AJC H2 MATH P1 soln

Anderson Junior College Preliminary Examination 2015 H2 Mathematics Paper 1 (9740/01) Qn Solution 1 2 2121 x xx     2 2 1 2 1 021 x x x x      243 021 xx x        22 1 4 3 0x x x       2 1 4 3 1 0x x x    31 42 x   or 1x (ans) 22 22 2 1 2 2 xx xx  2 2 2 12 2 12 1 x x x     2 3 1 1 42 x    or 2 1 1 x   2 110 2x  or 2 1 1 x   2 2x  or 2 1x   22xx  or or 1 1, 0xx    2 2 2 d 5 d 5dd y x y yxx y x     2 2 2 2 dd21 dd yyyy xx   3 2 2 3 2 2 2 3 d d d d d d2 2 2 0d d d d d d y y y y y yy y yx x x x x x        3 23 2 23 d d d d2 6 0d d d d y y y y yyx x x x      When x = 0, y = 5 d1 d5 y x 22 22 1 d d 310 25 125 d d 125 yy xx      3 33 33 1 1 3 d d 42 30 25 05 5 125 d d 625 yy xx                  23 23 1 3 45 5 125(2!) 625(3!) 1 3 2 5 5 250 1875 y x x x x x x            1 1/2 -3/4

3(i) 2arg( 4) 2 2arg( 4) 2 arg( 4)2              z z z 3(ii) From diagram, 8AB z AC   2 2 2 21 4 8 3 4 2z      17 8 7z   3(iii) maximum arg( 8)z 1132tan sin45 2.9096 2.91rad (3sf)       4 C: 12sinyx  L: 8 3yx   C & L intersect at 1 ,23   And y-intercept of L is -. Volume obtained when S is rotated 2 radians about the y-axis = 211 3 2 3 æ ö æ ö÷÷çç +÷÷çç÷÷ççè ø è ø - 2 3 0 sin 2 y dy   éù æö÷çêú ÷ò ç ÷çêú èøëû = 4 12 3 æö ÷ç ÷ç ÷çèø - 3 0 1 cos 2 y dy   -ò Re(z) Im(z) 2 (4,0) (4,3) 2 5 (8,0) A B C (8,0) 2 5 L C y x

= 2 9  - [ ]3 0sin2 yy   - = 2 9  - sin2 3 3   éùêú-êúëû = 2 9  - 2 3 6 2 2  éùêú+ êúêúëû = 23 4 18  - 5 By sine rule,     5 66 1 2 55 66 1 2 31 22 21 2 2 3 sin sin( ) 3 sin cos cos sin 3 cos sin 23 since is small 2 1 2 3 23 (shown) 2 2 3 AB AB                   Applying binomial expansion,       121 2 22211 22 221 2 2 3 1 3 3 1 3 3 3 1 3 3 7 3 7 33 3 3, 3, 22 AB a b c                              6(i) Given that 2sin xx   2 sin x xee  since xye is increasing  sin 2 110 xxe e   22 2 sin 00 x xe dx e dx    6(ii)    22 0 sinsin uxe dx e du     Let 1duux dx    =   2 sin 0 ue du    since sin (-u) = sin u C A B θ

6(iii) sin 0 xe dx    = 2 2 sin sin 0 xxe dx e dx     = 2 2 sin 0 xe dx    from the result in (ii) < 2 2 2 0 x e dx    from the result in (i) = 2 22 0 2 x e      = - [ e-1 – e0 ] =  1e e   7(i) 2 2 2 2p x y p where 1p 2 2 2 1yx p , , ie.   yx x y px p 7(ii) The transformation is that of a translation of 2 units in the direction of the positive x- axis. The equation of C2: 2 2 2 2( 2)  p x y p Sub (4,3) into C2: 2 2 2 2 22 2 2 (4 2) 3 49 39 3 3 (rej 3 1)          pp pp p p pp 7(iii) No. of roots = no. of intersection points between both graphs = 3 8(i) Let nP be the proposition:   2 2 11 2 9 ( 3)( 5) 30 4 5 n r n r r n n     , n Z , 2n . When n = 2, LHS = 22 (5)(7) 35 , y = px y = - px C1 y x ( 1, 0) ( -1, 0) C1 y x (1, 0) ( -1, 0)

