2014 H2 Maths Paper 1 solution vetted Mar 2018

2014 A Levels H2 Mathematics Paper 1 1 1. i 2 1 2 2 ff f f 1 f ( ) f(f( )) 1f 1 1 11 1 1 1 11 1f : 1 , , 1, 0. (Since D D ) R D ( ,0) (0,1) (1, ) Let f ( ) . 1 1 (1 ) 1 1 1 1 1 = 1 1 f : 1 , , 1, 0. Henc xx x x x x x x x x x x x x xy y x yx y xy xy y yx y y x x x x x                                       21e f ( ) f ( ).xx  ii 21 31 Since f ( ) f ( ), f ( ) f (f ( )) xx xx x     

2014 A Levels H2 Mathematics Paper 1 2 2 22 22 22 22 3 3 3 3 3 3 : 54 0 Differentiate w.r.t dd2 2 0 dd dWhen 1,d 2 2 0 Substitute into Substitute into 54 0 54 0 2 54 54 0 (N.A) 27 3 Coordinates of C x y xy x yyxy x y xy xx y x xy x y xy xy xy x y C x y C y y y y y y y                            the point at which the gradient is 1is ( 3, 3) Hence there is only one such point.    3 i a and b are parallel, or a or b is a zero vector. ii 2 2 2 1 Since 2 2 1 is parallel to the direction vector 2 . 2 1 1ˆ 2 1 2 ( 2) 2 1 1 23 2               n0 n n iii Let  be the required angle. 10 20 21cos 91 2 3                   

2014 A Levels H2 Mathematics Paper 1 3 4 Out of syllabus 5 i 2 33 1 2i (1 2 )(1 2 ) 1 4 4 34 11 (1 2i) 1 11 2 11 2 11 2 11 2 121 4 11 2 125 11 2 125 125 z z i i i i z i ii i i i                          ii From i   2 3 2 3 2 3 11 2( 3 4 ) ( ) 125 125 11 2 3 (4 )i 125 125 Since is real, 2(4 ) 0 125 250 113 250125 3 22 = 19 qpz p i q iz p q p q qpz z pq qp qpz p pz pp p                     

2014 A Levels H2 Mathematics Paper 1 4 6 ai Out of syllabus a ii     11 11 1 743 71 4 33 7 4 4 1 3 3 3 74 1439 nn r r rr nn r rr n n p n n             b i As n , S = approaches to 1, hence the seri es converges.n r r uu  S 1   b ii un = Sn - Sn-1 = 1- 1 (n +1)! - 1- 1 (n)! æ èç ö ø÷ = 1 (n)! - 1 (n +1)! = n +1-1 (n +1)! = n (n +1)! 7 i f(a ) = 0 a 6 - 3a 4 - 7 = 0 Using GC, a = 1.88524 » 1.885 (to 3.d.p) f(b) = -7 b 6 - 3b 4 - 7 = -7 b 6 - 3b 4 = 0 b 4(b 2 - 3) = 0 b = 0 or b 2 - 3 = 0 b = ± 3 Since b > 0, b = 3.

2014 A Levels H2 Mathematics Paper 1 5 ii 1.88524 3 1.88524 64 3 f ( ) ( 3 7) d 0.597272 (Using GC) 0.597 (to 3.d.p) x x x x       iii    3 0 3 64 0 3 64 0 375 0 75 22 77 22 7 2 3 2 ( 7 f ( ))d 7 ( 3 7) d 3d 3 75 3 3(3 ) 075 33 75 23 35 233 35 54 3 units35 xx x x x x x x xx                                iv 64 64 f ( ) ( ) 3( ) 7 3 7 f ( ) (shown) x x x xx x          

2014 A Levels H2 Mathematics Paper 1 6 Since f(x) = f(-x), the graph is symmetrical about the y-axis. From the graph above, it can be seen that the graph cuts the x-axis at 2 points; hence it has 2 real roots. Since the coefficients of the f(x) are all real, the equation f(x) = 0 has 2 real roots and 2 pairs of conjugate roots. In addition, since f(x) = f(-x), the roots form pairs within each which one root is the negative of the other. The four non-real roots are in the form iab , iab , iab and iab . 8 i 2 1 1 d 9 sin , where is an arbitrary constant3 x x x CC     ii (9 - x2) -1 2 = 9 -1 2 1- x2 9 æ èç ö ø÷ -1 2 = 1 3 1+ - 1 2 æ èç ö ø÷ - x2 9 æ èç ö ø÷ + - 1 2 æ èç ö ø÷ - 3 2 æ èç ö ø÷ 2! - x2 9 æ èç ö ø÷ 2 + - 1 2 æ èç ö ø÷ - 3 2 æ èç ö ø÷ - 5 2 æ èç ö ø÷ 3! - x2 9 æ èç ö ø÷ 3 + ... æ è ç ç ç ç ö ø ÷ ÷ ÷ ÷ = 1 3 1+ x2 18 + x4 216 + 15x6 34992 + ...æ èç ö ø÷ » 1 3 + x2 54 + x4 648 + 5x6 34992 iii From i,

2014 A Levels H2 Mathematics Paper 1 7 1 2 2 4 6 3 5 7 1 sin d 3 9 15 d3 54 648 34992 15 3 162 3240 244944 x x x x x x x x x xxc                Sub x=0: LHS=0; RHS=c Thus, c=0 3 5 7 1 5sin 3 3 162 3240 244944 x x x x x      9 i p 1 : 2 12 3      r Since p and q are perpendicular, the normal of p is parallel to q . Finding the vector equation of : 1 1 3Let , where 2 1 4 12 1 3 4 12 Hence : 1 1 , 34 l x y z x y z l                                       r Since q contains line l and the normal of p is parallel to q ,

2014 A Levels H2 Mathematics Paper 1 8 21 12 43 5 10 5 1 52 1 1 1 1 : 2 1 2 0 1 3 1 : 2 0 q q q n x y z                                                              r ii Using GC to obtain the line 64 : 3 1 , 02 m m                     r iii Since B is a general point on m,

2014 A Levels H2 Mathematics Paper 1 9   2 22 2 2 2 64 3 1 for some 02 6 4 1 3 1 23 54 = 4 23 5 4 (4 ) (2 3) 25 40 16 16 8 4 12 9 50 OB AB OB OA AB                                                                             2 2 2 36 21 50 36 21 Hence method: Let = Differentiate with respect to 2 36 42 when 0, 36 42 0 6 7 6 4 18 61Hence 3 1 15 . 770 2 12 AB y AB y dyy d dy d OB                                              

2014 A Levels H2 Mathematics Paper 1 10 iii (Otherwise Method) must be perpendicular to the line as i t is the shortest distance between and . 54 4 (see above) 23 0 5 4 4 4 1 0 2 3 2 2 AB m A B AB AB m                                0 16 4 4 6 0 18 21 0 6 7 6 4 18 61Hence 3 1 15 . 770 2 12 w OB                                           10 i 2d (1 )dt 1 d 1When and ,2 dt 4 1 1 1 14 2 4 1 (shown)5 x k x x xx k k          