HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
1.
( x8 y 7 )2 x16 y14
= 5 −6 = x11 y 20
x 5 y −6
x y
2.
Ax = (4 x + B )C
Ax = 4Cx + BC
Ax − 4Cx = BC
x( A − 4C ) = BC
BC
x=
A − 4C
3.
2
3
+
4x − 5 1 − 6x
2(1 − 6 x) + 3(4 x − 5)
(4 x − 5)(1 − 6 x)
2 − 12 x + 12 x − 15
=
(4 x − 5)(1 − 6 x)
−13
=
(4 x − 5)(1 − 6 x)
=
4.
(a)
5m − 10n = 5( m − 2n)
(b) m 2 + mn − 6n 2 = (m + 3n)(m − 2n)
(c)
m 2 + mn − 6n 2 − 5m + 10n
= (m + 3n)( m − 2n) − 5( m − 2n)
= (m − 2n)( m + 3n − 5)
5.
Let x be the number of female members
Then the number of male members = x (1 + 40%) = 1.4 x
x + 1.4 x = 180
2.4 x = 180
x = 75
Then the required difference is 1.4 x − x = 0.4 x = 30
1
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
6.
(a)
x + 6 < 6( x + 11)
x + 6 < 6 x + 66
−60 < 5 x
x > −12
Therefore, x > −12 or x ≤ −5
The compound solution is all real numbers
(b) The greatest negative integer is −1
7.
(a)
∠AOB = 135° − 75° = 60°
(b) OA = OB = 12 , AB = 12 2 + 122 − 2(12)(12) cos 60° = 12
The perimeter is 12 + 12 + 12 = 36
(c)
8.
∆AOB has 3 fold of rotational symmetry
(a) Let f ( x) = ax + bx 2 , where a and b are non-zero constants
f (3) = 48
3a + 9b = 48
a + 3b = 16......(1)
f (9) = 198
9a + 81b = 198
a + 9b = 22......(2)
(2) − (1)
b =1
a = 13
Therefore, f ( x) = 13 x + x 2
(b) f ( x ) = 90
x 2 + 13 x − 90 = 0
( x + 18)( x − 5) = 0
x = −18 or x = 5
9.
(a)
x = 2+4 = 6
y + 15 = 37
y = 22
z = 37 + 3 = 40
(b) c = y − 13 = 22 − 13 = 9
b = 13 − x = 13 − 6 = 7
The required probability =
9+7 2
=
40
5
2
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
10. (a) Let M ( x, y ) be the mid-point of A and B
x=
5 + 13
7 +1
=9, y =
=4
2
2
Slope of AB =
Slope of Γ =
7 −1
3
=−
5 − 13
4
4
3
Equation of Γ :
4
y − 4 = ( x − 9)
3
4 x − 3 y − 24 = 0
(b) put y = 0 into the equation of Γ
x=6
i.e. H = (6, 0)
put x = 0 into the equation of Γ
y = −8
i.e. K = (0, − 8)
as ∠HOK = 90° , HK is the diameter of the circumcircle of ∆HOK
diameter of the circumcircle = (6 − 0) 2 + ( −8 − 0) 2 = 10
The circumference of C = 10π > 30
Therefore, the claim is correct
3
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
11. (a) Let V cm3 be the original volume of milk in the vessel
3
V
 12 
= 
V + 444π  16 
V
27
=
V + 444π 64
64V = 27V + (27)(444π )
37V = (27)(444π )
V = 324π
Therefore, the final volume of milk in the vessel is
324π + 444π = 768π cm 3
(b) Let r cm be the radius of the final surface
1 2
π r (16) = 768π
3
r = 12
The final area of the wet curved surface
= π (12) 122 + 162
= 240π cm 2
< 800 cm 2
Therefore, the claim is disagreed
a + 11 = 11 + b + 4
a =b+4
If a = 12 , b = 8
If a = 13 , b = 9
(b) (i) The greatest possible median is attained when the extra ages are 7, 8,
9, 10 and a = 13 , b = 9
12. (a)
The median =
8+8
=8
2
(ii) The least possible mean is attained when the extra ages are 6, 7, 8, 9
and a = 12 , b = 8
The mean =
6 × 12 + 7 × 13 + 8 × 12 + 9 × 9 + 10 × 4
= 7.6
50
4
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
13. (a) In ∆ACD and ∆ABE
∠ADC = ∠AEB (given)
AD = AE (sides opp. eq. ∠s)
CD = CE + ED
= BD + DE
= BE
Therefore, ∆ACD ≅ ∆ABE (SAS)
(b) (i)
AD = AE and DM = EM = 9 cm
Therefore, AM ⊥ DE (prop. of isos. ∆ )
AM = 152 − 92 = 12 cm
(ii)
AB = 162 + 122 = 20 cm
BE 2 = (7 + 18)2 = 625
AB 2 + AE 2 = 202 + 152 = 625
i.e. AB 2 + AE 2 = BE 2
therefore, ∆ABE is a right-angled triangle (Converse of Pyth.
