2018-DSE MATH EP M2 Please stick the barcode label here. HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2018
MATHEM MA THEMA ATICS Exte Extended nded Part Candidate Number
Module 2 (Algebra and Calculus) Marking Scheme
/
8:30 am – 11:00 am (21 2 hours) This paper must be answered in English
INSTRUCTIONS 1. After the announcement of the start of the examination, you should first write your Candidate Number in the space provided on Page 1 and stick barcode labels in the spaces provided on Page
1, 3, 5, 7, 9 and 11. 2. This paper paper consists consists of TWO sections, sections, A and B. 3. Attempt ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins. Answers written in the margins will not be marked. 4. Graph paper and supplementary answer sheets will be supplied on request. Write your your Candidate Number, Number, mark the question number box and stick a barcode label on each sheet, and fasten them with string INSIDE this book. 5. Unless otherwise specified, all workings workings must be clearly clearly shown. 6. Unless otherwise specified, numerical answers must be exact. exact. 7. The diagrams diagrams in this paper are not necessarily necessarily drawn in scale. scale. 8. No extra time will be given to candidates for sticking on the barcode labels or filling in the question number boxes after the ‘Time is up’ announcement. © 香港考試及評核局
保留版權
Hong Kong Examinations and Assessment Authority All Rights Reserved 2018
Solution SECTION A
Marks
(0 marks)
(
2
1+h
) [(1 h) − 1]e (h 2h)e f (1 h ) − f (1) f (1) lim → h (h 2h)e − 0 lim
1. f 1 + h
=
+
2
=
+
=
0
h
2
=
h
(
2. x + 3
1+h
+
h
→0
lim h + 2 e1+h
=
=
1+h
+
h
→0
(
)
2 e
5
)
=
()
+
C 25 3 2 x 3 + C 35 3 3 x 2
()
()
x 5 + 15 x 4 + 90 x 3 + 270 x 2
=
5
x 5 + C 15 3 x 4
4
+
+
C 45 3 4 x + 35
()
405 x + 243
2
( x 3) x − x
5
4
3
2
( x 15 x 90 x 270 x 405 x coefficient of x (1)(16) (90)(−8) (405)(1) −299 3. (a) 2 sin( B − A ) 2 sin B cos A − 2 cos B sin A sin A sin B (2cot A − 2 cot B ) sin A sin B [cot A (cot A − 2 cot B )] sin A sin B (cot A (3 − 2) cot B ) sin A sin B (cot A cot B ) +
3
=
+
=
+
+
+
+
+
243 x 2
)
− 8
+
=
=
= =
+
=
+
=
+
=
sin B cos A + sin A cos B
=
sin A + B
(
)
(b) By (a), the equation is equivalent to sin
x +
4π 9
+
x +
sin 2 x +
5π
18
13 π 18
=
2 sin
=
2 sin
=
2
− 12
−1
=
2 x +
13π 18 x
4.
(a)
∫
u
( ) du
u5
=
u 5u
=
2018-DSE-MATH-EP(M2)-1
u
=
∫
( )− 1 ln 5 ln 5 5 u (5 ) − ln 5 (ln 5) u
=
2
3π 2 7π 18
5u du
+
C
1
x +
− π6
5π
18
−
x +
4π 9
+
16 x 2
Solution
Marks
( ) 0, x (5 ) 0
(b) When f x
=
2 x
x
= =
0
∫ ∫ 1
Required area Let u
=
x 52 x d x
( )
0
2 x . Then du = 2 d x . 2 1 Required area = u 5u du 4 0 =
=
=
=
5.
(a) Let u
1 4
( ) u(5 ) − u
2
ln 5
1 + x 2 . Then du
=
=
=
=
=
=
∫ ∫
1 2 1
∫ ∫ ∫ x 2 2 x
( ) 1 x (u − 1)√ u du
2 1 2
3 2
+
2
d x
1 2
(u − u ) du
u2
5
(1
3
− +
u2
+
3
x 2
5
C
5 2
) − (1
+
x 2
3
)
3 2
+
C
15 x 3 1 + x 2 d x x 3 1 + x 2 d x
15
5
3 1 + x 2 2 5 1 + x 2 Substitute 0, 2 , =
2
2 x d x .
5
=
0
2
=
y
2
2
x 3 1 + x 2 d x
(b)
(ln 5) 1 2(25) 25 1 0− − − 4 ln 5 (ln 5) (ln 5) 25 6 − 2 l n 5 (ln 5)
∫
=
5u
(
2=
) − ( ) ( ) 3 (1) − 5(1) C
3 2 +
C
+
C = 4
The equation of Γ is 6.
(a) When n
=
y
=
3 1 + x 2
(
5 2
) − 5(1
+
x 2
)
3 2 +
4.
