MATH 100Midterm Examination
Version 1
Date: October 24, 2014
Time: 18:0019:30
Surname:
Given name(s):
(Please, print!)
ID#:
Signature:
Please, check your section/instructor!
Section Instructor
EA1
V. Yaskin
EB1
F. Xanthos
EC1
E. Woolgar
ED1
H. Yahya
EE1
J
Written Assignment 2
Math 100 Fall 2012
Due: Friday, October 19
1. (8 pts) Evaulate the limit, if it exists:
(a)
4x8 5x + 2 + 3x2 + 2
lim
x (3x3 + 4x2 + 1) x2 + x + 1
(b)
lim x( x2 2x + 5 |x 1|)
x
Solution.
(a) We begin by factoring out x4 in the numerato
Welcome to Module 2: Contextualized Learning and Instruction or CLI.
Learning Outcomes for contextualized learning and instruction Part 1.1 include:
Describing contextualised learning and instruction
Defining the term scaffolding,
Describing the purpose o
DI Part 1.1
Learning outcomes for DI part 1.1a include:
Defining differentiated instruction and identifying the key elements DI.
Describing what differentiated instruction is and what it is not.
What is differentiated instruction? At its most basic leve
TB&AB1.1 In Canada exceptional are those students who are classified as either gifted or as
having a disability. Across the country a number terms are used interchangeably to describe
these students. For example, in Ontario and New Brunswick the term used
Math 100
Assignment 3
Fall 2014
Due: 5PM Friday, November 7, in Lab Group Drop-Boxes, 3rd oor CAB
Each question is worth 10 points.
1. Two runners at the same point begin running in opposite directions along a
circular track of radius 100m at a speed of 5
Math 100
Assignment 2
Autumn 2014
Due: 5PM Friday, October 17, in Lab Group Drop-Boxes, 3rd oor CAB
Each question is worth 10 points.
1. Find the following limits
(a)
(b)
2x sin(5x) + 3x
x2 1
3
lim ( x + 2 3 x).
lim
x+
x+
|f (x)|
if the
x2
function f sati
Math 100
Assignment 3
Fall 2014
Due: 5PM Friday, November 7, in Lab Group Drop-Boxes, 3rd oor CAB
Each question is worth 10 points.
1. Two runners at the same point begin running in opposite directions along a
circular track of radius 100m at a speed of 5
Math 100
Assignment 4 Solutions
Autumn 2014
1. (4 + 4 + 4 = 12 pts) Evaluate the following limits.
(a)
lim (tan x)cos x
x
2
1
1
2 ln
x x
(b) lim
x0+
1 + 3x
1 + 2x
(cos x) ln(x a)
ln(ex ea )
(c) lim
xa+
Solution.
(a) The limit has the form 0 . Let y = (t
Math 100
Assignment 2
Autumn 2014
Due: 5PM Friday, October 17, in Lab Group Drop-Boxes, 3rd oor CAB
Each question is worth 10 points.
1. Find the following limits
2x sin(5x) + 3x
x+
x2 1
(b) lim ( 3 x + 2 3 x).
(a)
lim
x+
Solution:
(a) Note that for each
Math 100
Assignment 1
Autumn 2014
Due: 5PM Friday, September 26, in Lab Group Drop-Boxes, 3rd oor CAB
Each question is worth 10 points.
1. Solve the inequalities
(a) x2 x < 2.
(b) (x + 1) x2 7x + 6 0.
(c) x < |x 1| 4 |x + 1|.
Report your solutions using i
Math 100 Lab 1A
1.
Find the set of all x that satisfy:
(i)
[1]
(ii)
x3 + 2 x2 3x = 0
[1]
(iii)
2.
x3 + 3x2 + 2 x = 0
2 x3 + x2 x > 0
[1]
Solve the following inequalities analytically and also
graphically (i.e., use a graph to illustrate your solution):
(i
Math 100 Lab 1B
1. Solve the following inequalities. Write the solutions in either
interval notation or set notation.
(i) x3 + 3x < 4 x2
[2]
(ii) 2 t 1
[1]
(iii) x6 = x3
[1]
Find the set of all x that solve the equation x 2 = 2 x .
[2]
2
2.
3. Let
(i)
(ii
MATH 100Final Examination
Date: December 11, 2010
Time: 2 hours
Surname:
Given name(s):
(Please, print!)
ID#:
Signature:
Please, check your section/instructor!
Section
Instructor
EA1/SA1 E. Osmanagic
EB1
L. Keener
EC1
E. Woolgar
ED1
V. Yaskin
EE1
O. Rivas
4. Find the derivative of the function f{I} = [sin{I”3)]|3 for all real I. (Hint: the
case I = I] needs special attention.)
Is the function f’[I} continuous on JR? Explain.
