NJC H2 MATHS P2 Solutions

Suggested Solution to 2015 SH2 H2 Mathematics Preliminary Examination Paper 2 1 (a) 2 3tan d 3secd x x     When 3,x tan 1 4    When 3x , 3tan 36          3 223 24 22 6 24 2 6 4 2 6 4 6 1 d 9 1 3sec d 9 tan 9 tan 9 11 3sec d9 tan 3sec 1 cos d9 sin 11 9 sin 1 1 1 9 sin sin46 22 9 x xx                                   1 (b)               2 2 2 2 2 2 2 2 2 21 ln 4 d 1 ln 4 d 2ln 4 d 4 2 4 8 ln 4 d 4 8ln 4 2 d 4 ln 4 2 4 tan 2 xx xx xx x x x x x x x x x x x x x xx x x c                                  2 (a) Amount saved after n months > 20000 1000 +1000(1.05) + +1000(1.05)n–1 > 20000

1000(1.05 1) 200001.05 1 n  20000(1.05 1) 20000 1.05 1 1 1.05 2 ln1.05 ln 2 ln 2 14.20669908ln1.05 n n n n n      Using GC, smallest n = 15 i.e. it takes 15 months for Joanne’s savings to exceed $20000. Joanne’s savings at the end of the Nth month 20000(1.05 1)N (from part (a)). Jim’s savings = 0 + 0 + 2000 + (2000+100) + ….+ [2000 +100(N – 3)] =  2 2 2000 ( 3)(100) , for 32 N NN     = 50( 2)( 37), for 3N N N   20000(1.05 1) 50( 2)( 37)N NN    By GC , we find that n = 50.

2 (b) Let x = 1 + r/100. For Joanne’s total savings to be at least $30000           2 24 24 24 24 25 1000 1000 ..... 1000 30000 1000 1 300001 1 301 1 30 1 1 100 1 31 30 0 x x x xx x xx x x x x xr xx                  Sketching 25 31 30y x x   , we see that 1.0174302x , Hence the interest rate must be at least 1.74% i.e. least value of r = 1.8 (correct to 1 decimal place)

3 (i) Converting the equations of p1 and p2 to Cartesian form, x + y – z = 0 --- (1) 2x – y – 2z = 6 --- (2) Using the PolySmlt2 app in the GC, 21 : 2 0 , 01 l                      r R 3 (ii) Equating the RHS of the two equations for  l and 1l , we have: 22 2 85                      2 2 ---- (3) 2 ---- (4) 8 5 ---- (5)         Substituting (4) into (3), we get 2 2 2 0 2      . Substituting 2 and 2  into (5), LHS = –2 = 8 + 5(−2) = RHS. Hence,  l and 1l intersect at exactly one point. Hence, 1 p , 2p and 1l have exactly one common point of intersection. 3 part after (ii)  l lies on 3p . So  l is perpendicular to the normal vector of 3p . 11 00 1 10 1 (shown) a b b b                   

3 (iii) Method 1: Finding angle between two planes Acute angle between 1 p and 2p = acute angle between 2p and 3 p       11 2 1 2 1 2 1 1 1 1 2 1 2cos cos 39 29                                              a a 2 2 43 3 2 3 2 3 4 a a aa             22 22 2 2 9 2 3 16 8 6 3 16 8 2 8 10 0 4 5 0 5 1 0 a a a a a a aa aa aa                  31rejected because and 5 or 1 are distinct and non-parallel ppa   Since  l lies on 3p , 21 25 01 2 2( 5) 12                         d 1 5 12 1     r

3 (iii) Method 2: Finding mirror image of point in plane Alternatively, a point on 1 p is the origin, (0, 0, 0) and a point on 2 p is (3, 0, 0). Denote the foot of perpendicular of O onto 2 p by P. Then 2 2 2 2 2 2 3 0 2 2 0 0 1 1 0 0 2 2 2 ( 1) ( 2) 2 ( 1) ( 2) 2 6 19 2 4 / 3 2 / 3 4 / 3 OP                                                            Let O be the mirror image of O in p2. Then 2 8 / 3 4 / 3 8 / 3 OO OP   Therefore a vector parallel to 3p 8 / 3 2 2 / 3 1 24 / 3 2 2 / 3 1 38 / 3 0 8 / 3 4                                             Hence normal vector to 3p = 1 1 1 1 0 5 4 1 1                              Hence 5a and an equation for 3p is 1 2 1 5 2 5 12 1 0 1                                  r

