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Department
Mathematics
Course
MATA30H3
Professor
Sophie Chrysostomou
Semester
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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