Exercises – Long Day#
Exercise 1: Sets on Roster Form#
Let \(A\) and \(B\) be finite sets given in the following roster forms:
Question a#
Which elements do the sets \(A\) and \(B\) contain?
Answer
Set \(A\) contains the following elements: \(A=\{1,4,9,16,25\}\).
Question b#
Which elements do the sets \(A \cap B\) and \(A \cup B\,\) contain?
Hint
To see how \(\cap\) (the symbol for set intersection) and \(\cup\) (the symbol for set union) are defined, see Equations (2.1) and (2.2) in the textbook.
Question c#
Let \(C\) and \(D\) be sets given in the roster forms:
Which elements do the sets \(C \cap D\) and \(C \cup D\,\) contain?
Answer
\(C \cap D\) contains all natural numbers that are both a multiple of \(2\) and of \(3\). Thus, \(C \cap D= \{n \in \Bbb{N} \, | \, n=6m \,\,\,\mathrm{where} \,\,\, m \in \Bbb{N}\}\).
\(C \cup D\) contains all natural numbers that are a multiple of \(2\) or a multiple of \(3\). Thus, \(C \cup D = \{2,3,4,6,8,9,10,12,\dots\}.\)
Question d#
Describe in your own words the sets \(\Bbb{R} \setminus \Bbb{Q}\) and \(\Bbb{Z} \setminus \Bbb{N}\,.\)
Exercise 2: Rules of Calculation for Set Operations#
Question a#
Some identities from set theory can be illustrated by drawing a circle diagram, also known as a Venn diagram. In the textbook the concepts of union, set intersection and set difference are illustrated in this way in Section 2.1. Also, the set identities in Theorem 2.1.2 in the textbook can be visualized by the help of such diagrams.
Draw circle diagrams that visualise the identities in Equations (2.8) through (2.11) in Theorem 2.1.2 of the textbook.
Question b#
Use propositional logic to prove the identities given in Equations (2.4) and (2.10) in Theorem 2.1.2 in the textbook.
Hint
In particular to prove the identity in Equation (2.10) you can get inspiration from the proof of the identity in Equation (2.11) in Theorem 2.1.2 in the textbook.
Question c#
Let \(A\) and \(B\) be sets with a finite number of elements. Let \(|A|\) denote the number of elements within a set \(A\). Explain the following identity:
\(|A \cup B| = |A| + |B| - |A \cap B|\).
Hint
To get the intuition right, first you can check that the identity holds true for the sets \(A\) and \(B\) from Exercise 1.
Exercise 3: Surjective, Injective, and Bijective#
Four sets are given by:
Four functions are given below. Determine for each if it is surjective, injective, and/or bijective.
Question a#
\(f_1 :A\rightarrow B\)
\(\ \ x \mapsto x^2\)
Hint
If you are in doubt about the exact meanings of the terms “injective”, “surjective”, and “bijective”, then you can refresh your memory by reading Section 2.2 in the textbook once more (more precisely the test after Lemma 2.2.1 and until Definition 2.2.1).
Hint
It can be helpful to first write out the sets \(A\) and \(B\) explicitely. Try for example to realise that \(A=\{1,2,3,4,5,6,7\}.\)
Answer
\(f_1\) is both injective and surjective. Hence it is also bijective.
Question b#
\(f_2 :D\rightarrow B\)
\(\ \ x \mapsto x^2\)
Hint
To show that a function is not injective, it is sufficient to find two different elements from the domain of the function that are mapped to the same element in the co-domain.
Answer
\(f_2\) is surjective but not injective, and hence also not bijective.
Question c#
\(f_3 :C\rightarrow B\)
\(\ \ x \mapsto x^2\)
Hint
To show that a function is not surjective it is sufficient to show that the image set of the function is not the entire co-domain.
Answer
\(f_3\) is injective but not surjective, and hence also not bijective.
Question d#
\(f_4 :\Bbb Z\rightarrow \Bbb Z\)
\(\ \ x \mapsto |x|\)
Answer
\(f_4\) is not injective nor surjective, and hence also not bijective.
Exercise 4: An Inverse Function#
We are given a function \(f: \mathbb{R} \rightarrow \mathbb{R}\) by its functional expression
Question a#
Let \(g: \mathbb{R} \rightarrow \mathbb{R}\) be a function with the functional expression
Show using Definition 2.2.1 from the textbook that \(g\) is the inverse function of \(f\).
Hint
If you draw the graphs of \(f\) and \(g\) you will be able to see a symmetric connection between them. What symmetry is that? Have a look at Example 2.3.1 in the textbook where a similar symmetry appears.
