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 and y=0. Students attempted to find the equation of PQ to intersect with the equation of its perpendicular bisector to equate the point of intersection with the midpoint of PQ – and got lost because they expanded (m-5)(n+3) and didn’t know how to factorize this back to simplify. Students proved (i) with hardly any problems, but (ii) became complicated when students changed x xtox 2 2 2 cos sintan Several students assumed that the RHS of (iii) is the same as the LHS of (ii) and thus equated 2secx to -1, instead of -4.

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 5 (ii) Solution: LHS = )costan)(sin1)(sin1(sin 222 xxxxx ++−+ = )tan1)(1(sin 22 xx +− = ))(seccos( 22 xx− = 1− = RHS (proven) (iii) Solution: )costan)(sin1)(sin1(sin4sin1 cos cos sin1 222 xxxxxx x x x ++−+=−+− )1(4sec2 −=x 2 1cos −=x °= 60α °°= 240120 orx 6 Solution: (a) (i) CBxxAxxxx ++−+≡+++ )1)((1 2224 When x = 0, 1= - A + C ------(1) When x = 1, CB+=4 -------(2) When x = -1, CB+−=2 -------(3) (method mark given for method of substitution or comparison of coefficients) (2) + (3): C = 3 (2): B = 1 (1): A = 2 (answer mark given for answers for A and B) (ii) 3)1)(2(1 2224 ++−+≡+++ xxxxxx Remainder = 3+x Solution: (b) 1 27 2 3 − ++ x xx = )1)(1( 28 +− ++ xx xx (by long division) Most students did long division to find the remainder instead of understanding that f(x)=(x-a)g(x) +R(x).

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 6 11)1)(1( 28 ++−=+− + x B x A xx x )1()1(28 −++=+ xBxAx Ax 210:1 == A = 5 Bx 26:1 −=−−= B = 3 (method of substitution or comparison of coefficients) 1 27 2 3 − ++ x xx = 1 3 1 5 ++−+ xxx 7 Solution: (a) x x 2 2 cos sin23+ = xx 22 tan2sec3 + = xx 22 tan2)tan1(3 ++ = 253 a+ Solution: (b) xx cotcos2 = x xx sin coscos2 = 0coscossin2 =− xxx 0cos)1sin2( =− xx 2 1sin =x or 0cos =x 2,6 ππα = 6 5,2,6 πππ=x 8 Scale: Many students chose a scale that is time- consuming to plot the points, and encounter the possibility of points being plotted wrongly. Several students split into partial fractions immediately without realizing that the degree of the polynomial numerator is higher than that of the denominator. Several students changed x xtox 2 2 2 sec tansin - the long way Several students changed x xtox 2 2 2 sec tansin - the long way Several students divided by cos x instead of factorizing it out and letting 0cos =x Many students are not aware the trig ratio of special angles as in this question and gave answers in radians but not in pi. Some even gave answers in degrees.

ACS(Independent)MathDept/Y3AddMathP2/2015/FinalExam 7 xy 352 = A few students still didn’t know to change this to 5.17= x y and draw the horizontal line to meet the line they drew. However, several forgot that the value on the horizontal axis does not give the x value but the xx value. The abnormal as well as the correct readings should be given in x and y, not x y and xx coordinates. For the graph of 2x y is plotted against 2x x , the student needs only re -arrange the given equation and see that q is the gradient of the new line and give its value as obtained in b(i). End of Paper