ACSI 2017 Y3EXP FYE AMath P2 Solutions

1. ( )x x 224 4 = ( )xx 224 2 1 1=− ( )xx 222 2 1 22 =− xx +=− 2 122 2 12=x 2. (i) ( ) 232 −−= xkx ( ) 0232 =+−− xkx Sum of roots = 41 +++   352 −=+ k 2 8−= k ----------------(1) Product of roots = ( )( )41 ++   2452 =++  --------(2) 242 852 8 2 =+     −+     − kk 0862 =−− kk =k 3 + 17 4. (i) )5(26522200 ke−+= ke 5265178 −= 265 178)5( =−ke 265 178ln5 =− k 0796.0=k [A1] (ii) )20(07959.026522 −+= eT

ACS(Independent)MathDept/Y3AddMathP2/2017/FinalExam 2 = 75.9oC [A1] (iii) 22=T 6(i) 24 113 +=−− xx 34 13 +=− xx 34 13 +=− xx or 34 13 −−=− xx 0=x 8=x 6(ii) Graph of 13 −−= xy (Red line) Graph of 24 1 += xy (Blue line) 6(iii) 1234 +− xx 343 +− xx 2413 +−− xx From the graph, 80  x . 7(i) Gradient of SR = 2 3 04 410 =− − Equation of SR: 42 3 += xy

ACS(Independent)MathDept/Y3AddMathP2/2017/FinalExam 3 7(ii) Gradient of SP = 3 2− Equation of SP: 43 2 +−= xy When y = 0, x = 6  P ( )0,6 7(iii) 41 = 4 0 10 4 00 6 4 0 2 1 q 2 24161041 −+= q 9=q Q (9, 0) 7(iv) 3 2 9 4 =− − x x xx 218123 −=− 305 =x 6=x 3 2 0 10 =− − y y yy 2330 =− 6=y The coordinates of U is ( )6,6 . 7(v) Gradient of SR = Gradient of ⊥ bisector = 2 3 . Mid-point = ( )2,32 04,2 60 =     ++ Equation of perpendicular bisector ( )32 32 −=− xy 2 5 2 3 −= xy When x = 6, ( ) 2 562 3 −=y = 2 16 Hence, the point U does not lie on the ⊥ bisector. [Alternatively, students can compare distance of U from S and P. If they are equal, Then U is on the perpendicular bisector.]

ACS(Independent)MathDept/Y3AddMathP2/2017/FinalExam 4 8(a) 3cos 3tan2 −= xx 3sec31sec2 −=− xx 02sec3sec2 =+− xx ( )( ) 02sec1sec =−− xx 1cos =x or 2 1cos =x oooo andx 360300,60,0= 8(bi) LHS = 1sec sin 1sec sin ++− x x x x = ( ) ( ) 1sec 1secsin1secsin 2 − −++ x xxxx = x xx 2tan secsin2 = x x 2tan tan2 = 2 Cot x 8(bii) ( ) ( ) 1 32 12sec2sin12sec2sin −=−++ xxxx ( ) ( ) 3212sec2sin12sec2sin −=−++ xxxx ( ) ( ) ( )12sec2cot212sec2sin12sec2sin 2 −=−++ xxxxxx ( )xx 2tan2cot2 2= x2tan2= 322tan2 −=x 32tan −=x 32 = xofref 2x = 3 − , 3 4− , 3 2 and 3 5 x = 6 − , 3 2− , 3  and 6 5 Otherwise,

ACS(Independent)MathDept/Y3AddMathP2/2017/FinalExam 5 ( ) ( ) 1 32 12sec2sin12sec2sin −=−++ xxxx ( ) ( ) 3212sec2sin12sec2sin −=−++ xxxx 322sin2tan2sin2tan −=−++ xxxx 322tan2 −=x 32tan −=x 32 = xofref 2x = 3 − , 3 4− , 3 2 and 3 5 x = 6 − , 3 2− , 3  and 6 5 Solutions: 9(i) Drawing of graph: Creation of Table