ACSI 2015 Y3EXP FYE AMath P2 Solutions

Anglo-Chinese School (Independent) FINAL EXAMINATION 2015 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 2 5 October 2015 1 hour 30 minutes PAPER REVIEW 1 Solution: xxx 2142 532 −−+ = ( )525 1 81 1432 x xx     =     × 4 8152532 ×=xxx 4 815150 ×=x 150lg 4 815lg      × =x 922.0=x (to 3 sf) Most students can get up to this line, but many could get no further. The mistake comes from xxx 532 = Some students started by logging both sides of the equation, but brought down the powers without bracketing them: 5lg213lg42lg2 xxx −=−++ And getting 3lg42lg215lg22 +−=+ xx

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 2 2 Solution: te kD 14.041 −+ = (i) When t = 0, D = 20: 041 20 e k + = 100=k (ii) When t = 10, 4.141 100 −+ = e D = 50 (nearest integer) ∴ There will be 50 deer after 10 years. (iii) When D = 70, te 14.041 10070 −+ = 170 1004 14.0 −=− te 14.0 28 3ln −=t 0.16=t (to 3 sf) The population will take 16 years to reach a population of 70. 3 (i) Solution: xx 42 =− M1 or xx 42 −=− 5 2=x 3 2−=x (NA) Students have either made 1) eº =0 thus getting k to be 20, and then cannot solve (iii) or 2) 4eº= 4e, thus getting k to be 237.46…. Students forget to check that x >0

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 3 (ii) Solution: xy −= 2 xy 4= –6 –4 –2 2 4 6 –4 –3 –2 –1 1 2 3 4 x y (0.4, 1.6) Accurate sketch of the graphs of xyxy 4 and 2 =−= Label of points of intersection      5 31,5 2 (iii) Solution: 5 2<x Several students didn’t label the point of intersection and did not realize that the range of values for x is obtained by referring to the sketch.

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 4 4 Solution: (i) When y = 0, x = m – 5 )0,5( −∴ misP . (ii) When x = 0, y = n + 3 )3,0( +∴ nisQ . Midpoint of PQ =      +− 2 3,2 5 nm (iii) Gradient of perpendicular bisector of PQ = 2 1 Gradient of PQ = 2− 2)5(0 03 −=−− −+ m n 1023 −=+ mn 132 −= mn -----(1) Since the midpoint of PQ lies on 3 2 1 += xy , 32 5 2 1 2 3 +−×=+ mn 12562 +−=+ mn (1): 67)132(2 −+=− mm 273 =m 9=m (1): 5=n 5 (i) Solution: LHS = x x x x sin1 cos cos sin1 −+− = )sin1(cos cos)sin1( 22 xx xx − +− = )sin1(cos cossinsin21 22 xx xxx − ++− = )sin1(cos sin22 xx x − − = )sin1(cos )sin1(2 xx x − − = 2secx = RHS (proven) Many students multiply by the LCM of the denominators and got lost in the expansion, before they let x=0 a