rr r nn n n = + += + + +∑ for .n +∈  For 1,n= ( )( ) ( )( )( ) 1 1 2 5 1 3 6 18 1 2161 1 1 3 31 74 1812 12 r LHS r r r RHS LHS = = + + =××= = + ++ = == ∑ Therefore, P1 is true. Assume kP is true for some k +∈  , i.e. ( )( ) ( ) ( ) 2 1 12 5 1 3 31 7412 k r rr r kk k k = + += + + +∑ We need to show 1kP+ is true, i.e. ( )( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 12 5 1 1 1 3 1 31 1 7412 k r rr r k k k k + = + += + ++ + + ++ ∑ For 1,nk= +

6 | Page ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) 1 1 1 2 2 32 2 32 2 2 25 25 11 21 5 1 1 3 31 74 1 3 612 1 1 3 31 74 12 3 612 1 1 3 31 74 12 108 21612 1 1 3 43 182 21612 1 1 2 3 37 10812 1 1 2 3 2 1 31 1 712 k r k r rr r rr r k k k kk k k k k k k kk k k k k k k kk k k kk k kk k k k k kk k + = = ++ = + + + + ++ ++ = + + + ++ + + = + +++ + + =+ ++ +++ =+ +++ = ++ ++ = + + + ++ ++ ∑ ∑ ( ) ( ) ( ) ( ) ( ) ( ) 2 4 1 1 1 1 3 1 31 1 7412 kk k k= + ++ + + ++ Therefore 1k kP P +⇒ is true. Since 1P is also true, nP is true for n +∈  by Mathematical Induction. (b)(i) ( )( ) 2 2 2 11 4 83 21 23 21 23rr r r r r = = −++ + + + + (ii) 2 11 2 11 4 8 3 2 12 3 11 35 11 57 11 79 ... 11 2 12 1 11 2 12 3 11 32 3 nn rr rr r r nn nn n = = = − ++ + +  = − +− +− +− −+ +− ++ = − + ∑∑

7 | Page (iii) 33111 1 10 103 23 3 23 498.5 smallest 499 nn n n −−− −< ⇒ <++ ⇒> =

8 | Page Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) Not possible to obtain the sampling frame, that is, not possible to get the list of all customers, with their age, who patronize the supermarket. (ii) Assume manager wants to get opinions of 10 customers from each of the following groups (by ages) Children: Below 12 years old, Teenagers: 12 to 18 years old, Young Adult: 18 to 35 years old, Mature Adult: 35 to 50 years old, Above 50 years old. Approach the first 10 customers who are willing to provide their opinions using a questionnaire. (iii) As mentioned in (ii), we are approaching the first 10 customers who are willing to provide their opinions. We will not obtain opinions of those who are not as vocal.

9 | Page Question 6 No. Suggested Solution Remarks for Student (i) Let X be the number of red sweets in a small packet of 10 sweets. ( )~ B 10,0.25X ( ) ( )P 4 1 P 3 1 0.775875 0.224XX≥= − ≤= − ≈ (ii) Let X be the number of red sweets in a large packet of 100 sweets. ( )~ B 100,0.25Y Since 100n= is large, and 25 5np= > ( )and 1 75 5,np −=> ( )25,~N 18.75Y approximately. ( ) ( )P 30 P 29.5 by continuity corrections 0.14935 0.149 YY≥= ≥ = ≈ Be sure to check the conditions and state the approximate distribution used. (iii) For 9740 syllabus: Let W be the number large packets, out of 15, containing at least 30 red sweets. ( )~ B 15,0.14935W ( )P 3 0.824655 0.825W ≤= ≈ For 9758 syllabus: Without using approximation for ( )P 30Y ≥ , ( )( ) ( )~ B 15, P 30 , that is, ~ B 15,0.14954WY W ≥ ( )P 3 0.82407 0.824W ≤= ≈ For 9740 syllabus: Exam report suggests that students need to use answer from (ii), as the only answer given is 0.825.

10 | Page Question 7 No. Suggested Solution Remarks for Student (i) Number of errors per page are assumed to remain constant uniformly. Errors occure independently between pages of the newspaper. Let X be the number of errors on one page. ( )~ Po 1.3X (ii) ( )123456 ~ Po 7.8Y XXXXXX=+++++ ( ) ( )P 10 1 P 10 1 0.83523 0.165YY>= − ≤= − ≈ (iii) ( )12 ... ~ Po 1.3nWX X X n= + ++ ( ) ( ) ( ) ( ) 1.3 1.3 1.3 P 2 0.05 P 0 P 1 0.05 e 1.3 e 0.05 e 1 1.3 0.05 0 ...(1) nn n W WW n n −− − << = += < +< + −< Solving (1) using GC graphically or table of values, least n = 4 To avoid careless mistake, we should also use the original inequality, ( )P 2 0.05,W << to check the answer.