\displaystyle \text{RELATION AND FUNCTION}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A general election of the Lok Sabha is a gigantic exercise. About }911\text{ million people were}
\displaystyle \text{eligible to vote and the voter turnout was about }67\%\text{, the highest ever.} \displaystyle \text{Let }I\text{ be the set of all citizens of India who were eligible to exercise their voting right in the}
\displaystyle \text{General Election held in }2019.\text{ A relation }R\text{ is defined on }I\text{ as follows:}
\displaystyle R=\{(V_1,V_2):V_1,V_2\in I\text{ and both exercised their voting right in the General Election--2019}\}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Two neighbours }X\text{ and }Y\in I.\ X\text{ exercised his voting right while }Y
\displaystyle \text{ did not cast her vote}  \ \text{in the General Election--2019. Which of the following is true?}
\displaystyle \text{(a) }(X,Y)\in R\qquad\text{(b) }(Y,X)\in R\qquad\text{(c) }(X,X)\notin R\qquad\text{(d) }(X,Y)\notin R
\displaystyle \text{Answer:}
\displaystyle \text{Since }X\text{ exercised his voting right but }Y\text{ did not, they cannot form an ordered pair in }R.
\displaystyle \therefore (X,Y)\notin R.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Mr. }X\text{ and his wife }W\text{ both exercised their voting right in the }
\displaystyle \text{General Election--2019. Which of the following is true?}
\displaystyle \text{(a) Both }(X,W)\text{ and }(W,X)\in R
\displaystyle \text{(b) }(X,W)\in R\text{ but }(W,X)\notin R
\displaystyle \text{(c) Both }(X,W)\text{ and }(W,X)\notin R
\displaystyle \text{(d) }(W,X)\in R\text{ but }(X,W)\notin R
\displaystyle \text{Answer:}
\displaystyle \text{Since both }X\text{ and }W\text{ exercised their voting right, }(X,W)\in R.
\displaystyle \text{Also, }(W,X)\in R.
\displaystyle \therefore \text{Both }(X,W)\text{ and }(W,X)\in R.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Three friends }F_1,F_2\text{ and }F_3\text{ exercised their voting right in the }
\displaystyle \text{General Election--2019. Which of the following is true?}
\displaystyle \text{(a) }(F_1,F_2)\in R,\ (F_2,F_3)\in R\text{ and }(F_1,F_3)\in R
\displaystyle \text{(b) }(F_1,F_2)\in R,\ (F_2,F_3)\in R\text{ and }(F_1,F_3)\notin R
\displaystyle \text{(c) }(F_1,F_2)\in R,\ (F_2,F_2)\in R\text{ but }(F_3,F_3)\notin R
\displaystyle \text{(d) }(F_1,F_2)\notin R,\ (F_2,F_3)\notin R\text{ and }(F_1,F_3)\notin R
\displaystyle \text{Answer:}
\displaystyle \text{Since }F_1,F_2\text{ and }F_3\text{ all exercised their voting right, every ordered pair formed by}
\displaystyle \text{any two of them belongs to }R.
\displaystyle \therefore (F_1,F_2)\in R,\quad(F_2,F_3)\in R\quad\text{and}\quad(F_1,F_3)\in R.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The above-defined relation }R\text{ is}
\displaystyle \text{(a) symmetric and transitive but not reflexive}
\displaystyle \text{(b) universal relation}\qquad\text{(c) equivalence relation}
\displaystyle \text{(d) reflexive but not symmetric and transitive}
\displaystyle \text{Answer:}
\displaystyle \text{The relation is not reflexive because every eligible voter did not necessarily exercise the right to vote.}
\displaystyle \text{Thus, for a person }x\in I\text{ who did not vote, }(x,x)\notin R.
