Paper G Sec 3 solutions for Student

  Paper G Sec 3 Solutions for students Qn Solutions 1(i) 2 32 2 42 x x x   =   32 2 22 2 x x xx    =   32 2 22 2 x xx x    =     32 2 2 22 xx xx    =   7 22xx 1(ii) 2 23 2 2 32 2 2 4 42 8 7( 2 4 ) ( 2)( 2) ( 2)( 2 4) 71 (2 ) (2 ) (2 ) 72 9 xx x xx x xx xx xx x xx x x x            2(a) 1125x x   11 2 5 25 522 5 21 51 0 21 51 0 21lg lg51 0 11 or log2 10lg 5 11 or lg 2 lg 5 lg 5 lg 2 xx x x x x x x xx xx                   

  2(b) 2 2 2 15 1( 5 ) 1( 5 ) 11 0 2 5 11 24 0 (8 ) (3 ) 0 8 (n.a.) or 3 xx xx xx xxx xx xx xx               2 (c)   Area of rectangle =length (breadth) 8+2 3 (5 12)(breadth) 82 3breadth 51 2 breadth 82 3 5 1 2 51 251 2 40 8 12 10 3 2(6) 25 12 52 16 3 10 3 13 42 3 breadth is 4 2 3 cm .                3(a)   2 31 0xk x kk  Two non-zero real roots, Discriminant,     2 2 2 2 40 34 1 0 13 4 0 13 4 0 401 13 ba c kk k kk kk k or k or k        

  3(b)(i)        1 4g3 2 g 43 2 4 32 32 4 326 4 31 0 2 10 2 3 x x Let x y gyx yx yx yx xy x y x yy yx y                1 10 2 : , 3 3 xgx x     Or  1 4 2 , 3 3gx x x    3(b)(ii) Domain of 1g : 3x . Range of 1g :   1 2gx  . 4(a) (i)      32 32 Let P 3 2 7 2 P1 31 21 7 2 0 yyyy      1y  is a factor of  P y . 4(a) (ii)      32 2 32 2 2 327 2 1 3 2 3 2 3 2 3 3 2 2 yyy y y b y yb y y yb y yb y b y               Comparing coefficient of 2y  32 5 b b          32 2327 2 1 35 2 1 3 1 2 yyy y yy yy y          Note : Long Division working acceptable

  4 (a) (iii) 64 22 481 42yy y  64 2 642 2 23 22 2 22 2 24 8 14 2 0 3 ( 8) 2 ( 4) 7 ( 2) 20 Replace " " by 2 3 ( 2) 2 ( 2) 7 ( 2) 20 (2 1)(6 1)(2 1) 0 11 or 26 yy y yyy yy yyy yyy yy             5 (i)    6 6 6 6 6 log 2 1 log log 2 1 loglog log 2 1 x x x (x ) xx x    5 (ii)         66 66 6 2 log log 2 1 log 1 log log 2 1 1 log 2 1 1 216 26 0 23 20 3 (n.a.), 22 xx xx xx xx xx xx xx xx            6(i)   2 2 2 1 42 1 2 82 1 = 1 92 91min. point : 1 , or 1 , 422              yx x xx x

  6(ii)      2 2 2 1 42 1 2 82 When 0 1 2802 1 42 02 4 or 2 Points : 4 , 0 or 2 , 0 yx x xx y, xx (x ) (x ) xx            6 (iii) 6 (iv) For 21 42 x xk to have 3 solutions, 104 2k . Or 4.5k  7 (a) 0: xye  Original graph I : 2 xye  Stretch graph of xye  parallel to y-axis, scale factor 2. II: 2 xye  Reflect graph of 2 xye  along x-axis.

  7(b) 8 (i)   22 2 22 22 22 31 7 621 0 4 9 0 623 9 0 22 0 3 1 39 xy xy xy xy xy x y ax by c a b c            8 (ii)  :3 , 6B 8 (iii) Equation of line BC : 22 22 2 mm 60m2 30 2( 1 ) 6 2 39 0 (2) Sub. (1) into (2) : 46 4 3 9 0 52 3 9 0 (5 13)( 3) 0 13 or 3 ( -coordinate of point )5 26 5 13 26 3 1: , or 2 , 555 5 5                           OB BC BC yx xy xy xxx x xx xx x xx B y C

  9 (a) 22 3 21 82 42 2 462332 11 64 3 3 20 4 31 20 41 23 1 10 2 31 22                         CB A C 9(b) (i) 400 250 50 450 200 140Q  Or 450 200 140 400 250 50  Q 9(b) (ii) 30 0 18 0 20 0 . P . .   9(b) (iii) 30 0400 250 50 1200 450 10018 0450 200 140 1350 360 28020 0 1750 1990               VQ P . . . 9(b) (iv)   Total Amount of profit 1750=1 1 1990 3740    OR   Total Amount of profit 1= 1750 1990 1 3740    Therefore, total amount of profit made was $3740.

  10 Plotting 2x y against x .  2 2 2 2 5( 1 )gradient = 26 3gradient = 2 352 2 21 0 3 6 31 6 2 OR 38 2 xy x xy x xy x y xx           11(i) Eqn AD : 21 6 1 82 yx y x    Gradient AD = gradient BC = 1 2 Equation of BC:  18 1 22 2 118 112 1 7 --- 32 y x yx yx      Equation of AB:  2 3 --- 1yx  1 72 x  = 23x  14 4 6 8 5 3 1 5 823 5 1 6 5 xx x y        31:1 , 655B 

  11 (ii) Gradient of AB = 2 Gradient of BC = 1 2  mm 12 2 1 ABB C    Hence line AB is  to line BC. Thus o90ABC . 11 (iii)    Let : , midpoint of = midpoint of 4.4 5.8 1.6 12 6.2 2,,22 2 2 4.4 5.22 6 5.8 2.12 10 :6 , 1 0 Ha b AH BD ab a a b b H                 11 (iv) Triangle AEC and Triang le ABC share common height, area of area of AEAEC ABC AB  22 2441 6 53 555 10404 5            AE units 22 23 4 141 5655 5 5 180            AB units

  area of area of 10404 area of 5 area of 180 area of 17 area of 5 AEAEC ABC AB AEC ABC AEC ABC          OR Using shoelace method  2 Area of triangle 4.4 22 16 441 5.8 18 35 5.82 1 1203.6 601.8 units2 AEC     2 Area of triangle 4.4 22 1.6 441 5.8 18 6.2 5.82 1 354 177 units2 ABC     area of 601.8 area of 177 area of 17 area of 5 AEC ABC AEC ABC       12 (i) 200 115 85PQR        2 2226 30 2 26 30 cos85 =1440.037041 RP = 37.94781998 37.9 (3 . .) RP km s f     12 (ii) 2 (North) 360 200 160 ( ) 180 160 20 sin sin 85 30 30sin 85sin 0.78755 51.957 52.0 (1 ) PQ QP North QPR PR QPR PR QPR dp                    20 51 957 072 0 1 bearing of R from P . = . ( dp )     