1,
( )( ) ( )( )
L.H.S. = 1 5 = 5 1 2 15 R.H.S. = = 5 = L.H.S. 6 The statement is true when n = 1. r r r + 1 2r + 13 Assume for some r k k + 4 = 6
k =1
(
2018-DSE-MATH-EP(M2)-2
)
(
)(
)
2
+
∈ Z
.
Solution When n
=
r + 1,
r +1
Marks
r
(
k k + 4
k =1
)
) (r 1)(r 5) r (r 1)(2r 13) (r 1)(r 5) 6 r 1 [r (2r 13) 6(r 5)] 6 r 1 (2r 19r 30) 6 (r 1)(r 2)(2r 15) 6 (r 1)[(r 1) 1][2(r 1) 13]
=
(
k k + 4
k =1
=
2
+
=
+
+
+
+
+
+
+
+
=
+
+
+
+
=
+
+
+
=
=
+
+
+
+
+
+
+
6
R.H.S.
The statement is true when n
=
k + 1.
555
(b)
k =333
k
k + 4
112
223
1 24 976
=
=
=
7.
(a)
555
24 976
332
+
k =1
1
+
∈ Z . k ( k 4) − k ( k 4) (555)(556)(1123) − (332)(333)(677)
By mathematical induction, the statement is true ∀ n
+
k =1
6
6
1813
M X = X M
7
3
a
6a
−1
5
b
c
7a + 3 b 42a + 3c
−
−
=
=
a
6a
7
3
b
c
−1
5
− 7b − c
7a 6 a
3a + 30a
5b a 5c 6 a 3b + 5c Comparing corresponding entries, 7a + 3b = a
42a + 3c
=
33 a
− 5c − 6 a
=
7 b c
5b a
−
3 b + 5c Solving, we have b = 2a and c (b) det X =
=
−3a.
=
6a
a
−2a −3a −3a 12a
=
−
2
+
2
9 a2 Since a 0 or otherwise X is a zero matrix, we have det X 0. =
Thus, X is non-singular.
2018-DSE-MATH-EP(M2)-3
3
(c) X −1
1
=
2a
a
−6
3
1
Marks
−3a −6a
9a 2
=
Solution
9a 2
1
( X )− ( X − ) T
1
=
8.
T
1
=
3
1
2
9a
−6 1 (a) f ( x ) − A( x − 4 x 7)− (2 x − 4) −2 A( x − 2) ( x − 4 x 7) f ( x ) −2 A( x − 4 x 7)− − 2 A( x − 2)(−2)( x − 4 x 7)− (2 x − 4) −2 A( x − 4 x 7)− [( x − 4 x 7) − 2( x − 2)(2 x − 4)] −2 A(−3 x 12 x − 9) ( x − 4 x 7) 6 A( x − 1)( x − 3) ( x − 4 x 7) When f ( x ) 0, we have x 2. 2 A f (2) − . 9 Note that A 0 or otherwise f ( x ) is a zero function. Therefore, f (2) Thus, f ( x ) attains its extreme at x 2. f (2) 4 2
=
=
2
2
+
2
=
3
+
3
+
2
2
+
+
2
=
2
3
+
2
=
2
+
2
=
2
+
3
+
=
=
=
0.
=
=
A
22
− 4(2)
=
7
+
4
A = 12
( ) ( x−24− 4( x x − 27)) . (b) Since (−4) − 4(1)(7) −12 < 0, the equation x − 4 x Thus, f x
=
2
2
+
2
2
=
+
7 = 0 has no real solutions.
Therefore, the graph has no vertical asymptotes.
As 0 = deg 12 < deg x 2
( )
no oblique asymptote.
( − 4 x 7) +
=
2, the graph has one horizontal asymptote
The graph has only one asymptote. The claim is disagreed. (c) By (a), 1)( x − 3) ( ) 72( x( x − − 4 x 7) When f ( x ) 0, x 1 or 3. f x
=
2
=
+
3
=
x
x < 1
1 < x < 3
x > 3
f x
−
+
−
()
f 1
()
=
()
3 and f 3
=
3.
( )
( )
Thus, the points of inflexions are 1, 3 and 3, 3 .
2018-DSE-MATH-EP(M2)-4
4
y
=
0 and
Solution SECTION B 9.
(a)
Marks
(0 marks) y
d y d x
1
ln x 2 1
=
=
2 x
Slope of tangent to C at P
1 . 2r 2r .
=
− Let the coordinates of Q be (q, 0). 0 − ln r −2r q − r ln r −4qr 4r Slope of normal to C at P 1 2
=
=
=
q
=
2
+
4r 2 + ln r 4r
The x -coordinate of Q is
4r 2
=
=
d A dr
=
=
When
d A dt
=
ln r
4r
(b) Let the area of PQR be A. A =
+
.
