MJC_JC2_H2_Maths_2012_Paper_1_Solutions

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 1 of 17 2012 MJC H2 MATH (9740) JC 2 PRELIMINARY EXAM PAPER 1 – SOLUTIONS Qn Solution 1 Inequalities                 2 2 2 32 2 32 2 2 2 22 464 , 1 1 4 1 4 6 01 4 4 4 6 01 45 01 45 01 since 4 5 2 1 0 for all real values of , 011 1 0 or 1 xxx x x x x x x x x x x x x x x x x x x x x x x x xx xx                              Qn Solution 2 SLE Let   32f x ax bx cx d    . Method 1:         f 0 0 0 f 1 3 (1) f 1 3 2 0 (2) Since the stationary point is a point of inflexion, f 1 6 2 0 (3) d abc a b c ab                 Using GC to solve, a = 3, b = – 9, c = 9   32f 3 9 9x x x x    Method 2:       f 0 0 0 f 1 3 (1) f 1 3 2 0 (2) d abc a b c               2f ' 3 2x ax bx c   Since there is only one stationary point,  f ' 0x  has only one solution. Hence, Discriminant = 0   2 2 4(3 )( ) 0b a c 1     1 0

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 2 of 17 24 12 0 (3)b ac Solving equations (1), (2) & (3), we have 3, 9, 9a b c     32f 3 9 9x x x x    Method 3:  f 0 0 0 d    f 1 3 (1)abc          2 2 32 1 is the only stationary point f 1 3 2 3 (2) Solving (1) and (2), 3, 9, 9 f 3 9 9 x x k x ax bx c k a b c a b c x x x x                   Qn Solution 3 Differentiation (Implicit) + Techniques of Integration (a) 2 ln 2x xy y    2 d 1 d20 dd d1 2d d 2 2 1d1 yyx x y x y x y x x yxy y x y xy y x xy x y            (bi)   2d 2d x x = 212 ln 2x (bii)    2 21 22 22 21 2 ln 2 d 1 2 ln 2 d2 1 2 2 d2 11 222 2ln 2 12 2ln 2 xx x xx xx x x xx xx xC xC              

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 3 of 17 Qn Solution 4 Complex 3 (include intersection of loci) (i) (ii) 41sin 8 2 6      smallest value of  arg 2 4iz 2 6 3    (iii) Method 1: AP =   224 4 4 4 2 1    w    2 4 2 1 cos 4 2 1 sin i44                        112 4 2 1 4 2 1 i 22 2 1 2 2 2 2 i                      Method 2: Equation of circle:     22 2 4 16xy    -------(1) Equation of half line:  0 tan 2 2, 24y x y x x        ----(2) Sub (2) into (1): ( 2,0)A  required region for (i) Im C (2, 4)  4 4   P Re O ( 2,0)A  Im C (2, 4)  8  4 4   P Re O (2, 4)

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 4 of 17 Qn Solution 5 Vec