NJC JC2 H2 Maths 2012 Solutions Paper 1

2012 NJC H2 Math Prelim P1 Solutions Page 1 of 16 Qn Suggested Solution 1   32f x ax bx cx d    where , , ,a b c d  R . Given  f 0 1 , 1d  .   2f ' 3 2x ax bx c   From the graph of  f 'yx ,    f ' 2 0 12 4 0 1 a b c          f ' 7 0 147 14 0 2a b c         f ' 2.5 9 18.75 5 9 3a b c      OR    f '' 2.5 0 15 2 0 4ab     Using GC to solve (1), (2) & (3): 4 10 56,,27 9 9a b c     324 10 56f1 27 9 9 x x xx     For  f x is concave downwards  f '' 0x , 2.5x . 2(i) 6 3 0 AB OB OA        1 2 1 2 2 6 12 3 1 3 32 2 0 0 AP                                                 Since 1 3AP AB for 1 and A is a common point, this shows that A, B and P are collinear.

2012 NJC H2 Math Prelim P1 Solutions Page 2 of 16 2(ii) Given that area of triangle OAP is 162 5 , 1 162 52 1 11 3 162 523 2 16 11 3 3 162 523 20 OA AP AB                                2 2 2 12 1 3 1 324 5 20 2 1 4 324 5 5 1 2 4 5 324 5 1 45 324 5 1 108 1 108 or 1 108 107 or 109                                            Since P is on BA produced, AP k AB for a negative value of k. Hence 109 . 3 1st part              2 2 3 2 2 2 3 2 3 2 3 23 1 1 1 1 3 1 2 2 2 2 21 1 (2 ) (2 ) (2 ) 2 2 ! 3! f ( ) e 1 2 1111 22 111 22 131 22 x x x x x xx x x x x x x x x x x x x                                   3(a) When 1 ,3x

2012 NJC H2 Math Prelim P1 Solutions Page 3 of 16   21 3 1 9 23 23 1 1 1 1 3 1e 1 2 1 3 3 2 3 2 3 5 1 1 1 3 1 13e1 3 3 2 3 2 3 9                                 11 99 5 13135e 9 e 9 1339   3(b)   29f ' 1 2x x x    Using   29f ' 1 , 2x x x      2 2 2 2 23 23 22 2 ef ( ) 2 e 1 2 12 ef ( ) 2 f ( ) 12 e f ( ) 2 f ( ) 12 d 1 31d 2 2 1321 22 91 2 2 2 51 2 x x x x x x x x x x x x x x x x x x xx x x x x x x x x xx                                Alternative method:      2 2 1 2 3 2 2 2 2 2 e e 1 2 1 2 12 131 1 2 422 11 2 4 2 2 51 2 x x xx x x x x x x x x x x x xx                       

2012 NJC H2 Math Prelim P1 Solutions Page 4 of 16 4(i)       2 2 22 2 2 2 4 2 4 2 dd 2dd d2 2 0 d d d d shownd x ut xu tu ttt ut tu t t ut ut t ut u tt u ut          4(ii) 2 2 2 1 d 1dt 1 ' ' where ' uu tCu t tCx tx C CtC         Since there was 0.2 milligrams of bacteria after 15 minutes, then   2 0.250.2 0.05 0.2 0.06250.25 1 16 CC C      216 16 1 tx t  When t = 4,     2 16 4 256 or 3.9416 4 1 65x    4(iii) As 216, .16 1 tt t  The particular solution of the DE suggests that the amount of bacteria in the Petri dish will grow indefinitely as time passes. Hence the model is not a realistic one.

2012 NJC H2 Math Prelim P1 Solutions Page 5 of 16 5(i) 5(ii) x = cos 2t, y = tan t d d x t = 2sin2t, d d y t = sec2t d d d d d d y y x x t t = 2sec 2sin 2 t t When t = 3  , P(0.5, 3 ) and d d y x = 2sec 3 22sin 3   =  4 3 So, 30 0.5 b   = 3 4 b = 4.5 5(iii) x = cos 2t = 0  t = 4   y = tan 4  = 1 1x x 0 1 ( 0, 1) y ( 0, –1) 1x x 0 1 ( 0, 1) y ( 0, –1) y = 4

