ACSI 2016 Y3EXP FYE AMath P1 Solutions

Solutions to AM Paper 1 EOY 2016 1. 9 3 13 2 1 2 1 = = − x xx When x=9 2 1 3 − = xy = 3(9)-0.5= 1 The point of intersection is ( 9 , 1)

2 (i) Area of triangle = )232 11( + ( )( ) 232 1145sin23232 1 0 +=++ b ( )( ) )232 11(2245sin2323 0 +=++ b Comparing both sides 1263 =+b 2=b ( shown) (ii) Perpendicular distance from B to AC = )23( )232 11(2 + + = ( ) )29( 23)2611( − −+ = 7 )2182111233( +−− = 237 )2721 +=+

3. (a) ktePP ο= From question given when t = 0 , P = 50 Therefore kteP 50= Also P = 960 when t = 8 ke850960= 5 968 =ke   = 5 96ln8k 369.05 96ln8 1 =  =k (b) 35100050 )24(369.0 == eP (c) kte50)50(15 = te 369.015= )15ln(369.0 1=t = 7.33 hours

4 Let ( )( ) 22 3 )1(1313 )3(2 −+−++= −+ − x C x B x A xx x )3()1)(3()1)(()3(2 22 ++−++−=− xCxxBxAx Let x=1 : -4 = 4C C= -1 Compare coefft of x2 : 2 = A + B _ Let x = -3 : 12 = A(-4) 2 A = 3/4 Sub in 1 B= 5/4 ( )( ) 22 2 )1( 1 )1(4 5 )3(4 3 13 )3(2 −−−++= −+ − xxxxx x

5 (a) let khxxxxf +++= 23 3)( By Factor Theorem : 0)2( =f 202 02128 −=+ =+++ kh kh By Remainder Theorem : 30)1( =−f 28 3031 =+− =+−+− kh kh Solving the equations simultaneously 202 −=+ kh --------------1 28=+− kh ---------------2 1 (-) 2 16 483 −= −= h h When 16−=h , 1628−=k =12 (b) (i) Remainder = 18)13)(63)(23()3( =−+−=f (ii) Let )6)(2()( 2 −+−= bxxxxf (By inspection) Compare coefft of x on both sides 5 2616 = −−=− b b Therefore )65)(2()( 2 −+−= xxxxf )1)(6)(2()( −+−= xxxxf

6 (a) 5cot3cos4 2 += xxec 5cot3)cot1(4 2 +=+ xx 01cot3cot4 2 =−− xx ( )( ) 01cot1cot4 =−+ xx ( ) ( ) 01cot01cot4 =−=+ xorx 1cot4 1cot =−= xorx 1tan4tan =−= xorx οοοο 225,450.284,0.104 == xorx (b) 6.0)12tan( =−z 9121 <−<− z (2z-1) lies in the I and III quadrant. Basic Reference angle = tan-1(0.6) Max (2z-1) = 2π+ ))6.0(tan 1− = 3.91 ( 3 sf)

7 (a) (ii) xx =−ππ 2cos2 ππ π xx =−12cos2 12cos2 += π xx The line to be inserted is y = (x/π)+1 (iii) (iv) from sketch there are 2 solutions to the given equation.

(b ) A lies in the II quadrant-------------------------M1 (i) 7 4sec −=A (ii) 4 7)cos( −=−A (iII) )2sin( A−π Acos= 4 7−=

8. (a) 2)3(log)1(log 42 =−−+ xx 2)3(log2 1)1(log 22 =−−+ xx 2)3(log)1(log 2 1 22 =−−+ xx 2 )3( )1(log 2 12 = − + x x 4 )3( )1( 2 1 = − + x x )3(16)1( 2 −=+ xx 049142 =+− xx 0)7( 2 =−x 7=x