VJC H2 MATH P2 ANS

1 2015 VJC JC2 Prelim Paper 2 Solutions Q1(i) Given  1e sin 2 (1)2 y x   Differentiating with respect to x,  de 2cos 2 (2)d y y xx   Differentiating again with respect to x,   22 2 dde e 4sin 2dd yy yy xxx   22 2 d d 1e e 4 ed d 2 y y y yy xx             22 2 22 2 dde e 4e 2dd dd 4 2e (3)dd y y y y yy xx yy xx         Q1(ii) Differentiating result in (i) with respect to x, 32 32 d d d d2 2e (4)d d d d yy y y y x x x x    At 0x , (1): 11e ln 22 y y     (2):  1 d d 2cos 0 42 d d yy xx   (3):   22 2 22 d 2 d 4 4 16 1dd 2 yy xx     (4):    33 33 dd 2 4 16 2(2)(4) 112dd yy xx     23 23 1 16 112ln 4 ...2 2! 3! 1 56ln 4 8 ...23 y x x x x x x                      Q1(iii)  1ln sin 22yx       3 3 21ln 2 ...2 3! 14ln 2 ...23 xx xx           318ln ln 1 4 ...23 xx      

2 233 3 31 8 1 8 1 8ln 4 4 4 ...2 3 2 3 3 3 x x xx x x                           3 32 23 1 8 1 1ln 4 16 ... 4 ... ...2 3 2 3 1 56ln 4 8 ...23 xx x x x x x               Alternative  1ln sin 22yx                     23 233 33 3 23 23 1ln ln 1 2sin 22 1 1 1ln 2sin 2 2sin 2 2sin 2 ...2 2 3 2214ln 2 2 ... 2 ...2 3! 2 3! 28 2 ... ...3 3! 1 8 8ln 4 2 4 ... 8 ... ...2 3 3 1 56ln 4 8 ...23 x x x x xxxx xx xx x x x x x                                                 Q2) ,,OA OB OC   a a c c Q2(i) 2 OB OXON      1 432 1 522 ON      a c c a ca Q2(ii) Area of triangle OAB = 1 2 OA OB     12 2 1 2 1 02            a a c a a a c a c a a 4  ac   4 4 43 OX OA AX OB AC       a c a ca

3 Area of triangle AXB = 1 2 AB AX     1 42 2 2 24 8           c c a c c c a ac Q2(iii) BXAX BX  is the length of perpendicular from A to BX . Area of triangle AXB 8 1 82 BXAX BX BX  16 16 16 43 16 34 BXAX BX BX OX OB          c a a c ca 16, 3, 4k m n    Q3(i) Q3(ii)  : 1 11aa    Re Im (5, –7) 6 (–1, –7) (11, –7) 2y O (5, –7) 6 Re Im O

4 Q3(iii) 2 4izk   is a circle with radius k and centre  2, 4 Distance between the two centres = 227 3 58 Possible exact values of k = 58 6 and 58 6 For 58 6k  58 6 42 77 58 58 58 6 18 33 58 58 a a b b         42 182 7 4 3 i 58 58 42 185 7 i 58 58 z                        42 18Similarly for 58 6, 5 7 i 58 58 kz                (5, –7) (–2, –4) 58 6 7 3 58 b a (–2, –4) (5, –7) 6 Re Im O

5 Q4(i)    2 44 1 ln 2d 2 ln d x x xy x x x x x x x    Let d 0 2 ln 0d y x x xx       10 or ln 2 NA 0 e xx x x      When ex ,   2 1 12 2ee y  Coordinates of A is 1e, 2e   Q4(ii) Required area 2 1 dyx 2 2 1 22 11 2 1 ln d ln 1 1 d 11ln 22 11 ln 222 x xx x xx x x x                        Q4(iii) Required volume   2 e 2 1 1 e 1 d2e yx      e22 2 1 1 ln e 1 d2e 0.0245 x xx               Q4(iv) 2 2eln xr x . From (i) and the curve, 2 2 ln 110 2e 2eln1 0 1 xx x xx x          Since 1r  , S does not exist. So I disagree with him.

