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(MATH021)[2009](f)final~kpleeaa^_10013.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 . 1).... + .( 1 . 1)....
1/.... 1 ..
= (ln .. . ..)O1 . (ln .. . ..)O..
1/.. 1
= .. +1/.. . 2.
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:
..(..+..)...(.....)
lim..0 .. .. (..+..)...(..) ..(.....)...(..)
= lim..0 +
.. ... .. (..+..)...(..) ..(.....)...(..)
= lim..0 + lim..0
.. ... = .. (..)+ .. (..)=2.. (..).
(b) Let .. be a function so that
....(..) =1 . ..(..) for al