\displaystyle \textbf{Question 1: }\text{Define a function as a set of ordered pairs.}
\displaystyle \text{Answer:}
\displaystyle \text{A function is a set of ordered pairs in which no two distinct ordered pairs}
\displaystyle \text{have the same first component. In other words, each element of the domain}
\displaystyle \text{is associated with exactly one element of the co-domain.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Define a function as a correspondence between two sets.}
\displaystyle \text{Answer:}
\displaystyle \text{A function is a correspondence between two sets, called the domain and}
\displaystyle \text{the co-domain, such that every element of the domain is assigned exactly}
\displaystyle \text{one element of the co-domain.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the fundamental difference between a relation and a function?}
\displaystyle \text{Is every relation a function?}
\displaystyle \text{Answer:}
\displaystyle \text{A relation is any subset of the Cartesian product of two sets.}
\displaystyle \text{A function is a special type of relation in which every element of the}
\displaystyle \text{domain is associated with exactly one element of the co-domain.}
\displaystyle \text{A relation may assign one, many or no images to an element of the domain,}
\displaystyle \text{whereas a function assigns exactly one image to each element of the domain.}
\displaystyle \text{Hence, every function is a relation, but every relation is not a function.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{-2,-1,0,1,2\}\text{ and }f:A\rightarrow Z\text{ be defined by}
\displaystyle f(x)=x^2-2x-3.\text{ Find: (i) the range of }f,\text{ i.e. }f(A),
\displaystyle \text{(ii) the pre-images of }6,\ -3\text{ and }5.
\displaystyle \text{Answer:}
\displaystyle A=\{-2,-1,0,1,2\},\qquad f(x)=x^2-2x-3
\displaystyle f(-2)=(-2)^2-2(-2)-3=5
\displaystyle f(-1)=(-1)^2-2(-1)-3=0
\displaystyle f(0)=0^2-2(0)-3=-3
\displaystyle f(1)=1^2-2(1)-3=-4
\displaystyle f(2)=2^2-2(2)-3=-3
\displaystyle \text{(i) }f(A)=\{-4,-3,0,5\}
\displaystyle \text{(ii) From the above values:}
\displaystyle \text{Pre-image of }6=\phi,\text{ since no element of }A\text{ has image }6.
\displaystyle \text{Pre-image of }-3=\{0,2\},\text{ since }f(0)=f(2)=-3.
\displaystyle \text{Pre-image of }5=\{-2\},\text{ since }f(-2)=5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If a function }f:R\rightarrow R\text{ is defined by}
\displaystyle f(x)=\begin{cases}3x-2,&x<0\\1,&x=0\\4x+1,&x>0\end{cases}
\displaystyle \text{Find }f(1),\ f(-1),\ f(0)\text{ and }f(2).
\displaystyle \text{Answer:}
\displaystyle f(1)=4(1)+1=5
\displaystyle f(-1)=3(-1)-2=-5
\displaystyle f(0)=1
\displaystyle f(2)=4(2)+1=9
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A function }f:R\rightarrow R\text{ is defined by }f(x)=x^2.
\displaystyle \text{Determine: (i) the range of }f\text{ (ii) }\{x:f(x)=4\}\text{ (iii) }\{y:f(y)=-1\}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }x^2\ge0\text{ for every }x\in R,
\displaystyle \therefore \text{Range}(f)=\{y\in R:y\ge0\}=[0,\infty).
\displaystyle \text{(ii) }f(x)=4\Rightarrow x^2=4
\displaystyle \Rightarrow x=\pm2
\displaystyle \therefore \{x:f(x)=4\}=\{-2,2\}.
\displaystyle \text{(iii) }f(y)=-1\Rightarrow y^2=-1
\displaystyle \text{Since }y^2\ge0\text{ for every }y\in R,\text{ there is no such }y.