4 (i), (ii) 4 (iii) Let W be the point representing 4 +10i. Then 22 min 4 10i = (4 ( 3)) (10 3) 5 2 7 2 5 2 2 2 i.e. 2 (shown) z WD WC CD m            4 (iv) Minimum value of  arg 6 w 11 22 11 11 5 2 3sin tan 939 50 3sin tan 990 51sin tan 93 1.978773429 rad 1.979 rad (to 3 d.p.)                                    P 5√2 D W(4, 10) C(–3,3) P 2239 α 3 6  –3 9 3 4  Im Re

5 (a) Assign a number from 1 to 2500 to each of the 2500 students OR obtain a list of the students from the administration office in order of their identification numbers or registration numbers. Use a random number generator to generate 5 00 numbers randomly and select the members who are assigned those numbers (based on the ordering of the list). 5 (b) Assign a number from 1 to 2500 to each of the 2500 students OR obtain a list of the students from the administration office in order of their identification numbers or registration numbers. Next, determine the sampling interval size 2500 5.500k  Randomly select any student from the list, say the 1 st student. Select every 5 th student thereafter (i.e. 6th, 11th, 16th…) until all 500 students are selected. 6 (i) Let X be the number of students, out of 20 randomly chosen students, who wore their House T-shirts to school. Then ~ B(20, 0.14).X P( 3) 0.24086 0.241 (3 s.f.) X   6 (ii) Let Y be the number of students, out of 60 randomly chosen students, who wore their school uniforms to school. Then  ~ B 60,0.78 .Y Since n = 60 > 30 is sufficiently large, 46.8 5np and (1 ) 13.2 5,np    therefore  ~ N , (1 )Y np np p  approximately i.e.  ~ N 46.8,10.296Y approximately.         P 45 P 45.5 by continuity correction 0.6573140131 0.657 3 s.f. YY    

7 (i) Number of ways = 5 54   7 (ii) Task 1 : Select the colour to appear once  5 1   ways Task 2 : Select the colour to appear thrice  4 1   ways Hence total number of ways 54 2011             or total number of ways 5 2 5 2! or P 202    7 (iii) Case 1 : All balls are distinct Number of ways = 5 (from part (i)) Case 2 : Exactly 3 identical balls are selected Number of ways = 20 (from part (ii)). Case 3: 2 pairs of identically-coloured balls Number of ways 5 102  Case 4: 4 identically-coloured balls Number of ways 5 51  Case 5: 1 pair of identically-coloured balls and 2 balls of different colours Number of ways 53 3031             or Number of ways 54 3012             or Number of ways 53 3021             or Hence total number of ways = 5 + 20 + 10 + 5 + 30 = 70

8 (i) Any two of the following three: 1. The demand for each single-room occurs singly i.e. does not occur at the same instant. 2. Each demand for a room in a day is independent of (or unaffected by) a demand occurring in any other part of the same day. 3. The mean number of demands over any time interval of the same duration within the day is constant. 8 (ii) Let X denote the random variable for the number of demands for a single -bed room in a day. Then Po(5.8).X P(X ≥ 10) = 1 – P(X ≤ 9) = 0.0708 8 (iii) Let Y denote the random variable for the number of demands for a double -bed room in a day. Then Po(37.1).Y   Po(5.8 37.1) Po 42.9 XY XY   P( 1 and 36 38) P( 1 and 36 | 38) P( 1 and 36 and 38) P( 38) P( 1 & 37) P( 2 & 36) P( 38) P( 1)P( 37) P( 2)P( 36) P( 38) 0.092653 0.0927