DHS H2 MATHS P2 Solution

1 2014 Year 6 Prelim Examination Paper 2 Suggested Solution Qn Suggested Solution 1(i) Let P(n) be the proposition:     22 1 11 1 2 112 1 n r rr n n rr n    , n  . LHS of P(1) =   33 12 2  RHS of P(1) =   2 11 1 2 121 1   3 2 LHS of P(1) Hence P(1) is true. Assume P(k) is true for some k  , i.e.     22 1 11 1 2 112 1 k r rr k k rr k    To show P(k + 1) is true, i.e.      221 1 11 1 2 2 112 2 k r rr k k rr k       LHS of P(k + 1) =   2 12 121 kk k   +     2 12 1 12 kk kk    =        22 12 2 2 2 12 2 121 2 kk k k k k kk        =     22 12 2 2 121 2 kk k kk     =    2 12 2 122 kk k    = RHS of P(k + 1) Hence P(k) is true P(k+1) is true. Since P(1) is true and P(k) is true P(k+1) is true, by mathematical induction,     22 1 11 1 2 112 1 n r rr n n rr n    for n  .

2 1(ii)    2 11 1 1 1 11 1 rr rrrr rr r r    By cover-up rule, 1, 1AB    2 11 11 1 1i.e. 11 nn rr rr rrr r r        2 11 1 1 1 11 1 1 11 11 2 11 23 11 1 11 1 nn n rr r n r n r rr rrr r r r nn r n                                  2 11 1 2 11 1 1 11 12 1 1121 1 11 1 1121 1 1 2 nn n rr r rrr rr r r nn nn nn nn nn                      

3 Qn Suggested Solution 2(i) 2 32 1 sin (3 )23 33 34 kx h x x xh x   3 22 13 3 3 34 3 4 khk x xxx     22 22 22 2 22 2 22 12s i n2 ( 3 ) 2 ( ) 2 ( 3 )23 3 682 33 6823 4 38 62 323 33 86 ( s h o w n )23 A xx h x h x xxh x kxxx x x kxx x x kxx x               2 d8 12 3 3d3 A kxxx x   For stationary values, 2 d8 12 3 3 0d3 Ak xxx x     2 3 3 1 3 812 3 3 0 3 9438 8 94 3 8 94 3 kxx x xk kx kx          2 23 d1 6 12 3 3d3 Ak x x  Since 3 0x  , 0k  , 12 3 3 0

4 Alternative   2 23 d1 6 12 3 3d3 94 31612 3 3 38 12 3 3 6 4 3 36 9 3 0 Ak xx k k             area A is a minimum. 2(ii) Using 360 and 300,kA 22 33 3 3 3 8(360)300 6 23 1800 36 9 3 5760 88.185 282.19 0 From GC, since 0, 3.8442 or 6.8587 xx x xx x xx x xx         2 2 When 3.8442, 360 3 (3.8442) 6.46.3(3.8442) 4 When 6.8587, 360 3 (6.8587) 0.419 (rej. 0)3(6.8587) 4 x h x hh          3.84, 6.46.xh 

5 Qn Suggested Solution 3(i) For 42 22 0ww  , 2 2 ( 2) ( 2) 4(1)(2) 1i2w      πi2 4 πi2 π4 1 i = 2e = 2e k w    π1 i π84 7 ππ11 ii 8844 7 π1 i 84 2 e , 0, 1 2e , 2e Since the coefficients of the equation are real, the conjugates of the above roots of are also roots for the equation. 2e k wk w w w                  ππ 7 π11 1ii i88 844 4,2 e ,2 e ,2 e              Alternatively For 42 22 0ww  , 2 2 ( 2) ( 2) 4(1)(2) 1i2w      22 ππii44 ππi2 π i2 π44 π1 i π84 1 i or 1 i = 2e = 2e = 2e = 2e 2 e , kk k ww w                π1 i π84 7ππ π 7π11 1 1ii i i 88 8 844 4 4 0, 1 or 2 e , 0,1 2 e ,2 e ,2 e ,2 e k kwk w                          [Note: Students ought to recognize that roots of w must be in conjugate pairs due to real coefficients in the equation.] 3(ii) [Note: Students ought to recognize that since the roots of w are in conjugate pairs, they will be reflections of one another in the x-axis.] Im π /8   1 42   π /8   Re π 8   w1 w4 w3 w2 Locus of z for (iii)

6 3(iii) Since the locus is a half-line from the point representing 1w and angled at π 8 from the positive Re-axis direction, it will pass through the origin and eventually the point representing 3w since  3arg π 8w      Alternative Method        i π 814 14 i π 8i 7 π 814 arg(2 e 2 cos 7 π 8i s i n7 π 8) =arg 2 e e              i π/8 i π14 i π/814 i π/854 =arg 2 e 1 e =arg 2 e 1 ( 1) =arg 2 e π 8      Alternatively, substitute 3zw into LHS of 1 πarg( ) = 8zw      31 11 3 1 1 1 arg arg (Since , rotation of π about origin) arg 2 arg 2 arg 7ππ 8 π 8 ww ww w w w w             *Locus drawn in (ii)               Cartesian Equation of locus: 1/4π 7πtan , 2 cos88yx x         

7 Qn Suggested Solution 4(i) 21 :10 1 lr a         Let the acute angle between l and p1 be .  13 01 12 1cos 90 sin 21 4 2 8 10.9                

4(ii) Method 1  0,5,0 is a point on p1. Perpendicular distance from the point A to p1 20 3 25 1 00 2 14 23 31 02 3 14 14                

Method 2 Perpendicular distance from the point A to p1 23 52 1 02 14 58 3 14 14   

Method 3 A  O   p1

8 Let F be the foot of perpendicular and it lies on both lAF and p1. 23 :2 1 , 02 AFlr          23 3 21 1 5 02 2 3 14              

23 321 1402 33 114 2 333 1 4114 142 OF AF OF OA AF                            4(iii) 21 :10 1 lr a         To find b: Method 1 Direction vector of l is perpendicular to normal vector of p3, 11 01 0 1 0 1 1 bb b         

Method 2 11 1 41 0 11 14 1 10 31 3, 1 k b b bk kb                        To find c:

9 Method 1 Since l lies on p2, 21 14 6 1a        241 6 0 aa  Since l lies on p3, 21 11 01 c         21 0 3 bc c     Method 2 Since l lies on p2, 21 14 6 1 66 0 a aa               Since l lies on p3,   21 11 31 0 c b cb                 Since the equation is always true regardless of , 30 3 & 10 1 cc bb   4(iv) Let M be the midpoint of A and B. 21 222 0 OA OBOM OB          231 21 52 02 23 3 21 1 1 0 02 2 3 11 0 8 2 2 A cartesian equation for the locus of is 32 2 . OB OB OB B xy z                    





10 Note: Locus is a plane parallel to p1