1 4r 2 + ln r 1 ln r r 2 4r 2 1 ln r ln r 2 4r 2 2 ln r 16r 2 1 ln r 2 ln r 16r r 16r 2 ln r 2 ln r
−
·
·
( )
( ) − ( ) ( − ) 16r 2
0,
( − )
ln r 2 ln r
=
0
ln r = 2 r = e 2
or or
0 1 (rej.)
r
1 < r < e2
r > e2
d A dr
+
−
A attains its greatest value when r = e 2 . ln e2 2 Greatest area = 16e2 1
(
=
)
4e 2
2018-DSE-MATH-EP(M2)-5
5
Solution (c) Let OP
=
p
d p dt
Marks
p.
=
r 2
=
r 2 +
1
=
+
r 2
2
2
ln r 2
2
(ln r ) 4
+
(ln r )
− 12
2
2r +
4
ln r dr 2r
When r = e , d p dt
1 2 e 2
=
1 1
· e
=
As
d p dt
e2 +
· e
=
+
e
2
− 12
1 4
2e +
− 12
1
1 2
4
dr dt
dr dt
≤ 32 e
2
≤ 32 e
3
+
1 dr 4 dt
e2 +
4 1
1 dr 2e dt
2
≤ 32e , 1
· e
e2 +
1 4
1 2
dr dt
d A dt
e
2
+
1 4
d A dr dr dt ln r 2 ln r
· ( − ) · dr dt 16r
=
=
2
When r = e , d A dt
=
1
dr · dt 16e 2
1
≤ 16e · 32e 2
=
2e
e2
<
+
e
2
+
1
− 12
4
1 2
2e 2
(e )
=
1 4
3
1 2
2
The claim is correct.
2018-DSE-MATH-EP(M2)-6
6
− 12
dt
10.
(a)
∫
(i)
4
sin x d x
=
Solution
∫
3
(sin x ) sin x d x 3
− cos x sin x
=
3
− cos x sin x
=
4
sin4 x d x
=
sin4 x d x
=
π
(ii)
4
sin x d x
−
=
β
−
∫
(
+
+
+
3
+
1 cos x sin3 x 4 +
3 8
∫
π
−
2
(1 − sin x ) sin x d x
3
sin2 x d x
3
sin2 x d x
3 4
sin2 x d x
π
+
0
π
3 4
2
0
0
∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫
()
x f x d x
=
β
β
( β − u) f (u) du β
=
β
() −
f u du
β
0
=
β
( ) −
f x d x
β
0
β =
()
x f x d x
0
()
f x d x
β
0
β =
β
β
2
()
f x d x
0
(ii) Since sin 4 x is continuous on R and sin 4 π
( − x )
π, we have
x sin4 x d x
=
0
=
=
y
=
π
π
2 0 π 3π 2 8 3π 2 16
sin4 x d x
·
0,
√ x sin x
=
0
sin x
=
0
x
=
π
2
()
x f x d x
0
β
()
x f x d x
0
(c) When
()
u f u du
0
β
π
sin2 x d x
=
0
=
sin4 x d x
− d x. − ( β − u) f ( β − u) du
=
and β
− 3
0
0
2
2
3
π
sin2 x
β
∫ ∫ ∫
)
(1 − cos 2 x) d x
0
x 8 3π = 8 x . Then du =
=
0
∫ ∫ ∫ ∫ ∫ ∫ ∫
cos x 3sin2 x cos x d x
+
3
− cos x4 sin x
=
(i) Let u
3
− cos x sin x
0
(b)
3
− cos x sin x
=
∫ ∫ ∫
Marks
2018-DSE-MATH-EP(M2)-7
or
2π
7
=
sin 4 x, by (b)(i), putting f x
()
=
sin 4 x
Solution
∫ ∫
2π
Volume of the solid
=
π
π
2π
=
Let u
=
x
∫ ∫ ∫
π
(√ x sin x ) 2
π
4
+
+
)
π du
0
π
π
0
(u
+
π
=
π
π sin4 u du
)
4
u sin u du + π
2
0
=
π
=
11.
