HKDSE Mock Paper

New Century Mathematics (Second Edition) 4B

Term Exam Paper S4

Paper 1 Marking Scheme

/ Solution / Marks /
1. / – 7
=i – 7
= –7 + 6i
∴ Real part = –7
Imaginary part = 6 / 1M
1A
1A
------(3)
2. / (a) / ∵ The value of the denominator x + 5 cannot be 0.
∴ x + 5 ¹ 0
x ¹ –5
∴ The domain is all real numbers except –5. / 1A
(b) / f(–3)
=
= 3 / 1M
1A
------(3)
3. / (a) / Substitute x = –4 and y = 0 into y = –x2 + 4x + k.
0 = –(–4)2 + 4(–4) + k
0 = –32 + k
k = 32 / 1A
(b) / Substitute x = 2 into y = –x2 + 4x + 32.
y = –(2)2 + 4(2) + 32
= 36
∴ The coordinates of the vertex are (2 , 36). / 1M
1A
------(3)
4. / ∵ The equation has two equal real roots.
∴ D = 0
k2 – 4(1)(–4k) = 0
k2 + 16k = 0
k(k + 16) = 0
k = 0 or –16 / 1M
1M
1A + 1A
------(4)
5. / (a) / f(–2) = –5
2(–2)3 – (–2)2 + k(–2) + 3 = –5
–16 – 4 – 2k + 3 = –5
–2k = 12
k = –6 / 1M
1A
(b) /
= 2– – 6+ 3
=–– 3 + 3
= 0
∴ 2x – 1 is a factor of f(x). / 1M
1A
------(4)
6. / (a) /
=
= / 1M
1A
(b) / Minimum value of
= minimum value of
=
= / 1M
1A
------(4)
7. / ∵ AM = BM
∴ OM ⊥ AB
AM =
=´ 24 cm
= 12 cm
In △OAM,
OA2 = OM2 + AM2
OM =cm
= 5 cm
∵ CD = AB
∴ ON = OM
= 5 cm / 1M
1M
1M
1A
------(4)
8. / ÐCDE =
=´ 110°
= 55°
∵ =
∴ ÐCED = ÐCDE
= 55°
In △CDE,
ÐECD + ÐCDE + ÐCED = 180°
ÐECD + 55° + 55° = 180°
ÐECD = 70°
ÐBAE = ÐECD
= 70° / 1M
1M
1M
1M
1A
------(5)
9. / (a) / Coordinates of B = (2 , –6) / 1A
(b) / Slope of L1 = –=
∵ L2 ⊥ L1
∴ Slope of L2 ´ slope of L1 = –1
Slope of L2 ´ = –1
Slope of L2 = –
The equation of L2 is
y – (–6) = –(x – 2)
2y + 12 = –3x + 6
3x + 2y + 6 = 0 / 1A
1M
1M
1A
------(5)
10. / (a) / When y = 0,
x2 + 2x – 24 = 0
(x + 6)(x – 4) = 0
x = –6 or 4
∴ The coordinates of A are (–6 , 0).
The coordinates of B are (4 , 0).
y-intercept = –24
∴ The coordinates of C are (0 , –24). / 1M
1A
1A
1A
------(4)
(b) / x-coordinate of the mid-point of AB
=
= –1
∴ The equation of the axis of symmetry is x = –1. / 1M
1A
------(2)
11. / (a) / 2cos q = –
cos q = –
q = 180° – 30° or 180° + 30°
= 150° or 210° / 1M
1A + 1A
------(3)
(b) / Take θ = 150°.
–
=–
= –+
= –+
= 0 / 1A + 1A
1A
------(3)
12. / (a) / The equation of L2 is
y = 2x – 4
2x – y – 4 = 0 / 1M
1A
------(2)
(b) /
(2) ´ 2 – (1): x – 1 = 0
x = 1
Substitute x = 1 into (2).
2(1) – y – 4 = 0
y = –2
∴ The coordinates of P are (1 , –2). / 1M
1A
------(2)
(c) / x-intercept of L2
= –
= 2
∴ The coordinates of B are (2 , 0).
Coordinates of the mid-point of AB
=
= (1 , –2)
From (b), the coordinates of P are (1 , –2).
∴ P is the mid-point of AB. / 1A
1M
1A
------(3)
13. / (a) /
∴ Quotient = 4x + 9
Remainder = 8x – 38 / 1M
1A
1A
------(3)
(b) / f(x)
= 4x3 – 7x2 – 20x + 36
= 4x3 – 7x2 – 12x – 2 – (8x – 38)
∴ f(x) = (4x + 9)(x2 – 4x + 4)
f(x) = 0
(4x + 9)(x2 – 4x + 4) = 0
(4x + 9)(x – 2)2 = 0
x =or 2 (repeated)
∴ The equation f(x) = 0 do not have three distinct real roots.
∴ The claim is disagreed. / 1M
1M + 1A
1A
------(4)
14. / (a) / Join BC.
ÐDBE = ÐABD
= 48°
∵ AB is the diameter of the semi-circle.
∴ ÐACB = 90°
ÐBED = ÐACB
= 90°
In △BDE,
ÐBDE + ÐDBE + ÐBED = 180°
ÐBDE + 48° + 90° = 180°
ÐBDE = 42° / 1M
1M
1M
1A
------(4)
(b) / ∵ ÐBED = 90°
∴ BD is a diameter of the circle.
