2019 JC1 H2 Promotional Examination Suggested Mark Scheme
Transcript of 2019 JC1 H2 Promotional Examination Suggested Mark Scheme
1
2019 JC1 H2 Promotional Examination Suggested Mark Scheme
1y
x
2
2 a(i) 2x b x b x b
So, x b and x b are asymptotes.Hence, 2 4b b .
2
2 2
ax bx a bx ab ay ax b x b
So, y a is an asymptote.Hence, 3a .
6m (ii)y
x
3y
2x 2x
( 0.578, 0.461) 3(0, )4
3
3
7m
(i)
Since any horizontal line intersects the graph of f at most once, f is one-one and it has an inverse. (shown)
(ii) fR [ 9, )
Since fR [ 9, ) Dg, therefore gf does not exists.
(iii) 2
2
fg : (e 1) 4(e 1) 5 e 2e 8, , 0
x x
x x
xx x
(e 1)x 1)1)1), 0,,,
fgR [ 9, 8)
y
x
y
x
y = – 8(0, – 9)
4
4
(a)
12 2
22
2 2
2
2
d 1 1 1 2d 2
1 1
1 1 22 1
1 1 01
11
y x xx
x
xx x
x xx x
x(b) 5 2 2d d sin
d dd d2 1 2 cosd d
d 2 cos 2d
d cos 2d 2
x xy y xx x
y yx x y y xx x
y x y x x yx
y x x yx x y
5
5 (i) 1sin cosec
4 4 2cosec
PQPQ
QR PQ RS2 cot 4 2 cosec 2cotPS QR
1 121 4 2 cosec 2cot 4 2 cosec 24 2 cosec cot shown
A PS QR
7m (ii) 2d 2 cosec cot cosecd
A
For maximum area, d 0d
A .2
2
2
2 cosec cot cosec 02 cosec cosec cos cosec 0
cosec 2cos 1 0
2cos 1 0 or cosec 0 No solution1cos2
3
3 3 3
dd
A0 0 0
Hence, A is a maximum.
Maximum Area 4 2 cosec cot3 3
4 143 3
343
4 3
mummmm....
a 4 24 24 24 24 224 24 2224 cosecececec cococococ t33
44444 14444 1114444444444444444444444444443
4 24 24 24 24 244 24 24 24 2444 24444 ccccccccosososososososoooososososoosooooosooooosoooososoooosoosoooooooooooooooo ecececec
4444444444444444
6
6 (i) 13
11 33
13
2
2
8 3
38 1 8
1 312 8
1 41 1 3 33 31 ... 2 3 8 2 8
1 1 21 ... 2 8 64
x
x
x
x x
x x
21 1 1 ... 2 16 64
x x
3 18
8 83 3
x
x
Method 2:13
4' 3
7'' 3
'
''
f ( ) 8 3
f ( ) 8 3
f ( ) 4 8 3
1f (0)21f (0)
161f (0)
32
x x
x x
x x
123
1 1 18 3 ... 2 16 64
x x x
3 8 818 3 3
x x
2222222222222222222222222222222222222211111111111111111111111111111111111111111111 ...... 2 111111111111111116 66 66 66 66 666 66 66 66 66 66 6666 666 66666 6666 666 6444444444444
x xxxx 222222xxxx xxx xxxxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxx xxxx
888888888888888888888 88881113333
1 xxxx1
7
7m (ii) 2
3
3
3
3
1 1 1 1 1 1 ... 2 16 16 64 1618 3
161 1 1 1 ...
2 256 163841251616 8192 64 15 16384
4128516 16384
8
7
8m
(a) 2
2
2
2
3 5 12
3 5 1 02
3 5 20
22 3 0
23 1
02
x xx
x xx
x x xx
x xx
x xx
Hence, 3 or 1 2x x .Method 2:
2
22
22
3 5 12
3 5 2 2
3 5 2 2 0
2 1 3 0
x xx
x x x x
x x x x
x x x
3 or 1 2x x .(b)
Hence, 2 1.99 or 4.94x x .
