华中中三E Math Practice B Solution

1 2024 Sec 3 EMath Practice Paper B Solution Qn Solutions 1(a) 236 4 41x xy y y  =   2 312 12x yy  =   12 3 12yx y  or   21 231yy x  1(b) 15 1227xx xx     = 22 21 5 71 2xx xx xx       =      53 34 xx xx   =   5 4 x x   1(c) 22 2 azx y z   2 22 2 azxy z   2 222x yz z a   2 22ax y z z   or  2 1az x y  2(a) 13 221x x   21 3 2 21xx x x   243 1 0xx    14 1 0xx  11 or 4xx   2(b) 2 6x yx 6xy  6 yx  sub into first eqn   2 66x xx 2 56 0xx    61 0xx  6 or 1xx   0 or 7yy 

2 3(a) P = 850 . 7 350 . 4   3(b)(i) C = 7.00 1.20 0.60    3(b)(ii) PC = 850 . 7 350 . 4   7 1.2 0.6    = 62.42 27.24   3(b)(iii) The top elem ent represents the total cost for handmade toys while the bottom element represents the total cost for machine-made toys. 4(i) 2() 2 8 1 5fx x x   =   22 224 2 2 1 5xx   =  2 22 2 3x  4(ii)  2 2 22 2 3 0 23(2 ) 2 23(2 ) 2 23 232 or 222 x x x x         4(iii) x = 2 5(i) Gradient =  75 13 10 1   = 8 8YX c  ; at (1, 13), 13 8(1) c   c = 5 3 85y xx    4385yx x  5(ii) 85YX   215 8 5 5   She’s incorrect.

3 6(i) ^ 1 102 cos 2.4630AO B   6(ii) Major Arc Length APB =  30 2π 2.4619 = 114.6  115 cm 6(iii) Percentage of area below water level =     22 2 1130 2.4619 30 sin 2.461922 100% π3 0    29.2% 7(a)(i)   1yx x k  At D,   21 20 1 20 k  19k  7(a)(ii) A(1, 0), B(19, 0), C(0, 19) Area of ABC = 1 20 192  = 190 units 2 7(a)(iii) Gradient of CD =  21 19 20 0   = 2 Equation of CD through A: 2 2yx  7(b)  22 22 22 22 2 244 24 4 0 (1 )1 ( 2 )4 4 0 (1 )( 2 )3 centre : 1, 2 3 units Gradient of tangent 1 422 5 21 5 3 4 Equation of tangent: 14 3 2 54 5 35 42 xy xy xx yy xy xy radius yx yx                    

4 8 (i)  22 22 2 2 67 0 ( 1 ) int : 1, 4 1:, 32 1 1 2 2 Equation of line : 42 ( 1 ) 22 ( 2 ) (2 2) 6 7 (2 2) 0 46 4 0 23 2 0 (2 1)( 2) 0 0.5 or 2 :( 2 , 2 ) yxx y Po P m m yx yx xx x x xx xx xx xx Q                            8(ii)  perpendicular bisector 1: ,3 , : 2, 22 1 2 31midpoint of PQ : , 42 Equation of perpendicular bisector : 11 3 22 4 11 28 PQ m yx yx               9(i) 1 12 23 :l n l n ( 3 ) Translate 3 units in negative -direction : ln( 3) 4ln( 3) Scale with scale factor 4 parallel to -axis : 4 ln( 3) 4ln( 3) Reflect about -axis Iy x y x x II y x y x y III y x y x x                 

5 9(ii) 10(i)       22 2 22 2 1 6 14.8 10cos 2 6 14.8 6 14.8 10cos 29.2 (1dp)2 6 14.8 DBC DBC         10(ii) 1 sin sin 6sin 29.194sin 17.0 (1dp)10 DCB DBC BD CD DCB         Bearing of B from C = 32 17.0 049.0   10(iii) Shortest distance from D to BC  sin 6sin 29.194 2.93 km (3sf) BD DBC   10(iv) TD = 6 tan 47.5 6.548 km 6548 m 10(v) 6.548tan 2.927  65.9 (1dp) 

6 11(i)     :, 91 3,1 , 922 9 12 11 13 92 5 :1 1 , 5 Sa b ab a a b b S              :, 31 4,1 , 922 1, 4 :1 , 4 Rc d cd cd R    11(ii)  2 Area of PQRS 11 1 9 3 111 54 1 3 1 452 1 44 13 126 15 (5 36 39 154)2 116 units          12(a) () ( ) (1 ) (2 ) 1, 2 yx xp x q yx x x pq     12(b) 2 32 32 32 41 ( 1 ) (1 ) (2 ) ( 2 ) 4 44 ( 2 ) 44 4 8 14 4 8 447 1 0 yx yx x x yx x x yx x x xx x x xxx                 