A Level 2016 P1 Solutions for new syllabus (edited CSC)

2016 A-Level H2 Maths Paper 1 Solutions 1     22 22 2 4 4 14 4 4 14 ( 4)( 3)344 4 4 14 12 4 3 5 2 4 (3 1)( 2) 4 x x x x x x xxx x x x x x xx x xx x                            2 2 4 4 14 3, 44 4 4 14 304 3 1 2 04 3 1 2 4 0, 4 xx xxx xx xx xx x x x x x                1 32 or 4xx   2(i) cos2 xy When x = 0, d 0d y x  When 2 d, 0.693 (3 s.f.)d yx x   2(ii)     cos0 cos 02 2 22 When 0, 2 2 Tangent at (0,2): 2 (1) When , 2 2 1 Tangent at ,1 : 1 0.693 0.693 2.09 (2) Substitute (1) into (2): 2 0.693 2.09 0.128 The tangents meet at (0.128, 2 xy y xy yx yx x x                         ). -2 x 4 1/3

2016 A-Level H2 Maths Paper 1 Solutions 3 44 f ( ) () x y x y k x l m     By the way (question doesn’t ask for it), this is the sequence of transformations: units44 in the positive -direction units44 in the positive -direction translate translatescale by factor parallel to -axis () ( ) ( ) l x m y k y y x y x l y k x l y k x l m               (0,0) ( , ) lm ( , ) ( , )l m a b , so la , mb . (0, )c lies on f ( )yx graph 4(0 )c k l m    44 c m c bk la  4(i) Given that the for the Arithmetic series: first term = a, common difference = d, Geometric series: first term = b , common ratio, r, ,, ,a b d r are non-zero, 4 7 14 74 43 14 7 77 77 43 37 3 3 10 3 10 3 3 (1) 8 (2) 11 (3) (2) (1) : 5 5 ( 1) (4) (3) (2): 3 3 ( 1) (5) (5) 3 ( 1):(4) 5 ( 1) 3 ( 1) 5 ( 1) 3 3 5 5 5 8 3 0 a d br a d br a d br d br br d br r d br br d br r br r br r rr r r r r rr                                Using GC, 0.74 or 1 (reject since <1)r r r Aim: Eliminate a, then eliminate d and b, leaving r. 4yx ( , )ab (0, )c 1( , )ba 1(0, )c 1 f ( )y x f ( )yx

2016 A-Level H2 Maths Paper 1 Solutions 4(ii) Required sum 12nnuu   1 2 1 2( ) ( ) (1 ) 11 1 (0.74) or 3.85 (0.74)0.26 n n n n n n u u u u u SS b b r rr br r b b              5(i)     2 1 , 0 2 22 1 , 1 22 2 2 2 1 1 4 4 2 2 2 a b aa bb a a b ba b b a                                                                               uv u v u v u v u v 5(ii) As the i- and k- components are equal, 22ba i.e. ba     21 8 2 4 21 a aa a                        u v u v    u v u v is a vector of length 1, given that it is a unit vector. 2 2 2 11 2 18 6 2 21 1262 1 2 4 1 1 1 2 4 1 1 2 1 4 1 1 2 18 1 (or ) a a a a a a               As usual, we do a dot product check to confirm this vector’s correctness. We can read it again: Yes, it’s i and k, not i and j.

2016 A-Level H2 Maths Paper 1 Solutions 5(iii)     22 22 2 2 2 0 0 0 2 ( 1) 2 9 3             u v . u v u.u v.v uv vu v 7(a) 2( 1 5i) ( 1 8i)( 1 5i) ( 17 7i) 1 10i 25 1 3i 40 17 7i = 10i 10i 23 23 0                        21 5i is a root of the equation ( 1 8i) ( 17 7 i) 0.ww         Let the second root be ipq .   Sum of roots 1 8i 1 5i ( i) 1 8i i 2 3i pq pq           7(b) As all the coefficients of the equation are real, 1i a is a root  1i a is a root. Let the third root be m .    32 22 5 16 (1 i) (1 i) ( ) = 2 (1 ) ( ) z z z k z a z a z m z z a z m               Comparing coefficients of z2 : 5 2 3mm     z : 2216 1 2 1 6a m a      2 9 3 ( 0)a a a    z0: 22(1 ) 3(1 3 ) 30k m a     8(i) 2 2 22 2 2 3 2 2 3 2 2 Given f ( ) tan( ) (1) f '( ) sec ( ) = 1 tan ( ) = (1 ) (shown!) f ''( ) 2 .f '( ) 2 ( ) 2 ( ) f '''( ) 2 (1 3 ).f '( ) 2 (1 )(1 3 ) y x ax b x a ax b a ax b a y a ay x ay x ay a ay a y y x a y x a y y                    8(ii) Substitute 4b  into (1):  4f ( ) tany x ax    2 2 3 2 3 2 2 3 2 2 3 3 23 4 2! 3! 8 3 f (0) tan 1 f '(0) (1) 2 f ''(0) 2 (1 1 ) 4 f '''(0) 2 (1 1 )[1 3(1 )] 16 f ( ) f (0) f '(0) f ''(0) f '''(0) 1 2 2 xx a a a aa aa xx ax a x a x                      If we try to use sum and product of roots (i.e. similar method to (a)), we find it doesn’t work as well this time – there are 3 unknowns to be found.

