1

Yishun Junior College +Mathematics Department

\
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+
= |
.
|

\
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2
1
ln
d
d
tan
x
e
x
y
y . (1)

Differentiate (1) wrt x:
1 d
d
tan
d
d
sec
2
2
2
2
+
=
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.
|

\
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+ |
.
|

\
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x
x
e
e
x
y
y
x
y
y . (2)
Differentiate (2) wrt x:

( )
2
2
3
3
2
2
2
2
2
2
3
2
1
1 d
d
tan
d
d
d
d
sec
d
d
d
d
sec 2
d
d
tan 2sec
+

+
=
|
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.
|

\
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+
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.
|

\
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.
|

\
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+
|
|
.
|

\
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|
.
|

\
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+ |
.
|

\
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x
x
x
x
e
e
e
e

x
y
y
x
y
x
y
y
x
y
x
y
y
x
y
y y


( )
2
2
3
3
2
2
2
3
2
1
1 d
d
tan
d
d
d
d
sec 3
d
d
tan 2sec
+

+
=
|
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.
|

\
|
+
|
|
.
|

\
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|
.
|

\
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+ |
.
|

\
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x
x
x
x
e
e
e
e
x
y
y
x
y
x
y
y
x
y
y y . (3)

At x =0: ( )
4
0 f
t
=

( )
( )
( )
4
1
0 ' ' ' f'
2
1
0 ' ' f'
0 0 f
'
=
=
=
Hence, Maclaurins series: ...
x x
y + + +
t
=
24 4 4
3 2


Maclaurins series of ( )
|
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.
|

\
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+
2
1
ln cot
x
e
y
|
|
.
|

\
|
+ +
t
=
24 4 4 d
d
3 2
x x
x 8 2
2
x x
+ =







2
Question 2

(a) ( ) ( ) ( ) 1 2
4
20
4
1
1 2 10 2
2
2
2
+ =
(

|
.
|

\
|
+ = n
n
n n
n
S
n

( ) ( ) ( ) 1
8
10
4
1
1 10 2
2
+ =
(

|
.
|

\
|
+ = n
n
n n
n
S
n




200
8 8
3
10
200
2
2
> +
>
n n
n
S S
n n


Using GC, 8 39. n < (rejected) or 4 13. n >

Hence, least number of terms is 14.


(b) i) Note: ... e e e ...
u lg u lg u lg
+ + + = + + +
3 2
3 2 1
T T T


1
=
n
n
T
T
r

u lg
u
u
lg
u lg
u lg
e
e
e
e
n
n
n
n
=
=
=
|
|
.
|

\
|

1
1
Since r is a constant, the series is a geometric progression.


For series to be convergent, 1 <
u lg
e
1 1 < <
u lg
e
Since 0 >
u lg
e , then 1 0 < <
u lg
e

1 0
0
< <
<
u
u lg










3

b) ii)

=
|
.
|

\
|
+
N
n
n
N
1
T
1


u lg
u lg
u lg
u lg u lg u lg
u lg
u lg u lg
N
n
u lg
e
e
e
e e e
e
e e
e
N
N
N
n

+
=

|
.
|

\
|

+ =
|
.
|

\
|
+ =
+
+

=
1
1
1
1
1
1
1
1
1
1
1




Question 3

(i) N is on H : 8
1
1
2
=
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.
|

\
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- ON
BN is perpendicular to H :
|
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.
|

\
|

=
1
1
2
k BN

|
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.
|

\
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=
= =
|
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.
|

\
|

-
|
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.
|

\
|

+
+
|
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|
.
|

\
|

+
|
|
|
.
|

\
|
=
4
2
5
1 8
1
1
2
5
1
2 3
1
1
2
5
1
3
ON
k
k
k
k
k ON


Notice that =
|
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.
|

\
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-
|
|
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.
|

\
|

0
1
5
2
1
1
2
H and l
1
are parallel to each other.
Distance between H and l
1
= 6
1
1
2
=
|
|
|
.
|

\
|

= BN
4

(ii)
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=
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\
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=
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.
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\
|

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.
|

\
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2
0
1
6
12
0
6
1
1
2
1
5
2


Equation of H
2
=
|
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|
.
|

\
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-
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.
|

\
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=
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.
|

