华中中三E Math Practice C Solution

2024 Sec 3 E.Math Practice Paper C Solution Qn Solutions 1(i) y < 0 1(ii) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 10 1 3 100 1 9 100 9 100 100 9 100 100 (reject +ve since 0) 9 100 10 (since 0) 9 100 a by y a by y a y a by a y a b ayy ab aya ab += += =− =− =−  − = − 2(a) 3( 8)( 4) 7( 8) 3( 8)( 4) 7( 8) 0 ( 8)[3( 4) 7] 0 ( 8)(3 5) 0 528 or or 1 33 x x x x x x xx xx xx − + = − − + − − = − + − = − + = = =− − 2(b) 21 43 3 2(2 1) 4(4 3) 4 2 3 16 12 279 81 33 3 33 4 2 15 16 20 13 13 20 x x x x xx xx x x + − + − + − + = = = + = − = =

3 AND AND AND 2 2 7 5 2 113 2 2 7 5 7 5 2 113 23 7 9 63 92 > 23 3 92i.e. 23 3 x x x x x x x x x x xx x − +  + − + − +  + + − + −  +  − −   4(a) 32 3 2 2 2 2 2 22 2 1 ( 2 1) ( 1)( 1) ( 1) ( 1)( 1) ( 1) ( 1) ( 1) u u u u u u u u u u uu u u u u u u u oru u u u −+ − −+= − + + −= − + + −−= + + + + 4(b) 2 2 2 4 2 4 (4 4 ) ( 2 ) (2 ) (2 ) (2 )(2 1) x x x y y y x x y y x y x y x y x y x y − − + + = − + + − + = − − − = − − −

5 26xy−= 22 8xy+= Sub. 26yx=− into 22 8xy+= 22 22 2 (2 6) 8 4 24 36 8 5 24 28 = 0 (5 14)( 2) 0 2 or 2.8 2 or 0.4 xx x x x xx xx xx yy + − = + − + = −+ − − = == =− =− Points of intersection: A ( )2, 2− and ( )2.8, 0.4B − or : B ( )2, 2− and ( )2.8, 0.4A − 6(i) Centre of circle, C (6, 8)− Radius 226 ( 8) 525r = + − + 25= Alternatively: 22 22 2 2 2 12 16 525 0 ( 6) 36 ( 8) 64 525 0 ( 6) ( 8) 25 x y x y xy xy + − + − = − − + + − − = − + + = Centre of circle, C (6, 8)− and Radius = 25 6(ii) Equation of : 28 28 8 21 21 6 28 4 21 3 3 4 4 3 CQ y x y x yx yx + − +=−− +− =− =− =− As (0,0) lies on 34yx= , CQ passes through O.

6 (iii) Centre of new circle = Midpoint of OQ = (10.5, 14)− Radius of circle = Half of diameter OQ 221 21 282 17.5 =  + = 2 2 2 22 22 Hence, equation of the circle: ( 10.5) ( 14) 17.5 21 110.25 28 196 360.25 21 28 0 21 28 xy x x y y x y x y k q − + + = − + + + + = + − + = =− = 7(i) Since 90 , thereforeBCO =  cos(0.9273) 6 cos(0.9273) 6cos(0.9273) 6cos(0.9273) 1 cos(0.9273) 9.00 (shown) x x xx x x = + += = − = 7(ii) 2 2 2 22 1Area of Sector (8.999 6) (0.9273)2 = 104.307 cm 1Area of (9)(14.999)sin(0.9273)2 = 53.997 cm Shaded Area = 104.307 cm 53.997 cm OBD OBC =+ = − 2 = 50.3 cm (3s.f) 8(i) 7616 43 112 96 64 48 =  = B

8(ii) 45 65 = C 8(iii) 112 96 45 64 48 65 11280 = 6000   =       D 8(iv) The elements represent the total amount Peter earns from the 16-week block of sessions on weekdays and on weekends respectively. 8(v) 36 52   8(vi) ( ) 10 8 3616(1 1) 7 4 52 19776   =      = T 9(i) Equation of line Y = 2.5X + c and (4, 8) lies on the line, 8 = 10 + c c = –2 Hence, the equation of the line: 2.5 2 2.5 2 ( 2.5) 2 2 2.5 2 and 2.5 xy y xy y yx y x ab =− − =− − =− −= − =− =− 9(ii) tan 2.5 1.19 (3 s.f)   = =

10(i) 180 (6 2)Interior 6 120 AFE −= = 10(ii) 1 Bearing of from : = 180 120120 (180 170 ) 2 120 (10 ) (30 ) 080 DF AFE AFN DFE − − − = − −  −  = −  −  = 10(iii) 225 5 2(5)(5)cos120 5 3 AE= + −  = OR 2 5cos30 5 3 AE =   = 2tan , where of elevation 2tan 53 13.0 (1d.p) AE   = =  = = A B C D E F N

11(i) y-coordinate of P 12 1.52 +== 3 4 1.5 27 3 33 11 x x x −  = = = x-coordinate of P = 11 hence, 5 11 2 17 k k + =  = 11(ii) Gradient of line AB = Gradient of line CD = 4 2 1 1 5 2 − =−− Since (17, 1) lies on line CD, 1 2 1(1) (17)2 9.5 1 9.52 y x c c c yx =− + =− + = =− + 11(iii) Equation of is 3 4 27PQ x y −= (Given) Equation of 1 is 9.5 2CD y x =− + Solving simultaneously: 13 4( 9.5) 272 3 2 38 27 13 3 (13,3) xx xx x y Q − − + = + − = = = 11(iv) has coordinates (11,1.5)P Area of the quadrilateral ABPQ 2 1 5 11 13 11 4 2 1.5 3 42 1 [(2 7.5 33 52) (20 22 19.5 3)]2 1 [94.5 64.5]2 15 unit = = + + + − + + + =− =

12(a) 23( ) 1 2 3f x x x x= − + − 23( ) 7 4 6 2h x x x x= − + − 23 23 2 3 2 3 1 23 1 2 3 2 3 12 Let 1 2 3 ( ) 7 4 6 2 I : 1 2 3 2(1 2 3 ) 2 4 6 2 Scale with scale factor 2 parallel to -a xis. II : 2 4 6 2 7 4 6 2 Translate 5 units in the positi y x x x h x x x x y x x x y x x x y x x x y y x x x y x x x = − + − = − + − = − + − −−− = − + −  = − + − = − + − −−− = − + − ve -direction.y 12(b)(i) ( ) 2 2 41 13 1 1 4 111 4 13 44 gradient m y x x y x x −= − = + − = − = − + 12(b)(ii) 13(a) ( ): 1 4newC − 13(b)(i) 1 ( 2)y f x=− x y ()y f x= B ( 3,2)A − (1, 2)C −

13(b)(ii) 2 1 ( )y f x=− x y ()y f x= B ( 3,2)A − (1, 2)C −