acjcp2sol
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Anglo-Chinese Junior Col legeH2 Mathematics 9740
2010 JC 2 PRELIM Marking Scheme
Paper 2:
1( )
( ) ( )
( ) ( )
( )
1
2
2 1
1
112
2 2
2 1 2 111
2
1 12
2 1 2 12 2
112
4 2 2
1
1 1
1 1 1 1
2 2 2 2
14 1
2
x
x
x x
x x
x xx x
e dxe
e dx e dxe e
e e e e
e e e e
= +
= + + +
= + + +
2 (i)
2 (ii)
2 (iii)
Least value of z w = 1
Greatest ( )arg 3z + = 1 11 3
tan sin 0.826 (3 dp)5 26
+ =
Least ( )arg 3z + = 1 11 3
tan sin 0.432 (3 dp)5 26
=
3ln 2 0
ax
x
where 1a > .
2
2 1
20
1 1 8 ) 1 1 8 )2
4 40
1 1 8 1 1 8
04 4
ax
x
x x a
x
a ax x
x
a a
x or x
+ + +
+ + +
<
4 (a)( ) ( )3 2 3cos 3 sin
dx x x
dx=
x
y
(2,1)
O3
(5,3)
o
2
3
26
3
z
w
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2
5 3
3 2 3
3 3 3 2
3
3 2 3
33 3
sin
( sin )
1 1cos cos 3
3 3
cos cos3
1cos sin
3 3
x x dx
x x x dx
x x x x dx
xx x x dx
xx x c
=
=
= +
= + +
4 (b)5 sec 10,
4
5 sec tan 2 5,3
x when x
dxwhen x
d
= = =
= = =
( )
( )
[ ]
2 5 32 2
10
332 2
4
3
2
4
3 3
2
4 4
3
3
4
4
5
5sec 5 5 sec tan
1 sec
5 tan
1 cos 1 cot cos5 sin 5
1 1 1cos
5 sin 5
1 2 1 22 3 . .
5 15 5 15
x dx
d
d
d or ec d
ec
i e a b
=
=
=
= =
= = =
5
(i)
(ii)
1
1
: 2 3
1
p r
=
1
1 2
: 0 1 ,
1 4
l r
= +
2 1
1 . 2 2 2 4 0
4 1
= + =
1 1
0 . 2 1 0 1 3
1 1
= +
l1is parallel top1. l1is not contained inp1.
Alternative method:
1 2 1
2 2 3
1 4 1
+ = +
Since no solution for , l1is parallel and not contained top1
1 2 32 1 3 2
1 4 1
=
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No , (i) The weight of a husband and wife may not be independent
Or (ii) Randomness is not there ( a randomly chosen women but spouse is not
randomly chosen)
Or (iii) Distribution of weight of married woman is different from distribution of
adult woman.
Etc
9 Telephone costs are assumed to be normally distributed.
To test H0: = 72
against H1: > 72 at 5% level of significance
Under H0,
ns
x 0
_
t(5-1)
Test statistics : T=ns
x 0
_
=
82.6 72
10.6677 / 5
= 2.2219
pvalue = P(T > 2.2219) = 0.0452 < 0.05
Reject H0at the 5% level of significance. We conclude that there is sufficient
evidenceat the 5% level of significancethat there is evidence of an increase in
mean monthly costs.
To test H0: = 0
against H1: > 0 at 5% level of significance
Under H0,n
x
89.9
0
_
N(0,1)
Test statistics : Z =n
x
0
_
= 082.6
9.89 / 5
= (82.6 - 0 )89.9
5
Do not reject H0if P(Z > (82.6 - 0 )89.9
5) > 0.05
(82.6 - 0 )89.9
5< 1.64485...........(1)
0> 75.310
(i) 161x= (from calculator or computation)
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(ii)
(iii)
when 161x= , 103.6 0.726x y= +
(161 103.6) / 0.726y=
79.06336088=
using y y n= 1
79.06336088 (65.1 73.2 85 80.9 89.9)6
k= + + + + +
80.3k= Use G.C. to find regression line ofyonx:
97.593 1.097y x= +
Useyonxline to predict weight.
When 165x= , 97.593 1.097(165)y= +
83.4y= (1 d.p.) using 3 d.p. of aand bto compute.
or 83.5y= - using full accuracy of aand bto compute.
Using G.C., 0.893r=
C is unusually overweight.
11
(i)
Breakages occur randomly or
Breakages occur independently or
Meannumber of breakages is a constant
Let A be the r.v for the number of broken cups per day
A Po(2.1)
P(A 3) = 1 P(A 2) = 0.350369 = 0.350 (3 sig figs)
Let X be the r.v for the number of days with a least 3 broken cups out of n cups.
X
y
65.1
73.2
80.380.9
85.0
89.9
150 157 160 162 167 170
x
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(ii)
P(X 2) > 0.9991 P(X 1) > 0.999............(1)P(X 1) < 0.001n 22least n is 22
T Po(2.1x7 + 1.6 x 7)
T Po(25.9)Method 1:
Since n is large,_
T (25.9,100
9.25) approx by central limit theorem
P(_
T 26) = 0.578 (to 3 sig fig)Method 2:
100 weeks, Y Po(2590)
= 2590 > 10 . Normal approx to Poisson
Y N(2590,2590) approx
P(Y 2600) = P( Y < 2600.5) (With cc)= 0.578 (to 3 sig fig)
12
(i)
(ii)
Let X be the r.v for the number of fish which measures less than 8 cm long.
9 5 4
5
1 4 1( ) ( ) ( )5 5 5
C = 0.00330 ( to 3 sf)
X B(n,0.2)
Since n large and p= 0.2 , X N(0.2n,(0.2)(0.8)n) approx
P(X 10) 0.0227P(X < 10.5) 0.0227
P(Z