NJC_H2_MATHS_P1_Solutions

Suggested Solution to 2015 SH2 H2 Mathematics Preliminary Examination Paper 1 1 Let PN be the statement 11 0 3 2 1 1( 1)!3 ( 1)!3 N nN n n nN     for all N ≥ 0. When 0,N  0 1 0 1 0 01 3 2 3(0) 2 2LHS ( 1)!3 (0 1)!3 3 12RHS 1 LHS(0 1)!3 3 n n n n             Thus, 0P is true. Assume Pk is true for some 0k , i.e. 11 0 3 2 1 1( 1)!3 ( 1)!3 k nk n n nk     . Consider 1Pk : To show 1 12 0 3 2 1 1( 1)!3 ( 2)!3 k nk n n nk      1 1 1 0 12 0 12 12 22 32LHS of P ( 1)!3 3 2 3( 1) 2 ( 1)!3 ( 2)!3 1 3 5 1 ( 1)!3 ( 2)!3 ( 2)(3) 3 5 1 ( 1)!( 2)3 3 ( 2)!3 3 6 3 51 ( 2)!3 ( 2)!3 1 k k n n k nk n kk kk kk n n nk nk k kk kk k k k kk kk                                  2 12 3 5 3 6 ( 2)!3 1 1 RHS of P( 2)!3 k kk kk k k           1P is true P is truekk  . Since 0P is true, and 1P is true P is truekk  , by mathematical induction, PN is true for all N ≥ 0.

2 (i) 2 (ii)

3 (i) Method 1:                  12 2 24 24 4 2 2 4 4 2 4 2 4 4 2 2 4 4 4 2 cos 2 cos 2 11 221 ... 1 ...2 4! 21 2 ... 1 ... 3 21 2 2 ... 3 2or 1 2 2 ... 3 1 3 x xxx xx xx xx x x xx x x x xx x x x xx                                   3 (ii) 24 2 1cos 2 3 1 1 1 1 1 21511 3 3 3 9 243 24311 3 2 215 8cos 3 243 9 1720 2187                     

4 (i) Either a = 0 OR b = 0 OR a and b are perpendicular to each other. 4 (ii) Projection vector of a – b onto a     2 2 ˆˆ() ˆ ˆ ˆ ˆ0 ( 0 since 0) (since )                     a b a a a a b a a aaa b a b aaa aa a a a a a a Alternatively, From the diagram, since ,ab the projection vector of vector of a – b onto a is a itself. 4 (iii) Method 1: Geometrical Definition of Cross Product Since ,ab the angle between a and b is 90°. Hence,     ˆsin 90 ˆsin 90 ( 1) (1) (shown)       ab a b n a b n ab ab ab Method 2: Area of Triangle or Rectangle

Area of triangle with sides and 1 2 11 22 (shown)    ab a b a b a b a b ab Method 3: Length of Perpendicular Component Since ,ab the component of a perpendicular to b is a itself. Hence, ˆ (shown)      a b a baa b ab ab a b a b

5 (i)                      11 1 1 1 2 1 2 3 4 2 1 4 2 3 1 1 1 4 1 5 11 37 11 59 ... 11 2 2 1 2 2 3 11 2 1 1 2 1 3 11 2 1 2 3 1 1 1 114 3 2 1 2 3 1 4 2 3 2 1 4 3 2 1 2 3 11 3 2 1 2 3 nn rr r r r r nn nn nn nn nn nn n nn                                        5 (ii) (a)               5 4 11 1 2 1 2 3 11 2 1 2 3 2 1 2 3 1 1 1 5lim 3 2 1 2 3 3 (9) 11 1 1 503 3 99 5 99 r rr n rr r r r r n nn                           