\
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-
2
0
1
5
1
3
2
0
1
r

13 2
13
2
0
1
= +
=
|
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|
.
|

\
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-
z x
r



Question 4


i)























ii)
x
t
t
y
x
y
d
d
d
d
d
d
=
x
y
- a
a
1
t x
t y
cos a
sin
2
=
=
5

2
2
2
cos 2
sin
1
cos sin 2
a
x
a
x
a
a
t
t a
t t
=
|
.
|

\
|
=
=

=


When units/s
3 d
d a
t
x
= : |
.
|

\
|
= |
.
|

\
|
t
x
a
x
y
t d
d 2
d
d
d
d
2


a
a
a
x
y
t 3
2
3
2
d
d
d
d
2
= |
.
|

\
|
= |
.
|

\
|


iii)
2 2
2
2
d
d
a x
y
=

Since 1 0 < < a : < < 1 0
2
a 1
1
2
>
a

2
2
2
<
a
2
d
d
2
2
<
x
y



iv) At
6
t
= t :
2
3 a
x = ,
4
1
= y

3 2
1
2
3
4
1
gradient
a a
= =

Hence, equation of line:
3 2a
x
y =


Volume V
R
( ) ( )
|
.
|

\
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|
|
.
|

\
|
t
+ t =
}
t
t
4
1
2
3
3
d cos sin 2 cos
2
2
6
2
a
t t t t a
6

( )
32
11
16 64
9
0 2
16 4
cos
2
16
d cos sin 2
2
2
2
2
2
6
4
2
2 2
6
3 2
a
a
a
a t
a
a
t t t a
t
=
t
+
(

t =
t
+
(
(

t =
t
+ t =
t
t
t
t
}






YISHUN JUNIOR COLLEGE
H2 MATHEMATICS
JC2 (2007) PRELIMINARY EXAM STATISTICS
MARKING SCHEME

Qn Solution
5a The number of :
one-digit nos. 4 =
two-digit nos. 12 =
three-digit nos. 24 =
four-digit nos. 24 =
The required number is 64 24 24 12 4 = + + +
5b No. of three-digit nos. that are greater than 300 18 2 3 3 = =
No. of four-digit nos. that are greater than 300 24 1 2 3 4 = =
Required probability
32
21
64
24 18
=
+
=
6a Let X be the r.v. no. of student (out of 40) eating western food at the canteen at a
particular instant.
) 33 . 0 , 40 ( ~Bin X
) 14 ( 1 ) 15 ( s = > X P X P
326 . 0 = using ) 14 , 33 . 0 , 40 ( 1 binomcdf
6b Let Y be the r.v. no. of students (out of 150) who felt that the prices at the western food
stall are reasonable.
( )
7
3
, 150 ~Bin Y
Since 150 = n is large,
7
3
= p is neither large nor small, 5 3 . 64 > = np and 5 7 . 85 > = nq ,
) 73 . 36 , 3 . 64 ( ~N Y approx.
) 5 . 60 5 . 24 ( ) 60 25 ( < < s s Y P Y P
cc

7
Qn Solution
265 . 0 = or 266 . 0 (3 sfs)
7ai Let X be the r.v. no. of dengue fever cases occurring in a 7-day period (from April to
J uly).
) 5 / 7 ( ~Po X
) 0 ( 1 ) 1 ( = = > X P X P
753 . 0 7534 . 0 = = (3 sfs)
using ) 0 , 5 / 7 ( 1 poissonpdf
7aii Let Y be the r.v. no. of weeks (out of 5), each of which is free of dengue fever cases
(from April to J uly).
) , 5 ( ~ p Bin Y where 753 . 0 1 = p
985 . 0 ) 3 ( = s Y P (3 sfs) using ) 3 , 753 . 0 1 , 5 ( binomcdf
7b Let W be the r.v. no. of dengue fever cases occurring in a period of n days (from April
to J uly).
) ( ~ x Po W where
5
n
x =
9 . 0 ) 3 ( s s W P


9 . 0
6 2
1
3 2
s
|
|
.
|

\
|
+ + +

x x
x e
x

Using GC to sketch graphs of
|
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.
|

\
|
+ + + =
6 2
1 9 . 0
3 2
x x
x e y
x
and solve for 0 > x and
0 > y to obtain 7447696 . 1 > x
72 . 8 7447696 . 1
5
> > n
n

Least 9 = n days
7c Let R be the r.v. no. of dengue fever cases occurring from April to J uly and S be the
r.v. no. of dengue fever cases occurring for the rest of the year.
Then ) 5 / 122 ( ~Po R and ) 15 / 243 ( ~Po S .
) 6 . 40 ( ~Po S R +
( )
) 15 (
) 10 5 (
15 | 5
= +
= =
= = + =
S R P
S R P
S R R P
0241 . 0 024086 . 0
) 15 (
) 10 ( ) 5 (
= =
= +
= =
=
S R P
S P R P

