RI 9740/02 2014

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student 23xt= , 6yt= (i) d 6d x tt = , d 6d y t = d1 d y xt⇒= d1 0.4 0.4 2.5d y txt= ⇒= ⇒= (ii) Equation of tangent at point P(3p2, 6p) on C: ( ) 2163yp xp p−= − When x = 0, ( ) 216 3 36 3yp p y pppp− = − ⇒= − + = Coordinates of D are (0, 3p) Mid-point of PD is 239 ,22 pp  23 2 px= , 9 2 py= 2 232 2 2 9 27xy y ⇒= =  ∴Cartesian equation is 22 27xy= Question 2 No. Suggested Solution Remarks for Student ( ) ( ) 2 22 9 13 25 925 9 x x A Bx C xxxx +− += + −+−+ ( ) ( )( ) 229 13 9 2 5x x A x Bx C x+− = + + + − 22 5 55 5: 9 13 9 32 22 2x AA  = +−= + ⇒=    Raffles Institution H2 Mathematics Solution for 2014 A-Level Paper 2

2 | Page ( ) ( )0 : 13 3 9 5 8x CC= −= + −⇒= ( ) ( )( )1 :911 3 3 19 825 3x BB= + −= +++ −⇒= ( ) ( ) ( ) 222 2200 222 220 00 2221 0 0 1 1 9 13 3 3 8dd 25 925 9 3 38ln 2 5 d d2 99 33 8ln5 ln 9 tan22 33 33 382ln5 ln13 ln9 tan22 233 3 13 8 2ln tan2 45 3 3 xx xxx xxxx xx xx xx xx − − − +− += + −+−+ =  −+ + ++ = − + ++   = −+ −+ = + ∫∫ ∫∫ 31 382, ,,2 45 3 3ab cd∴= = = = You may use the GC to verify your answers. Question 3 No. Suggested Solution Remarks for Student (i) (a) Distance run by athelete who completes first 10 stages [ ] ( ) ( )( )( ) ( ) 2 4 8 12 ... 40 102 2 4 10 1 42 10 8 36 440m ++ ++ = +− = += (b) Distance run by athelete who completes first n stages ( ) ( )( )( ) ( ) 2 24 1 42 44 n n nn = +− = + ( )For 4 4 5000,nn +≥ consider the following n ( )44nn + 34 4760 < 5000 35 5040 > 5000 Least n is 35.

3 | Page (ii) OA1 = 4 A1A2 = 4 OA2 = 8=4(2) A2A3 = 8=4(2) OA3 = 16 = 4(2)2 A3A4 = 16= 4(2)2 OA4 = 32 = 4(2)3   An−1An = 4(2) n−2 OAn = 4(2) n−1 Required distance ( ) ( ) 12 4 8 16 ... 4(2) 212(4) 21 82 1 n n n −= ++ ++  −=  − = − Let B be the point where he completes 10km exactly. Need n such that ( )8 2 1 10000 1000021 8 10.289 n n n −≥ ⇒ ≥+ ⇒≥ ⇒Athlete ran a total of 10km after completing the 10th stage but not the 11th stage. Distance covered between O and B = 10 000 − ( ) 1082 1 − = 1816 OA11 = 4(2) 11−1 = 4096 Since OA11 > 1816, athlete is running away from O towards A11 when he reaches B and his distance from O is 1816 m.

4 | Page Question 4 No. Suggested Solution Remark s for Student (a)(i) ( )5i 4 5i 4zz+− = ⇒ −−+ = (ii) ( )6i 10 4i 6i 10 4izz zz−= ++⇒ −= − −− ( ) ( )1 5 4 c o s i 1 4 s i n 5 22 i 1 2244z ππ  =−− + + =−− + +     ( ) ( )2 5 4 c o s i 14 s i n 5 22 i 12244z ππ  =−+ + − =−+ + −     −5 −9 −1 1 −3 5 Im Re 0 4 −5 −9 −1 1 -3 5 Im Re 0 (0,6) (-10,-4) z1 z2 4

5 | Page (b)(i) i 63 i 2ew π−= −= ( )2 and arg 6ww π= =− ( ) ( ) 666 6 2 and arg 6arg 2 2ww w w π π ππ⇒ = = = += − += 6i 64ew π⇒= Or ( ) 6 i i6 6i 62e 2 e 64ew π π π − −  = = = (ii) For * nw w to be real, arg 0 or * nw w π = ( ) 1 0, , 2 , 3 ,66 6 nn πππ πππ+⇒− − =− = ± ± ±  3 smallest positive whole number values of n are ( ) 1 , 2 , 3 5,11,176 n nπ πππ+− = −− − ⇒= Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) Re-order the list of customers by alphabetical order of their names. We need a sample size of 10000 0.05 500×= Note that 10000/500 20= Thus, first customer is selected by r andomly generating a number from 1 to 20 inclusive, say 4. Then every 20 th customer from the 4 th customer will be included in the sample. In other words, the sample of 500 will consist of the 4th, 24th, 44th , … 9984th customers. (ii) Advantage: customers of different first names (which has relation with religion and race as well) have equal chance of being represented. Disadvantage: Any customer who is not able to make it for the survey would result in re -sampling which may be time

