MJC H2 MATH P2 Solution

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 1 of 16 2015 H2 MATH (9740/02) JC 2 PRELIM EXAMINATION – MARKING SCHEME Qn Solution 1 Mathematical Induction 1(i) 2 1 3 2 2 4 3 3 213. 21 2 7 2 13. 3 3 2 1 2 15 2 13. 7 7 2 1 3 u u u             Hence 2a . 1(ii) Let 1 21P be the statement that for 1.21 n nn nun    1 2 When 1, LHS = 3 2 1 3RHS = 3 LHS (shown).2 1 1 n u      1P is true. Assume 121P is true some ,i.e., . 21 k kk kku     To prove 2 11 1 21P is also true,i.e. . 21 k kk ku        1 1 1 11 1 1 1 2 1 LHS 23 2= 3 21 21 2 2 1 3 21 3.2 3 2 2 21 2.2 1 21 21 21 RHS k k k k k k kk k k k k k u u                            1 P is true P is true.kk  1 1 n Since P is true, and P is true P is true, by Mathematical Induction, P is true for all . kk n    

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 2 of 16 1(iii) 2 3 4 1 1 2 3 23 1 1 2 1 2 1 2 1 2 1... ... 2 1 2 1 2 1 2 1 21 21 2 1 as . n n n n n u u u u n                                 Hence, the limit does not exist.

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 3 of 16 Qn Solution 2 Differentiation (Maxima and Minima) 2(i) Base of isosceles triangle 10 2 x 2 20yh By Pythagoras Theorem,   222 2 2 2 2 5 25 10 25 10 x h x x h x x hx                    2 2 32 4 3 2 1 10 22 1 2510 2 20 22 10 1 10 255 1 10 25 2505 1 10 25 250 (shown)5 V x hy h hh h h h h h h h h h h h                   (ii) Differentiate wrt x,   32d1 4 30 50 250d5 V h h hh     For maximum V, d 0.d V h    32 32 d1 4 30 50 250 0d5 4 30 50 250 0 V h h hh h h h           Using GC, 8.0902 cm (rejected 2 18.090 10), 3.0902 cm (rejected 0) or 2.5 cm. hx hh h       Differentiate wrt x,   2 2 2 d1 12 60 50d5 V hhh    For      2 2 2 d12.5 cm, 12 2.5 60 2.5 50d5 Vh h    25 0 is maximum.V   Alternatively Using GC, 2 2 2.5 d 25 0d h V h    is maximumV .

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 4 of 16 Alternative (By 1st derivative test) h 2.5 2.5 2.5 d d V h is maximum.V          4 3 2 3 1 maximum 2.5 10 2.5 25 2.5 250 2.55 70.3125 cm V     

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 5 of 16 Qn Solution 3 Complex 1 & 2 (ai) Method 1: Since 1 2i is a root to 32 50x ax bx    ,             321 2i 1 2i 1 2i 5 0 11 2i 3 4i 1 2i 5 0 ab ab                 Comparing real part, 11 3 5 0 3 16 (1) ab ab          Comparing imaginary part, 2 4 2 0 4 2 2 (2) ab ab        Solving (1) and (2), 3, 7ab  . 32 3 7 5 0x x x    Using GC, the other roots are 1 2i and 1 . Method 2: Since the coefficients of the polynomial are all real, 1 2i is a root implies 1 2i is also a root.            22 2 1 2i 1 2i 1 2i 25 x x x xx           By comparing coefficients,   3 2 2 5 2 5 1x ax bx x x x       By comparing coefficient of 2x , 1 2 3a   By comparing coefficient of x , 2 5 7b   Therefore, the other roots are 1 2i and 1 . (bi)             21 44 11 424 33 111 1 1 1 44 444 4 4 4 4i i i ii ii 2e 2e 2 e 2, 1,0,1 2 e , 2 e , 2 e , 2 e k k z z k                (ii) i 2ew   or iw

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 6 of 16 (iii) Hence method:    1 4 2 2 3 3 4 2 2 * 22 42 wz w z z z     Otherwise method:       1 4 1 2 22 3 3 3 2 2 3 2 ** 2i 42 4 2 4 2 wz w z z w w z       B1 B1 B1 B1

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 7 of 16 Qn Solution 4 Vectors 2 & 3 (i) 11 10 11 2cos 3 2 6 0.615rad or 35.3                     (ii) 1 2 x y z xz     By GC, 21 : 1 0 , 01 l                      r (iii) 11 1 1 1: 2 1 22 2 2 k k k kp x y z                         Since pk tends to q as k and p is 1p , the limit of the acute angle is 0.615rad. (iv) 1 1 1 10 1 1 021 1 k            Thus pk is parallel to l for any k  . 11 1 2 1112 220 1 kk                     Thus the point  2, 1,0 lies in pk for any k  . l lies in all planes in P. (v) Since ac and 1 00 1 a b a c c                     ,  is not parallel to l,  and any two planes in P must intersect at only 1 point which lies on l. (vi) 1 1:1 2 1      r Since  is parallel to the p2 in P, and  2, 1,0 lies in p2, the perpendicular distance is also the shortest distance between  and p2. 3 1 12required distance 3111 4   

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 8 of 16 Alternatively Note that (1,0,0) lies in  . 121 110 200 1 1required distance 3111 4                        

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 9 of 16 Qn Solution 5 Permutation and Combination 5(a) Number of arrangements = 5! 4! 2880 (b)(i) Number of arrangements = 9 8 8! 453608C  (ii) Number of arrangements = 7 6 6! 6 5 252006C     Alternatively: Number of arrangements = 76 62 6! 2! 252006CC   

MJC/2015 JC2 Prelim Examination Marking Scheme/H2 Math (9740/02)/Math Dept Page 10 of 16 Qn Solution 6 Binomial Distribution (a) Let X be the number of students, out of 10, who score distinction  ~ B 10,0.01Xp                   0 10 1 9 10 9 10 9 P( 2) 0.95 1 P( 0) P( 1) 0.95 10 101 0.01 1 0.01 0.01 1 0.01 0.9501 1 1 0.01 10 0.01 1 0.01 0.95 1 0.01 10 0.01 1 0.01 0.05 Using GC, 39.416 39.4 X XX p p p p p p p p p p p                               (b) Let W be the number of students, out of 10, who score distinction  ~ B 10,0.4W Since n = 50 is large, by Central limit theorem, 2.4~ N 4, approximately50W    P 3.5 0.011239 0.0112W    Alternatively: (not for students) Let Y be the number of students, out of 500 , who score distinction  ~ B 500,0.4Y    P 3.5 50 P 175 P( 174) 0.0095558 0.00956 (3 s.f ) YY Y      