IJC H2 MATH P2 (9740) Solutions

1 (i)   5 i0 i0 2 π i2 π 243 243e 243e 243e k k w      2 πi 5 4π 2π 2π 4πii i i55 5 5 3e , where 2, 1, 0,1, 2 3e , 3e , 3, 3e , 3e k wk       (ii) 4πi 5 1 3ew   and 2πi 5 2 3ew   2 (i) 1f : 3 2x x  , ,2xx Let f( )yx . 13 2 12 3 12 3 y x x y x y       1 1 f ( ) 2 3x x    , ,3xx (ii)   f f D2 , R3 ,   Since ffRD , the composite function 2f exists. 3 Im(z) O Re(z)

(iii)  2 2 f 1f3 2 13 132 2 13 1 2 3( 1) ( 2) 1 45 (1 ) 55 0 xx xx x x xx x xx xx xx x xx                 Using GC, f 1.38 (rej 1.38 D ) or 3.62xx         11 1 ff ff f f ff xx x x xx      Therefore 3.62x  satisfies    1ff x x . 3 (i) When x = 0,  10 0 o r 1 1 1 or 2 tt t t yy      Coordinates are (0,1) and 10, 2   . x y O

(ii)     22 22 22 dd 2 21 , dd 1 d2 1 d2 1 1 2 12 1 xy tttt t yt xt t t tt         When tangent is parallel to y-axis,    2222 112 10 1 0 2tt t t         Equation of tangent: 1 4x (iii) Area of the required region   0 1/ 4 1 21/2 1 221/2 121 1/2 1 1 d 1 21 d 1 21 d 11 ln 1 tan 51ln 2 ln tan44 2 81ln tan54 2 yx tt t t t tt tt                            4 (a)(i) Area of unsown ploughed land  2 0.4 0.4 300 100 88 m    11When , 42 When 0, 1 xt xt  

(a)(ii) n Beginning of week End of week 1 300  0.4 300 2 0.4 300 100    2 0.4 0.4 300 100 0.4 300 0.4 100     3  20.4 300 0.4 100 100        2 32 0.4 0.4 300 0.4 100 100 0.4 300 0.4 100 0.4 100    .. … … n …     1 21 0.4 300 0.4 100 ... 0.4 100 0.4 100 nn    Area of land unsown ploughed land at the end of nth week     1 12 0.4 1 0.4 0.4 300 100 10 . 4 2000.4 300 1 0.4 m3 n n nn          the value of k is 200 3 . (a)(iii) Method 1      1 1 1 40 2000.4 300 1 0.4 703 200 2000.4 300 (0.4) 0.4 7033 400 100.433 10.4 40 ln ln 0.4 4.02588 nn nn n n n n           Hence the number of complete weeks required is 5. Method 2    12000.4 300 1 0.4 703 nn  

Using GC, when n = 4, unsown ploughed land = 70.08 (> 70) when n = 5, unsown ploughed land = 68.032 (< 70) when n = 6, unsown ploughed land = 67.213 (< 70) Hence the number of complete weeks required is 5. (b)(i) n Beginning of week End of week 1 300 300 80 2 300 100 80  300 100 80 100  3 300 2 100 80 100   300 2 100 80 100 120   .. … … n …     300 1 100 80 100 80 20 1 n n      Area of unsown ploughed land at the end of nth week      2 2 300 100 1 2 80 20 12 300 100 100 140 202 300 100 100 70 10 10 30 200 nnn nnn nn n nn                 (b)(ii) For the farmer to finish sowing all the ploughed farmland, 210 30 200 0nn   Method 1: Solving the inequality, 6.21699 or 3.21699 (rejected)nn  Hence the number of complete weeks is 7. Method 2: Using GC to set up a table, When 6n  , area unsown 20 When 7n  , area unsown 80 When 8n  , area unsown 200 Hence the number of complete weeks is 7. In week 6, the area of unsown ploughed land 210(6) 30(6) 200 20    m2  area of ploughed land to be sown in week 7 (the final week) 20 100 120  m2

