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    