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 \(x=\dfrac{4}{3}\).
Is \(g=\{(1,1),(2,3),(3,5),(4,7)\}\) a function? Justify. If this is described by the relation \(g(x)=\alpha x+\beta\), then what values should be assigned to \(\alpha\) and \(\beta\)?
Yes. A relation is a function if each input (first coordinate) is associated with exactly one output (second coordinate). Here the first coordinates \(1,2,3,4\) are distinct and each appears exactly once, so \(g\) is a function.
Assume \(g(x)=\alpha x+\beta\). Use any two ordered pairs to determine \(\alpha,\beta\). From \(g(1)=1\) we get \(\alpha(1)+\beta=1\Rightarrow \alpha+\beta=1\). From \(g(2)=3\) we get \(2\alpha+\beta=3\). Subtract the first from the second: \(\alpha=2\). Then \(\beta=1-\alpha=1-2=-1\).
Hence \(\alpha=2,\;\beta=-1\).
Find the domain of each of the following functions:
(i) \(f(x)=\dfrac{1}{\sqrt{1-\cos x}}\)
(ii) \(f(x)=\dfrac{1}{\sqrt{x+|x|}}\)
(iii) \(f(x)=x|x|\)
(iv) \(f(x)=\dfrac{x^3-x+3}{x^2-1}\)
(v) \(f(x)=\dfrac{3x}{2x-8}\)
(i) Require denominator \(\sqrt{1-\cos x}\) to be defined and nonzero. So \(1-\cos x>0\Rightarrow\cos x\ne1\). Therefore \(x\ne 2n\pi\) for integer \(n\). Domain: \(\mathbb{R}\setminus\{2n\pi\mid n\in\mathbb{Z}\}\).
(ii) Inside square root: \(x+|x|\ge0\) and denominator nonzero. Note \(x+|x|=\begin{cases}2x,&x\ge0\\0,&x<0\end{cases}\). For \(x<0\) the expression is \(0\) giving division by zero, so excluded. For \(x\ge0\), \(x+|x|=2x\ge0\) and positive for \(x>0\). Hence domain: \(x>0\). That is the positive reals \((0,\infty)\) (often denoted \(\mathbb{R}^+\)).
(iii) The expression \(x|x|\) is defined for all real \(x\) (absolute value always defined). Domain: \(\mathbb{R}\).
(iv) Rational function with denominator \(x^2-1\). Exclude zeros of denominator: \(x^2-1=0\Rightarrow x=\pm1\). Domain: \(\mathbb{R}\setminus\{-1,1\}\).
(v) Rational function with denominator \(2x-8\). Exclude \(2x-8=0\Rightarrow x=4\). Domain: \(\mathbb{R}\setminus\{4\}\).
Find the range of the following functions:
(i) \(f(x)=\dfrac{3}{2-x^2}\)
(ii) \(f(x)=1-|x-2|\)
(iii) \(f(x)=|x-3|\)
(iv) \(f(x)=1+3\cos2x\)
(i) Let \(y=\dfrac{3}{2-x^2}\). Rearranging gives \(x^2=2-\dfrac{3}{y}\). For real \(x\) we need \(\dfrac{2y-3}{y}\ge0\). This inequality holds for \(y\ge\tfrac{3}{2}\) or \(y<0\). Also \(y\ne0\). Thus the range is \((-\infty,0)\cup[3/2,\infty)\).
(ii) Since \(|x-2|\ge0\), maximum of \(1-|x-2|\) is \(1\) (when \(|x-2|=0\)). As \(|x-2|\to\infty\) the expression tends to \(-\infty\). Hence range: \(( -\infty,1]\).
(iii) Absolute value yields nonnegative values. So range: \([0,\infty)\).
(iv) As \(\cos2x\in[-1,1]\), we have \(1+3\cos2x\in[1-3,\;1+3]=[-2,4]\). Hence range: \([-2,4]\).
Redefine the function \(f(x)=|x-2|+|2+x|\) for \(-3\le x\le3\) as a piecewise function (i.e. give \(f(x)\) on intervals determined by the absolute-value expressions).
We partition \([-3,3]\) by the points where expressions inside absolute values change sign: \(x=-2\) and \(x=2\). Evaluate on intervals:
For \(-3\le x<-2\): \(x-2<0\) and \(2+x<0\). So \(|x-2|=-(x-2)=-x+2\) and \(|2+x|=-(2+x)=-x-2\). Thus \(f(x)=(-x+2)+(-x-2)=-2x\).
For \(-2\le x<2\): \(x-2<0\) but \(2+x\ge0\). So \(|x-2|=-x+2,\;|2+x|=x+2\). Hence \(f(x)=(-x+2)+(x+2)=4\) (constant).
For \(2\le x\le3\): both \(x-2\ge0\) and \(2+x\ge0\). So \(|x-2|=x-2,\;|2+x|=x+2\). Then \(f(x)=(x-2)+(x+2)=2x\).
