IJC H2 MATH P1 (9740) Solutions

1 Let nP denote     0 ! 1 ! 1 n r r r r     for , 0nn .     0 When 0, LHS 0 0! 0 RHS 0 1 ! 1 0 LHS Therefore P is true. n               0 1 1 0 Assume P is true for some , 0, i.e. ! 1 ! 1 Want to prove that P is true, i.e. ! 2 ! 1 k k r k k r kk r r k r r k                                    1 0 0 LHS ! ! 1 1 ! 1 ! 1 1 1 ! 1 ! 1 1 1 1 ! 2 1 2 ! 1 RHS k r k r rr r r k k k k k kk kk k                              1 01 Thus P is true P is true. Since P is true, and P is true P is true, by mathematical induction, P is true for all , 0. kk kk n nn      2 (i)                   2 24 e cos d e cos e sin( ) d e cos e sin e cos( ) d e cos sin e cos d x n xx nn x x x n n n xx nn nx x n nx n n nx x n nx n n nx n n nx x n nx n nx n nx x                               42 2 4 1 e cos d e cos sin e cos d e cos sin 1 xx nn xx nn n nx x n nx n nx nnx x nx n nx C n          

(ii)                        2 22 2 4 4 4 4 4 e cos d e cos sin 1 e cos 2 e cos 1 e e cos 1 e e 1 if is even 1 e e 1 if is odd 1 xx nn nn nn nn nn nnx x nx n nx n n nn n n n n n n n n n n                        3 (i) (2 5 ) (2 5 ) 4 10 10 25 20 2 20 1 1 0 20 2 b a a ab b                                   p q p q p p p q q p q q pq Alternative:   22 (2 5 ) (2 5 ) 2 1 5 1 2 1 5 1 00 4 5 4 5 37 22 6 14 8 10 8 10 28 35 12 15 20 20 40 20 bb aa bb aa aa a ab a ab bb a ab b                                                                                           p q p q 20 2 a ab b           

Given that the i- and j- components of the vector 20 2 a ab b     are equal, 0 ( 1) 0 Since 0, thus 1 a ab ab a ab ab      (ii) 2 2 2 (2 5 ) (2 5 ) 80 20 80 2 4 21 2 9 4 2 9 16 7 2 7 14 22 a ab b a a a a a a or                 p q p q (iii) Since 25pq and 25pq are perpendicular,        22 22 2 2 2 5 2 5 0 4 25 0 25 4 25 114 25 2 52 2           p q p q pq pq p Alternative: 2 2 4 5 4 5 (2 5 ) (2 5 ) 3 7 22 16 25 21 4 4 30 aa a a                           p q p q Since 25pq and 25pq are perpendicular,

2 2 (2 5 ) (2 5 ) 0 4 30 0 15 2 a a      p q p q 22 15 25 5 22 1 5 2 2 2a      p 4 (a) Method 1 Since the coefficients are real, 2iw is another root of the equation.        22 2 2 2 i 2 i 2 i 4 4 1 45 w w w ww ww               32 30 0w pw qw      2 4 5 6 0w w w    (By inspection) Comparing coefficients of 2w , 6 4 2p   Comparing coefficients of w , 24 5 19q   Method 2 Substitute 2iw ( or 2iw ) into the given eqn, 32(2 i) (2 i) (2 i) 30 0 (3 4i)(2 i) (3 4i) (2 i) 30 0 (6 3i 8i 4) (3 4i) (2 i) 30 0 (32 3 2 ) ( 11 4 )i 0 pq pq pq p q p q                                Comparing the real parts, 3 2 32pq  --- (1) Comparing the imaginary parts, 4 11pq  ---- (2) (1)  (2)  2: 3 8 32 11 2 5 10 2 pp p p        From (2): 11 4 2 19q    2p , 19q (b) Substitute 3izu into the given eqn,         2 2 2 3 i ( 5 2i) 3 i (21 i) 0 9 6 i 15 5 i 6i 2 21 i 0 15 2 5 i 0 uu u u u u u u u                       Compare imaginary coefficient: 50 5 u u   3 5iz  