6 | Page consuming. Question 6 No. Suggested Solution Remarks for Student Available are 3 goalkeepers, 8 defenders, 5 midfielders, 6 attackers. Total:22 Team consists of 1 goalkeeper, 4 defenders, 2 midfielders, 4 attackers. Total:11 (i) 38 5 6 142 4 31500CCC C = different teams can be formed. (ii) Case 1: Brother(midfield) not in but Brother(attacker) in 38 4 5 14 23 12600CC CC = possible teams Case 2: B rother(attacker) not in but Brother(midfield) in 3845 1414 4200CC CC = possible teams ∴Total number of possible teams is 12 600 + 4200 = 16800 (iii) Let A be the midfielder who can play as either midfielder or defender. After the brothers left, there are 3 goalkeepers and 5 attackers. Case 1: A serves as a possible midfielder There are 4 midfielders and 8 defenders 38 4 5 14 24 6300CC CC = possible teams Case 2: A serves as a possible defender There are 3 midfielders and 9 defenders 39 3 5 1424 5670CCCC = possible teams Number of possible teams without A = 38 3 5 1424 3150CCCC = Total: 6300 + 5670 – 3150 = 8820 Note that teams formed without A is counted twice in Cases 1 and 2. Question 7 No. Suggested Solution Remarks for Student (i) Let X be the number of 6’s out of 10 rolls of the fair die. 1~ B 10, 6X   ( )P 3 0.155X = ≈ (ii) Let Y be the number of 6’s out of 60 rolls of the fair die.

7 | Page 1~ B 60, 6Y   Since 60n= is large, and 10 5np= > ( )and 1 50 5,np −=> 2510, 3~NY   approximately. ( ) ( )P 5 8 P 4.5 8.5 by continuity correction 0.273 YY≤≤ = ≤≤ ≈ Be sure to check the conditions and state the approximate distribution used. (iii) Let W be the number of 6’s out of 60 rolls of the biased die. 1~ B 60, 15W   Since 60n= is large, and 1 15p= is small such that 4 5,np= < ( )4~ PoW approximately. ( ) ( ) ( )P5 8 P 8 P 4 0.350 W WY≤≤ = ≤− ≤ ≈ Be sure to check the conditions and state the approximate distribution used. Question 8 No. Suggested Solution Remarks for Student (a)(i) (ii) (b)(i) Product moment correlation coefficient between m and P = –0.9470 (4dp) Product moment correlation coefficient between lnm and P = –0.9749 (4dp) y x O y x O

8 | Page (ii) –0.9749 is nearer to –1 compared to –0.9470. The scatter diagram of P on m also suggest a non-linear relationship between the 2 variables. Thus P = c lnm + d is the better model. 195693.5593 33659.72805ln 196000 33700ln (3 s.f.) Pm Pm = − = − (iii) From GC, estimated price is $64015.93 ≈ $64000 (3 s.f.) Question 9 No. Suggested Solution Remarks for Student (i) Null hypothesis, 0H : 4.3µ = Alternative hypothesis, 1H : 4.3µ < µ is the population mean number of minutes the bus is late. (ii) Perform a one-tailed test at 10% significance level. Under H0, ( )4.3 ~ 1, / XT tn Sn −= − From sample, 10,n= with x t= and ( ) 210 10 3.299sk= = Given that null hypothesis is not rejected, p-value > 0.1, that is, ( ) 4.3P 0.1 10 3.29 10 tT    − <>   ( )3 4.3P 0.1 3.2 tT − <>  ( )3 4.3 1.38303 3.2 t − >− 3.4753 3.48t >≈ ∴Set of values of t is (3.48,∞). (iii) Perform a one-tailed test at 10% significance level. Under H0, ( )4.3 ~ 1, / XT tn Sn −= − From sample, 10,n= with 4.0x = and 210 9sk=

9 | Page Given that null hypothesis is rejected, p-value < 0.1, that is, ( ) 2 4.0 4.3P 0.1 10 9 10 T k    − <<   0.9P 0.1T k − << 0.9 1.38303k − <− 0.65075k < 220 0.42347, that is, 0 0.423kk<< << ∴Set of values of k2 is (0,0.423). Question 10 No. Suggested Solution Remarks for Student (i)(a) Required probability = 121 1 10 10 10 500    =     (b) Required probability = 989 4 41 10 10 10 125   −=    