Theorem)
5
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
14. (a)
p ( x) = (lx 2 + 5 x + 8)(2 x 2 + mx + n)
= 2lx 4 + (10 + lm) x3 + (16 + 5m + nl ) x 2 + (8m + 5n) x + 8n
Comparing coefficients
2l = 6
l =3
10 + lm = 7
3m = −3
m = −1
p (−2) = p (2)
[3(−2) 2 + 5(−2) + 8][2(−2) 2 − (−2) + n] = [3(2) 2 + 5(2) + 8][2(2) 2 − (2) + n]
10(10 + n) = 30(6 + n)
n = −4
(b) p ( x ) = 0
(3 x 2 + 5 x + 8)(2 x 2 − x − 4) = 0
3 x 2 + 5 x + 8 = 0 or 2 x 2 − x − 4 = 0
For 3 x 2 + 5 x + 8 = 0
∆ = 52 − 4(3)(8)
= −71
<0
There is no real root
For 2 x 2 − x − 4 = 0
∆ = (−1) 2 − 4(2)(−4)
= 33
>0
There are two distinct real roots
As a result, there are 2 distinct real roots
6
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
15. The required probability =
5! × C46 × 4! 5
=
9!
42
16. Let σ be the standard deviation of the distribution and x be the score of Mary
22 − 61
σ
= −2.6
σ = 15
x − 61
= 1.4
15
x = 82
The range of the distribution
≥ 82 − 22
= 60
> 59
Therefore, the claim is not correct
17. (a) Let d be the common difference
666 + (38 − 1)d = 555
d = −3
Therefore, the common difference is −3
(b)
n
[2(666) + (n − 1)(−3)] > 0
2
n(1332 − 3n + 3) > 0
n(n − 445) < 0
0 < n < 445
Therefore, the greatest value of n is 444
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HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
18. (a)
f ( x) =
−1 2
x + 12 x − 121
3
1
= − ( x 2 − 36 x) − 121
3
1
= − ( x 2 − 36 x + 324 − 324) − 121
3
1
= − ( x − 18)2 − 13
3
Therefore, the coordinates of the vertex is (18, − 13)
(b) The vertex of the graph of y = g ( x ) is (18, 0)
The graph of g ( x ) is obtained by translating the graph of y = f ( x )
upwards by 13 units
1
Therefore, g ( x) = f ( x) + 13 = − ( x − 18) 2
3
−1 2
x − 12 x − 121
3
h( x ) = f ( − x )
(c) Let h( x) =
i.e.
