For sets and let and be functions such that for each . State which of the statements are true.
(a) The function must be one-to-one.
(b) The function must be one-to-one.
(c) The function must be onto.
(d) The function must be onto.
Solution
False-True-True-False.
If then so is one-to-one. Also is onto because each is in the image of namely The other two statements are false, for example by constructing an that is a larger finite set than .
Composition
B1a, 2021
is a function defined on the domain and codomain and is a function defined on the domain and codomain . Fill in the blanks with with one of the four options. If is one-one then, is and is . If is onto then, is and is .
Options: A: One-one and onto. B: One-one but need not to be onto. C: Onto but need not to be one-one. D: Need not to be one-one and need not be onto.
Solution
Ans. B-D-D-C. Explanation. If is one-one, the must be one-one. If not, there are two elements and with such that , which implies that . A contradiction. However, need not be onto and need not be onto or one-one as the example below shows.
If is onto, then must be onto since every element in must be reached by some element from via . As the example below shows, this the only constraint.
Write whether the functions have horizontal asymptotes, vertical asymptotes and removable discontinuities.
Solution
How to find a horizontal asymptote?
Let us consider the case when the given function is of the form:
If , then there is no horizontal asymptote.
If , then a horizontal asymptote.
If , then a horizontal asymptote.
We can apply this directly to the given functions. The first function has no horizontal asymptotes. The second and third functions have and as their horizontal asymptotes, respectively.
How to find a vertical asymptote?
Vertical asymptotes occur at those points where the denominator is zero and the numerator is non-zero.
The first two functions have and as their vertical asymptotes. The denominator is always positive for the third function, so there are no vertical asymptotes for this function.
Removable discontinuities
In all the functions, the term can be factored out from the numerator and the denominator. Hence, is a removable discontinuity for all the functions.
Constrained function
B4, 2020
i) A continuous function has the property that If the domain of is and then show that is unique and find
Solution
Since is non-zero, . Since , the range of is non-negative. We will show that .
Let be an arbitrary point and . Since the sequence converges to 0 as . Since the function is continuous: The sequence must converge to 1. This is possible only if . By continuity, too. Therefore, the conditions imply that is unique and that .
ii) Consider the same property of but the domain of the function being Show that either or
Solution
The proof is similar to the previous proof. For we have: So is either 0 or 1.
Lemma. If , then . Proof. For a contradiction, let us say there exists a point such that , where .
Lemma. If , then . Proof. For a contradiction, let us say there exists a point such that .
iii) Show that there exist infinitely many continuous functions with the same property and with domain such that
Solution
For any , the following function satisfies the conditions:
Surjective if and only if injective
B11, 2011
A function from a set to itself satisfies for fixed positive integers and with . The notation stands for times). Show that is one-to-one (injective) if and only if is onto (surjective).
Solution
Claim 1: If is one-to-one, then it is onto.
Proof: Let . The second implication follows since implies that for a one-to-one function. Since is an identity function, its range must be . So the range of must also be , which proves the claim.
Claim 2: If is onto, then it is one-to-one.
Proof: We prove by contradiction. Assume that is onto but not one-to-one. So there must be an such that for some . Since is onto we should be able to find two numbers and such that Also, . It follows that: In general, for every natural number , we should be able to find two different numbers with the property that and . In particular, consider the numbers and , where . Due to the condition , we have: This implies that , which is a contradiction.
Hence, the function is one-to-one if is onto.
Continuity of two functions
A10, 2017
Consider the following function:
(a) Find the value of for which is continuous.
(b)
(c)
Solution
(a) is sandwiched between -1 and so makes continuous.
(b) Now which is similarly 0.
(c) For nonzero calculate so does not exist and does not exist.