\displaystyle \text{If }(x,y)\in R,\text{ then both }x\text{ and }y\text{ voted. Hence }(y,x)\in R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{If }(x,y)\in R\text{ and }(y,z)\in R,\text{ then }x,y\text{ and }z\text{ all voted.}
\displaystyle \therefore (x,z)\in R,\text{ and hence }R\text{ is transitive.}
\displaystyle \therefore R\text{ is symmetric and transitive but not reflexive.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Mr. Shyam exercised his voting right in the General Election--2019. }
\displaystyle \text{Then Mr. Shyam is related to which of the following?}
\displaystyle \text{(a) All eligible voters who cast their votes}\qquad\text{(b) Family members of Mr. Shyam}
\displaystyle \text{(c) All citizens of India}\qquad\text{(d) Eligible voters of India}
\displaystyle \text{Answer:}
\displaystyle \text{By the definition of }R,\text{ two eligible voters are related if both exercised their voting right.}
\displaystyle \text{Since Mr. Shyam voted, he is related to every eligible voter who also cast a vote.}
\displaystyle \therefore \text{Mr. Shyam is related to all eligible voters who cast their votes.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the }
\displaystyle \text{die, Sherlin's sister Raji observed and noted that the possible outcomes of}
\displaystyle \text{every throw belong to the set} \ \{1,2,3,4,5,6\}.\text{ Let }A\text{ be the set of players and }
\displaystyle B\text{ be the set of all possible outcomes.} \ A=\{S,D\},\qquad B=\{1,2,3,4,5,6\}. \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Let }R:B\to B\text{ be defined by }R=\{(x,y):y\text{ is divisible by }x\}.\ R\text{ is}
\displaystyle \text{(a) reflexive and transitive but not symmetric}
\displaystyle \text{(b) reflexive and symmetric but not transitive}
\displaystyle \text{(c) not reflexive but symmetric and transitive}\qquad\text{(d) equivalence}
\displaystyle \text{Answer:}
\displaystyle \text{For every }x\in B,\ x\text{ divides }x.\text{ Hence, }(x,x)\in R.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{If }(x,y)\in R\text{ and }(y,z)\in R,\text{ then }x\mid y\text{ and }y\mid z.
\displaystyle \therefore x\mid z,\text{ so }(x,z)\in R.\text{ Hence, }R\text{ is transitive.}
\displaystyle \text{Also, }(1,2)\in R\text{ but }(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \therefore R\text{ is reflexive and transitive but not symmetric.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Raji wants to know the number of functions from }A\text{ to }B.\text{ How }
\displaystyle \text{many functions are possible?}
\displaystyle \text{(a) }6^2\qquad\text{(b) }2^6\qquad\text{(c) }6!\qquad\text{(d) }2^{12}
\displaystyle \text{Answer:}
\displaystyle n(A)=2\quad\text{and}\quad n(B)=6.
\displaystyle \text{Number of functions from }A\text{ to }B=[n(B)]^{n(A)}.
\displaystyle =6^2=36.
\displaystyle \therefore \text{There are }36\text{ functions from }A\text{ to }B.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }R\text{ be a relation on }B\text{ defined by}
\displaystyle R=\{(1,2),(2,2),(1,3),(3,4),(3,1),(4,3),(5,5)\}.\text{ Then }R\text{ is}
\displaystyle \text{(a) symmetric}\qquad\text{(b) reflexive}\qquad\text{(c) transitive}\qquad\text{(d) none of these three}
\displaystyle \text{Answer:}
\displaystyle (1,2)\in R\text{ but }(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle (1,1)\notin R.
\displaystyle \therefore R\text{ is not reflexive.}
\displaystyle (1,3)\in R\text{ and }(3,4)\in R,\text{ but }(1,4)\notin R.
\displaystyle \therefore R\text{ is not transitive.}
\displaystyle \therefore R\text{ is neither symmetric, reflexive nor transitive.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Raji wants to know the number of relations possible from }A\text{ to }B.
\displaystyle \text{ How many relations} \ \text{are possible?}
\displaystyle \text{(a) }6^2\qquad\text{(b) }2^6\qquad\text{(c) }6!\qquad\text{(d) }2^{12}
\displaystyle \text{Answer:}
\displaystyle n(A)=2\quad\text{and}\quad n(B)=6.
\displaystyle n(A\times B)=n(A)\times n(B)=2\times6=12.
\displaystyle \text{A relation from }A\text{ to }B\text{ is any subset of }A\times B.
\displaystyle \therefore \text{Number of relations from }A\text{ to }B=2^{12}=4096.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }R\text{ be a relation on }B\text{ defined by}
\displaystyle R=\{(1,1),(1,2),(2,2),(3,3),(4,4),(5,5),(6,6)\}.\text{ Then }R\text{ is}
\displaystyle \text{(a) symmetric}\qquad\text{(b) reflexive and transitive}
\displaystyle \text{(c) transitive and symmetric}\qquad\text{(d) equivalence}
\displaystyle \text{Answer:}
\displaystyle \text{Since }(x,x)\in R\text{ for every }x\in B,\ R\text{ is reflexive.}
\displaystyle \text{Also, }(1,2)\in R\text{ but }(2,1)\notin R.
\displaystyle \therefore R\text{ is not symmetric.}
\displaystyle \text{The only non-diagonal pair is }(1,2).\text{ Its possible compositions with pairs in }R
\displaystyle \text{produce }(1,2),\text{ which already belongs to }R.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \therefore R\text{ is reflexive and transitive but not symmetric.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{An organization conducted a bike race under two different categories, boys and girls.}
\displaystyle \text{There were a total of }250\text{ participants. Finally, three from Category 1 and two from}
\displaystyle \text{Category 2 were selected for the final race. Ravi forms two sets } B\text{ and }G
\displaystyle \text{ with these participants for his}  \ \text{college project.} \displaystyle \text{Let }B=\{b_1,b_2,b_3\}\text{ and }G=\{g_1,g_2\},\text{ where }B\text{ represents the set of boys selected and }G
\displaystyle \text{represents the set of girls selected for the final race.}
\displaystyle \text{Ravi decides to explore these sets for various types of relations and functions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Ravi wishes to form all the relations possible from }B\text{ to }G.
