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(MATH1018)[2009](f)final~=u_bitzkj^_46938.pdf
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HKUST
MATH021 Concise Calculus
Final Examination Name:
17th Dec 2009 Student I.D.:
Tutorial Section:
Seat Number:

Directions:
.
DO NOT open the exam until instructed to do so.

.
You may write on both sides of the examination papers.

.
You must show the steps in order to receive full credits.

.
Electronic calculators are not allowed.


Question No. Points Out of
1 3
2 4
3 4
4 4
5 10
6 10
7 10
8 10
9 10
10 10

Some formula:
.
sin(. + ) = sin . cos . + cos . sin .

.
sin(. . ) = sin . cos . . cos . sin .

.
cos(. + ) = cos . cos . . sin . sin .

.
cos(. . ) = cos . cos . + sin . sin .


tan +tan
. tan(. + )=
1.tan . tan . tan .tan
. tan(. . )=
1+tan . tan .
.
sin 2. = 2 sin . cos .

.
cos 2. = cos2 . . sin2 . = 2 cos2 . . 1=1 . 2 sin2 .


1. (a) What is an even function?
(b) If . is an even di.erentiable function (so that . does have a derivative), eval-uate . (0).
Solution:
A function . is evn if
()= (.) for all .

Taking derivatives on both sides of this equation yields
. ()= .. (.) for all .
Evaluating at 0 yields
. (0) = .. (0)

or . (0) = 0.
2. (a) For each >0, evalualte

.. . =0
(b) Is your result in part (a) valid if 0?
Solution: We identify the given series as the geometric series with .rst term 1 and common ratio .. . Since >0, 0 <.. <1. The given series converges to
1
.
.
<=
1 . .

However, if 0, .. 1 and the given series diverges.

3. Let
1
()= . 1 for all . =0M.
Evaluate the area of the region bounded by the graph of ,the -axis and the two vertical lines de.ned by . =1/. and . = .
Solution: Observe that () > 0 when 1/. < . < 1, and () < 0 when 1 <<. The area of the given region is
1
. 1). + . =(ln . . )O= . +1/. . 2.
1
1
.(

1
. 1).

(

1/
11
..O(ln )1
4. For each of the following in.nite series and improper integral, determine if it con-verges or diverges
(a) =1 .. ,
+
(b) 0 sin . .
Solution:
(a) Since the given series has positive terms and
(+1)..1
lim
.. (+1).
= 1 . lim . = 1 . < 1,
the given series converges by ratio test.
(b) For each > 0,
.
sin . =1 . cos .
0
But the limit of 1 . cos . does not exist as . +, the given improper integral diverges.
5. (a) Let . be a function. . is a number. Evaluate

(. + ) . (. . )
lim .
0 .
Solution:
(+).(.)
lim0 . (+).() (.).()
=lim0 . + .. (+).() (.).()
=lim0 . +lim0 .. = . ()+ . ()=2. ().
(b) Let . be a function so that
() =1 . () for all .
Evaluate . (0).
Solution:
Take derivatives on both sides of the given relation,

. ()() = .. ()
so that
. (0)((0) +1) = 0.

However, (0) +1 > 1 for all . Therefore, . (0) = 0.
(c) Evaluate

1 . cos .
lim .
0 . sin .
Solution:
1.cos
lim0
. sin .
(1.cos )(1+cos )
=lim0
. sin (1+cos ) sin