i.e. BD = 6 cm
In △BCD,
sin ÐCDB =
BC = 6 sin 74° cm
In △ABD,
ÐBAD + ÐADB + ÐABD = 180°
ÐBAD + 74° + 48° = 180°
ÐBAD = 58°
In △ABC,
sin ÐBAD =
AB = cm
= 6.80 cm, cor. to 3 sig. fig.
∴ The diameter of the semi-circle is 6.80 cm. / 1M
1M
1M
1M
1A
------(5)
15. / ¸´
=´´
=´´
= / 1M + 1M
1M
1A
------(4)
16. / (cos q + 1)(sin q + cos q) = 0
cos q + 1 = 0 or sin q + cos q = 0
cos q = – or + = 0
cos q = – or tan q + 1 = 0
cos q = – or tan q = –1
When cos q = –,
q = 180° – 45° or 180° + 45°
= 135° or 225°
When tan q = –1,
q = 180° – 45° or 360° – 45°
= 135° or 315°
∴ q = 135° or 225° or 315°
∴ The equation has three distinct roots. / 1M
1M
1M
1A
------(4)
17. / (a) / Sum of roots = –a
(3 + 5i) + (3 – 5i) = –a
6 = –a
a = –6
Product of roots = b
(3 + 5i)(3 – 5i) = b
9 – 25i2 = b
9 – 25(–1) = b
b = 34 / 1M
1A
1M
1A
------(4)
(b) / x2 + ax + b = k
x2 – 6x + 34 – k = 0
∵ The equation has real root(s).
∴ D ³ 0
(–6)2 – 4(1)(34 – k) ³ 0
36 – 136 + 4k ³ 0
4k ³ 100
k ³ 25 / 1M
1A
------(2)
18. / (a) / f(x)
= 32x – x2
= –(x2 – 32x)
= –
= –(x2 – 32x + 256) + 256
= –(x – 16)2 + 256
∴ The coordinates of the vertex are (16 , 256). / 1M
1A
------(2)
(b) / (i) / When P = 0,
3 log2 = 0
= 1
32x – x2 = 60
x2 – 32x + 60 = 0
(x – 2)(x – 30) = 0
x = 2 or 30 / 1M
1M
1A
(ii) / From (a),
maximum value of 32x – x2
= 256
∴ Maximum value of
=
=
∴ Maximum value of 3 log2
= 3 log2
= 6.28, cor. to. 3 sig. fig.
> 6
∴ The annual profit of the company in selling the toys can be greater than 6 million dollars.
∴ The claim is disagreed. / 1M
1M
1A
------(6)
19. / (a) / Let ÐGBF = x.
ÐCBF = ÐGCF
= 22°
ÐBAF = ÐGBF
= x
∵ M is the circumcentre of △ABC.
∴ M is the centre of the circle.
∴ ÐHBG = 90°
ÐEBH = ÐHBG – ÐCBF – ÐGBF
= 90° – 22° – x
= 68° – x
∵ H is the orthocentre of △ABC.
∴ ÐAEB = 90°
In △ABE,
ÐBAF + ÐAEB + ÐABE = 180°
x + 90° + ÐABH + 68° – x = 180°
ÐABH = 22° / 1M
either one
1M
1M
1A
------(4)
(b) / Extend BH to meet AC at T.
∵ H and M are the orthocentre and the circumcentre of △ABC respectively.
∴ AT = CT and ÐATB = ÐCTB = 90°.
BT = BT
∴ △ATB @ △CTB (SAS)
∴ ÐABH = ÐCBH
i.e. BH is the angle bisector of ÐABC.
∴ The in-centre of △ABC lies on the line passing through B and H.
∴ The claim is agreed. / 1M
1M
1A
------(3)
(c) / Join FM.
ÐCMF = 2ÐCBF
= 2 ´ 22°
= 44°
In △CMF,
∵ CM = FM
∴ ÐFCM = ÐCFM
ÐFCM + ÐCFM + ÐCMF = 180°
2ÐCFM + 44° = 180°
2ÐCFM = 136°
ÐCFM = 68°
In △BDH,
∵ H is the orthocentre of △ABC.
∴ ÐBDH = 90°
ÐBHD + ÐBDH + ÐABH = 180°
ÐBHD + 90° + 22° = 180°
ÐBHD = 68°
∵ ÐBHD = ÐCFM
∴ M, H, C and F are concyclic. / Ð at centre twice Ð at circumference
radii
base Ðs, isos. △
Ð sum of △
Ð sum of △
ext. Ð = int. opp. Ð
Marking Scheme:
Case 1 Any correct proof with correct reasons. / 3
Case 2 Any correct proof without reasons. / 2
Case 3 Incomplete proof with any one correct step and one correct reason. / 1
------(3)
(d) / ∵ BH is the angle bisector of ÐABC.
∴ ÐCBH = ÐABH
= 22°
ÐMBF = ÐCBH + ÐCBF
= 22° + 22°
= 44°
∵ MB = MF
∴ ÐMFB = ÐMBF
= 44°
In △BMF,
ÐHMF = ÐMFB + ÐMBF
= 44° + 44°
= 88°
¹ 90°
∴ FH is not a diameter of the circle passing through M, H, C and F. / 1M
1M
1A
------(3)