Practice Paper B Sec 3 IP Math Answers (for students)

1 Practice Paper B Sec 3 Math Answers Qn Solutions 1(a) 33 22 222 2 ( 3) ( 1) ( 3 1)[( 3) ( 3)( 1) ( 1) ] (2 4)( 6 9 4 3 2 1) 2( 2)( 4 7) xx x x x x x x x x x x x x x x x x                            1(b) 21 2 21 2 xx xx x x     or 2 2 1 2 12 2 ( 1) 4 2( 1) 22 ( 1) 2( 1) 2 xx x x x x x xx xx x xx x x          2 Length of the side BC 1 3 2 28 (1 3 2)(2 2 2) (2 8)(2 2 2) 2 2 2 6 2 12 4 10 4 2 4 5 2 m 2          3(a) 2 ( 1) 22 ( 1) ( 1) 22 2 log (3 7 2) 2 log (3 7 2) log ( 1) 3 7 2 2 1 2 5 3 0 (2 1)( 3) 0 1 (rejected) or 3 2 x xx xx x x x x x x x xx xx x                     

2 3(b) 2 2 1ln 2 ln 0, where 02 ln 2 ln 0 ln 2 0 21 2 1 or 1 (Two cases) Case 1: 2 1 21 2 1 4 6 1 0 3 2 2 (reject 3 2 2) OR 5.83 or 0.172 (rejected, as it does n ot satisfy x x x xx xx xx xx xx xx x x x xx x x                         the Case 1 equation) 2 2 Case 2: 2 1 21 2 1 4 ( 1) 0 1 xx xx x x x x x         or 2 2 1 0 ( 1) 0 1 xx x x      Hence 3 2 2 or 1 x 4(i) P(R Q) or 100 120 5 70 85 6 x y          or 100 120 5 70 85 6 x y                  4(ii) 1 100 120 5 1710 70 85 6 1205 5 100 120 1710 6 70 85 1205 5 85 120 17101 6 70 100 1205100 5 7.5 68 12.5, 14 x y x y x y x y xy                                                           The price of pandan cake is $12.50 each, and the price of chocolate cake is $14 each. 5 2 ( 1)y mx m x m    is always positive 0 and 0m  

3 22 22 2 ( 1) 4 0 2 1 4 0 3 2 1 0 (3 1)( 1) 0 1 or 1 3 is positive, 1 mm m m m mm mm mm mm                  6(i) 2 2 2 2 2 1 ( 2) ( 2) (2 1) 0 ( 2) 4(2 1) 0 40 ( 4) 0 0 or 4 x k k x x k x k kk kk kk kk                  6(ii) 22 2 2 2 2 21 11 ( ) 2 11 ( 2) 2(2 1) 11 9 3 k k kk k k                        , are real, range of values for : 0 or 4 3 k k k k     7(a) 32 32 32 2 2 5 2 2 5 2 0 Let f( ) 2 5 2 f(1) 0, ( 1) is a factor of f( ). Using long division, ( 1)(2 3 2) 0 ( 1)(2 1)( 2) 0 11, or 2 2 x x x x x x x x x x xx x x x x x x x                       

4 7(b) 200 100 2 2 200 100 Let f( ) ( 3) ( 2) . When divided by 5 6, f( ) ( 5 6)Q( ) , where and are real constants. f( ) ( 2)( 3)Q( ) By Remainder Theorem, f(2) (2 3) (2 2) 1 2 1 f(3) (3 3 x x x xx x x x x ax b ab x x x x ax b ab                         200 100) (3 2) 1 3 1 Solving the above two equations, 2, 5 The remainder is 2 5. ab ab x            8 2 2 When 2 and 3, 1, 4 1 ...(1) 3 1When 4 and 6, 7 , 2 116 7 ...(2) 62 Solvng equations (1) and (2) simultaneously, 1 ,32 13 2 bS ax y x y S ba x y S ba ab Sx y                  

