Let \(A = \{-1,2,3\}\) and \(B = \{1,3\}\). Determine
(i) \(A \times B\)
(ii) \(B \times A\)
(iii) \(B \times B\)
(iv) \(A \times A\)
(i) \(\{(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)\}\)
(ii) \(\{(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\}\)
(iii) \(\{(1,1),(1,3),(3,1),(3,3)\}\)
(iv) \(\{(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\}\)
If \(P=\{x:x<3,\;x\in\mathbb{N}\}\), \(Q=\{x:x\le 2,\;x\in W\}\). Find \((P\cup Q)\times(P\cap Q)\), where \(W\) is the set of whole numbers.
\(P=\{1,2\},\;Q=\{0,1,2\}\). Hence \(P\cup Q=\{0,1,2\}\) and \(P\cap Q=\{1,2\}\).
So \((P\cup Q)\times(P\cap Q)=\{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}\).
If \(A=\{x:x\in W,\; x<2\}\), \(B=\{x:x\in\mathbb{N},\;1 (i) \(A\times(B\cap C)\) (ii) \(A\times(B\cup C)\)
\(A=\{0,1\},\;B=\{2,3,4\},\;C=\{3,5\}\).
(i) \(B\cap C=\{3\}\) so \(A\times(B\cap C)=\{(0,3),(1,3)\}\).
(ii) \(B\cup C=\{2,3,4,5\}\) so \(A\times(B\cup C)=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}\).
In each of the following cases, find \(a\) and \(b\):
(i) \((2a+b,\;a-b)=(8,3)\)
(ii) \(\left(\dfrac{a}{4},\;a-2b\right)=(0,\;6+b)\)
(i) From \(a-b=3\) we get \(b=a-3\). Substitute in \(2a+b=8\): \(2a+(a-3)=8\Rightarrow3a=11\Rightarrow a=\dfrac{11}{3},\; b=\dfrac{2}{3}\).
(ii) From \(\dfrac{a}{4}=0\) we get \(a=0\). Then \(a-2b=6+b\Rightarrow -2b=6+b\Rightarrow -3b=6\Rightarrow b=-2\).
Given \(A=\{1,2,3,4,5\}\), \(S=\{(x,y):x\in A,\;y\in A\}\). Find the ordered pairs in \(S\) which satisfy:
(i) \(x+y=5\)
(ii) \(x+y<5\)
(iii) \(x+y>8\)
(i) \(\{(1,4),(2,3),(3,2),(4,1)\}\).
(ii) \(\{(1,1),(1,2),(2,1),(1,3),(3,1),(2,2)\}\) (all pairs with sum \(<5\)).
(iii) \(\{(4,5),(5,4),(5,5)\}\) (pairs with sum \(>8\)).
Given \(R=\{(x,y):x,y\in W,\;x^2+y^2=25\}\). Find the domain and range of \(R\) (\(W\) = whole numbers).
Integer whole-number solutions of \(x^2+y^2=25\) are \((0,5),(5,0),(3,4),(4,3)\) so
Domain \(=\{0,3,4,5\}\) and Range \(=\{0,3,4,5\}\).
If \(R_1=\{(x,y)\mid y=2x+7,\;x\in\mathbb{R},\;-5\le x\le5\}\) is a relation, then find the domain and range of \(R_1\).
Domain \(= [-5,5]\).
Range: for \(x=-5\) we get \(y=-3\), for \(x=5\) we get \(y=17\). Hence Range \(=[-3,17]\).
If \(R_2=\{(x,y)\mid x,y\in\mathbb{Z}\text{ and }x^2+y^2=64\}\) is a relation, then find \(R_2\).
Integer solutions of \(x^2+y^2=64\) are \((\pm8,0)\) and \((0,\pm8)\). Thus
\(R_2=\{(8,0),(-8,0),(0,8),(0,-8)\}\).
If \(R_3=\{(x,|x|)\mid x\text{ is a real number}\}\) is a relation, then find the domain and range of \(R_3\).
Domain \(=\mathbb{R}\) (all real numbers).
