A Level 2012 P1 Solutions for new syllabus (edited EKS)

2012 A-Level H2 Math (9740) P1 1 Let the ticket prices for “under 16 years”, “between 16 and 65 years” and “above 65 years” be x, y and z respectively. 9 6 4 162.03 7 5 3 128.36 10 4 5 158.50 x y z x y z x y z          Using GC, $7.65, $9.85, $8.52.x y z   2(i)   3 3 4 4 4 1 4d d1 4 1 1 ln 1 , where is an arbitrary constant4 x x x xx x x C C        (ii) 1 2 2 1 2d 1 d 2 u x x u x uu        1 2 4 2 1 12 2 2 2 1 1 2 dd d1 1 d 1 d1 2 1 1 d2 1 1 tan2 1 tan , where is an arbitrary constant2 x u xx ux u u u u uu uu u C x C C                           (iii) 2 1 40 d 0.186 (3 d.p.)1 x xx    

4(i)       3 4 4 1 4 42 cos sin cos sin 2 2 2 by sine rulesin sin( ) 1 sin cos cos sin 1 1 22 cos sin 1 (shown!)cos sin AC AB AC AC AC                           (ii) Given that  is sufficiently small, 2 112 2 22 2 2 2 2 sin , cos 1 2 1 1 2 2 1 2 2 1 2 31 2 AC                                                         

5(i) ) ) 1Area of triangle 126 2 1 ( 1262 2 126 ( 2 126 2 126 ( 0 1 1 1 1 2 1 0 2 1 ) 3 4 2 126 6 OAC OA OC                                                 a a b a b a a 0       (ii) When 4  , 1 1 1 4 4 82 1 0 OC                                  Given 5 3OC  ,     2 2 2 2 4 8 5 3 3 8 80 75 (3 5)( 1) 0 5 or 13                       17 19 5, ,3 3 3C    or  5, 7,1C 6(i)         2 33 2 3 1 i 1 3 i 3 i i 1 3 i 3 z c z c c c c c c           (ii) Given 3z is real, 3 2 3 0 (3 ) 0 c c c c     0 (rejected 0) or 3 1 3i c c c z        (iii) Given 0c , 2 2 2 2 ( 8 16) ( 16 64) 3 8 80                

i 3 i3 1 3i 2e 2 e n n n z z         1000 2 1000 Smallest 10 n nz n       10 10 10 2 1024 10 2arg 3 3 z z       7(i) Let g( )y x     1 , , 11 ( 1) 1 g g 1 x ky x xx xy y x k xy x y k x y y k y kx y x kx x x                      g is self-inverse (shown!). (ii)

(iii) A line of symmetry is y x . 111 1x k ky x x     1 0 0 1 Scaling by factor 1 parallel to the -axi Translate Translate s 1 1 1 1 11 1 1 ky k ky y y yx x x x                          8(i)   2 (0)x y x y    Differentiate (0) w.r.t. x:     d d1 2 1d d d1 2 2 2 2 1 d d 1 2 2 d 2 2 1 d (2 2 1) (1 2 2 )1 d 2 2 1 2 (shown!) (1)2 2 1 y y x yx x yx y x y x y x y x x y y x y x y x x y x y                             (ii) Differentiate (1) w.r.t. x:     2 22 2 2 3 d 2 d 2 2d d 2 2 1 1 4 d 2 12 d 2 2 1 d d1 1 d d d1 (2)d y y x x x y y xx y y y x x y x                                    (iii) At the turning point, d 0dyx  . From (2), 2 2 d 1 0d yx    The turning point is a maximum point.

9(i) 8 1 16 8 2 8 1 7 1 Line : 8 2 , 9 1 AB OB OA AB                                          r        (ii) 0CN AB CN AB          0 7 1 1 1 8 2 8 2 0 ( is on line ) 9 1 3 1 6 1 1 0 2 2 0 6 1 1 12 6 0 2 ON OC AB N AB                                                                                                7 1 5 8 ( 2) 2 4 9 1 7 ON                                 2 4 2 NA OA ON                while 612 36BN ON OB NA                  : 1: 3AN NB  (iii) The reflection of line AC in line AB is line 'AC . By observation, '(9,0,11)C , A(7,8,9) B(–1,–8,1) C(1,8,3) N

so 2 1 ' ' 8 2 4 2 1 AC OC OA                            . 7 1 Line ': 8 4 , 9 1 AC t t                    r  A Cartesian equation is 7 8 91 4 1x y z    . 10(i) Let V and S be the volume and surface area of the model respectively. Given V k , 3 2 3 2 2 2 3 2 3 2 (1)3 r r h k rr h k k rh r            2 2 2 2 2 3 2 (2) S r r rh r rh            Sub (1) into (2): 2 2 2 2 2 23 2 3 2 43 3 5 2 (5)3 k rS r r r k rr r r k r                   Given S is minimum, d 0dSr  2 2 3 10 2 03 10 2 3 3 (3)5 r k r r k r kr         From (1), 3 2 (4)3h kr r    Sub (3) into (4):  35 2 5 2 13 3 3kh kr      1 33 5 kh r        (from (3)) Here I’m using the method of finding hr first (i.e. the ratio of h to r) instead of finding h directly. It works well here as we see the RHS of (4) has just one term in r – and it has 3r , which links well with (3).

(ii) Sub 200k  and 180S  into (5): 2 3 5 2(200)180 3 5180 400 3 r r rr       35 180 400 03 r r    6.758 (rejected as 0), 3.722 or 3.037r r  When 3.722, 2.1156r h  , so h r . When 3.037, 4.8775r h  , so r h . 3.04, 4.88r h   11(i) d 1 cosdx    , d sindy   2 d d d d d d sin 1 cos 2sin cos2 2 2sin 2 cot (shown!)2 y y x x               As 0  and 2  , ddyx  .  The tangents approach vertical lines. (ii) [Note: Graph is vertical at 0, 2  and horizontal at   .] (iii) Required area 2 0 2 0 d d dd y x xy            When 0, 0 ;2 , 2 .xx      x y C

   2 0 2 2 0 2 0 2 0 2 0 2 1 cos 1 cos d 1 2cos cos d 1 cos 21 2cos d 2 3 1 2cos cos 2 d2 2 3 1 2sin sin 22 4 3 units                                            (iv) Gradient of pn tan 2p   2 2 2 2 2 2 2 2 2 2 0 (1 cos ) tan ( sin ) 2 cos 1 tan ( sin ) 1 2sin 1 tan ( sin ) sin2sin ( sin ) cos 2sin cos sin sin sin sin sin p p p p p p p p p p a p p p a p p a p p a p p a p p p a p p a p p p p                                 pn crosses the x-axis at ( ,0)A p (shown!) x A double dose of double-angle formula is needed for this method. y C ( sin ,1 cos )P p p p  pn ( ,0)A a