f ( x ) is reflected in the y-axis
8
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
19. (a) In ∆ABD
10
15
=
sin ∠ADB sin 86°
10sin 86°
sin ∠ADB =
15
∠ADB = 41.68560132°
∠ABD = 180° − 86° − 41.68560132°
= 52.3°
In ∆BCD
52.31439868
CD = 82 + 152 − 2(8)(15) cos 43°
= 10.7 cm
(b) In ∆ABC
10.65246974
AC 2 + BC 2 = 62 + 82 = 100 = 102 = AB 2
Therefore, ∠ACB = 90° (converse of Pyth. Theorem)
In ∆ABD
AD = 102 + 152 − 2(10)(15) cos ∠ABD = 11.89964475 cm
In ∆ACD
AC 2 + CD 2 = 149.4751116
AD 2 = 141.6015451
I.e. AC 2 + CD 2 ≠ AD 2
Therefore, ∠ACD ≠ 90°
Hence, C is not the projection of A on BCD
∠ABC is not the angle between AB and the face BCD
The claim is disagreed
9
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
20. (a) I is the in-centre
Then ∠IPO = ∠IPQ = x
J is the circumcenter
Then JO = JP = JQ
∠JOP = ∠JPO = x
( base ∠s, isos. ∆ )
( base ∠s, isos. ∆ )
∠JQP = ∠JPQ = x
( base ∠s, isos. ∆ )
∠JOQ = ∠JQO = y
∠POQ = x − y = ∠PQO
Therefore, PQ = PQ ( sides opp. eq. ∠s )
P
x
x
I
x
x
O
y
y
J
(b) (i)
Let P = ( a, 19)
PO = PQ
(a − 0) 2 + (19 − 0) 2 = (a − 40) 2 + (19 − 30) 2
a 2 + 361 = a 2 − 80a + 1600 + 121
a = 17
i.e. P = (17, 19)
Let the equation of C be x 2 + y 2 + Dx + Ey + F = 0
Put O (0, 0) , F = 0
Put (17, 19) ,
17 2 + 192 + 17 D + 19 E = 0
17 D + 19 E = −650......(1)
Put (40, 30) ,
402 + 302 + 40 D + 30 E = 0
4 D + 3E = −250......(2)
Solving (1) and (2)
D = −112 , E = 66
Therefore, the equation of C: x 2 + y 2 − 112 x + 66 y = 0
10
Q
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
(ii) Method I
Let y =
3
x + c be the equation of the tangent
4
Put into C
3
3
x 2 + ( x + c) 2 − 112 x + 66( x + c) = 0
4
4
9
3c
99
x 2 + x 2 + x + c 2 − 112 x + x + 66c = 0
16
2
2
25 2 3c − 125
x +
x + c 2 + 66c = 0
16
2
2
25 x + 8(3c − 125) x + 16c(c + 66) = 0
∆=0
64(3c − 125) 2 − 4(25)(16c)(c + 66) = 0
9c 2 − 750c + 15625 − 25c 2 − 1650c = 0
16c 2 + 2400c − 15625 = 0
(4c − 25)(4c + 625) = 0
25
625
or c = −
4
4
3
25
3
625
i.e. L1 : y = x +
and L2 : y = x −
4
4
4
4
c=
S = (−
25
25
625
625
, 0) , T = (0, ) , U = (
, 0) , V = (0, −
)
3
4
3
4
ST = (−
25
25
125
− 0)2 + (0 − )2 =
3
4
12
625
625
3125
− 0) 2 + (−
− 0) 2 =
3
4
12
Height of the trapezium = Diameter of circle
UV = (
−112 2 66 2
) + ( ) = 130
2
2
Area of the trapezium
= 2× (
1 125 3125
= ×(
+
) × 130
2 12
12
1
= 17604
6
> 17000
Therefore, the claim is correct
11
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
Method II
In the figure, M and N are the feet of the perpendiculars from T and S
to UV respectively
Let θ be the inclination of L1 and L2 ,
Then tan θ =
3
3
4
, sin θ = , cos θ =
4
5
5
∠SUV = θ and ∠TVU = 90° − θ
PR = diamater of the circumcircle
=2 (
−112 2 66 2
) +( )
2
2
= 130
= TM = SN
In ∆SUN ,
SN
SU
130 650
SU =
=
3
3
5
sin θ =
In ∆TVM ,
12
HKDSE2016 Mathematics (Compulsory Part) Paper 1 - Solution
TM
TV
130 650
TV =
=
4
4
5
sin(90° − θ ) =
Area of STUV
1
= × SU × TV
2
1 650 650
= ×
×
2 3
4
1
= 17604
6
> 17000
Therefore, the claim is correct
13