8
Let X be the r.v. Life (hours) of a randomly chosen light bulb and ( )
2
, ~ o N X
Given 123 . 0 ) 800 ( = < X P and 719 . 0 ) 1100 ( 281 . 0 ) 1100 ( = < = > X P X P
123 . 0
800
= |
.
|

\
|
<
o

Z P and using GC,
1601 . 1
800
=

o

---------- (1)
719 . 0
1100
= |
.
|

\
|
<
o

Z P and using GC,
8
Qn Solution
5799 . 0
1100
=

o

--------- (2)
Solving (1) with (2), 172 = o and 1000 =
8i
( )
2
3 2 1
172 3 , 1000 3 ~ + + N X X X
i.e. ( ) 88752 , 3000 ~
3 2 1
N X X X + +
16 . 0 ) 3300 (
3 2 1
= > + + X X X P (2 d.p.)
8ii ) 1100 1100 1100 (
3 2 1
> > > X X X P
( ) 0222 . 0 281 . 0
3
= =
The two answers are different because the two events are related as:
{ } hrs life has bulbs the of each 1100 3 > is a proper subset of
{ } 3300 3 > bulbs the of life total
9ai The sample chosen would be biased as it might not represent the quality of those produced
throughout the shift and also the different flavours of milk that were being produced.
9aii A stratified ramdom sampling would be suitable. This would require a selection of 14
cans of chocolate flavoured milk, 5 cans of strawberry flavoured milk, 20 cans of vanilla
flavoured milk and 11 cans of unflavoured milk. These would be randomly selected from
all the cans produced from each shift.
9b Let X be the r.v. no. of patients admitted into a private hospital in a day.
Given 85 ) ( = X E and 35 = o
Let ( )
30 2 1
30 ...
1
X X X
n
X + + + =
Since 30 = n is large, by Central Limit Theorem,
( )
30
35
, 85 ~
2
30 N X approximately
941 . 0 ) 95 ( 30 = < X P (3 sfs), using ) 30 35 , 85 , 95 , 99 ( E normalcdf

Let ( )
n
n X X X
n
X + + + = ...
1
2 1

Assuming that n is large, by Central limit Theorem, ( )
n
N X n
2
35
, 85 ~
Given 75 . 0 ) 90 80 ( > < < n X P
( ) 75 . 0 80 2 1 > < n X P
( )
2
25 . 0
80 < < n X P
125 . 0
7
<
|
|
.
|

\
|
<
n
Z P
Using ) 125 . 0 ( Invnorm , 15035 . 1
7
15035 . 1
7
> <
n n

84 . 64 > n
Least 65 = n
10a Let X be the r.v. the mass, in g, of a component.
Assumption: The masses of components produced are normally distributed.
:
0
H 6 =
9
Qn Solution
:
1
H 6 =
Under
0
H , the test statistic is ) 1 , 0 ( ~N
n
X
Z
o

= where 6 = , 8 . 0 = o , 10 = n
Level of sifnificance: % 5 = o
We reject
0
H if 96 . 1 > z and do not reject the claim if 96 . 1 < z
i.e. 96 . 1
10
8 . 0
6
96 . 1 <

<
x

496 6 504 5 . x . < <
i.e. 50 6 50 5 . x . < <
10b :
0
H 6 =
:
1
H 6 <
Under
0
H , the test statistic is ) 1 ( ~

= n t
n
s
X
T

where 6 = , 10 = n ,
( )
4 . 5 4
10
4
= +

=

x
x ,
( ) ( )
8882 . 0
4 4
9
10
2
2
=
(
(

|
|
.
|

\
|

=

n
x
n
x
s
Level of sifnificance: % 5 = o
We reject H
0
if T < 1.833
t =
n s
X
/

= 2.1362 < 1.833


Alternative
Using GC, the 05 . 0 0307 . 0 < = value p
Conclusion:
Hence, we reject H
0
and conclude that at 5% significance level, there is sufficient
evidence that mass of the components produced by the new machine is less than 6g.
11a The scatter diagram gives a good overview of how the two variables are related.
11ai












10
Qn Solution
11ai








11bi 98 . 0 = r . High positive linear correlation between X and Y.
11bii r remains unchanged. It is because r is a measure of the degree of scatter and it is
unchanged when there is a change of scales.
11c Regression line of x on y :
5 . 5
8
9
+ = x y
9
44
9
8
+ = y x
Given
25
16
2
= r , b (
9
8
) =
25
16

b =
25
18

Since ( ) y x, =(4,1) lies on the line of regression line of y on x ,
Hence the required regression line is
) 4 (
25
18
1 = x y
25
97
25
18
+ = x y .