2016 A-Level H2 Maths Paper 1 Solutions 8(iii) Substitute 2 and 0ab into (1): f ( ) tan(2 )y x x 2 23 3 2 2 f (0) 0 f '(0) 2 2(0 ) 2 f ''(0) 2(2 )(0 0 ) 0 f '''(0) 2(2 )(1 0 )[1 3(0 )] 16              33 8 3! 3tan 2 2 16 2 xx x x x     9(i)(a) Let x be the distance the stone has fallen through the water at t seconds. Let d (1)d xy t  2 2 dd (2)dd yx tt   Substitute (1) & (2) into the given D.E.: dd 2 10 10 2 (shown!)dd yy yytt     9(i)(b) 2 2 2 2 1 2 1 d 1 d10 2 ln 10 2 where is an arbitrary constant ln 10 2 2 2 10 2 e e where et C t C yty y t C C y t C y B B                  2 2 2 2 2 5 2 5 2 55 22 dWhen 0, 0 10 d 10 2 10e 5 5e d 5 5ed 5 e where is an arbitrary constant When 0, 0 5e t t t t t xt y B t y y x t x t A A t x A xt                         9(ii) 2 2 2 1 2 1 2 1 2 d 10 5sind d 10 10cos (1)d 5 20sin (2) ( , are arbitrary constants) x tt x t t Ct x t t Ct D C D           Sub d0, 0d xt t into (1): 10C Sub 0, 0tx into (2): 0D 2 1 25 20sin 10x t t t    9(iii) The stone reaches the bottom of the pond when 5x . For model in (i): 255 225 5 e 1.47 (2 d.p.)ttt      For model in (ii): 2 1 25 5 20sin 10 1.05 (2 d.p.)t t t t     Just as we made use of part (i) to do part (ii), we should ask ourselves if we can make use of earlier parts to do part (iii). Did you? Don’t. Waste. Time. Be methodical: Integrate twice first, then sub in the initial conditions.

2016 A-Level H2 Maths Paper 1 Solutions 10(a)(i) Let f ( ) 1 , for , 0y x x x x     . 12 12 1 f ( ) ( 1) f ( ) ( 1) , , 1 xy y x y x x x x           10(a)(ii) If ff ( )xx , then f (1 ) xx . 22 2 2 2 4 3 2 1 1 (1) 11 1 ( 1) 2 1 2 ( 2 ) 4 4 xx xx x x x x x x x x x x x x x                    4 3 24 4 0 (shown!)x x x x     By GC, 0, 0.383,1 or 2.62x . Substitute each value into (1) to check: When 0, 0.383,1x , LHS RHS .  They do not satisfy ff ( )xx . (Note: LHS > 1 for all x ) When 2.62x , LHS = RHS.  2.62x is the only solution to ff ( )xx . As 2.62x is inside both fD and 1fD  , it satisfies -1f ( ) f ( )xx . 10(b)(i) For a start… As usual, we check if any value(s) needs to be rejected. Also, how many values is the question looking for? A good approach for this question is to draw a table like in the Sec Sch days. This question is a test of perseverance & focus 

2016 A-Level H2 Maths Paper 1 Solutions n 0 1 2 3 4 5 6 7 8 9 10 11 12 g(n) 1 2 4 At the end: n 0 1 2 3 4 5 6 7 8 9 10 11 12 g(n) 1 2 4 5 6 7 7 8 8 9 9 10 9 g(4) = 6, g(7) = 8, g(12) = 9 10(b)(ii ) g(5) = g(6) = 7, so g is not one-one and 1g does not exist. 11(i)(a ) When a = 0, 1 1 0 : 3 2 4 , , 2 0 2 12 : 0 1 , 12 p l t t                                                        r r As 21 1 2 2 2 0 20                     and 20 1 4 4 4 0 22                     , l is perpendicul