Mathematics 2003 Paper 1 Marking Scheme

HOK YAU CLUB HONG KONG MOCK EXAMINATION 2003 MATHEMATICS PAPER 1 (MARKING SCHEME)

3

1.

(2 mn) 6m 2 n 5

= = =

8m3 n 3 6m 2 n 5 4m 3 − 2 3n 5 − 3 4m

3n 2

1A 1M for a m ÷ a n = a m −n 1A ----(3)

2.

(a)

y = mx + t 2 mx = y − t 2 x=

(b)

x=

y −t2 m 10 − ( −2) 2 =3 2

1M 1A 1A ----(3)

3.

(a) (b)

x 2 + 6 x + 9 = ( x + 3) 2 x 2 + 6 x + 9 − 4 y 2 = ( x + 3) 2 − ( 2 y) 2 = ( x + 3 − 2 y ) ( x + 3 + 2 y) = ( x − 2 y + 3) ( x + 2 y + 3)

1A 1M 1A ----(3)

4.

(a)

2 a = 3b = 7c a b c = = 21 14 6

1M

a : b : c = 21 :14 : 6

1A

a b c = = = k , where k is a constant (常數). 21 14 6 We have a = 21k , b = 14k , c = 6k

(b) Let

1M

a 2 + b 2 ( 21k ) 2 + (14 k ) 2 = 2 ac 2 ( 21k ) ( 6k ) =

91 36

1A ----(4)

5.

(a)

f ( x ) = x 2 + ( k + 1) x + k Since f ( x ) = 0 has equal roots (等根), we have ∆ = ( k + 1) 2 − 4(1)( k ) = 0 ( k − 1) 2 = 0

k =1 (b) Sub. k = 1 into f (x) , we have f ( x ) = x 2 + 2 x + 1 . When f (x) is divided by x – 5, Remainder (餘數) = f (5) = 5 2 + 2 × 5 + 1 = 36

6.

1M 1A

1M 1A ----(4)

∠TSR = ∠ STR = ∠TRS = 60o

(Property of equilateral triangle / 等邊 ∆ 性質) 1A

∠PST = 90 − 60 = 30 As SP = ST, ∠PTS = ∠TPS

(base ∠ s of isos. ∆ / 等腰 ∆ 底角)

1A

( ∠ sum of ∆ / ∆ 內角和)

1A

o

o

o

180 o − 30 o = 75 o 2 o ∠UTR = 180 − 75 o − 60 o = 45o ∠PTS =

1A ----(4)

7.

( ∠ s in the same segment / 同弓形內的圓周角)

∠BED = ∠BAD o

= 40 ∠ACD = ∠ CAD (base ∠ s of isos. ∆ / 等腰 ∆ 底角) = 40o ∠ADE = 40o + 40o (ext. ∠ of ∆ / ∆ 外角) = 80o ∠ABE = ∠ADE ( ∠ s in the same segment / 同弓形內的圓周角) o = 80

8.

8 2 + 10 2 − 9 2 cos ∠ACB = 2 × 8 × 10 ∠ACB = 58.7 o a = 40o (alt. ∠ s, 2 lines //) b = 58.7o – 40o = 18.7o The bearing of C from B = N 18.7o W

(a) For L1 : 2 x + ky = 5

8 cm

1A 1A ----(4)

N

1M a

40°

b 10 cm

1A

N

A 9 cm

1M 1A ----(4)

passes P(1 , 3),

2(1) + 3k = 5 k=1 (b) For slope (斜率) of L1 = -2 and L1 ⊥ L2 , we have Slope of L2 =

1A

C

B

9.

1A

1 2

y −3 1 = x −1 2 x − 2y + 5 = 0

1M 1A

1A

Equation of L2 :

1A ----(4)

10. (a) Area of A1 =

1 1 ×8× 6 − × 4 ×3 2 2

= 18 square units (b) Area of A2 = =

1M

9 square units 2

1A

9 2

18

=

1 4

18 1 1− 4

P = −30 + k 1 x + k 2 x 3

1A 1M

= 24 square units

11. (a)

1A

1 1 3 × 4× 3− × 2× 2 2 2

(c) Common ratio = Total area =

1M

(where k 1 , k2 are constant. (常數))

When x = 1, P = -48, we have − 48 = −30 + k1 × 1 + k 2 × 13 k 1 + k 2 = −18 … … … … … … ..(1) When x = 6, P = 72, we have 72 = −30 + k1 × 6 + k 2 × 6 3 k 1 + 36k 2 = 17 … … … … … … ..(2)

1A ----(7)

1A 1M (substitution)

1A

(2) – (1) :

35k 2 = 35 k2 = 1 Sub. k 2 = 1 into (1), we have k 1 = -19. ∴ P = x 3 − 19 x − 30 (b) Let f ( x ) = x 3 − 19 x − 30

1

R = f ( −2) = ( −2) 3 − 19 × ( −2) − 30 = 0 ∴ (x – 2) is a factor of f(x). Hence, f(x) = ( x + 2 )( x 2 − 2 x − 15)

1M + 1A

= ( x + 2)( x + 3)( x − 5)

1A

When f(x) = 0, x = -2 or –3 or 5. So, 5 thousand copies (reject negative) should be sold.

