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    