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 Vectors 1 & 2 (Ratio Thm, application of dot & cross product) (i) Using ratio theorem, 4 5 5 4 11 1 5 3 14 22 OM OAOB OB OAOM                           3 1 82 4    (ii) OA OB OB  represents the shortest/perpendicular distance from point A to the line OB. (iii) 11 13 22 8 4 2 4 22 1 OA OB                                        22 2 2 2 16 28 2 2 2 2 2 2 or 2 2 2 . 0 4 2 2 2 1 2 2 2 2 i xx x x x x rej x y w                   Method 3: 4cos 2 24 4sin 2 24 PN CN         2 2 2 4 2 2 2 1 2 2 2 2 i x y w        C P 4 4  N

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 5 of 17 Normal to the plane containing , and O A B is 4 2 1   . Method 1: 14 32 81shortest distance 21 2 21 21 21 2 21 units21                    Method 2: Let N be the foot of perpendicular from C to the plane. Equation of line CN is 14 3 2 , 81                      r Equation of plane: 4 20 1     r Since N lies on the plane, 1 4 4 3 2 2 0 81 2 21                          2 1314 21 21 2 5932 21 21 17028 2121 ON                  So 13 8 21 21 1 59 4 321 21 8170 2 21 21 CN ON OC                                      2 2 2 8 4 2 4 21 21 21 21 2 21 units21 CN                      

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 6 of 17 Qn Solution 6 Maclaurin + Binomial Series (include concept of approximation)       22 2 33 (3) (3) 3 f ( ) ln e 2 f (0) ln 2 eef ( ) f (0)e 2 2 eef ( ) f (0) 4e2 2e ef (0) f (0) 4e2 xx x x x x x             (i)               23 (3) 2 2 3 3 2 2 3 3 f f 0 f 0 f 0 f 0 ... 2! 3! e e ef ln 2 ... 2 2! 4 3! 4 e e ef ln 2 ... 2 8 24 xxxx xxxx x x xx                             (ii)   2 2 3 3 2 3 2 22 e e eln e 2 ln 2 ... 2 8 24 Differentiate both sides wrt : e e e e ...e 2 2 4 8 2 e e 1 ...e 2 2 4 x x xx x xx x xx x               (iii)          1 1 1 2 22 2 2 e 2e2 e2 2 1 2 12ee1 1 ... 2 2! 2 ee1 ...24 xx x xx xx                         Qn Solution 7 MI + Summation (i)                2 3 4 1 1 3 1 1 112 2 1 2 4 4 2 2 3 1 5 1 1 114 2 2 3 6 6 2 3 5 1 7 1 1 116 2 3 4 8 8 2 4 a a a                      11 , 22 nac n   

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 7 of 17 (ii) Let Pn be the statement 11 for all 2 nan n    . When 1n , LHS = 1 1 2a  RHS =   111 2 1 2 1P is true Assume Pk is true for some k  , i.e. 11 2 ka k . To prove 1Pk is true. i.e.   1 11 21 ka k                11 1LHS of P : 21 111 (from assumption)2 2 1 111 2 1 2 1 1 21 11 RHS21 k k k aa kk k k k k k k k k k kk k                   Pk is true 1Pk is true Since 1P is true and Pk is true 1Pk is true, by Mathematical Induction, Pn is true for all n  . (iii)              2 1 1 1 1 1 ...2 6 12 1 1 1 1 1 ...1 2 2 3 3 4 1 1 N n NN N N n n                       11 1 1 21 32 43 1 1 11 21 2 1 2 2 NN nn N nn n NN NN n n n n aa aa aa aa aa aa                    112 1121 2 1 2 Naa N     

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 1/Math Dept Page 8 of 17   11 1N  Qn Solution 8 Application of Differentiation (Max/Min, Rate of Change) (i) Method 1: Area of triangle PQR = 1 ( )( )2 QR PQ since 90PQR (angle inscribed in a semicircle is a right angle). Let QPR  . Then 2 sinQR r  and 2 cosPQ r  Area of PQR = 1 (2 sin )(2 cos )2 rr  = 222 sin cos sin 2rr    . For maximum area of triangle PQR,   2d sin 2 0d r   22 cos 2 0r   cos2 0 2 2   4    2 2 2d 2 cos 2 4 sin 2 4 0d r r r     when 4    Area is maximum. 2 cos 24PQ r r  2 sin 24QR r r  Method 2: Let POQ  . Area of PQR = Area of POQ + Area of ROQ =     2211 sin sin note:sin sin22rr          = 2 sinr  . For maximum area of triangle PQR,   2d sin 0d r   2 cos 0r   cos 0 2     2 2