Range \(=\{y\mid y\ge0\}= [0,\infty)\).
Is the given relation a function? Give reasons for your answer.
(i) \(h=\{(4,6),(3,9),(-11,6),(3,11)\}\)
(ii) \(f=\{(x,x)\mid x\in\mathbb{R}\}\)
(iii) \(g=\{(n,\tfrac{1}{n})\mid n\text{ is a positive integer}\}\)
(iv) \(s=\{(n,n^2)\mid n\text{ is a positive integer}\}\)
(v) \(t=\{(x,3)\mid x\in\mathbb{R}\}\)
(i) Not a function (the first component \(3\) maps to both \(9\) and \(11\)).
(ii) Function (identity mapping: each input has exactly one output).
(iii) Function (each positive integer \(n\) maps to a unique \(1/n\)).
(iv) Function (each \(n\) maps to a unique \(n^2\)).
(v) Function (constant function: each \(x\) maps to \(3\)).
If \(f\) and \(g\) are real functions defined by \(f(x)=x^2+7\) and \(g(x)=3x+5\), find each of the following:
(a) \(f(3)+g(-5)\)
(b) \(f(\tfrac{1}{2})\times g(14)\)
(c) \(f(-2)+g(-1)\)
(d) \(f(t)-f(-2)\)
(e) \(\dfrac{f(t)-f(5)}{t-5},\; t\ne5\)
(a) \(f(3)=16,\;g(-5)=-10\Rightarrow f(3)+g(-5)=6\).
(b) \(f(\tfrac{1}{2})=\tfrac{29}{4},\;g(14)=47\Rightarrow f(\tfrac{1}{2})\times g(14)=\dfrac{1363}{4}\).
(c) \(f(-2)=11,\;g(-1)=2\Rightarrow f(-2)+g(-1)=13\).
(d) \(f(t)-f(-2)=t^2+7-11=t^2-4\).
(e) \(f(5)=32\Rightarrow\dfrac{f(t)-f(5)}{t-5}=\dfrac{t^2-25}{t-5}=t+5\) for \(t\ne5\).
Let \(f\) and \(g\) be real functions defined by \(f(x)=2x+1\) and \(g(x)=4x-7\).
(a) For what real numbers \(x\) is \(f(x)=g(x)\)?
(b) For what real numbers \(x\) is \(f(x)
(a) Solve \(2x+1=4x-7\Rightarrow 2x=8\Rightarrow x=4\).
(b) \(2x+1<4x-7\Rightarrow 8<2x\Rightarrow x>4\).
If \(f\) and \(g\) are two real valued functions defined as \(f(x)=2x+1\), \(g(x)=x^2+1\), then find:
(i) \(f+g\)
(ii) \(f-g\)
(iii) \(fg\)
(iv) \(\dfrac{f}{g}\)
(i) \((f+g)(x)=x^2+2x+2\).
(ii) \((f-g)(x)=2x+1-(x^2+1)=2x-x^2\).
(iii) \((fg)(x)=(2x+1)(x^2+1)=2x^3+x^2+2x+1\).
(iv) \(\dfrac{f}{g}(x)=\dfrac{2x+1}{x^2+1}\) (where \(x^2+1\ne0\), always true for real \(x\)).
Express the following function as set of ordered pairs and determine its range. \(f:X\to\mathbb{R},\; f(x)=x^3+1\), where \(X=\{-1,0,3,9,7\}\).
Compute values: \(f(-1)=0,\;f(0)=1,\;f(3)=28,\;f(9)=730,\;f(7)=344\).
So \(f=\{(-1,0),(0,1),(3,28),(7,344),(9,730)\}\) and Range \(=\{0,1,28,344,730\}\).
Find the values of \(x\) for which the functions \(f(x)=3x^2-1\) and \(g(x)=3+x\) are equal.
Solve \(3x^2-1=3+x\Rightarrow3x^2-x-4=0\). Discriminant \(\Delta=1+48=49\).
Roots: \(x=\dfrac{1\pm7}{6}\Rightarrow x=-1\) or \(x=\dfrac{4}{3}\).