华中中三E Math Practice A Solution

2024 Sec 3 E Math Practice Paper A Solution Qn Solutions 1(i) x is negative 1(ii) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 81 64 1 1 64 64 64 (reject +ve since 0) OR1 64 8 (since 0) 1 64 x a bx x a bx a b x a axx ab a a ab =+ =+ −= =−  − = − 2(a) 22 2 9 12 4 6 4 (3 2 ) 2(3 2 ) (3 2 )(3 2 2) a ab b a b a b a b a b a b − + − + = − − − = − − − 2(b) ( ) ( ) ( ) 2 1 1 3 3 2 2 2 2xx x −−−+ − ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 22 2 2 2 2 3 2 18 2 6 2 2 2 8 3 12 12 18 36 6 2 2 x x x x xx x x x x xx − + − − − += −+ − − + − − −= −+ ( ) ( ) 2 2 6 56 6 2 2 xx xx − − −= −+ 2(c) ( ) ( ) 233 3 2 2a b a b+ = + 6 3 3 6 6 4 2 2 4 6 4 2 2 4 3 3 22 22 2 3 3 3 3 2 3 3 2 2 3 a a b b a a b a b b a b a b a b a b ab ab ab + + = + + + += += + =

3 22 933 1 684 33 9 1 81 27 k m x x x    =    942 33 1 4 64 8 3 3 3 3 k m x x x −    =    12 6 4 48 8 3 33 k k mx xx −= 36 2 433 k k mxx− − − = 9; = 4.5km=− 4 3410 3 2 AND 3 2 5 3 8 4 AND 9 6 15 3 4 1 > 2 AND 2 12 1i.e. 2 2 12 xx x x x x x xx x −− −  − −  − −  −  − + − −   Thus, largest integer is 2. 5 23xy−= 223 5x xy y− + = Sub. 23xy=+ into 223 5x xy y− + = 22 2 2 2 2 2 3(2 3) (2 3) 5 27 36 12 3 2 5 0 11 + 33 + 22 = 0 + 3 + 2 = 0 ( 1)( 2) 0 = 2 or = 1 = 1 or = 1 y y y y y y y y y yy yy yy yy xx + − + + = + + − − + − = + + = −− − Points of intersection: ( )1, 2−− and ( )1, 1−

6a 1 0 1 4 2 1 0 0 41 x yy x y    − = −   +    −= + QP 2 11 00 1 xx yy xy x y xy   =     += P 6b(i) 10 24 40 3012 20 60 508 18 1840 1500 1680 1360 1400 1140      = 6b(i) Alternatively: 40 60 10 12 8 30 50 24 20 18 1840 1680 1400 1500 1360 1140         = 6b(ii) 1840 1500 61680 1360 101400 1140 26040 23680 19800      = 6b(ii) Alternatively: ( ) ( ) 1840 1680 14006 10 1500 1360 1140 26040 23680 19800   = 7a(i) 2 2 240 35 70cos 2 40 35 137.8 (1 d.p) ABC ABC +−=   =  7a(ii) 2 1Area of 35 40 sin137.8222 = 470 m (3 s.f) ABC =    

7(b) Let x be the difference of height between the two poles, therefore tan 3.6 40 x= 40 tan 3.6x= Height of pole at A = 5 40 tan 3.6+ m 7.5 m (1 d.p)= 7(c) BH is the shortest distance from B to path AC 11 70 35 40 sin137.82222 13.4285 m BH BH   =     = Let largest angle of depression be  , 3tan 13.4285 12.6 (1 d.p)   = = 7(d) 2 3 Area 2 Area 3 ZC AC ZCB ABC =  = Probability = 2 3 8(i) Centre of circle, C = (2, 3) Radius 222 3 23r = + + 6= units 8(ii) 22 2 2 2 4 6 23 0 At 4, 16 16 6 23 0 6 9 0 ( 3) 0 Repeated root of 3 line 4 is a tangent to the circle, . or Discriminant 36 4(1)(9) 0 4 is a tangent to the circle, . x y x y x yy yy y y xC xC + − − − = =− + + − − = − + = −= = = = − = = Alternatively, the distance from 4 x=− to (2,3) is 6 units, which is equal to the radius. Therefore 4 x=− is a tangent line.

8 (iii) 22 50xx+= (Pythagoras Theorem) 5x= Centre of new circle = (7,8) 22 22 22 ( 7) ( 8) 36 14 49 16 64 36 0 14 16 77 0 xy x x y y x y x y − + − = − + + − + − = + − − + = 9(i) Gradient of line = 3 Equation of line Y = 3X + c and (1, 2) lies on the line, 2 = 3 + c c = –1 Hence, the equation of the line: lg y = 3 lg (x + 1) – 1 lg y = 3 lg (x + 1) – lg 10 y = 9(ii) tan 3 1.25 (3 s.f)   = = 10 (i) 3.11cos 4 0.684 (3 s.f ) OXY −=  = 10(ii) Area of sector XYZ ( ) 2 2 2 1 3.1 16.2 (cos 2) cm24 26.3 cm (3 s.f) −=  = (2, 3) x x

10(iii) 3.112 2 cos 4YOZ − =    at centre of circle 2 at circumference =  ( ) 3.114 4cos cm 4 10.9 cm YQZ r =  −=   = 11(i) Mid-point PQ = ( )9 1 9 3, 5,322 − − − =− Gradient PQ = 9 3 3 9 1 2 + =− + − 2grad bisector= 3⊥ Eqn of bisector⊥ is ( ) 32 53 235 3 2 19 33 y x yx yx − =+ − = + =+ 11(ii) Eqn of QR is 85yx=+ Let R ( ),8 5xx + 2 1985 33 24 15 2 19 22 4 42 22 11 xx xx x x + = + + = + = == 2 168 5 511 11 71 11 5611 y  = + = + = = 25,611 11R 

11(iiii) Let ( ),S x y Mid-pt RS=mid-pt PQ 2 71 9 1 9 311 11,,2 2 2 2 xy ++ − − −=   2 71 10 and 611 11 2510 and 11 11 xy xy + =− + = =− =− Hence, S is 2510 ,11 11 −− . 11(iv) Eqn of line l is 33 22 867 0yx−−= 33 22 867 0 22 867 33 33 2 289 3 11 yx yx yx −−= =+ =+ Gradient of l = Gradient of of bisector⊥ and the lines have different y-intercept, hence the two lines do not intersect. (two lines are parallel and distinct lines) 12 (a) 4 2(2 1) 2 2 22 1 2 1 2 1 xx x x x +−= = −+ + + 12 (b) 12 (c) 1 12 : 6lg( 2) 3lg( 2) : 3lg( 2) 3lg( 2 4) ( ) 3lg( 2) II y x y x I y x y x h x x = − −−−− = − = − −−−− = − + =+