\displaystyle \therefore \{y:f(y)=-1\}=\phi.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }f:R^+\rightarrow R,\text{ where }R^+\text{ is the set of all positive}
\displaystyle \text{real numbers, be defined by }f(x)=\log_e x.\text{ Determine:}
\displaystyle \text{(i) the image set of the domain of }f\text{ (ii) }\{x:f(x)=-2\}
\displaystyle \text{(iii) whether }f(xy)=f(x)+f(y)\text{ holds.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:R^+\rightarrow R,\qquad f(x)=\log_e x
\displaystyle \text{(i) As }x\text{ varies over }R^+,\ \log_e x\text{ assumes every real value.}
\displaystyle \therefore \text{Image set of the domain }=R.
\displaystyle \text{(ii) }f(x)=-2\Rightarrow \log_e x=-2
\displaystyle \Rightarrow x=e^{-2}
\displaystyle \therefore \{x:f(x)=-2\}=\{e^{-2}\}.
\displaystyle \text{(iii) }f(xy)=\log_e(xy)
\displaystyle =\log_e x+\log_e y
\displaystyle =f(x)+f(y).
\displaystyle \therefore f(xy)=f(x)+f(y).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the following relations as sets of ordered pairs and find}
\displaystyle \text{which of them are functions:}
\displaystyle \text{(i) }\{(x,y):y=3x,\ x\in\{1,2,3\},\ y\in\{3,6,9,12\}\}
\displaystyle \text{(ii) }\{(x,y):y>x+1,\ x=1,2,\ y=2,4,6\}
\displaystyle \text{(iii) }\{(x,y):x+y=3,\ x,y\in\{0,1,2,3\}\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }y=3x,\ x\in\{1,2,3\}
\displaystyle \text{For }x=1,2,3,\text{ the corresponding values of }y\text{ are }3,6,9.
\displaystyle \therefore R=\{(1,3),(2,6),(3,9)\}
\displaystyle \text{Each element of the domain has exactly one image.}
\displaystyle \therefore R\text{ is a function.}
\displaystyle \text{(ii) Given }y>x+1,\ x=1,2,\ y=2,4,6.
\displaystyle \text{For }x=1,\ y=4,6;\qquad \text{for }x=2,\ y=4,6.
\displaystyle \therefore R=\{(1,4),(1,6),(2,4),(2,6)\}
\displaystyle \text{Since }1\text{ and }2\text{ each have more than one image, }R\text{ is not a function.}
\displaystyle \text{(iii) Given }x+y=3.
\displaystyle \Rightarrow y=3-x
\displaystyle \text{For }x=0,1,2,3,\text{ the corresponding values of }y\text{ are }3,2,1,0.
\displaystyle \therefore R=\{(0,3),(1,2),(2,1),(3,0)\}
\displaystyle \text{Each element of the domain has exactly one image.}
\displaystyle \therefore R\text{ is a function.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Let }f:R\rightarrow R\text{ and }g:C\rightarrow C\text{ be two functions}
\displaystyle \text{defined by }f(x)=x^2\text{ and }g(x)=x^2.\text{ Are they equal functions?}
\displaystyle \text{Answer:}
\displaystyle f:R\rightarrow R,\qquad g:C\rightarrow C
\displaystyle \text{Although both functions are defined by the same rule }x\mapsto x^2,
\displaystyle \text{their domains and co-domains are different.}
\displaystyle \text{Domain}(f)=R,\qquad \text{Domain}(g)=C
\displaystyle \text{Co\mbox{-}domain}(f)=R,\qquad \text{Co\mbox{-}domain}(g)=C
\displaystyle \therefore f\neq g.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }f,\ g,\ h:R\rightarrow R\text{ are defined as follows:}
\displaystyle \text{(i) }f(x)=x^2\qquad\text{(ii) }g(x)=\sin x\qquad\text{(iii) }h(x)=x^2+1,
\displaystyle \text{find the range of each function.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }x^2\ge0\text{ for all }x\in R,
\displaystyle \text{Range}(f)=\{y\in R:y\ge0\}=[0,\infty).
\displaystyle \text{(ii) Since }-1\le\sin x\le1\text{ for all }x\in R,
\displaystyle \text{Range}(g)=\{y\in R:-1\le y\le1\}=[-1,1].