3 1 2
9
8 (i)2 2
2
12
4
1 114 4
4 114
1 11 15 2
n
nr
r
n
10m (ii) 22 1 2 3 2 1 2 3
1 2when , 112 2 32
3 2when , 132 2 12
2 1 12 1 2 3 2 1 2 3
A Br r r r
r A
r B
r r r r
1 1
1
1 1 22 1 2 3 2 2 1 2 3
1 1 1 2 2 1 2 31 1 12 3 5
1 15 7
...1 1
2 1 2 11 1
2 1 2 31 1 12 3 2 31 16 2 2
n n
r r
n
r
r r r r
r r
n n
n n
n
3n
222222222222211111111111 166666666666666666666666666666666 222222222
1 11113333333333333333333333333333333333333333333333333333333 222222 333333333333333333333333333 222222222222222 33
22
10
(iii)
4
2
1
2( 1)
1
2
11
4
2( 1)
2
1
4
142 1 2 3
4
4 4
1 1 116 6 2 2 3
1 1 15
12 1 2 3
1 16
42
2
u
2
16 6 2 2 31 1 1 1=
16(15) 6 2 2 2 31 1 (ded ced) since
240
3
4
1 111 2
nr
r
nr
r
n
n
r
r
r
n
r
n
n
r
r r
n
r
n
n
n
r
10 and 02 2 3n n
11
9 (i) Solving the cartesian equations representing planes 1 and 2 (on the GC) produces
3 2 3 22 0 2
0 1
x xy yz z
3 2 : 0 2 ,
0 1l r 0
000
(ii) 3 : ( 2 2 3) ( 2 2 6) 0 ( 2) (2 1) (2 2) 3 6
p M x y z x y zM x M y M z M
2
2 1 3 62 2
Mr M M
Mr 2222
2 2222 2, A normal vector to plane 3p is
22 12 2
MMM
.
9m (iii) (Elegant) method :
Let ( , , )x y z be a point on line l.Line l is the common line of intersection between planes 1p and 2p , ( , , )x y z
is on plane 1p as well as plane 2p .2 2 3x y z ,
2 2 6x y z .
The expression
( 2 2 3) ( 2 2 6)
( 3 3) ( 6 6)
(0) (0) 0
M x y z x y z
M
M
( , , )x y z lies on plane 3p .Any point on line l lies on plane 3p , line l lies on plane 3p .
Alternative standard method :
Direction vector for line l,22
1mm 2
111.
Normal vector for plane 3p , 3
22 12 2
Mn M
M3n3 M22
222.
pla
dard dddddd ddddddddddd memmememmemeeeethththththththththththththtthththththhththththththththtthhthththhhthhththththhhhhththhhhthhhtthhthhhhhhthhhhhththhhhhthhhthhhhhhhhhhhhhhhhhhoddododododododoododoodoooodododododododoododoododododoododoodoododododdodododdodododdododoododododododdodododododdoooooo ::
ioooooooooon n n n nnnn nn vvvevevvevvvvvvvvvvvv ctctctctctctctctctctctctctctttc orooooooooooooooooooo fffffffffffffffffffffoorororororororoororororoorooorooo lllllllllllllllllllllllinininininininninininininininininininnnnnnninnne e e e ll, m22
m
12
3
2 22 2 1
1 2 22( 2) 2(2 1) (2 2) 0 ,
Mm n M
MM M M
m n 22111
3m n3n
line l is parallel to plane 3p .Point (3, 0, 0), a point on line l.
The expression
3 2(0) 2(0) 3 2(3) 0 2(0) 6(3 3) ( 6 6)(0) 0 0 ,
MMM
.
this point is also on plane 3p .
Line l // plane 3p , and l contains a point (3, 0, 0) that’s also on 3p ,
line l lies on plane 3p .
(iv) Normal vector for plane 1 , 1
122
n1n 2222
.
Normal vector for plane 3 , 3
22 12 2
Mn M
M3n3 M222
222Let be the angle formed between normal vectors 1n1n and 3n3n .
