Download may 2008 tz1 p3 ms - HLMath-PreAP/IB Precal-blog

1.
u n 1
Consider

10 n  1
un
1  1
 1  
10  n 
10

n  1 10 n


M1A1
n10
10
A1
1
as n  
10
A1
1
1
10
R1
So by the Ratio Test the series is convergent.
R1
[6]
2.
(a)
lim
x 
x
1
 lim x
x
x


e
e
M1A1
=0
(b)
AG
Using integration by parts

a
0

  e
 e 
x e  x dx   xe  x
 ae  a
= 1  ae
(c)
M1
a
a
0
0
x
dx
x a
0
a
a
e
Since ea and aea are both convergent (to zero), the integral is
convergent.
Its value is 1.
A1A1
A1
A1
R1
A1
[9]
IB Questionbank Mathematics Higher Level 3rd edition
1
3.
(a)
Rewrite the equation in the form
dy 2
x2
 y 2
dx x
x 1
M1A1
2
  x dx
M1
= e 2ln x
A1
1
x2
A1
Integrating factor = e
=
Note: Accept
(b)
1
as applied to the original equation.
x3
Multiplying the equation,
1 dy 2
1
 3 y 2
2
x dx x
x 1
(M1)
d  y 
1
 2  2
dx  x  x  1
(M1)(A1)
y
dx
 2
2
x
x 1

M1
= arctan x + C
A1
Substitute x = 1, y = 1.
M1
1

4
 C  C 1 

A1
4
 

y  x 2  arctan x  1  
4 

A1
[13]
4.
(a)
The area under the curve is sandwiched between the sum of the
areas of the lower rectangles and the upper rectangles.
Therefore
 dx
1
1
1
1
1
1
1 3  1 3  1 3  ...
 1 3  1 3  1 3  ...
3
3
4
5
6
x
3
4
5
which leads to the printed result.

IB Questionbank Mathematics Higher Level 3rd edition
M2
A1
2
(b)
We note first that


3

dx  1 
1
  2  
3
x  2 x  3 18
M1A1
Consider first

n
1
n 1
3
1 
1 1  1 1 1

 3   3  3  3  ...
3
2 3 4 5 6

1 1 1
 1  
8 27 18

n
n 1
1
3
1.22 which is an upper bound 
263
216
=
1 
1 1
1 1

  3  3  3  ... 
3
2 3
4 5

1 1
 1 
8 18
85  255 

 1.18 (which is a lower bound)
72  216 
=
M1A1
M1A1
A1
M1A1
M1A1
A1
[15]
5.
(a)
(b)
Constant term = 0
f  ( x) 
1
1 x
f  ( x) 
f ′′′ (x) =
A1
A1
1
1  x 2
2
1  x 3
f (0) = 1; f (0) = 1; f (0) = 2
A1
A1
A1
Note: Allow FT on their derivatives.
f ( x)  0 
1  x 1 x 2 2  x 3


 ...
1!
2!
3!
= x
x2 x3

2 2
IB Questionbank Mathematics Higher Level 3rd edition
M1A1
AG
3
(c)
1
1
 2 x 
1 x
2
(A1)
1 1 1
ln 2   
2 8 24
M1
=
(d)
2
0.667 
3
Lagrange error =
A1
f n  1 (c)  1 
 
(n  1)!  2 
n 1
1 1
=
  
4
1  c  24  2 
6

6
 1
1  
 2
4

(M1)
4
1 1

24 16
giving an upper bound of 0.25.
(e)
Actual error = ln 2 
2
 0.0265
3
The upper bound calculated is much larger that the actual
error therefore cannot be considered a good estimate.
A1
A2
A1
A1
R1
[17]
IB Questionbank Mathematics Higher Level 3rd edition
4