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Mathematics

MATA30H3

Sophie Chrysostomou

Winter

Description

University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT A30F
Solution to Assignment #1
A. Homework problems from the lectures:
1. Show that:
(i) the product of two even functions is an even function.
(ii) the product of two odd functions is an even function.
(iii) the product or an odd and an even function is an odd function.
solution : (i) Let f and g be even functions. Then f(−x) = f(x) and g(−x) = g(x) for
all x ∈ R. Let h(x) = f(x)g(x). Then:
h(−x) = f(−x)g(−x) = f(x)g(x) = h(x)
Thus h, is an even function.
(ii) Let f and g be odd functions. Then f(−x) = −f(x) and g(−x) = −g(x) for all
x ∈ R. Let h(x) = f(x)g(x). Then:
h(−x) = f(−x)g(−x) = (−f(x))(−g(x)) = f(x)g(x) = h(x)
Thus h, is an even function.
(iii) Let f be an even function and g be an odd function. Then f(−x) = f(x) and
g(−x) = −g(x) for all x ∈ R. Let h(x) = f(x)g(x). Then:
h(−x) = f(−x)g(−x) = f(x)(−g(x)) = −f(x)g(x) = h(x)
Thus h, is an odd function.
2. Use the factor theorem to show that
(a) x − 2x + 4x − 5x − 6 is divisible by x − 2.
(b) x + 4x + 2x − 2x + x − 6 is divisible by x + 2x − 3.
(c) x + (2 − a)x + (5 − 2a)x − 5a is divisible by x − a.
solution : Factor Theorem: If p is (a nonzero) polynomial, then p(a) = 0 if and only if
x − a is a factor of p.
(a) Let p(x) = x − 2x + 4x − 5x − 6, then p(2) = 0. Thus, by the factor thm (x − 2) is
a factor of p.
1 (b) First factor x + 2x − 3: x + 2x − 3 = (x + 3)(x − 1).
Substitute x = −3 and x = 1 into p(x) = x + 4x + 2x − 2x + x − 6 to obtain :
5 4 3 2
p(−3) = (−3) + 4(−3) + 2(−3) − 2(−3) + (−3) − 6 = 0 ⇒ (x + 3),(x − 1)
5 4 3 2 are both factors of p.
p(1) = 1 + 4(1 ) + 2(1 ) − 2(1 ) + 1 − 6 = 0
2
Thus, (x + 3)(x − 1) = x + 2x − 3 is a factor of p.
(c) Let p(x) = x + (2 − a)x + (5 − 2a)x − 5a.
3 2 3 2 3 2
p(a) = a + (2 − a)a + (5 − 2a)a − 5a = a + 2a − a + 5a − 2a − 5a = 0.
Thus (x − a) is a factor of p.
3. Use the rational root theorem to ﬁnd if p(x) = x +8x +19x +22x+10 has any rational
roots. If there are such roots, factor p(x) as far as possible.
m
solution : If p(x) has any rational roots they must be of the form n where m,n are
integers and m divides 10, where n divides 1.
Possibilities for m: 10,−10,5,−5,2,−2,1,−1.
Possibilities for n: 1,−1.
m
Possibilities for : 1,−1,2,−2,5,−5,10,−10.
n
As each coeﬃcient in p is positive, the roots cannot be positive. So we need only test the
negative candidates. We get:
p(−1) = 0, p(−5) = 0, whereas p(−2) ▯= 0,p(−10) ▯= 0. Therefore x + 1 and x + 5 are
two factors of p(x).
Divide p(x) by (x + 1)(x + 5) = x + 6x + 5 to get the next factor x + 2x + 2.
2
x + 2x + 2
2 4 3 2
x + 6x + 5 | x + 8x + 19x + 22x + 10
x + 6x + 5x 2
3 2
2x + 14x + 22x + 10 2 2
2x + 12x + 10x so p(x) = (x + 5x + 6)(x + 2x + 2)
2
2x + 12x + 10
2
2x + 12x + 10
0
2 2 2
Hence p(x) = (x + 5)(x + 1)(x + 2x + 2). Note x + 2x + 2 = (x + 1) + 1 has no factors
over the real numbers.
2 P(x) R(x)
4. For the following functions f(x) = express f(x) in the form f(x) = Q(x) + .
2 D(2) D(x)
x x + 1
i) f(x) = 2 ii) f(x) =
x − 1 x − 2
x3 x − b
iii) f(x) = iv) f(x) = 3
x − 2 x + a
x − 3 x − 1
v) f(x) = vi) vi) f(x) =
x + x + 1 x + x + 1
solution :
x2
i) f(x) = 2
x − 1
1
2 2
x − 1 | x 1
x − 1 so f(x) = 1 + x − 1
1
2
ii) f(x) = x + 1
x − 2
x + 2
x − 2 | x2 + 1
2
x − 2x 5
2x + 1 so f(x) = x + 2 + x − 2
2x − 4
5
x 3
iii) f(x) =
x − 2
2
x + 2x + 4
x − 2 | x3
3 2
x − 2x
2x 2 2 8
2 so f(x) = x + 2x + 4 +
2x − 4x x − 2
4x
4x − 8
8
x − b
iv) f(x) =
x + a
1
x3 + a | x − b a + b
3 so f(x) = 1 − 3
x + a x + a
−a − b
3 3
x − 3
v) f(x) = x + x + 1
x − 1
x + x + 1 | x 3 − 3
x + x + x 2
2 so f(x) = x − 1 −
−x − x − 3 x + x + 1
−x − x − 1
−2
vi)
3 2
x − 1 (x − 1)(x + x + 1)
f(x) = x + x + 1 = (x + x + 1) = x − 1
C. Problems from topics from CRM:
1. Let A = {2,5,6,8,11,15,18,19,24,29}, B = {1,3,7,11,13,18,22,25} and
C = {3,4,6,13,14,19,22,26,29}. Find the following:
(a) A ∩ B = {11,18}, A ∩ C = {6,19,29}, B ∩ C = {3,13,22}, A ∩ B ∩ C = { } or Ø.
(b) A ∪ B = {1,2,3,5,6,7,8,11,13,15,18,19,22,24,25,29},
B ∪ C = {1,3,4,6,7,11,13,14,18,19,22,25,26,29}
A ∪ C = {2,3,4,5,6,8,11,13,14,15,18,19,22,24,26,29}.
(c)
A ∩ (B ∪ C) = A ∩ {1,3,

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