Therefore the piecewise definition on \([-3,3]\) is:
\(f(x)=\begin{cases}-2x,&-3\le x<-2\\4,&-2\le x<2\\2x,&2\le x\le3.\end{cases}\)
If \(f(x)=\dfrac{x-1}{x+1}\), then show that
(i) \(f\bigl(\dfrac{1}{x}\bigr)=-f(x)\)
(ii) \(f\bigl(-\dfrac{1}{x}\bigr)=-\dfrac{1}{f(x)}\)
(i) Compute \(f\bigl(\dfrac{1}{x}\bigr)=\dfrac{\tfrac{1}{x}-1}{\tfrac{1}{x}+1}=\dfrac{1-x}{1+x}\cdot\dfrac{1}{x}\div\dfrac{1}{x}=\dfrac{1-x}{1+x}\). Simplify: \(\dfrac{1-x}{1+x}=-\dfrac{x-1}{x+1}=-f(x)\). Thus \(f(1/x)=-f(x)\).
(ii) Compute \(f\bigl(-\dfrac{1}{x}\bigr)=\dfrac{-\tfrac{1}{x}-1}{-\tfrac{1}{x}+1}=\dfrac{-(1+ x)}{-(1-x)}\) after multiplying numerator and denominator by \(x\). Simplify: \(\dfrac{-(1+x)}{-(1-x)}=\dfrac{1+x}{1-x}\). On the other hand \(f(x)=\dfrac{x-1}{x+1}\) so \(\dfrac{1}{f(x)}=\dfrac{x+1}{x-1}\) and hence \(-\dfrac{1}{f(x)}=-\dfrac{x+1}{x-1}=\dfrac{1+x}{1-x}\). Therefore \(f(-1/x)=-\dfrac{1}{f(x)}\).
Let \(f(x)=\sqrt{x}\) and \(g(x)=x\) be two functions defined in the domain \(\mathbb{R}^+\cup\{0\}\). Find
(i) \((f+g)(x)\)
(ii) \((f-g)(x)\)
(iii) \((fg)(x)\)
(iv) \(\bigl(\dfrac{f}{g}\bigr)(x)\)
Domain for both: \([0,\infty)\).
(i) \((f+g)(x)=f(x)+g(x)=\sqrt{x}+x\).
(ii) \((f-g)(x)=\sqrt{x}-x\).
(iii) \((fg)(x)=f(x)g(x)=x\sqrt{x}=x^{3/2}\).
(iv) \(\bigl(\dfrac{f}{g}\bigr)(x)=\dfrac{\sqrt{x}}{x}=\dfrac{1}{\sqrt{x}}\) for \(x>0\). (Note: at \(x=0\) this quotient is not defined.)
Find the domain and range of the function \(f(x)=\dfrac{1}{\sqrt{x-5}}\).
Inside the square root we need \(x-5>0\) (strictly positive because denominator cannot be zero). Hence domain: \(x>5\) i.e. \((5,\infty)\).
For \(x>5\) we have \(\sqrt{x-5}>0\) so \(f(x)>0\). As \(x\to5^+\), \(\sqrt{x-5}\to0^+\) so \(f(x)\to+\infty\). As \(x\to\infty\), \(\sqrt{x-5}\to\infty\) so \(f(x)\to0^+\). Thus range is \((0,\infty)\) (all positive real numbers).
If \(f(x)=y=\dfrac{ax-b}{cx-a}\), then prove that \(f(y)=x\).
We are given \(y=\dfrac{ax-b}{cx-a}\). We solve this relation for \(x\) in terms of \(y\) and then show that replacing the variable by \(y\) in the formula for \(f\) returns \(x\).
Start with \(y=\dfrac{ax-b}{cx-a}\). Cross-multiply:
\(y(cx-a)=ax-b\Rightarrow cxy-ay=ax-b\).
Collect terms in \(x\): \(cxy-ax = ay-b\Rightarrow x(cy-a)=ay-b\).
Provided \(cy-a\ne0\), we obtain
\(x=\dfrac{ay-b}{cy-a}\).
But the right-hand side is exactly the same functional form as \(f\) with \(x\) replaced by \(y\):
\(f(y)=\dfrac{ay-b}{cy-a}=x\).
Therefore \(f(y)=x\), as required (under the nondegeneracy condition that denominators are nonzero).