Question b#
Justify that \(f\) is bijective.
Hint
Rather than starting out by showing that \(f\) is injective and surjective, try applying Lemma 2.2.2.
Exercise 5: From Angle Measure to Radian Value and Vice Versa#
Bijections are involved in the conversion from one unit to another. In this exercise we will recap on our ability to convert angle measures from the degree unit to the radian unit, and vice versa.
Question a#
State the radian value that corresponds to the angle measures: \(30, 60, 120, 135\), and \(300\,\mathrm{degrees}\).
Answer
\(\frac{\pi}{6}\), \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(\frac{3\pi}{4}\), \(\frac{5\pi}{3}\). In general: \(x\,\mathrm{degrees}\) corresponds to the radian value \(x \pi / 180.\)
Question b#
Draw the unit circle in a \((x,y)\) coordinate system centered at the origin. Draw points on the unit circle corresponding to the arc lengths
Which angle measures in degrees do these correspond to?
Answer
\(180\), \(60\), \(330\), \(330\), \(105\), \(90\), \(315\). In general: the arch length \(x\) corresponds to \(180x/\pi\) degrees if \(x \in [0,2\pi[\). If \(x \not\in [0,2\pi[\), then we must first add a multiple of \(2\pi\) such that the result is within the interval \([0,2\pi[\).
Exercise 6: Cosinus and Sinus Repetition#
Question a#
Use the figure (the blue triangle) for geometric determination of the exact values of \(\,\displaystyle{\cos\left(\frac{\pi}{4}\right)}\,\) and \(\,\displaystyle{\sin\left(\frac{\pi}{4}\right)}\,.\)
Hint
Remember that the Pythagorean Theorem implies that \(\cos^2(x)+\sin^2(x)=1\) for all real numbers \(x\).
Answer
Both have the value \(\frac{\sqrt{2}}{2}\,. \)
Question b#
Determine via symmetry considerations the numbers
Answer
\((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}),(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}),(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}),(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}),(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}), (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}),(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}).\)
Question c#
We are informed that \(\,\displaystyle{\cos\left(\frac{\pi}{6}\right)}=\frac{\sqrt 3}{2}\,\) and \(\,\displaystyle{\sin\left(\frac{\pi}{6}\right)}=\frac{1}{2}\,.\) Draw the point
on a unit circle and determine via symmetry considerations the values:
Exercise 7: \(\mathrm{arccos}\), \(\mathrm{arcsin}\) and Trigonometric Equations#
In this exercise we will consider the inverse trigonometric functions \(\mathrm{arccos}\) and \(\mathrm{arcsin}\) as well as some trigonometric equations. If you need a refresher or would like to see the graphs of these functions then have a look in Section 2.3 in the textbook.
Question a#
State the values of \(\,\displaystyle{\mathrm{arccos}\left(\frac{1}{2}\right),\,\mathrm{arcsin}\left(-\frac{\sqrt 3}{2}\right)}\) and \(\displaystyle{\mathrm{arcsin}(1)}\,.\)
Answer
\(\frac{1}{3} \, \pi\), \(-\frac{1}{3} \, \pi\), \(\frac{1}{2} \, \pi\).
Question b#
Let \(x \in \mathbb{R}\) and \(y \in [-1,1]\) be real numbers. Let \(P\) be the logical proposition \(\mathrm{arccos(y)}=x\) and \(Q\) the logical proposition \(y=\cos(x)\). Show that \(P \Rightarrow Q\) is true but that \(Q \Rightarrow P\) is not necessarily true.
Hint
For \(P \Rightarrow Q\): What happens if you apply the \(\cos\) function on both sides of the equal sign in \(P\)?
For \(Q \Rightarrow P\): Can you find \(x \in \mathbb{R}\) and \(y \in [-1,1]\) such that \(Q\) is true while \(P\) is not true?
Question c#
We are now given the sets \(\,A=\left[\,0\,,\,2\pi\,\right]\,\) and \(\,B=\left[\,-\pi\,,\,\pi\,\right]\,.\)
Solve the equation \(\,\displaystyle{\cos(x)=\frac{1}{2}}\,\) within each of the sets \(\,A,\,B\,\) and \(\,\Bbb R\,.\)
Hint
A solution to the equation is achieved directly from Question a because according to Question b, \(\mathrm{arccos}(y)=x\) implies that \(y=\cos(x).\) Now, find all solutions within \(\mathbb{R}\) by drawing a sketch of the graph of the \(\cos\) function.
