ACSI 2018 Y3EXP FYE AMath P2 Solutions

AM Paper 2 2018 (1i) 2 12 2x kx += −  2 2 2 kx kx+= − 2 20 2 kx kx− ++= 2 40b ac−> ( ) 2 41 2 0 2 kk − +>  2 82 0kk−− > 2 2 80kk− −> ( )( )4 20kk− +> 24k or k<− > (ii) Since k > 4 here, the roots are real and distinct. (2) 2 16yx= --------------(1) 2 4y x= ---------------(2) 2 2 4 16xx  = y x 2 16yx= 2 4y x=

2 4 16 16xx = 5 1x = 1x= 2 4 1y= = 4 The point of intersection is (1, 4) [A1] (3i) When t = 0, M = 30 ( )30 6 ln 0 1k= −+ k = 30 (3ii) ( )15 30 6ln 1 t= −+ ( )6ln 1 30 15t+=− 15 61te+= 11.18 4 22 37300xy+= - - - - (1) 3 3 ( ) 860x y xy++−= 4 2 860xy+= 430 2yx= − - - - - (2) (2) in (1) : ( ) 22 430 2 37300xx+−= 25 1720 147600 0xx−+ = ( ) 2 1720 1720 4(4)(147600) 2(4)x ±− −= 180 164x or= 70 102y or= The tables are of lengths 180 cm and 70 cm, respectively or 164 cm and 102 cm. (5a) 3log 2 3log 3 0 xx+− = 3 3 1log 2 3 0 logx x +− =  

3 Let 3logyx= 320y y+− = 2 2 30yy+ −= ( )( )3 10yy+ −= 31y or y= −= 33log 3 log 1x or x= −= 1 327x or x= = (5b) 1 22 24 4 4 17 xxx + +−=− ( ) 1 224 4 4 4 4 17xx x −= −  ( ) ( ) 24 2 4 16 4 17xx x−= − Let y = 4x 2 18 17 0yy− += ( )( )1 17 0yy− −= 41x = or 4 17x = x = 0 lg17 lg 4x= = 2.04 6 (a) If ( ) 2 3 54ab−= − , find the value of a and of b. [4] 23 3 5 4aa b− +=− 323 54a ab+− =− 35a+= 2a= 23 4ab= 12 16ab= 12 16ab= 3 2b= (6b) A cuboid of volume ( ) 318 11 2 cm+ stands on a square base of side ( )12 cm+ . Find the height of the cuboid. [4]

4 Height of cuboid = ( ) 2 18 11 2 12 + + = 18 11 2 322 + + = 18 11 2 3 2 2 322 322 +− × +− = ( )10 3 2 cm− (7ai) 1cos 5 A= Using Pythagoras Theorem, opp = 2 2sin 5 A= = 25 5 (aii) ( ) 122cos 26 1tan 90 tan o A A π = − 2= (7bi) cos 1 tan1 sin cos x xxx−=− cos 1 1 sin cos xLHS xx= −− ( ) ( ) 2cos 1 sin cos 1 sin xx xx −−= − ( ) 21 sin 1 sin cos 1 sin xx xx − −+= − ( ) 2sin sin cos 1 sin xx xx −= − ( ) ( ) sin 1 sin cos 1 sin xx xx −= − tan x=

5 (7bii) cos 1 51 sin cos x xx= −− cos 1 51 sin cos x xx−= −− tan 5x=− Reference ∠ x = 78.7o 101.3 281.3oox and= (8i) Equation of AD is 12 6 615yx −= + 2 65yx= + ------------------------(1) 8 165yx= −+ ----------------------(2) Solving simultaneously, 28 6 1655xx+= − + 5, 8xy= = ( )5,8B (8ii) when y = 0, ( )10, 0E . When 10x= , ( )2 10 65y= + =10 ( )10,10C (8iii) Area of ∆ABE = 0 10 5 01 6 0862 = 25 units2 226 10BE = + 136= units Shortest distance = 2 25 136 × = 4.29 units

6 (8iv) Gradient of ⊥ bisector = 5 4− Mid-point of OD = 0 15 0 12,22 ++  = 17 ,62   Equation of ⊥ bisector is 5167 42yx −= − −  5 123 48yx= −+ When 0,y= 12.3x= . Hence, it does not pass through E. ---------------------------------------------------------The End ----------------------------------------------------------