2012 NJC H2 Math Prelim P1 Solutions Page 6 of 16 Area = 4 1 dxy or  =  1tan 4 2 4 cos2 sec dt t t   =    1tan 4 22 4 2cos 1 sec dt t t   =  1tan 4 2 4 2 sec d tt   =    1tan 4 4 2 tantt    = (2tan1 4 4) + ( 2   1) = 3 + 2   2tan1 4 6 1x --- (1) 25x y az   --- (2) Substitute (1) into (2): 25 3 y az y az     Let z  , Hence, 1 3 x ya z                      r , where  is a real number. (Shown) 6(a) The angle between l and 3p is 60 . This implies    2 2 2 2 01 2 11sin 60 1 1 2 1 a a                               2 22 22 2 2 213 2 16 4 4 4 1 18 1 16 16 4 18 18 2 16 14 0 8 7 0 7 1 0 7 or 1 a a a a a a a a aa aa aa aa                            4 1 dxy

2012 NJC H2 Math Prelim P1 Solutions Page 7 of 16 6(b) Since ON is parallel to the normal vector of 3p , 2 2 2 1 61 23 1 2 1 1 ON         1 61 23 6 1 1/ 3 1/ 3 2 / 3 or 2 / 3 1/ 3 1/ 3                          Alternative Method: Let N be the foot of perpendicular from the origin to 3p . Since ON is parallel to the normal vector of 3p , and 2 2 2 2 2 6 3 64 3 66 3 66 9 1 3 ON              , 1/ 3 1/ 3 2 / 3 or 2 / 3 1/ 3 1/ 3 ON                   6(c) Given that 1p , 2p and 3p do not have common point, then line l must be parallel to 3p . Hence 01 20 11 a                 

2012 NJC H2 Math Prelim P1 Solutions Page 8 of 16 2 1 0 1 2 a a     Also, a point (1, 3, 0) in l must not lie in 3p . Hence 11 32 01 7 b b                    7 (a)(i) 2 2 2 2 2 2 3 38 ( )( 8 ) ( 3 ) 9 8 6 9 30 ( 3 ) 0 0 (rej.) or 3 (shown) a d a dr a d a d a d a d a d a ad d a ad d d ad d d a d d a                       Alternatively, let b be the first term of the geometric series. Then 2 2 2 52 2 2 5 5 5 7 2 0 (5 2)( 1) 0 2 or 1 (rej because otherwise 0)5                  b br br brd b br br br rr rr r r d Hence 2 335 ( 8 )5 25 25 25 3 24 3 (shown)         bb d b a d d a d da 7 (a) (ii) 3 4 2 .3 9 10 5     a d a a ar a d a a a Since 2 1,5  the geometric series is convergent.

2012 NJC H2 Math Prelim P1 Solutions Page 9 of 16 8Sum to infinity 1 24 21 5 5 (25 )3 125 3       ad r aa a a 7(b) (i) The distance the mountaineer climbs for each hour follows an arithmetic progression with first term 300 metres and common difference (– 10) metres. Total distance travelled after n hours ≤ x   2 2(300) ( 1)( 10)2 (600 10 10)2 (610 10 )2 (305 5 ) 5 305 (shown) n nx n nx n nx n n x n n x             5, 305pq  7(b) (ii) If x = 2500, then 2 2 5 305 2500 5 305 2500 0 nn nn        9.757n or 51.24n Hence 9.n 8         32 p 3 0 3 3 7 3 15 0 27 9 21 15 0 1 m m m                 8(i)           32 2 7 15 0 3 4 5 0 4 16 4 1 53 or 21 44 2 = 2 i z z z z z z zz                1 2 3 2 i, 2 i, 3z z z     

2012 NJC H2 Math Prelim P1 Solutions Page 10 of 16 8(ii) Since 12 and zz are complex conjugates,     * * 1 1 2 = nnnz z z  . 12 and nnzz are complex conjugates as well. Thus       * 1 2 1 1 12i Im n n n n n z z z z z     which is purely imaginary. Alternative method:                         11 11 12 i tan 0.5 i tan 0.5 i tan 0.5 i tan 0.52 11 2 11 12 2 i 2 i 5e 5e 5 e e cos tan 0.5 isin tan 0.5 5 cos tan 0.5 isin tan 0.5 2 5 isin tan 0.5 nnnn nn n nn n n zz nn nn n                             Since  12Re 0 nnzz  , 12 nnzz  is purely imaginary. 8(iii) Since 1 5,z  3 3z  , 13zz . The locus of complex numbers satisfying the equation wa , for some positive constant a, will not pass through all the points representing the complex numbers