Answer
Within \(A\) the solutions are \(\,\frac{\pi}{3}\,\) and \(\,\frac{5\pi}{3}\,.\)
Within \(B\) the solutions are \(\,\frac{-\pi}{3}\,\) and \(\,\frac{\pi}{3}\,.\)
Within \(\mathbb{R}\) the solutions are \(\,\frac{-\pi}{3}+p\cdot 2\pi\,\) and \(\,\frac{\pi}{3}+p\cdot 2\pi\,\) where \(\,p \in \Bbb Z\,.\)
Question d#
Solve the equation \(\,\displaystyle{\sin(x)=-\frac{\sqrt 3}{2}}\,\) within each of the sets \(\,A,\,B\,\) and \(\,\Bbb R\,.\)
Answer
Within \(A\) the solutions are: \(\,\frac{4\pi}{3}\,\) and \(\,\frac{5\pi}{3}\,.\)
Within \(B\) the solutions are: \(\,\frac{-2\pi}{3}\,\) and \(\,-\frac{\pi}{3}\,.\)
Within \(\mathbb R\) the solutions are: \(\,\frac{4\pi}{3}+p\cdot 2\pi\,\) and \(\,\frac{5\pi}{3}+p\cdot 2\pi\,\) where \(\,p \in \Bbb Z\,.\)
Exercise 8: Second-Degree Polynomial Functions#
We consider a function \(h: \mathbb{R} \rightarrow \mathbb{R}\) with the functional expression
Question a#
Bring the function \(h\) to the form \(h(x)=2(x-k_1)^2+k_2\), and state the constants \(k_1\) and \(k_2\). Use this form to determine the image set of \(h\).
Hint
A possible method that can be used here is what is known as “completing the square”.
Answer
\(h\) has the image set \(\{ x \in \mathbb{R} \, | \, x \ge 7\}.\) This set can also be described as \(\mathbb{R}_{\ge 7}\).
Question b#
State the largest possible interval \(J \subseteq {\Bbb R}_{\geq 0}\) on which the restriction of \(h\) becomes injective.
Hint
Use Question a to draw a sketch of the graph of \(h\).
Answer
\(J=\{ x \in \mathbb{R} \, | \, x \ge 5\}.\) This set can also be described as \(\mathbb{R}_{\ge 5}\).
Question c#
We now consider the restriction of \(h\) to the interval \(J\) from Question b and delimit the co-domain of \(h\) to the set \(\mathbb{R}_{\ge 7}\) from Question a. The resulting function is bijective and is denoted by \(h_1\). More directly, \(h_1\) is the function \(h_1: J \rightarrow \mathbb{R}_{\ge 7}\) that is given by \(h_1(x)=2x^2 -20x +57.\)
Provide a functional expression for the inverse function \({h_1}^{-1}\)
Hint
A good start is to solve the equation \(2x^2 -20x +57=y\) for \(x\).
Question d#
State the domain and the image set of \({h_1}^{-1}\).
Answer
The function \(h_1^{-1}\) has the domain \(\mathbb{R}_{\ge 7}\) and the co-domain \(\mathbb{R}_{\ge 5}\). Since \(h_1^{-1}\) is surjective (in fact bijective), then the image set of \(h_1^{-1}\) is the same as its co-domain, so \(\mathbb{R}_{\ge 5}\).
Exercise 9: Bijection#
We are given the function \(f :\Bbb N\rightarrow \Bbb Z\) defined by
\( x \mapsto \left\{ \begin{array}{ll} \frac{x}{2} & \text{when }x \text{ is even,} \\ -\frac{x-1}{2} & \text{when }x \text{ is odd.} \\ \end{array} \right. \)
Question a#
Is \(f\) a bijection?
Hint
To get an idea of the behaviour of the function, first compute \(f(1)\), \(f(2)\), \(f(3)\), \(f(4)\), and \(f(5)\).
Answer
Yes.
Exercise 10: Hyperbolic Functions#
In this exercise we will introduce two new functions that are formed from already known functions. The two functions are called hyperbolic sinus and hyperbolic cosinus and they are defined by:
We assume for both functions that both their domains and co-domains are equal to \(\mathbb R\).
Question a#
Justify that \(\mathrm{sinh}(x)\) is injective and that \(\mathrm{cosh}(x)\) is not injective.
Hint
Are the functions monotone?
Question b#
Determine a functional expression for \(\mathrm{sinh}^{-1}\) by isolating \(x\) in the equation \(y=\mathrm{sinh}(x)\).
Hint
Multiply your equation through with \(\mathrm e^x\) and solve the quadratic equation that appears.
Hint
Why can we ignore one of the solutions?