Exam 1                              Friday, March 23, 2001

 
 

NAME:

Student Number:------            No.----           Marks ----
 
 


Instructions:

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 3 questions and has a total of 100 marks.

3. Show all of your work. Your work must justify the answer that you give.
 

1.    (20 Marks)    Find the derivative of the following function
 
 
f(x) =  æ
è
x2+1
x4+1
ö
ø
[1/3]
 

Solution. Taking ln, we obtain
 
 
lnf(x)
ln æ
è
x2+1
x4+1
ö
ø
[1/3]
 
1
3
ln æ
è
x2+1
x4+1
ö
ø
1
3
(ln(x2+1)-ln(x4+1)).

Taking the derivative, we have
 
 
f¢(x)/f(x) =  1
3
é
ë
2x
x2+1
- 4x3
x4+1
ù
û
.

This implies f¢(x) = [1/3]([(x2+1)/(x4+1)])[1/3][[2x/(x2+1)]-[(4x3)/(x4+1)]].
 
 

2.    (20 Marks)    Evaluate the following limit
 
 

lim
x® 0+
æ
è
sinx
x
ö
ø
[1/(x2)]
 

Solution: Let f(x) = ([sinx/x])[1/(x2)]. Then we have lnf(x) = [(ln([sinx/x]))/(x2)] = [(lnsinx-lnx)/(x2)].
 
 

lim
x® 0+
lnf(x)

lim
x® 0+
lnsinx-lnx
x2

lim
x® 0+
(lnsinx-lnx)¢
2x

lim
x® 0+
(xcosx-sinx
2x2sinx

lim
x® 0+
cosx-xsinx -cosx
2(2xsinx+x2cosx)

lim
x® 0+
-sinx 
2(2sinx+xcosx)

lim
x® 0+
-cosx 
2(3cosx-xsinx)
-1/6.

Hence limx® 0+lnf(x) = -1/6. Since f(x) = elnf(x) and g(x) = lnx is continuous, we have
 
 

lim
x® 0+
f(x) = 
lim
x® 0+
elnf(x) = elimx®0+lnf(x) = e-1/6.

 

3.    (60 Marks)    Evaluate each of the following indefinite integrals.

(a) ò[(x+1)/((3x+1)1/3)] dx                   (b) òsin3x cos4x dx

(c) ò ln dx                                           (d) òexcos2x dx 

Solution: (a) Let u = (3x+1)1/3. Then u3 = 3x+1 and 3u2du = 3dx. This implies
 
 
dx = u2du   and x = (u3-1)/3.

 
ó
õ
x+1
(3x+1)1/3
 dx
ó
õ
(u3-1)/3+1
u
.u2 du =  1
3
é
ë
ó
õ
(u3-1)u+3 u du ù
û
1
3
æ
è
ó
õ
(u4+2u) du ö
ø
1
3
æ
è
1
5
u5+u2 ö
ø
+C
1
15
u5+ 1
3
u2+C
1
15
(3x+1)5/3+ 1
3
(3x+1)2/3+C

(b)
 
 
ó
õ
sin3x cos4x dx
ó
õ
sin2xcos4x sinx dx = - ó
õ
sin2xcos4x dcosx
- ó
õ
(1-cos2x)cos4x dcosx =  ó
õ
(cos6x-cos4x) dcosx
1
7
cos7x- 1
5
cos5x+C.

(c) ò lnx dx = x ln xx.[1/x] dx = x ln xdx = x ln x-x+C.

(d)
 
 
ó
õ
excos2x dx
ó
õ
cos2x d ex = excos2x- ó
õ
ex dcos2x
excos2x-2 ó
õ
ex(-sin2x) dx
excos2x+2 ó
õ
sin2x  d ex
excos2x+2ex sin2x-2 ó
õ
ex d sin2x
excos2x+2ex sin2x-2 ó
õ
ex (2cos2x) dx
excos2x+2ex sin2x-4 ó
õ
ex cos2x dx

This implies 5òexcos2x dx = excos2x+2ex sin2x+C1 and
 
 
ó
õ
excos2x dx =  1
5
æ
è
excos2x+2ex sin2x+C1 ö
ø
1
5
æ
è
excos2x+2ex sin2x ö
ø
+C

 


File translated from TEX by TTH, version 2.79.
On 2 Apr 2001, 15:15.