2009 ACJC P2/Q3

3(i)
3(ii) /
Required area =
=
= 2
=
Therefore,
3
(iii) / Required volume =
= 5.75 square units

2009 CJC P1 Q3

3 /



2009 CJC P2 Q2

2(a) / Hint: Exact value not required, so can use GC.
Sketch the graphs and find the intersection points.
By G.C, Intersection points
(1.05395, -0.947453), (4.3919, 0.47976)
=1.68
2b / Hint: When asked to rotate about lines other than x and y- axis, we need to consider a translation of the graph, and rotate about x or y-axis considering the new graph.
Therefore, we consider rotating the graph about the x-axis instead of about the line y=1.
Answer: b=5/2

2009 PJC P1 Q9

9. Objective:Use of sum of area of rectangles to estimate area under graph

Total area of four rectangles

Total area of n rectangles

Area under graph

Since Sum of Area of Rectangles < Area under graph

------(1)

Consider rectangles as seen in the diagram,

Total area of n rectangles

Since Sum of Area of Rectangles > Area under graph

------(2)

Considering (1) and (2),

. (deduced)

2009 JJC P1 Q5

1(i)
(ii)
(iii)
2(i)
(ii)
(iii)
3
4
(i)
(ii)
5(a)
(b)
6(a)
(b)
7(a)
(b)
8
8(i)
(ii)
9(i)
9(ii)
(iii)
10(a)
(b)
(c)
11(a)
(b) / Graph of


Graph of
Graph of




since is a subset of
exists. (shown)


















From the diagram,

(Shown)



When t = 0, P = 1000, A = 270
Hence
For P > 2000,

]




when ; when






Let x be no. of chickens.
Let y be no. of horses.
Let z be no. of sheep.
------(1)
2x + 4y + 4z = 1250------(2)
Case 1:
If ------(3)
By GC, x = , y = 625, z = (rejected)
(Alternative):
Reject , because any combination of 250 animals will never have 1250 legs.
(Maximum no of legs = 250 x 4 = 1000)
Case 2:
If ------(3)
By GC, x = 75, y = 125, z = 150




or 0 < x < 5


For , replace x with ln x
or
0 < or
Let be the statement that
when , L.H.S. = (given)
R.H.S. =
is true.
Assume that is true, i.e.
When , R.H.S.=
L.H.S. =
= (given)
=

is true
Since is true and is true is true,
By the principle of induction, is true for all all non-negative integral values of n.
L.H.S. =

=
=
=
=
= = R.H.S. (Shown)

=
=
=
Since



As
(shown)


Sub into (1)


Since , the geometric series is convergent. .

From the GC, the least value of N =12.









Subst (1) into (2),




Since is real, =0
, where
Since is real and negative, .
Given that ,
has to be odd

, when t = 2
, and the gradient of the normal at P is 4.
The equation of the normal is …………….(1)
Subst , into (1)
we have

as


……….(1)


………….(2)
(1) = (2)


Since r > 0 (deduced)

2009 AJC P2 Q3

3(a) / ------(1)
Given z = x + y  ------(2)
Sub (2) into (1) :



3(b)(i) / Speed = positive constant
At t = 0, x = 0 , = 10
 10 = 8k  k =
 (shown )
3(b)(ii) / Integrating wrt t ,


At t = 0 , x = 0 , A = 8 
At x = 6, t = 1.11hrs
3(b)(iii) /
When x = 8 ( i.e. he completes his 8 km jog) ,
Model predicts infinite amount of time to complete jog, therefore model is NOT suitable.

2009 CJC P1 Q5

5 /
Substituting,








2009 ACJC P1 Q9

9 (i) /
d.e. becomes after replacing and
Therefore,
9 (ii) /

where D =

When x = 0, y = 2 . Therefore, .
The particular solution is
9 (iii) /
=
When x = 0, y = 2 therefore,

2009 DHS P1 Q6

6(i) /
(ii) /
(iii) / Any positive value of k
But A must be a positive value STRICTLY LESS than
e.g. k=1, A=, then
m
2/3

2/3-A

0 t