MATH 1200 A02 Fall 2012 Problem Workshop 11 Solutions
1. Put the following complex numbers into Polar form
(a)
r=
1
1
( )2 + ( )2 = 1 = 1
2
2
tan =
1
2
1
2
= 1
which happens at 3/4, 7/4. Since y < 0 and x > 0 we have 7/4. Hence the
polar form is
7
.
1cis
MATH 1200 A02 Fall 2012 Problem Workshop 9 Solutions
1. All of these can be found using the special triangles for sine and cosine along unit the
unit circle to put it into dierent quadrants.
(a)
1
= .
4
2
sin
(b) Quadrant 2 gives a negative value for cosi
MATH 1200 A02 Fall 2012 Problem Workshop 2 Solutions
1. (a) The numerator is 12 , 22 , 32 , . . . , 152 and the rst term of the denominator is 2i
and the second term is one more than the rst term, so we get the general term
i2
.
(2i)(2i + 1)
Hence the sig
Updated: August 29, 2013
MATH 1200 Elements of Discrete Mathematics, Fall 2013
(MWF 1:30-2:20pm, ARMES 208)
Instructor:
Office hours:
K. Kopotun, [email protected], 422 Machray Hall, 474-9789
MWF 12:30-1:20pm (or by appointment)
Textbook:
Option
MATH 1200 A02 Fall 2012 Problem Workshop 4
1. Find the general term for the sequences, and nd the limits if they exist.
(a) c1 = 2, cn+1 = 3cn + 1 for n 1.
cn
for n 1.
2
cn + 21 for n 1.
n
2cn + 21 for n 1.
n
(b) c1 = 1, cn+1 = 1 +
(c) c1 = 1, cn+1 =
(d)
MATH 1200 Fall 2012 Quiz 2B
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Student ID: ttttttttttttttttttttttttttttttttttttttttttttttttttttttttt
1. Assume the follow
MATH 1200 Fall 2012 Quiz 2B Solutions
1. We can assume the sequence is bounded, so we just need to show it is monotonic. The
rst few terms are
c1 = 1, c2 =
9 = 3, c3 =
5 + 4(3) =
17 4.12
so it appears that the terms are increasing. We show this by inducti
MATH 1200 A02 Fall 2012 Problem Workshop 1
Prove the following using induction:
1. 3 + 6 + 12 + + 3(2n ) = 3(2n+1 1) for all n 1.
2. 1(2) + 2(3) + 3(4) + + n(n + 1) = n(n + 1)(n + 2)/3 for all n 2.
3. 8n 3n is divisible by 5 for all n 0.
4. n! > 2n for al
MATH 1200 A02 Fall 2012 Problem Workshop 3
1. Show the following sequences are monotonic and bounded, then nd the limit.
(a) c1 = 2, cn+1 = cn + 4 for all n 1.
1
(b) c1 = 2, cn+1 = 3cn for all n 1. (Hint: Use indution to show that the bounds
are 0 cn 2 at
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MATH 1200 A02 Fall 2012 Problem Workshop 8 Solutions
Just to spice things up, and because I dont want to re-type in the functions multiple
times, I will do all 3 questions for each function at once. Itll be a wild ride. Better then the
scariest haunted ho
MATH 1200 A02 Fall 2012 Problem Workshop 4
1. Find the general term for the sequences, and nd the limits if they exist.
(a) If we start listing out a few terms
c1 = 2
c2 = 3c1 + 1 = 3(2) + 1
c3 = 3c2 + 1 = 3(3(2) + 1) + 1 = 32 (2) + 3(1) + 1
c4 = 3c3 + 1
MATH 1200 A02 Fall 2012 Problem Workshop 6 Solutions
1. (a) To use the rational root theorem, we need to make the coecients integers. So
change the polynomial to
Q(x) = 2x3 + 3x2 + 9x + 4
which has no sign changes. Hence there are no positive roots. Letti
1
MATH 1200: Test #4 (Fall 2011)
B04.
Solution; marking scheme
Important: In all three questions use the function f (x) = x 3 3x + 1 .
[7] 1. [3] (a) Show that there is a solution c of f (x) = 0 such that 0 < c < 1 .
[4] (b) The first two steps of approxi
1
MATH 1200: Test #2 (Fall 2011)
B02.
Solution; marking scheme
[6] 1. Consider the sequence cfw_cn defined by c0 = 2 , cn +1 =
(a) Show that cfw_cn is monotonic.
cn
for n 1 .
2
(b) Show that the sequence cfw_cn is bounded.
(c) Find lim cn .
n
Solution.
Mathematics 1200
Lab 9
Winter 2014
3
+1
1. Sketch a graph showing the line y = x and the curve y = g(x) = x 3 showing a
3 3x + 1 = 0 in
region around their intersection point x = , the root of f (x) = x
the interval [2, 1]. (NOTE: this is the root we did
MATH 1200 A02 Fall 2012 Problem Workshop 5 Solutions
1. For the following polynomials, P (x), solve for P (x) = 0.
(a) We test some numbers to see if they make P (x) = 0.
P (1) = 13 2(1) + 1 = 1 2 + 1 = 0.
Hence we know that x 1 is a factor of P (x). Usin
MATH 1200 A02 Fall 2012 Problem Workshop 12 Solutions
1. Rearranging to nd the characteristic equation
cn+2 = 5cn+1 6cn cn+2 5cn+1 + 6cn = 0.
Therefore the characteristic equation is
2 5 + 6 = 0 ( 3)( 2) = 0 = 3, 2.
Hence we have two dierent real solution
Mathematics 1200
Lab 10
Winter 2014
1. (a) A wheel rolls 3m while rotating through an angle of
diameter of the wheel?
11
6 .
What is the (exact)
(b) A circular pizza has radius 30cm and a slice from the pizza has area 1000 cm2 .
What (exact) angle (at the
Mathematics 1200
Lab 11
Winter 2014
1. Show that
sin 3 sin
= tan
cos 3 + cos
1 + tan x
(b)
= tan x +
1 tan x
4
(a)
Remember you are expected to produce a chain of equalities.
2. Find all solutions
(a) cos 2x = sin x
(b) sin 3x + sin 5x = cos 3x cos 5x