ACSI 2015 Y3EXP FYE AMath P1 Solutions

Solutions 1 ) 3 135()323( 2 +×+=π V )3 335()123129( +×++= )3 335()31221( +×+= 12180373105 +++= 1923112 += 2(i) 2 1 2 35 xx = 3 52 5 =x 3 255 =x 23.1=x 32.3=y (ii) x y (1.23, 3.32) 2 1 3xy= 2 5 xy =

3 06152 =+− yxyx 0)3 1(6)3 1(152 =+− xxxx 0252 =+− xx 0252 2 =+− xx 0)12)(2( =−− xx 2=x or 2 1=x 6 1=y 3 2=y Grad of AB 3 1−= Grad of perpendicular 3= Midpoint of AB )12 5,4 5(= Equation: 3 103 −= xy 4 0632 2 =−− xx (i) 2 3=+βα 3−=αβ (ii) 33 αββα + )( 22 βααβ += )]3(2)2 3[(3 2 −−−= 4 99−= 4433 ))(( βααββα = 4)3(−= 81= 0814 992 =++ xx 0324994 2 =++ xx 5(i) amplitude = 4 Period = °180 (ii) 022sin4 =+x 2 12sin −=x

330,2102 =x 165,105=x 6(i) kkxx >+ )( 02 >−+ kkxx 0<D 0)(42 <−− kk 0)4( <+kk 04 <<− k (ii) 05)2( 22 =−++ pxx 0544 222 =−+++ ppxxx 0545 22 =−++ ppxx 0>D 05)(5(4)4( 22 >−− pp 01002016 22 >+− pp 0252 <−p 0)5)(5( <+− pp 55 <<− p 7(i) 01561022 =−−−+ yxyx 03515)3()5( 3222 =−−−−+− yx M1 49)3()5( 22 =−+− yx Centre = (5, 3) Rad = 7 (ii) Distance = 22 16 + 737 <= It lies in the circle. (iii) New centre = (-7, 3) 49)3()7( 22 =−++ yx 0961422 =+−++ yxyx 8(i) )2(log2log)718(log 333 −=−− xx

3log )2(log2log)718(log 3 3 33 −=−− xx 2 33 )2(log2 718log −=− xx 2)2(2718 −=− xx 0102 2 =−− xx 0)2)(52( =+− xx 2 5=x or 2−=x (NA) (ii) 1 1 1214 − −     =+ x x ee 0214 1)1(2 =−+ −− xx ee Let 1−= xey 02142 =−+ yy 0)7)(3( =+− yy 31 =−xe 10.2=x 71 −=−xe (NA) 9(a)(i) Let 2723)( 23 −−−= xxxxf )23)(1( 2 −++= axxx 2)2()3(3 23 −−+++= xaxax 23 −=+a 5−=a )253)(1()( 2 −−+= xxxxf 0)2)(13)(1( =−++= xxx 2,3 1,1 −−=x (ii) 2sin7)2sin3(sin2 +=− θθθ 02sin7sin2sin3 23 =−−− θθθ 1sin −=θ °= 270θ 3 1sin −=θ °°= 5.340,5.199θ )(2sin NA=θ (b) 614)( 3 ++= xxxf 614)( 3 ++= aaaf

3082)( 23 −−−= xxxxg 3082)( 23 −−−= aaaag 6143082 323 ++=−−− aaaaa 036222 2 =++ aa 0)9)(2( =++ aa 92 −−= ora