Solution: For each I aé El, by the diﬁerential rules we have that
WIN!
mm} = 3(sin
Solutions for Math 100 April 2007 Final
1. (a) Introducing the substitution u =
du =
1=x, leads to
1
dx and that x ! 1 () u = 0 and x = 0+ () u !
x2
=)
1
Z
1
e x
dx =
x2
0
Z
0
eu du = eu ]
1
1
0
1;
= 1:
(b) Observing that the integrand is an odd function
Solutions for Math 100 April 2006 Final
1. (a) Observing that
1
Z
ex
dx =
2 + 2ex + e2x
1
1
Z
1
ex
2
1 + (1 + ex )
dx;
and then introducing the substitution u = 1 + ex , which implies
du = ex dx and x = 1 () u = 1 and x =
1 () u = 1;
leads to
1
Z
ex
1
1 +
MATH 100Midterm Examination
Version 1
Date: October 24, 2014
Time: 18:0019:30
Surname:
Given name(s):
(Please, print!)
ID#:
Signature:
Please, check your section/instructor!
Section Instructor
EA1
V. Yaskin
EB1
F. Xanthos
EC1
E. Woolgar
ED1
H. Yahya
EE1
J
MATH 100Midterm Examination
version 1
Date: October 27, 2012
Time: 90 minutes
Surname:
Given name(s):
(Please, print!)
ID#:
Signature:
Please, check your section/instructor!
Section Instructor
EA1
V. Yaskin
EB1
E. Leonard
EC1
D. Hrimiuc
ED1
H. Yahya
EE1
V
MATH 100Midterm Examination
version 1
Date: October 27, 2012
Time: 90 minutes
Given name(s):
Surname:
(Please, print!)
ID#:
Signature:
Please, check your section/instructor!
Section Instructor
EA1
V. Yaskin
EB1
E. Leonard
EC1
D. Hrimiuc
ED1
H. Yahya
EE1
V
MATH 100: Solutions for Problem Set 6
Page 720, #8 : To use the integral test for the series
1
X
n3
n2 e
;
n=1
we compute the integral
Z
1
x3
x2 e
x3
e
dx =
=
3
1
1
1
1
:
3e
Hence the integral is convergent and so is the series.
Page 726, #22 : The series
MATH 100: Solutions for Problem Set 5
Page 371, #20 : We have
f (x) = x2 +
Hence
4x =
7
4
=
n
so that the area is given by
Z
7
p
2
x +
4
p
1 + 2x, 4
x
7:
3
3i
, xi = 4 + i 4x = 4 + ;
n
n
n
3X
1 + 2x dx = lim
n!1 n
i=1
"
s
2
3i
4+
n
#
3i
1+2 4+
n
+
Page 38
MATH 100: Solutions for Problem Set 4
Page 281, #48 : We have
2x3 , x 2 [0; 4] :
f (x) = 5 + 54x
We see that
f (0) = 5 and f (4) = 93:
Also
f 0 (x) = 54
6x2 = 6 9
x2 = 0 =) x = 3 =) f (3) = 113:
Hence the absolute minimum occurs at x = 0 with value 5 and
MATH 100: Solutions for Problem Set 1
Page 98, #30 : We have
lim
x! 3
x+2
= +1, since x + 2 < 0 and x + 3 ! 0
x+3
as x !
3 .
Page 98, #36 : We have
lim
x!2
= lim
x!2
x
x
2
=
x2
2x
x (x 2)
= lim
4x + 4 x!2 (x 2)2
x2
1, since x > 0 and x
2!0
as x ! 2 .
Page
MATH 100: Solutions for Problem Set 3
Page 223, #26 : We have
sec x tan x + sec2 x
= sec x:
sec x + tan x
y0 =
y 00 = sec x tan x:
Page 235, #26 : The population is given by
n (t) =
a
1 + be
We are told that
n (0) = 20 and
Thus
0:7t
:
dn
(0) = 12:
dt
a
=
MATHEMATICS 100 - FINAL EXAMINATION
Date: April 26, 2007
Time: 120 Minutes
Instructor: Professor G. E. Swaters
1. Evaluate the following integrals:
(a)
(b)
(c)
(d)
Z
1
1
e x
dx
x2
0
Z tan 1 ( 4 )
Z
Z
tan
1+e
sin (2 )
p
2
sin3 ( ) d
1( 4)
1 + cos2 ( )
d
ta
Written Assignment 3
Math 100 Fall 2016
Due: Friday, November 18 at 11 pm
Total: 100 points
Please complete each problem on a separate piece of paper. You will need to scan your work
and submit it electronically through Crowdmark. The link will be sent to
Written Assignment 1
Math 100 Fall 2016
Due: Friday, September 30 at 11 pm
Total: 100 points
Please complete each problem on a separate piece of paper. You will need to scan your work
and submit it electronically through Crowdmark. The link will be sent t
Written Assignment 2
Math 100 Fall 2016
Due: Friday, October 21 at 11 pm
Total: 100 points
Please complete each problem on a separate piece of paper. You will need to scan your work
and submit it electronically through Crowdmark. The link will be sent to
Written Assignment 4
Math 100 Fall 2016
Due: Wednesday, December 7 at 11 pm
Total: 100 points
Please complete each problem on a separate piece of paper. You will need to scan your work
and submit it electronically through Crowdmark. The link will be sent
MATHEMATICS 114 (B1)
MIDTERM EXAMINATION
VERSION I
Date: October 19, 2016
Time: 50 minutes
Instructor: E. K. Leonard
I
Last Name: m
First Name: ID#:
INSTRUCTION S:
Eli-NH
. Justify your work to receive full credit.
. No books, notes, formula sheets, calcu