ACSI 2018 Y3EXP FYE AMath P1 Solutions

Anglo - Chinese School (Independent) FINAL EXAMINATION 2018 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 1 Friday 5 October 2018 1 hour 30 minutes

ACS(Independent)MathDept/Y3EXP/AM P1/2018/Final Exam 1 SOLUTION Let P be (a, b). Let R be (c, d) 2 02 a− = and 1 22 b− = 2 62 c− = and 1 02 d− = P = (2, 5) R = (14, 1) C = ( )2 14 5 1,,22 hk++ = = (8, 3) 2 SOLUTION (a) 2 180 245 3 125+− = 12 5 7 5 15 5+− = 45 (b) 1 213 x x = − 2 13xx= − 24 13xx= − 24 3 10xx+ −= ( )( )41 10xx− += 1 14x or= − (NA) 1 4x∴= 3 SOLUTION (i) gradient = 24 51 −− − = 3 2− ( )3( )4 1 2xy x y+ −= − − 33 422xy y x+ =−+ ( 23 ) 8 23yx x += −+ 11 2 23 xy x −= +

ACS(Independent)MathDept/Y3EXP/AM P1/2018/Final Exam (ii) 11 2 23 xy x −= + 2(2 3) 11 2xx+=− 65x= 5 6x= 4 SOLUTION (i) 22 53αβ+= 22 196αβ = 2( ) 2 53α β αβ+− = 14αβ = (α and β are lengths and are positive) 2( ) 2(14) 53αβ+= + 9αβ+= Equation whose roots are α and β is 2 9 14 0xx−+= (ii) Since α and β are roots of 2 9 14 0xx−+= and ( )( )7 20xx− −= , 2α = or 7 Dimensions of the rectangle = 4 cm by 14 cm. 5 SOLUTION (i) Centre of 1C = (1, 4)− Gradient of PQ = 1 ( 5) 4 ( 2) −−−−−− = 3 (negative reciprocal of radius at P) Equation of PQ is: ( )23 5yx+= + 3 13yx= + (ii) Radius of 1C = ( ) 221 4 23+− + = 40 Distance between centre of 1C and R = ( ) ( ) 22 13 41+ +−− = 41 > radius Hence, R lies outside of 1C . (iii) Centre of 2C = ( 7, 4)−− Equation of 2C : ( ) ( ) ( ) 222 7 4 40xy+ ++ = 22 14 8 49 16 40xy xy++ +++= 22 14 8 25 0xy xy++ ++= [Note, not necessary to have this form]

ACS(Independent)MathDept/Y3EXP/AM P1/2018/Final Exam 6 SOLUTION See graph paper (ii)(a) lg lg lgPbT a= + Gradient = b = 1.52. Intercept = lg a = 2.2 a = 158.5 (b) Abnormal Reading: P = 10000 Correct Reading: lg P = 3.75 P = 5623 (nearest whole number) (c) After 4 days, T = 4 lg T = 0.60 lg P = 3.15 P = 1413 (nearest whole number) The population after 4 days is 1413.

ACS(Independent)MathDept/Y3EXP/AM P1/2018/Final Exam 7 SOLUTION (a)(i) 23x−= 23x−= or 23x−= − 5x= or 1x=− (ii) 7(a)(iii) 54 2 xx+= − 234 2 0 xx++ −− = 24yx= −− is the equation of the line to be inserted. 7(b) 2 2 cos 2xxππ−= 11 2yx π = −  is to be inserted into the sketch. There are 2 solutions.

ACS(Independent)MathDept/Y3EXP/AM P1/2018/Final Exam 8 SOLUTION (a) 2 2log log log nm m mn m × = lg 2lg 2lg lg lg lg mnm nm m×× = 4 (b) 26 2xy = 27 2xy = ( ) 2 6242rs = ( ) 2 724 2rs×= 226rs+= - - - (1) 47rs+= - - - (2) 2(2) - (1): 68s= 4 3s= (1): 5 3r = (b) (5 ) 1 1 lnlog xe xe− = + ( )ln 5 ln lnxe e x−= + ( )5x e ex−= ( )5x ee−= 5 ex e= − END OF PAPER ONE