PJC H2 MATH P2 SOLUTIONS

PJC2011 [Turn Over] 1 PJC 2011 JC2 H2 Mathematics End of Year Paper 2 Solution 1 32Let sequence, then 1 8 4 2 2 27 9 3 4 64 16 4 8 11 4Using GC: , , , 062 3 rTa rb rc r d abcd abc d abc d ab c d ab cd             32114 623 rTrrr 2(i) 3 21 14 x x   21 104 5 04 4 5 ---------------(1) x x x x x      (2 e ) e 1(1 4e ) e 2e 1 1e4 xx xx x x       1 2 Let lim i.e as , and then 3 30 11 1 2 2 11 3 11 3 or (NA since >0)22 nn nn n Xn X X x               Note: Graph is not a straight line 3y xx  1 1 from graph of 3 for 0 , 0 30 0 nn nn nn y xx xy xx xx xx           – 4 5 + + –

PJC2011 [Turn Over] 2 Replace in (1) with e then 4 e 5 since e 0 for all 0e 5 ln 5 ln 5 x x x x x x x x             4(i)  04 2 11 OL 4 2 322 06 3 OC OD                   4(ii)  11area sin 22 24 1 302 30 0 1 122 12 1 288 6 22 OAL OL OA AOL OL OA            

4(iii) 0 using dot 22 1 1 14 14 cos 33 6cos 14 64.6 ML OD                      5(i) 2 22 22 22 2 12 1 1 4422 4 1 2 44 44 24 xxx x x xx xx x         

PJC2011 [Turn Over] 3 5(ii) 22 1 0 0 21 11 1 4d 4 4 s i n22 2 2 1 4 4sin 42 4 kk xxx x x kkkk a                   5(iii) 2244yx 2 2 2 12 xy  2 0 1 4d2 k R xx  5(iv) 1 Required area 4 with 1 1 3 4sin 2 3 4 6 2 3 3 Rk          6(i) 31 zz 6(ii) arg 3 i arg66 arg66 6 arg 03 0a r g 3 31 lies on excluding point ' ' z z z z zz zO A C O                 6(iv)  23 3 3 tan 3 1 3 Hence arg 2 3i23 AB AB z            2 1 k R 3  1 A B  23 O C Locus of Q Locus of P Locus of z Im(z) Re(z) 3 −1

PJC2011 [Turn Over] 4 7(i) N = 90, n = 10, so k = 9. List all the faculty staff by their names in alphabetical order and ch oose a random number from 1 to 9. Say we have chosen the number 4, then we select the 4 th, 13th, 22nd, 31st, 40th,49th, 58th, 67 th,76th, 85th names from the list to form the sample of 10. 7(ii) Choose the number of Professor, Seni or Lecturers and Lecturers in the following: Professor Senior Lect urer Lecturer Total Population 9 18 63 90 Sample 1 2 7 10 Compile 3 lists of names – Professo r, Senior Lecturers and Lecturers. Perform simple random sampling from the separate lists of Professor, Senior Lecturers and Lecturers to select 1 Professor, 2 Senior Lecturers and 7 lecturers to form a sample of 10. Stratified sampling is preferred as the sample selected is more representative of the faculty. 8(i) r = 0.9785 8(ii) r = 0.9785 indicates that there is a strong positive linear correlation but from the scatter diagram, there is non-linear relation between x and y. So a linear model may not be appropriate. 8(iii) From GC, ln y = 1.35553 + 0.132498 x 8(iv) From the description, x is the controlled variable while y is the response. So we will use ln y on x to estimate the amount: ln15 = 1.35553 + 0.132498x, so x = 10.2 mg 9(i) Let A ~ students own IPod B~ students own IPhone Given n(A ) = 25, n(B) = 40 and n(A  B) – n(A  B) = 35 n(A  B) = n(A) + n(B) – n(A  B) 35 + n(A  B) = 25 + 40 – n(A  B), so n(A  B) = 15 P(A  B) = 20 3 100 15  x y

