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 