Let \( n(A)=m \) and \( n(B)=n \). Then the total number of non-empty relations that can be defined from A to B is
\( m^n \)
\( n^m - 1 \)
\( mn - 1 \)
\( 2^{mn} - 1 \)
If \( [x]^2 - 5[x] + 6 = 0 \), where \([.]\) denotes the greatest integer function, then
\( x \in [3,4] \)
\( x \in (2,3] \)
\( x \in [2,3] \)
\( x \in [2,4] \)
The range of \( f(x)=\dfrac{1}{1-2\cos x} \) is
\( \left[\dfrac{1}{3},1\right] \)
\( [-1,\tfrac{1}{3}] \)
\( (-\infty,-1] \cup [\tfrac{1}{3},\infty) \)
\( [-\tfrac{1}{3},1] \)
Let \( f(x)=\sqrt{1+x^2} \). Then
\( f(xy)=f(x)f(y) \)
\( f(xy)\ge f(x)f(y) \)
\( f(xy)\le f(x)f(y) \)
None of these
The domain of \( \sqrt{a^2-x^2} \) where \( a>0 \) is
\( (-a,a) \)
\( [-a,a] \)
\( [0,a] \)
\( (-a,0] \)
If \( f(x)=ax+b \), where \( a,b \) are integers, \( f(-1)=-5 \) and \( f(3)=3 \), then \( a \) and \( b \) are
\( a=-3, b=-1 \)
\( a=2, b=-3 \)
\( a=0, b=2 \)
\( a=2, b=3 \)
The domain of the function defined by \( f(x)=\sqrt{4-x}+\dfrac{1}{\sqrt{x^2-1}} \) is
\( (-\infty,-1) \cup (1,4] \)
\( (-\infty,-1] \cup (1,4] \)
\( (-\infty,-1) \cup [1,4] \)
None of these
The domain and range of the real function \( f(x)=\dfrac{4-x}{x-4} \) is
Domain = \( \mathbb{R} \), Range = \{ -1, 1 \}
Domain = \( \mathbb{R}-\{1\} \), Range = \( \mathbb{R} \)
Domain = \( \mathbb{R}-\{4\} \), Range = \{ -1 \}
Domain = \( \mathbb{R}-\{4\} \), Range = \{ -1,1 \}
The domain and range of the function \( f(x)=\sqrt{x-1} \) is
Domain = \( (1,\infty) \), Range = \( (0,\infty) \)
Domain = \( [1,\infty) \), Range = \( 0,\infty) \)
Domain = \( [1,\infty) \), Range = \( [0,\infty) \)
Domain = \( [1,\infty) \), Range = \( [0,\infty) \)
The domain of the function \( f(x)=\dfrac{x^2+2x+1}{x^2-x-6} \) is
\( \mathbb{R}-\{3,-2\} \)
\( \mathbb{R}-\{-3,2\} \)
\( \mathbb{R}-[3,-2] \)
\( \mathbb{R}-(3,-2) \)
The domain and range of the function \( f(x)=2-|x-5| \) is
Domain = \( \mathbb{R}^+ \), Range = \((-\infty,1] \)
Domain = \( \mathbb{R} \), Range = \((-\infty,2] \)
Domain = \( \mathbb{R} \), Range = \((-\infty,2) \)
Domain = \( \mathbb{R}^+ \), Range = \((-\infty,2] \)
The domain for which the functions \( f(x)=3x^2-1 \) and \( g(x)=3+x \) are equal is
\( \{-1,\dfrac{4}{3}\} \)
\( \left\{-1,\dfrac{4}{3}\right\} \)
\( \{-1,\dfrac{4}{3}\} \)
\( \{-1,\dfrac{4}{3}\} \)
Let f and g be two real functions given by
f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}
then the domain of f · g is given by ____
{2, 3, 4, 5}
Let \(f = \{(2, 4), (5, 6), (8, -1), (10, -3)\}\) and \(g = \{(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\}\) be two real functions.
| Column A | Column B |
|---|---|
\(f - g\) | (i) \(\{(2, \tfrac{4}{5}), (8, -\tfrac{1}{4}), (10, -\tfrac{3}{13})\}\) |
\(f + g\) | (ii) \(\{(2, 20), (8, -4), (10, -39)\}\) |
\(f \cdot g\) | (iii) \(\{(2, -1), (8, -5), (10, -16)\}\) |
\(\dfrac{f}{g}\) | (iv) \(\{(2, 9), (8, 3), (10, 10)\}\) |
| Column A | Matched Item from Column B |
|---|---|
\(f - g\) | (iii) \(\{(2, -1), (8, -5), (10, -16)\}\) |
\(f + g\) | (iv) \(\{(2, 9), (8, 3), (10, 10)\}\) |
\(f \cdot g\) | (ii) \(\{(2, 20), (8, -4), (10, -39)\}\) |
\(\dfrac{f}{g}\) | (i) \(\{(2, \tfrac{4}{5}), (8, -\tfrac{1}{4}), (10, -\tfrac{3}{13})\}\) |
The ordered pair \((5, 2)\) belongs to the relation \(R = \{(x, y) : y = x - 5,\ x, y \in \mathbb{Z}\}\).
False
If \(P = \{1, 2\}\), then \(P \times P \times P = \{(1,1,1),\ (2,2,2),\ (1,2,2),\ (2,1,1)\}\).
False
If \(A = \{1,2,3\}\), \(B = \{3,4\}\) and \(C = \{4,5,6\}\), then \((A \times B) \cup (A \times C) = \{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\}\).
True
If \((x - 2,\ y + 5) = (-2,\ \tfrac{1}{3})\) are two equal ordered pairs, then \(x = 4\) and \(y = -\tfrac{14}{3}\).
False
If \(A \times B = \{(a,x),\ (a,y),\ (b,x),\ (b,y)\}\), then \(A = \{a,b\}\) and \(B = \{x,y\}\).
True