6 Statistics Q5(i) A stratified random sample of 60 students can be obtained by sampling 2 groups from the 2 levels as shown: Level Number to be sampled Year One 540 60 27540 660  Year Two 660 60 33540 660  Simple random sampling is used for the selection of students in each stratum. To select the 27 students from the Year One group, we first randomly assign each Year One student a distinct number from 1 to 540. Then generate 27 numbers using a random number generator and select the students corresponding to the 27 numbers generated. Repeat the procedure for the selection of the Year Two group. Q5(ii) A better sample could be achieved by increasing the number of strata, example 4 strata instead of 2 : Female JC1, Male JC1, Female JC2 and Male JC2. Q6(i) number of ways that 6 couples are seated together in a circle 65! 2 7680   Q6(ib) no. of ways that 12 people are seated in three distinct rows of four seats without restriction 12! no. of ways that 12 people are seated in a circle with 0 female in one of the rows      36 12 8! 4!CC Hence required no. of ways      36 1212! 8! 4!CC 435456000 Q7(i)   5 3 13 3 5P wins on her turn 143 CA C     5 3 13 3 8 3 13 3 138P didn't win on her turn 1 143 115P didn't win on her turn 1 143 CA C CB C         2 5 138 115 5 138 115 5P wins ... 143 143 143 143 143 143 143A                             5 143 138 1151 143 143 715 0.156 (3 s f)4579          

7 Q7(ii)  P wins on 1st draw | wins the gameAA     P wins on 1st draw P wins the game 5143 7154579 0.224 A A   No. of possible draw sequence 3 3 3 1 27     Q8(i) (1) Meteors occur at a constant average rate. (2) Meteors occur independently. other meteors. Q8(ii) Let X be the number of meteors seen by Amy in 4 minutes  ~ Po 8X P( 6) 0.122X  Q8(iii) Let Y be the number of meteors seen by Amy in 2 minutes  ~ Po 4Y Required probability 12P( 3) P( 3) 0.0382YY     Q8(iv) Since part (iii) is one of the possible solutions for part (ii), hence answer to (ii) is greater than answer to (iii). Q8(v) Let W be the number of meteors seen by Amy in n minutes  ~ Po 2Wn Since 10 2 20nn   ,  ~ N 2 , 2W n n approx    P 3 0.005 P 3 0.5 0.005W n W n      (continuity correction) 3 0.5 2P 0.005 2 0.5P 0.995 2 nnZ n nZ n       From the GC,  P 2.3263 0.99Z  0.5 2.57583 2 n n  2.5758 3 z 0 0.995 1a Either 0,1, 2 2a Either 0,1, 2 3a Either 0,1, 2 4a Only 3

8 3.64277 0.5 0nn   Consider 3.64277 0.5 0nn        2 3.64277 3.64277 4 1 0.5 0.14286 or 3.49992n    0.14286 3.4999 0nn     0.14286 or 3.4999 NA 10 12.249 nn n n    smallest value of 13n Q9) Let 160wx , then 480w and 2 8837w    2 2 480480 1 3.2 and 8837 49150 149 150 wws         22160 100 and xwx w x w s s      3.2 160 163.2x    , 22 49xwss Let  cm be the population mean of X. H0 : 162 H1 : 162 Level of sig : 5% Test Statistic : When H0 is true, 162 49 150 XZ     Computation : 163.2x  , 2 49s  , p – value = 0.0179 Conclusion: Since p – value = 0.0179  0.05, H0 is rejected at 5% level of significance. Hence there is sufficient evidence that the mean height of 162 cm is an understated value. At 5% significance level” means there is 0.05 probability that the test will conclude that the mean height of 162 cm is an understated value when actually it is not an understated value. Q9(i) A t-test is carried out instead. We need to assume that the height of a female student follows a normal distribution. 3.4999 0.14286

9 Q9(ii) Yes. For an upper tail test, Ho is rejected if 162m s n   c, where c is the