[Note: if using 215 2 0uu   , need to reject 3u ] Method 1 Let the other root be w.   2 ( 5 2i) (21 i) 3 5iz z z z w         Comparing coefficients of z, 5 2i 3 5i 2 3i w w       Method 2 Let the other solution be iab ,   2 2 2 ( 5 2i) (21 i) ( (3 5i))( ( i)) ( i) (3 5i) (3 5i)( i) 3 ( 5)i (3 5i)( i) zz z z a b z a b z z a b z a b z a b                           Compare the z term: ( 3) 5 2aa     ( 5) 2 3bb     2 3iz   is another root. 5 (i)                           22 2 1 2 23 34 45 1 1 21 22 22 22 2 11 .... .... 11 2 2 2 1 2 1 1 1 11 8 21 N n N nn n NN NN N n n n uu uu uu uu uu uu uu NN NN                          (ii) As N  ,   22 1 0 21NN  

    22 2 21 811n n n n     which is a constant, hence it is a convergent series.     22 2 21 0811 1 8 n n n n       (iii) Method 1                 2222 11 1 22 2 22 22 22 ( 1) 2 ( 1) 2 2 ( ) 1 1 11 8 2 1 2 418 12 NN nn N n N N n n n n n n N n n n N NN N NN                     Method 2 By listing the terms     22 2 2 2 2 2 2 2 2 11 2 2 2 2(1) (3) 3(2) (4) ( 1) ( 1) N n n n n N N N          22 1 2 2 2 2 2 2 2 ( 1) 2 2 2 2 2(1) (3) 3(2) (4) ( 1)( ) ( 2) N n N n n n N N N N                  1 22 2 22 22 2 11 11 8 2( 1) ( 2) 418 12 N n N n n n N NN N NN            

6 (i) 2 2 22 2 d( ) 2 (1)d Differentiating (1) w.r.t. : d d d d( ) 1 0 d d d d d d d( ) (1 ) 0 (2)d d d yx y ky x x y y y yx y k x x x x y y yx y k x x x               3 2 2 2 3 2 2 2 32 32 Differentiating (2) w.r.t. : d d d d d d( ) 1 (1 ) 2 0d d d d d d d d d( ) 2 3 0d d d x y y y y y yx y k x x x x x x y y yx y k x x x                          2 2 3 22 3 d0, 1: 2 d d 36 d d 6 36 48 6 6 8 d yx y k x y k x y k k k k x                2 23 2 2 3 6 6 8361 (2 ) ... 2! 3! 361 (2 ) 6 8 ... 2 kkky k x x x kk x x k k x                     (ii) sin 2 sin 2 cos cos 2 sin cos 22 2 2x x x x         22 22 22 2 11 cos 2sin 2 (2 )1 2 12 1 4 ... xx x x x            

3642 2 10 3 k k    7 (i)   22d 100 , 0d M k M kt    Since  22d 0 and 0, 100 0 and 0 100d M M M Mt          22 1 dd 100 1 100ln200 100 M k t M M kt CM       200 ' 100ln 200 '100 100 e , where e100 kt C M kt CM M AAM      When 0, 5tM   105 21 95 19A When 1000 1000 3 215, 20 e 2 19 19 1 19e 200 ln 14 5 14 k k tM or k       Thus     51000 5 5 55 100 21 21 19 e100 19 19 14 21 19100 100 19 14 21 19 21 191 100 119 14 19 14 tt k t tt M M MM M                                   55 55 21 19 19100 1 100 21 1919 14 14 21 19 19 1 21 1919 14 14 tt ttM OR                                   OR 5 5 19 19100 14 21 19 19 14 21 t t      (ii) When t = 15, 3 3 21 19100 119 14 46.847 21 19 119 14 M       47M  (nearest whole number)

(iii) Method 1: Graphical Method Sketch the graphs of M=f(t) and M=80 From the graph, when 34.336397, 80tM Least number of days required is 35. Method 2: Use GC table When 𝑡 = 34, 𝑀 = 79.627 < 80 When 𝑡 = 35, 𝑀 = 80.718 > 80 When 𝑡 = 36, 𝑀 = 81.756 > 80 Thus least number of days required is 35. Method 3: 5 5 55 5 5 21 19100 119 14 80 21 19 119 14 5 21 19 21 19 114 19 14 19 14 1 21 19 9 4 19 14 4 19 57 14 7 575ln 7 34.33639719ln 14 t t tt t t t                               Least number of days required is 35. 8 (i) Range of f is [−1, 3] or  1,3fR  or  : 1 3fR