5 9(a) 2 2 2 0 720 30 30 390 2 6 4cos 30 9sin 3022 6 4 4sin 30 9sin 30 22 4sin 30 9sin 30 2 0 22 4sin 30 1 sin 30 2 0 22 x x xx xx xx xx                                                                              1sin 30 or 2 (rejected) 24 B.A. = 14.48 30 165.52 or 374.482 271.0 or 689.0 x x x              9(b) 2 2 22 2 22 2 2 2 22 tan cotLHS 2sin tan cot sin cos cos sin2sin sin cos cos sin sin cos2sin sin cos 2sin (sin cos ) sin cos 1 RHS xxx xx xx xxx xx xx xxx xx x x x xx             10(i) e ln ln bxy a y a bx    x 1.5 2.5 3.5 4.5 5.0 6.0 y 9.9 6.0 3.6 2.2 1.7 1.4 ln y 2.29 1.79 1.28 0.79 0.53 0.34

6 10(ii) ln 3.04 20.9 (accept 20.0 to 21.8) 0.500 (accept 0.47 to 0.53) a a b     10(iii) The abnormal reading is 1.4. When x = 6.0, ln y is estimated to be 0.04. Hence, the correct value of 0.04e 1.04y . (accept 1.00 to 1.04) 11 1 1 1 f( ) 1 Let f ( ) f( ) 1 1 f ( ) , 1 1 2f( 1) f (2) 4 2 1 4 1 2 1 2 2 4 2 ax x yx yx a xy ay x axx x aa aa a                         

7 12 Hence, the range of the function is 1 f( ) 5x   . 13(a) Hence, the range of values of x such that 110 (2 1)( 2)2 x xx   is 0x . 13(b) e 1 e 1 ln( 1)xxy y x y        First transformation: reflection in the line y = x. Second transformation: stretching / scaling parallel to the y-axis by a factor of 2. 14(i) Height of the flagpole XY 30 tan 25 13.989 14.0 m (3 s.f.)   14(ii) The bearing of T from S and the bearing of T from Y are 042 and 075 respectively.

8 42 , 180 75 105 180 42 105 33 Alternatively, by stating that 75 , 75 42 33 sin105 sin 33 30 sin105sin 33 53.205 53.2 m (3 s.f.) YST SYT YTS NYT YTS ST SY ST                              14(iii) Let the midpoint of S and T be M. 1 26.60252SM ST   Using Cosine Rule, 2 2 2 22 2 cos 30 26.6025 2 30 26.6025 cos 42 20.531 MY SY SM SY SM YSM MY                 Angle of elevation of X from the man 11 13.989tan tan 20.531 34.3 XY MY   15 2 2 22 1 2cot 1cos 1 tan 1sin 1 2cot cos sec cot 1 2cot cos sec cot lies in the 4th quadrant, sec 0, cot 0, sec sec , cot cot 1 2cotHence, the expression cos sec cot 1 2 1                                     16(i) 22 22 6 4 21 0 ( 3) ( 2) 34 x y x y xy          Hence, the coordinates of P are (3, 2) , and the radius of 1C is 34 units.

9 16(ii) 22 22 2 2 4 6 ...(1) 6 4 21 0 ...(2) From (1), 4 6 ...(3) Sub (3) into (2): (4 6) 6(4 6) 4 21 0 17 68 51 0 4 3 0 ( 1)( 3) 0 yx x y x y xy y y y y yy yy yy                        1 or 3 2 or 6 ( 2,1), (6,3) y x MN       16(iii) Equation of : 4 6 13 42 Gradient of the perpendicular bisector 1= 4 14 6 2 3 1Midpoint of , (2, 2) 22 MN y x yx MN         Equation of the perpendicular bisector: 2 4( 2) 4 10 yx yx       16(iv) Since Q is the centre of 2C , it lies on the perpendicular bisector of chord MN. Hence, the coordinates of Q satisfy the equation 4 10 ...(1)yx  Also, since the radius of 1 2212 units, 1 2212QM  . 22 22 1( 2) ( 1) 221 2 221( 2) ( 1) ...(2) 4 xy xy         Substitute (1) into (2):

10 22 2 2 221( 2) ( 4 9) 4 11917 68 0 4 4 16 7 0 (2 1)(2 7) 0 17 or 22 8 or 4 xx xx xx xx x y                 Since Q lies above the x-axis, only the pair of solution 1 ,82   is accepted. Hence, coordinates of Q are 1 ,82   . 16(v) Area of PMQN 2 13 6 2 31 22 2 3 8 1 2 142 units 2   