\displaystyle \text{(iii) Since }x^2\ge0,\ x^2+1\ge1\text{ for all }x\in R,
\displaystyle \text{Range}(h)=\{y\in R:y\ge1\}=[1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }X=\{1,2,3,4\}\text{ and }Y=\{1,5,9,11,15,16\}.
\displaystyle \text{Determine which of the following sets are functions from }X\text{ to }Y:
\displaystyle \text{(i) }f_1=\{(1,1),(2,11),(3,1),(4,15)\}
\displaystyle \text{(ii) }f_2=\{(1,1),(2,7),(3,5)\}
\displaystyle \text{(iii) }f_3=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}
\displaystyle \text{Answer:}
\displaystyle X=\{1,2,3,4\},\qquad Y=\{1,5,9,11,15,16\}
\displaystyle \text{(i) }f_1=\{(1,1),(2,11),(3,1),(4,15)\}
\displaystyle \text{Each element of }X\text{ has exactly one image in }Y.
\displaystyle \therefore f_1\text{ is a function from }X\text{ to }Y.
\displaystyle \text{(ii) }f_2=\{(1,1),(2,7),(3,5)\}
\displaystyle \text{Here }4\in X\text{ has no image in }Y.
\displaystyle \text{Also, }7\notin Y.
\displaystyle \therefore f_2\text{ is not a function from }X\text{ to }Y.
\displaystyle \text{(iii) }f_3=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}
\displaystyle \text{Here }2\in X\text{ has two images, }9\text{ and }11.
\displaystyle \therefore f_3\text{ is not a function from }X\text{ to }Y.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Let }A=\{12,13,14,15,16,17\}\text{ and }f:A\rightarrow Z\text{ be}
\displaystyle \text{defined by }f(x)=\text{the highest prime factor of }x.\text{ Find the range of }f.
\displaystyle \text{Answer:}
\displaystyle A=\{12,13,14,15,16,17\}
\displaystyle f(12)=3,\qquad f(13)=13,\qquad f(14)=7
\displaystyle f(15)=5,\qquad f(16)=2,\qquad f(17)=17
\displaystyle \therefore f=\{(12,3),(13,13),(14,7),(15,5),(16,2),(17,17)\}
\displaystyle \therefore \text{Range}(f)=\{2,3,5,7,13,17\}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }f:R\rightarrow R\text{ is defined by }f(x)=x^2+1,\text{ find}
\displaystyle f^{-1}\{17\}\text{ and }f^{-1}\{-3\}.
\displaystyle \text{Answer:}
\displaystyle \text{For any }y\in R,\qquad f^{-1}\{y\}=\{x\in R:f(x)=y\}.
\displaystyle \text{Now }f(x)=17
\displaystyle \Rightarrow x^2+1=17
\displaystyle \Rightarrow x^2=16
\displaystyle \Rightarrow x=\pm4
\displaystyle \therefore f^{-1}\{17\}=\{-4,4\}.
\displaystyle \text{Also, }f(x)=-3
\displaystyle \Rightarrow x^2+1=-3
\displaystyle \Rightarrow x^2=-4
\displaystyle \text{Since }x^2\ge0\text{ for all }x\in R,\text{ there is no real solution.}
\displaystyle \therefore f^{-1}\{-3\}=\phi.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Let }A=\{p,q,r,s\}\text{ and }B=\{1,2,3\}. \text{ Which of the following}
\displaystyle \text{relations from }A\text{ to }B\text{ is not a function?}
\displaystyle \text{(i) }R_1=\{(p,1),(q,2),(r,1),(s,2)\}
\displaystyle \text{(ii) }R_2=\{(p,1),(q,1),(r,1),(s,1)\}
\displaystyle \text{(iii) }R_3=\{(p,1),(q,2),(p,2),(s,3)\}
\displaystyle \text{(iv) }R_4=\{(p,2),(q,3),(r,2),(s,2)\}
\displaystyle \text{Answer:}
\displaystyle A=\{p,q,r,s\},\qquad B=\{1,2,3\}
\displaystyle \text{(i) }R_1=\{(p,1),(q,2),(r,1),(s,2)\}
\displaystyle \text{Each element of }A\text{ has exactly one image in }B.