PJC2011 [Turn Over] 5 OR P(A  B) = P(A) + P(B) – P(A  B) 100 35 + P(A  B) = 100 40 100 25  − P(A  B) P(A  B) = 20 3 100 15  9(ii) From (i) n( A  B) = 50 So the prob required = 2 1 100 501  9(iii) P( A | B) = 8 3 40 15 )( )(  BP BAP Alternative: Use venn diagram to consider the numbers: 10(i) Number of way required = 10 4 = 100 00 10(ii) Number of ways for last digit = 5 Number of ways required = 9  8  7  6  5  5 = 75600 10(iii) Number of ways for 3 odd digits = 3 5C = 10 Number of ways for 3 even digits = 3 5C = 10 Number of ways required = 10  10  6! = 72000 11(i) X ~ number of defective pen, out of 10 X ~ B(10, 0.01) P( X > 1) = 1 – P(X  1) = 0.0042662 0.00427 11(ii) Y~ number of boxes being rejected, out of 1000 Y~ B(1000, 0.0042662) Since n is large, np = 4.2662 < 5, Y~Po(4.2662) At least 990 boxes being accepted is th e same as at most 10 boxes being rejected P( Y  10) = 0.995 IPod IPhone x y z Given x + y = 25 ----- (1) y + z = 40 ----- (2) x + z = 35 ----- (3) (1)+(2): x + 2y + z = 65, so y = 15 (1) + (2) + (3): 2x+2y+2z = 100 So x + y + z = 50 (i) Prob req = 20 3 100 15  (ii) Prob req = 2 1 100 50 

PJC2011 [Turn Over] 6 11(iii) W~ number of boxes being rejected out of 2000 W~B(2000, 0.0042662) Since n is large, np = 8.5324 > 5 and n(1 – p) = 1991.4676>5 W~N(8.5324, 8.495999) approx P( W  12) = P( W < 12.5) = 0.913 12(i) Unbiased estimate for mean ( 30) 126630 30 82.7524 24 xx     Unbiased estimate for variance = s2=   391.49024 12665.7806023 1 24 )30()30(23 1 22 2                 xx 12(ii) H0:  = 80 vs H1:  > 80 Level of Sig = 3% Test statistics t = 82.75 80 490.391/ 24  =0.60837 p-value = 0.27445 > 0.03 Since p-value is more than level of significance, do not reject H 0 and conclude that there is insufficient evidence to suggest that the mean speed is greater than 80 km/h at 3% level of significance. The unbiased estimate of mean remains unchanged as 82.75 The unbiased estimate of pop variance  2 2 2 ( 30)1 1 12660( 30) 780605 471.925239 240 239 240 x x           but the distribution 471.925~( 8 0 , ) 240XN approximately under CLT test statistics 82.75 80 1.9611 471.925 / 240 z  p-value is 0.02493 < 0.03 Since p-value is less than the level of significance, we reject H0 and conclude that there is sufficient evidence that the mean speed is greater than 80 km/h, so there is a change in conclusion.

PJC2011 [Turn Over] 7 13(i) X~ time taken by route A X~ N(14, 2.52) P( X > 12) = 0.788 13(ii) Y ~ time taken by route B Y~ N(12, 7.3) X – Y ~N(2, 2.5 2 + 7.3) X – Y ~N(2, 13.55) P(| X – Y|  3) = 1  P(3 X – Y  3) = 0.480 Let T = X1 + X2 + X3 + Y1 + Y2 ~ N(42 + 24, 3(2.52)+2(7.3))=N(66,33.35) T ~ N(66,33.35) P ( T < 70) = 0.756 14 X~no of fish caught by angler in 1 hour X~Po(1.5) P( X = 3) = 0.126 14(i) Y~no of fish caught by angler’s son in 1 hour Y~Po(1) X + Y ~Po(2.5) P( X + Y  2) = 1 – P(X + Y  1) = 0.713 14(ii) U~no of fish caught by angler in 2 hours V~ no of fish caught by his son in 2 hours U~Po(3) V~Po(2) U + V~Po(5) Prob required = ( 4 )( 0 ) ( 3 )( 1 )(| 4 ) (4 ) PU PV PU PVPU V U V PU V        = 0.475 Let F and S be the number of fish caught by angler and his son in n minutes respectively. F + S~Po( 60 5.2 n ) P(F + S  1) > 0.9 P(F + S = 0) < 0.1 From the GC, n = 55: P(F + S = 0) = 0.1011 >0.01 n = 56: P(F + S = 0) = 0.09697 <0.01 so least n = 56 Since 50n is large, By Central Limit Theorem 1.5~( 1 . 5 , )50XN approx P( X > 2) = 0.00195