1A 1A ----(9)

12. (a) For tan 30 o =

10 x 2

1M

x = 20 3

1A

Shaded area (陰影面積) =

1 ( 20 3 ) 2 sin 60 o − π × 10 2 2

1M

(300 3 − 100π ) cm2 1A 300 3 − 100π 100π The required probability (所求概率) = 2 × × 1M 300 3 300 3 = 0.478 1A =

(b) (i)

(ii) The required probability = 0.478117768 +

300 3 − 100π 300 3

= 0.634

×

300 3 − 100π 300 3 1M+1A ----(8)

13. (a) x = 100 – (12 + 18 + 16 + 24 + 10) = 20 y = 360 × 24% = 86.4 (b) Number of students who go to school by bus

1M + 1A 1A = 800 × 20% 1M = 160 1A (c) a = 20 × (1 + 10%) = 22 1A b = 18 × (1 + 33? %) = 24 1A % of student by private car = 100 – (22 + 12 + 24 + 16 +24) = 2% 1M o c = 360 × 2% = 7.2 1A ----(9)

1 1 14. (a) [ × 10 2 × θ − × 10 2 × sin θ ] × 3000 = 150000 2 2 50θ − 50 sin θ = 50 θ − sin θ − 1 = 0 Let f (θ ) = θ − sin θ − 1

For f(1) < 0 and f(2) > 0, there exist a root between 1 and 2

1M + 1A

1 1A

Range

mid-value c

f(c)

1<α<2

1.5

-

1.5 < α < 2

1.75

-

1.75 < α < 2

1.875

-

1.875 < α < 2

1.9375

+

1M (for next interval)

1A (for correct sign)

1M

1.875 < α <1.9375 So, θ = 1.9

1A

(b) Area of the road surface (路面面積) = 3000 × 10 2 + 10 2 − 2 × 10 × 10 cos 1.9 1M = 48804.9 m2 Cost of painting (油漆的費用) = $20 × 48804.9 = $976 000

15. (a)

1A 1A ----(11)

1 1 108 × 3 × 4 × sin θ + × 4 × 15 × sin θ = 2 2 5 36 sin θ =

108 5

sin θ =

(b) AA’= 3 sin θ = 3 ×

3 9 = cm 5 5

BB’= 15 sin θ = 3 ×

3 = 9 cm 5

3 5

A’X= 3 cos θ = 3 × So,

A’B’= 12 −

4 12 = cm 5 5

12 48 = cm 5 5

1A 1A

A

1A

3 5 −3 4 For sin θ = , cos θ = = 5 5 5 4 XB’= 15 cos θ = 15 × = 12 cm 5 2

1M + 1A

2

B’

θ

Y A’

θ

X

1M

B

1A

A’B =

tan ∠ABA ' =

( A' B' ) 2 + ( BB ' ) 2 = 9.6 2 + 9 2

1M

= 13.159 cm

1A

AA' 1 .8 = A' B 13.159

1M

∠ABA ' = 7.79 o Hence, the angle of elevation of A from B

is 7.79o

1A ----(11)

16. (a) z = 100 – x – y The constraints are : For z ≥ 0, 100 – x – y ≥ 0 x + y ≤ 100 … … … … … .(1) x ≥ 3 y … … … … … … . (2) z ≤ 6 y , 100 – x – y ≤ 6 y x + 7 y ≥ 100 … … … … .. (3) x ≥ 0, y ≥ 0 … … … … . (4)

1A

1A 1A 1A 1A

y 60

x + y = 100 40 x = 3y

20 x + 7y = 100

x 0

20

40

60

80

100

2A (straight lines) 1A (shaded region) (b)

P = 5x + 10 y + 15 z

= 5 x + 10 y + 15(100 − x − y ) = 1500 − 10 x − 5 y

1A

Draw the line 2x + y = 0 , P attains maximum at (30, 10) So, x = 30, y = 10 , z = 100 – 30 – 10 – 60

1M 1A ----(11)

2y + 5 … … … ..(1) 3 Sub. (1) into C : x 2 + y 2 − 4 x − 14 y + 27 = 0 , we have

17. (a) From L : 3x – 2y = 5 , x =

(

1M

2y + 5 2 2y + 5 ) + y 2 − 4× ( ) − 14 y + 27 = 0 3 3 y 2 − 10 y + 16 = 0

1A

(y – 2) (y – 8) = 0 y = 2 or y = 8 x = 3 or x = 7 So,

P = (3, 2) , Q = (7, 8)

Mid-point (中點) of PQ = ( Radius (半徑) =

1A 3+7 2 +8 , ) = (5, 5) 2 2

(5 − 3) 2 + (5 − 2) 2 = 13

So, equation of the required circle : (x – 5)2 + (y – 5)2 =

1M 13

2

1M+1A

x 2 + y 2 − 10 x − 10 y + 37 = 0 (b) O = ( −

− 4 − 14 ,− ) = (2 , 7) 2 2

Sub. (2, 7) in the circle, L.S. = 22 + 72 – 10(2) – 10(7) + 37 = 0 So, (2, 7) lies on the circle. As PQ is a diameter (直徑) of the circle, ∠POQ = 90 o ( ∠ in semicircle / 半圓上的圓周角) If A

1A

1M 1

lies on the major arc (優弧) PQ, ∠PAQ = 45o ( ∠ at centre twice ∠ at circumference / 圓心角兩倍於圓周角) 1A

If A

lies on the minor arc (劣弧) PQ, ∠PAQ = 180o - 45o = 135o (opp. ∠ , cyclic quad. / 圓內接四邊形對角) 1A ----(11)