\displaystyle \therefore R_1\text{ is a function.}
\displaystyle \text{(ii) }R_2=\{(p,1),(q,1),(r,1),(s,1)\}
\displaystyle \text{Each element of }A\text{ has exactly one image in }B.
\displaystyle \therefore R_2\text{ is a function.}
\displaystyle \text{(iii) }R_3=\{(p,1),(q,2),(p,2),(s,3)\}
\displaystyle \text{Here }p\in A\text{ has two images, }1\text{ and }2,\text{ and }r\in A\text{ has no image.}
\displaystyle \therefore R_3\text{ is not a function.}
\displaystyle \text{(iv) }R_4=\{(p,2),(q,3),(r,2),(s,2)\}
\displaystyle \text{Each element of }A\text{ has exactly one image in }B.
\displaystyle \therefore R_4\text{ is a function.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Let }A=\{9,10,11,12,13\}\text{ and let }f:A\rightarrow N\text{ be defined}
\displaystyle \text{by }f(n)=\text{the highest prime factor of }n.\text{ Find the range of }f.
\displaystyle \text{Answer:}
\displaystyle A=\{9,10,11,12,13\}
\displaystyle f(9)=3,\qquad f(10)=5,\qquad f(11)=11
\displaystyle f(12)=3,\qquad f(13)=13
\displaystyle \therefore f=\{(9,3),(10,5),(11,11),(12,3),(13,13)\}
\displaystyle \therefore \text{Range}(f)=\{3,5,11,13\}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The function }f\text{ is defined by}
\displaystyle f(x)=\begin{cases}x^2,&0\le x\le3\\3x,&3\le x\le10\end{cases}
\displaystyle \text{The relation }g\text{ is defined by}
\displaystyle g(x)=\begin{cases}x^2,&0\le x\le2\\3x,&2\le x\le10\end{cases}
\displaystyle \text{Show that }f\text{ is a function and }g\text{ is not a function.}
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}x^2,&0\le x\le3\\3x,&3\le x\le10\end{cases}
\displaystyle \text{At }x=3,\qquad 3^2=9\qquad\text{and}\qquad 3(3)=9.
\displaystyle \text{Thus }f(3)\text{ has the same value from both parts.}
\displaystyle \text{Hence every element of the domain has exactly one image.}
\displaystyle \therefore f\text{ is a function.}
\displaystyle g(x)=\begin{cases}x^2,&0\le x\le2\\3x,&2\le x\le10\end{cases}
\displaystyle \text{At }x=2,\qquad 2^2=4\qquad\text{and}\qquad 3(2)=6.
\displaystyle \text{Thus }g(2)\text{ has two different values, }4\text{ and }6.
\displaystyle \text{Hence }g\text{ does not assign a unique image to }2.
\displaystyle \therefore g\text{ is not a function.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }f(x)=x^2,\text{ find }\frac{f(1.1)-f(1)}{1.1-1}.
\displaystyle \text{Answer:}
\displaystyle f(x)=x^2
\displaystyle f(1.1)=(1.1)^2=1.21
\displaystyle f(1)=1^2=1
\displaystyle \therefore \frac{f(1.1)-f(1)}{1.1-1}=\frac{1.21-1}{1.1-1}
\displaystyle =\frac{0.21}{0.1}=2.1
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Express the function }f:X\rightarrow R\text{ given by }f(x)=x^3+1
\displaystyle \text{as a set of ordered pairs, where }X=\{-1,0,3,9,7\}.
\displaystyle \text{Answer:}
\displaystyle X=\{-1,0,3,9,7\},\qquad f(x)=x^3+1
\displaystyle f(-1)=(-1)^3+1=0
\displaystyle f(0)=0^3+1=1
\displaystyle f(3)=3^3+1=28
\displaystyle f(9)=9^3+1=730
\displaystyle f(7)=7^3+1=344
\displaystyle \therefore f=\{(-1,0),(0,1),(3,28),(9,730),(7,344)\}
\displaystyle \\

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