(a)
d x
=
π
=
2
x sin4 x d x
− π . Then du d x . (u π) sin (u π
Volume =
Marks
∫
π
sin4 u du
0
2
· 316π
· 38π
π2
+
9π3 16
1
(i) (1) 2 1
( ) a − 1 2(a − 1) −1 −12 a
4 a+1
1
−a − 1 −6a − 10 −a − 1 −4a − 16
0
=
4a + 4
a
0 1
4a + 4
a
−a − 1 −6a − 10 0 0 2a − 6 (1)(−a − 1)(2a − 6) 2 (a 1)(a − 3) ( E ) has unique solution if and only if 2(a 1)(a − 3) 0, i.e., a −1 and a Thus, the required range is a < −1 or −1 < a < 3 or a > 3. 1 4(a 1) 18 1 4a 4 18 a a (2) 2 a − 1 2(a − 1) 20 ∼ 0 −a − 1 −6a − 10 −16 1 −1 −12 b 0 −a − 1 −4a − 16 b − 18 0
=
= =
+
+
+
∼
+
1 0 0
4a + 4
a
18
−a − 1 −6a − 10 −16 0 2a − 6 b − 2
− and 2(a − 3) (−a − 1) (−6a − 10) z −16 −3ab 22a − 5b − 38 (a − 3)(a 1)
Therefore, z
b 2
=
y
+
(
)
x + a y + 4a + 4 z
(
Thus, x, y, z
2018-DSE-MATH-EP(M2)-8
)
=
3.
=
y
and
x a2 b +
=
+
=
+
18 a2 b + ab + 10a
− 2b − 50 (a − 3)(a 1) ab 10a − 2 b − 50 −3ab 22a − 5 b − 38 b − 2 , , 2(a − 3) (a − 3)(a 1) (a − 3)(a 1) =
+
+
+
+
8
+
Solution (ii) (1) When a
=
Marks
( )
3, by (a)(i)(2), E is equivalent to
1
3
16
18
−4 −28 −16 0 0 0 b − 2 If ( E ) is consistent, then b − 2 0
1
(2)
0 0
3
16
18
−4 −28 −16 ∼ 0 0 b − 2 ∼ ∼
Let z
=
∈ R.
r , where r
Then x
=
6 + 5r and
y
=
=
1 0 0
0, i.e., b
3
16
−4 −28 −16 0
0
0
16
18
0 1
7
4
0 0
0
0
1 0
−5
6
7
4
0
0
0 1 0 0
− 7r .
− ) ∈ R}. (b) The first three equations are equivalent to ( E ) at a 3. Thus, the solutions are
{(6
2.
18
1 3
4
=
+
5r , 4 7r , r : r =
By (a)(ii)(1), the first three equations are consistent when s By (a)(ii)(2), the solutions are equation,
(
{(6
) − 5(4 − 7r ) − 45r
2 6 + 5r
+
=
−
)
5r , 4 7r , r : r
t
−8
t =
12.
(a)
−→ (−i −−→ AC 3i
(i) AB
= =
+
+
− ) − (4i − 3 j k) −5i
3 j 3k
2 j + 4k i
−→ × AC −−→ −5
AB
=
3
j
k
6
−4
2
4
=
+
−
6 j 4k
(24 8)i − (−20 12) j (−10 − 18)k 32i 8 j − 28k −→ × AC −−→) · AD −−→ (32i 8 j − 28k) · (−i j − 6k) ( AB −32 8 168 =
=
(ii)
+
+
=
=
+
+
=
+
=
+
=
Volume =
+
+
+
144
1 AB 6 144
−−→) · AD −−→| |(−→ × AC
6 24
2018-DSE-MATH-EP(M2)-9
9
=
2.
∈ R} .
Substitute it into the forth
Solution (iii) Let n
Marks
−→ × AC −−→. −−→ · n − DE −→ DA (n) n·n −→ × AC −−→) · AD −( AB (n) n·n −144 (32i 8 j − 28k) =
AB
=
=
=
=
(b)
322 4 13
+
+
82 + 282
(−8i − 2 j
)
7k
+
−−→ −−→ −→ BD · BC − − − → (i) BF − ( BC ) − → − − → BC · BC (4i − 5 j − 2k) · (8i − 4 j =
=
82
+
) (8i − 4 j
8k
42 + 82
− −−→ −−→ −−→ DF BF − BD −2i 4 j 4k −−→ −−→ −−→ (ii) E F DF − DE 1 (i 10 j 4k) 13 −−→ −−→ 1 (i 10 j E F · BC 13 =
+
+
)
8k
2i j + 2k
=
=
+
+
=
=
+
+
=
=
+
+
) · (8i − 4 j
4k
+
)
8k
0
−−→ −−→ Thus, BC ⊥ E F . ⊥ BC and E F ⊥ BC .
(c) Note that BC is the line of intersection of two planes, D F The angle required is equal to ∠ DF E .
(−2i
+
4 j +
−−→ · −−→ ( )( ) √ 1 1 ( 4k) · 2 4 4 · i 10 j 4k) 13 √ 54 6 · 3 13 cos ∠ DF E √
DF E F = DF E F cos ∠ DF E
+
+
=
2
+
2
+
2
=
cos ∠ DF E =
3 13 13
−1 ∠ DF E = cos The required angle is cos−1
√
3 13 13
.
END OF PAPER
2018-DSE-MATH-EP(M2)-10
10
√
3 13 13
2
+
102 + 42 cos ∠ DF E 13