1 3
1 3cos n n
n n1 3n n1
n n1 3n1 3n n1
1 3
1 22 2 12 2 2
( 2) 2(2 1) 2(2 2)9
Mn n M
MM M MM
1 3n n1 3n3 222222
2 2 21 1 2 2 3n1 1n1 1
2 2 23
2 2 2
2
2
( 2) (2 1) (2 2)
( 4 4) (4 4 1) (4 8 4)
9 9
3 1
n M M M
M M M M M M
M
M
3 (n3 ((
2(22
2 333333333322 22
22222222
2
2222)))))))))) (2222 1)))) ((((22 22
( 444444444 4444))2
922
222222222222))))))))))) ((222222222 1) ((((22 2
44444444 44444))))222222
2222222222222222222)))))))))))) (((((((((((((((((((((2222222222222222 1) (2122222222 2
( 4444444444444444444444 4444))))222
22
(((
13
1 3
1 3
22
cos
9 0 if 013 3 1
n nn n
M M MMM
1 3n n1
n n1 3
9n1
M3n n1
The cosine of the acute angle between planes 1p and 3p is 2 1
M
M.
14
10
(i)
1Area of triangle 2
OAB OA OBOA OBOA
01 1 12
2 1
1 21 0 22
0
11 22
2 21 1 22
21 1 52
10m (ii)By the Ratio Theorem, 3
4OA OCOB A OC3OAOB OA
011 3 14
1 2OCOCOCOC
4 44 3 14 6 2
OCOC4
4444
OBOB is perpendicular to ,OCOC 2
41 1 4 1 01 2
1 02
(iii) Exact perpendicular distance from B to OA = BNBN
From (ii), 12
21 1 1Area of triangle 1 5 1 52 2 4
OAB
1 9 12 4 2
BN OABN OA
3 1 42
BNBNBN
32 5
BN 3BN
2 2 2
21 91 14 4 5
OB ON BN
ON
2 2 2OB ON BNON
92ONON
of trrrriaiaiaiangnnn
OAAAAAAAAAAAA
4444444444444444N 11111111111111N 1 41111 41111N 111111111111NNNNNNNNN
3
15
2 95
ON2 9ON
35
65
01
023
5 1 4ON 3ON
16
11
(i)
As 21, 1x t y e e
12m (ii)3
1 d 1d1 2 1
xxtt t
2 2d 2
dt tyy e te
t
2
3
d 21d
2 1
ty tex
t
2 34 1 (shown)tte t
(iii)Gradient of normal at 5
4t is
54
1 1 0.012421d 80.50628d t
yx
When 54
t , the point on C is 25162 ,
3e or 2 , 4.7707
3
The normal at 54
t meets the y-axis at the point P:
At y-axis: 2516 20.012421 0
3y e
4.76245yP: 0, 4.76245
tsts tthe
:
on CC isis 22 ,33233
,33
ormrmmmrmmmmmmrmrmmrrrmrmalaaaaaaaaaaaaa atatatatatatatatatatatatataaaa 555555555555555555554444444444444444
ttttttttttttt
01
17
Given Q: 21 ,1
qeq
PQ =2
221 4.76245
1qe
q
Min PQ = 0.667 units when q= 1.25
ORFor PQ to be a minimum, PQ must be perpendicular to the tangent of Q [B1],
thus Q must be 54
q [B1]
Thus Min PQ =
2
225161 4.76245
5 14
e =0.667 [B1]
ORFor PQ to be a minimum, PQ must be perpendicular to the tangent of Q [B1]Thus
2 3
5 34
1 4 1 1554 14 4
qqe q
e
Solving gives q = 1.25 [B1]
Thus Min PQ =
2
225161 4.76245
5 14
e =0.667 [B1]
ORFor PQ to be a minimum, gradient of the line PQ must 0
[B1]1]1]1]
in PPPPPPPPPPPPPPPQQQQQQQQQQQQQQQQQQ ==============
22222222222222222222222222
1111111111111111 252511111111111111111111111 e555555555555555555555555555 1555555555555555 144
18
P: 0, 4.76245 , Q: 21 ,1
qeq
, thus 2
d 4.76251d
1
qy ex
q
Solving for 2
d 4.7625 01d1
qy ex
q1.25q (to 3 sf) [B1]
Thus Min PQ = 22 25
161 4.762451.2493 1
e =0.667
19
12 (a) (i) Month Outstanding amt. owed
1st day of month aft. instalment Last day of month aft. interestOct-2019 12000 12000(1.005)Nov-2019 12000(1.005) P 212000(1.005 ) (1.005)PDec-2019 212000(1.005 ) (1.005)P P
Amount owed upon paying the 2nd instalment (in $)= 212000(1.005 ) (1.005)P P
14m (a)(ii) Instal Outstanding amt. owed
-ment 1st day of month aft. instalment Last day of month aft. interest0 12000 12000(1.005)1 12000(1.005) P 212000(1.005 ) (1.005)P2 212000(1.005 ) (1.005)P P 3 212000(1.005 ) (1.005 ) (1.005)P P3 3 212000(1.005 ) (1.005 ) (1.005)P P P …… … …n 112000(1.005 ) (1.005 )... (1.005)n nP P P
Amount owed upon paying the nth instalment (in $)1
Geometric series, termsStarting from right : first term , common ratio 1.005
12000(1.005 ) (1.005 )... (1.005)
(1.005 1)12000(1.005 ) 1.005 1
12000(1.005 ) 200 (1.005 1
n n
nP
nn
n n
P P P
P
P
tric series terms
(1.005 )... (1.005) 1(1 005 (1.005)(1.005)(1 005)
) (shown)
(a)(iii)
Intended last instalment on 1st December 2021.- 2019 : 2 instalments- 2020 : 12 instalments- 2021 : 12 instalmentsTotal no. of instalments = 26
Amount owed upon paying the 26th instalment (in $)26 2612000(1.005 ) 200 (1.005 1) 0 if loan is repaid completelyP
formulate eqn.26 26
26 26 60
493.3395...
