MATH104 Exam May 2009, Solutions
All questions except q.1 are standard examples similar to classwork or homework.
1. delta, zeta ,ω. (1 mark each.)
1 + 1 + 1 + 1 = 4 marks
2.
(a) f(n) = −3n.
(b) f(n) = n(n+ 1)
(c) {−1,1}.
(d) 1 ≤x
(2 marks for (a), 4 marks for (b) and 3 marks for each of c) and d).)
2 + 4 + 3 + 3 = 12 marks
3. The negations are as follows
(a) x≤10 or x≥11 (1 mark)
(b) There exists x∈Rsuch that x2−1 = 0 and x6= 1.(2 marks)
(c) For all x∈Rwith x≥2, we have f(x)<1(2 marks)
(d) There exists a function f(x)which is both even and odd. (2 marks)
For the original statements (or equivalently, for their negations),in a), xis a free variable,
and in c) the function f(x) is a free variable (4 marks)
b) is false and its negation is true, taking x=−1 (2 marks)
d) is false and its negation is true, taking f(x) = 0 for all x(2 marks)
1 + 2 + 2 + 2 + 4 + 2 + 2 = 15 marks
4.
(a) fis not strictly increasing if there exist x, y in the domain of fsuch that x < y but
f(x)≥f(y). (2 marks)
(b) f(1) = |1|=| − 1|=f(−1), and so fis not injective.(2 marks)
(c) f(0) = 0 and f(2) = −2. So fis not strictly increasing. (2 marks)
To show that fis injective:
f(x) = f(y)⇒x
1−x=y
1−y⇒x(1−y) = y(1−x)⇒x−xy =y−xy ⇒x=y. (4 marks)
(d) It suffices to show that if x6=ythen f(x)6=f(y). If x6=ythen either x < y or y < x. We
can assume that x < y, renaming if necessary. Then f(x)< f(y) and so f(x)6=f(y) (5
marks)
2 + 2 + 2 + 4 + 5 = 15 marks
5.
a) Sclosed under addition means that, for all x∈Sand y∈S, it follows that x+y∈S.
(1 mark)
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