CJC For Sharing 2012 JC2 H2 Prelim Paper 2 Solutions

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 1 SLE Part Assessment Objectives Mark Scheme Able to form system of linear equations Able to solve a system of linear equations using GC Let dcxbxaxx  23)f( . Also, cbxaxx  23)('f 2 6248)2f(  dcba 15)1f(  dcba 023)1('f  cba 0412)2('f  cba Solving, 21232)f(2,12,3,2 23  xxxxdcba 2 Vectors Part Assessment Objectives Mark Scheme (i) Find intersection between line and plane At pt of intersection, 4 2 0 1 22 1 31                             9 So intersection pt is (28, -8, 16). (ii) Form equation of plane given a line and a point A direction vector on p2 is                                   0 2 1 2 1 0 2 1 1

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 2 Vectors Part Assessment Objectives Mark Scheme A vector normal to p2 is                                  7 2 4 2 1 3 0 2 1 So an equation of p2 is 16 7 2 4 2 1 0 7 2 4                                     r i.e. 16 7 2 4             r (iii) Find intersection of 2 planes using GC 42  zx and 16724  zyx . Solving, an equation of line is                           , 2 1 4 0 8 4 r (iv) Understand relationship between normal of a plane and a line // to it For p3 // l2, 0 2 1 41                       b a 042  ba

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 3 FUNCTIONS Part Assessment Objectives Mark Scheme (i) Draw graph of a function - (ii) Understand condition for inverse to exist For inverse to exist, function f must be one-one and thus, largest k is 0. (iii)  Know how to find inverse function and select the appropriate function depending on domain of f  Understanding relationship between f and its inverse HOT 1 1 2  xy 12  yyx y yx  12 y yx  1 x xxf  1)(1 Solving    xx ff -1  is equivalent to solving   xx f : xx  1 1 2 013  xx Solving and taking the real root: a =  0.682 (3sf) (iv)  Understand the condition for composite function to exist  Know how to restrict domain of a function to For composite function to exist, )0,1[1  fg DR Restricted domain of g = [– 4, – 3) -1

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 3 FUNCTIONS Part Assessment Objectives Mark Scheme achieve required range HOT

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 4 Definite Integral Part Assessment Objectives Mark Scheme Feedback (i) Know how and when to use double angle formula Know how to apply integration by parts     C2cos2 12sin4 1 d 2sin2 1 2 2sin 2 1 22 1d sin 2 2sin 1d d 2cosd d d 2cos2 1 22 1 d 2cos2 1d 2 1 d 2 2cos1 d sin 2 2 22 2 2                                xxxx xxxxxxxx xvx u xx vxu xxxx xxxxx xxx xxx Less than half the student cohort correctly used the double angle as part of the problem solving procedure. Only a handful of students used the alternative methods and lead to a longer solution. They were conceptually correct and many did well.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 4 Definite Integral Part Assessment Objectives Mark Scheme Feedback (ii) Know how to use definite integral to find volume of solid of revolution Know how to use previous answer in evaluation of definite integral 75.72 12cos2 12sin4 2cos2 12sin4 d sin 2 0 2 0 2                  V xxxxV xxxV This part is generally well done. (iii) Know how to find required volume by using volume of previous solid found     3.23 7516.7Volume 2    Most students could not even state the volume of cylinder correctly. Some do not even know how to subtract the volume from one another. This part is badly done. I suspect the students don’t have a visual approach to solving this part. They need to use a diagram to aid their problem solving.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 5 DIFFERENTIAL EQUATIONS AND APPLICATIONS Part Assessment Objectives Mark Scheme Feedback  Able to solve DE using substitution 22 d d yxx yxy  uxy xxux y d du d d    22 d du)( uxxxxuuxx         22322 d du uxxxuxxu  uxx 1 d du  xxu 1 d du  xxu d1du   cxu  ln2 2 dxu  ln22 222 ln2 dxxxy  This part is generally well done. A handful of students forgot to include the modulus sign.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 5 DIFFERENTIAL EQUATIONS AND APPLICATIONS Part Assessment Objectives Mark Scheme Feedback HOT  Able to find particular solution, given conditions  For the curve that passes (1, 2): 222 )1(1ln)1(22 d 4d 222 4ln2 xxxy  When y = 3 (from GC): x = 1.39, – 1.39 Quite a number of students stated 1.39 as the only value for x.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 6 PERMUTATION AND COMBINATION Part Assessment Objectives Mark Scheme Feedback (i) HOT Solve counting problems by applying the idea of combinations. Total possible groupings 2 27 2 29 2 9 CCC  5130216 The question was very badly done with many kinds of mistakes including using permutation instead, adding instead of multiplying and also many who simply gave 4 29 2 9 CC  , not realising that the 4 not involved in DI must still be split into pairs. (ii) HOT Total possible groupings     22 27 1 28 2 8 2 27 2 29 1 8  CCCCCC 1690416 Out of those who completed (i) correctly, only a few went on to get this part correct. The best solutions used the idea of complement instead. Many made the error of using  2 27 1 28 2 9 CCC  instead, not realising that Jill cannot be among the ones selected for DI component. (iii) HOT Total possible groupings   22 27 1 28 2 9  CCC 707616 Out of those who completed (i) correctly, only a few went on to get this part correct. The best solutions used the idea of complement instead. Many made the error of using  2 27 1 29 2 9 CCC  instead, not realising that Jack cannot be among the ones selected for the component which they had assumed that Jack is already in. (iv) HOT Calculating probabilities using P&C. Relating probabilities to the idea of fairness. From the previous parts, if the arrangement is at random, the probability of a CI student on duty 330.05130216 1690416  the probability of a SS student on duty 138.05130216 707616  Hence, it is unfair as the probability that a CI student is put on This part was also badly attempted as most could not get the previous parts correct. Majority of the students realised that fairness has to do with equal chance and probability, but did not fulfil the ‘hence’ requirement, which is to use the arrangements to calculate probability.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 2 SOLUTIONS 2012 6 PERMUTATION AND COMBINATION Part Assessment Objectives Mark Scheme Feedback duty is much higher than a SS student. Arguments made on the number of students in each wing and/or the special requirement of the DI comp