12000(1.005 ) 1.005200(1.005 1) 1.005 1
P
Minimum monthly instalment needed = $493.34 (nearest cent)
(b)(i)
Let a hertz be the frequency of the tone produced by the 1st key on the piano, and r be the common ratio of the frequency between successsive keys.
y ng tttthehehehe 2622626222222000000000000000000000000000 (((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((111111111111111111111111111111111111111111111111111111.....0000000055552626
26226226226626626666666
262626262262622222 601222222220000000000000000000000000000000000 0(0(0(00(0(0(0(0(0(0(00((0(0(00(000(1.1.1.111111.11.111.111.00000000000000000000000000000000000000005 )5 )5 )5 )5 )5 )5 )5 )5 )5 )5 )))5 ))5 )5 )5 )5 )5 ))5 )55 ))55 )2666660
20202020020200002000202000002022000(0(0(0(0(0(00(0(0(0(00((((((11.111.111.111 00000000000000000000000000000000000000000005 15555555555555555 111))))2226222222222222 11)26262666266((((((((((( )))))))))))))
20
the 49th key produces 440 hertz, 48 440ar (1)the 61st key produces 440 hertz, 60 880ar (2)
Eliminating a, 12(2) : 2(1)
r , 1
122r
112
st
48 48
4 (frequency of 1 key in hertz)
440 440From (1) : 2
440 440 = 27.5 162
ar
(b)(ii)
Frequency produced by the nth key on the piano (in hertz),1
12
112
1 127.5 (2 )
27.5 (2 )n
n nnu ar
Sequence of logarithm of the frequencies produced by successive piano keys :
1 2 3ln( ), ln( ), ln( ), ... , ln( ), ... nu u u u
The difference between consecutive terms of the sequence
112
112
1 2 3
1
1
, , , ... , , ... is in a G.P., with 2
ln(2 ) , a constant
ln( ) ln( )
ln , using properties of the ln function
ln( ) ,
n n
n
n
nu u u
u u
u
uu
r r1 21 22, ,, 1 22
1 2 3ln( ), ln( ), ln( ), ... , ln( ), ... nu u u u is in an A.P.
(b)(iii)
Ratio of ascending frequencies produced in a major triad chord, sounded by the nth, ( 4n )th and ( 7n )th keys is
1 112 12
1 73 12
1 ( 4) 1 ( 7) 1
1 1 4 1 7
4 7
4 7
: :
(1) : ( ) : ( )
1 : :
1 : (2 ) : (2 )
1 : 2 : 2
n n n
n n n
ar ar ar
ar ar r ar r
r r
,
which is independent of n. (shown)
11111 11111
11111111
( 7)7
1111111111111111 7777
4 7
44 774 77444444 1
( ) ::: (((( )))) 4 1111111111111111111111111111111 7777
: 4
1 ::::::::::: ((((((((((((((222222222222222222222 )))))))))))))))))))) : (((((((2222 )12212212121212121211212112221 12124444444444444 1212
: 2222 33
ar
( ))))))) ::::::: ((((
rrrrrrrrrr ::::::
( 7)7