RHS =      2 2 911 2 30 6 7 35  . Since LHS = RHS, 2P is true. Assume kP is true for some k Z , 2k i.e.   2 2 11 2 9 ( 3)( 5) 30 4 5 k r k r r k k     . Need to show that 1kP is also true. i.e.         1 2 2 1 92 11 11 2 11 ( 3)( 5) 30 1 4 1 5 30 5 6 k r k k r r k k k k               . ( ) 1 1 2 2LHS of 3 ( 5) k k r P rr + + = = ++å ( ) ( )( )2 22 3 ( 5) 4 6 k r r r k k= =+ + + + +å       11 2 9 2 30 4 5 4 6 k k k k k                  2 9 6 2 511 30 4 5 6 k k k k k k            211 2 21 54 2 10 30 4 5 6 k k k k k k            211 2 19 44 30 4 5 6 kk k k k            4 2 1111 30 4 5 6 kk k k k        11 2 11 30 5 6 k kk   Since 2P is true, and kP is true  1kP is true , by mathematical induction, nP is true for all n Z , 2n . 8(ii) 4 4 2 ( 2) n r rr    = 1 1 2 ( 3)( 5) n r rr       1 2 22 ( 3)( 5) 4 6 n r rr        11 2 11 2 30 5 6 24 n nn        9 2 11 20 5 6 n nn   8(iii) 44 22 44 1 1 2 ( 1) 2 ( 1) nn rr rr    4 4 12 2 ( 2) n r rr     (Since     2 221 2 1 2 2r r r r r r r        )

   1 9 2 11 2 20 5 6 n nn     9 40 (since    2 11 056 n nn   for all n  ) 9(i) 𝑑 𝑑𝑥 ( 1 √𝑥2 − 3 ) = 𝑑 𝑑𝑥 ((𝑥2 − 3)−1 2) = (−1 2) (𝑥2 − 3)−3 2(2𝑥) = −𝑥 (𝑥2 − 3) 3 2 9(ii) 1 22 d 1 1 1sind 11 x x x x             = 1 √𝑥2−1 𝑥2 (− 1 𝑥2) = 𝑥 √𝑥2−1 (− 1 𝑥2) ( √𝑥2 = 𝑥 𝑎𝑠 𝑥 > 1) = 2 1 1xx   9(iii) Area = 3 1 3 dyx = 2 3 2 dln dd xtt t =   2 3 32 22 ln( ) d 1 ttt t    from(i) =   2 3 22 2 3 ln . d 1 ttt t   (𝑎𝑠 ∫ 𝑓(𝑥)𝑑𝑥 = 𝑏 𝑎 − ∫ 𝑓(𝑥)𝑑𝑥 𝑎 𝑏 ) 2 2 22 2 2 3 3 1 1 1ln d 11 t t tt ttt        2 2 2 3 2lnln 2 1 3 d 31 1 3 t tt           2 1 2 3 ln 2 2 13 ln sin 33 t      (by part (ii)) = − 1 √3 𝑙𝑛2 + √3𝑙𝑛2 − √3 ln(√3) − [𝑠𝑖𝑛−1 ( 1 2) − 𝑠𝑖𝑛−1 (√3 2 )] = (√3 − 1 √3)𝑙𝑛2 − √3 2 𝑙𝑛3 − ( 𝜋 6 − 𝜋 3) = 2√3 3 𝑙𝑛2 − √3 2 𝑙𝑛3 + 𝜋 6 10(i) Let  fyx 2 1 2y x  2 12x y  

12x y   since 0x Therefore -1 1f : 2x x  ,   1,0 , 2x      10(ii)          -1 2 3 2 f =f f = 1 2 2 1 0 1 1 0 1 1 4 1 51, 22 xx xx xx xx x x x xx                         -1f f 15 22 xx x     10(iii) For f -1 g to exist, range of g  Domain of f-1  11 ,1 , 2e       11 2 1ln 2 e      Greatest value of  = 1ln 2     1f10, ,1 1,0 2 g       . Range of f -1 g =  1,0 y x y =-2 x = -2 y = x y = f -1 (x) y = f(x) y = f(x) y = f -1 (x) 0.5 0.5 y =1 y = g(x)

11(i) i) From triangle APQ, 12tan 30 23 ax ax hh     2 3 x a h     2 2 22 2 Volume, base area height = 6 area of PST 16 sin 602 23 3 233 23 32 3 (shown) 2 o V h xh h a h h a h h a h            Alternative method: find the height RS of triangle PST From triangle PSR, 122tan 30 3 xx RS RS   32 2 3 RS a h       222 Volume, base area height 16 2 2 3 23 233 3 3 2 3 3 =2 3 (shown) 2 2 2 3 V x RS h h a h a h h a h h a h                                    11(ii) 2 323 2V h a h   2 d 3 3 2 3 2 ( 1)d 2 2 V a h h a hh                  B C D E F a h 120º 30º x P Q R S A T

332 3 2 22 332 3 3 22 a h h a h a h a h   