5 (i) Number of arrangements = 66! 2 46080 (ii) Required probability   6 5 12 10 51 !2 10 1 ! 288 23950080 0.0000120 (3 sig fig) C C     6 (i) P(Clark wins in 3rd draw) 77772 99999 0.081322 0.0813    (ii) P(Kara wins) 35 72 7 2 7 2 ....99 9 9 9 9       35 2 27 7 7 ...99 9 9 7 2 9 9 71 9 70.4375 16or              7 (i) Let Y be the number of calls received by the office in a t-minute period. ~P o(0 . 4 )Yt Given: P( 0 ) 0.1Y  Using GC, 6 (nearest minute)t  (ii) Let T be the number of calls received by the office in a 2-hour period. ~P o(4 8)T Since E( ) 4 8 1 0T  , therefore ~N(4 8 , 4 8 )T approximately. P( 50 ) P( 50.5 ) (with continuity corrections) 0.641 (3 sig fig) TT 

(iii) The average number of calls may not increase at a constant rate over longer periods of time interval. The calls that arrived may not be independent of one another, because the calls might be made by the people who witness the same accident at a particular location. 8 (i) Whether a randomly chosen patient turns up for an appointment is independent of any other patient. (ii) Let X be the number of patients who turn up for their appointments, out of 20 appointments. ~ B 20 ,0.845X P( 15)X  1P ( 1 5 )X  0.812 (3 sig fig) (iii) Required probability P( 1 7 | 1 2)XX  P (12 17 ) P( 1 2) P( 1 7) P( 1 1) 1P ( 1 1 ) 0.618 (3 sig fig) X X XX X       (iv) Let A be the number of appointments for which the patients fail to turn up, out of 300 appointments. ~ B 300 ,0.155A Since n = 300 is large, 46.5 5np and 253.5 5nq  , therefore ~ N 46.5 ,39.2925A approximately. P( 40 50 ) P( 39.5 50.5 ) (by continuity corrections) 0.606 (3 sig fig) A A     9 (i)(a) Given:  2~N 3 5 . 2 , 5 . 2L   2~ N 24.6 ,3.8P   2~ N 29.3 , 4.3C Let 32TL P . E ( ) 3 35.2 2 24.6 154.8T    22 22Var ( ) 3 5.2 2 3.8 301.12T    ~ N 154.8 ,301.12T

Let a be the required score exceed by 1% of the candidates. P( ) 0.01Ta  P( ) 0.99Ta Using GC, 195.2 (1 dec pl)a (i)(b) Required probability   32 5!P( 150 ) P( 140 ) 2!3!TT     0.0875 (3 sig fig) (ii) Consider 325A LPC E ( ) 154.8 5 ( 29.3 ) 8.3A    22Var ( ) 301.12 5 4.3 763.37A    ~ N 8.3 , 763.37A Required probability    P2 5 P2 5 2 5 0.613 (3 sig fig) A A    Required percentage = 61.3% 10 (i) (ii) The product moment correlation coefficient between t and m is 0.94597r   (5 d.p.). A value of 0.94597 for r suggests that there is a strong positive linear correlation between t and m. However, the points on the scatter diagram show a curvilinear relationship . Therefore this value of r does not necessarily mean that the linear model is best model for the relationship between t and m. m t

(iii)  ln ln ln ln ln b b ma t ma t mbt a      The product moment correlati on coefficient between ln t and ln m is 0.98967 0.990 (3 sig fig)r  Reason 1: From the scatter diagram, as t increases, the weight of the foetus increases at an increasing rate. Reason 2: The value of r between ln t and ln m is 0.98967 , which is closer to 1 as compared to that between t and m, hence indicating a stronger positive linear correlation between ln t and ln m. Hence bma t is a better model. (iv) From GC, ln 8.3764 4.5938 ln (5 sig fig)mt  4 ln 8 3764 23 0 1 0 a. a.    and b = 4.59 (v) When 26, ln 8.3764 4.5938ln 26 728 (nearest grams) tm m       Since the value of 26 is with in the range of values of t and the value of r is close to 1, this estimate is reliable. 11 (i) Let X be the random variable denoting the mass of stra wberry jam, in grams, in a randomly chosen jar. Unbiased estimate of population mean 66 200 197 830x . Unbiased estimate of population variance s2 =  2661 958 28 0275929 30 .     H0 :  = 200 H1 :  < 200 Test at 2% significance level Assume H0 is true. 28 02759N 200 30 .X~ ,  

Test statistic: 200 N0 1 28 0275930 XZ ~, .  Using GC, p-value = 0.011420121 < 0.02 Reject H 0 and conclu