\displaystyle \text{ How many such relations} \ \text{are possible?}
\displaystyle \text{(a) }2^6\qquad\text{(b) }2^5\qquad\text{(c) }0\qquad\text{(d) }2^3
\displaystyle \text{Answer:}
\displaystyle n(B)=3\quad\text{and}\quad n(G)=2.
\displaystyle \therefore n(B\times G)=n(B)\times n(G)=3\times2=6.
\displaystyle \text{A relation from }B\text{ to }G\text{ is any subset of }B\times G.
\displaystyle \therefore \text{Number of relations from }B\text{ to }G=2^6=64.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }R\text{ be a relation on }B\text{ defined by }R=\{(x,y):x\text{ and }y
\displaystyle \text{ are students of the same sex}\}. \ \text{Then this relation is}
\displaystyle \text{(a) equivalence}\qquad\text{(b) reflexive only}
\displaystyle \text{(c) reflexive and symmetric but not transitive}
\displaystyle \text{(d) reflexive and transitive but not symmetric}
\displaystyle \text{Answer:}
\displaystyle \text{Every student is of the same sex as himself. Hence, }R\text{ is reflexive.}
\displaystyle \text{If }x\text{ is of the same sex as }y,\text{ then }y\text{ is of the same sex as }x.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{If }x\text{ and }y\text{ are of the same sex and }y\text{ and }z\text{ are of the same sex, then }x\text{ and }z
\displaystyle \text{are also of the same sex. Hence, }R\text{ is transitive.}
\displaystyle \therefore R\text{ is reflexive, symmetric and transitive.}
\displaystyle \therefore R\text{ is an equivalence relation.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Ravi wants to know, among these relations, how many functions can be} \\ \text{formed from }B\text{ to }G.
\displaystyle \text{(a) }2^2\qquad\text{(b) }2^{12}\qquad\text{(c) }3^2\qquad\text{(d) }2^3
\displaystyle \text{Answer:}
\displaystyle n(B)=3\quad\text{and}\quad n(G)=2.
\displaystyle \text{Number of functions from }B\text{ to }G=[n(G)]^{n(B)}.
\displaystyle =2^3=8.
\displaystyle \therefore \text{There are }8\text{ functions from }B\text{ to }G.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }R:B\to G\text{ be defined by }R=\{(b_1,g_1),(b_2,g_2),(b_3,g_1)\}.\text{ Then }R\text{ is}
\displaystyle \text{(a) injective}\qquad\text{(b) surjective}
\displaystyle \text{(c) neither surjective nor injective}\qquad\text{(d) surjective and injective}
\displaystyle \text{Answer:}
\displaystyle R(b_1)=g_1,\qquad R(b_2)=g_2,\qquad R(b_3)=g_1.
\displaystyle \text{Since }R(b_1)=R(b_3)=g_1\text{ for }b_1\ne b_3,\ R\text{ is not injective.}
\displaystyle \text{Also, both }g_1\text{ and }g_2\text{ have pre-images in }B.
\displaystyle \therefore \text{Range of }R=G.
\displaystyle \therefore R\text{ is surjective but not injective.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Ravi wants to find the number of injective functions from }B\text{ to }G.
\displaystyle \text{ How many} \ \text{injective functions are possible?}
\displaystyle \text{(a) }0\qquad\text{(b) }2!\qquad\text{(c) }3!\qquad\text{(d) }0!
\displaystyle \text{Answer:}
\displaystyle n(B)=3\quad\text{and}\quad n(G)=2.
\displaystyle \text{For an injective function, distinct elements of the domain must have distinct images.}
\displaystyle \text{Since }n(B)>n(G),\text{ an injective function from }B\text{ to }G\text{ is not possible.}
\displaystyle \therefore \text{Number of injective functions from }B\text{ to }G=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 4}

\displaystyle \text{Students of Grade 9 planned to plant saplings along straight lines, parallel to each other and to}
\displaystyle \text{one side of the playground, ensuring that they had enough play area. Let us assume that they}
\displaystyle \text{planted one of the rows of saplings along the line }y=x-4.\text{ Let }L\text{ be the set of all lines}
\displaystyle \text{on the ground and }R\text{ be a relation on }L. \displaystyle \text{Answer the following using the above information.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Let the relation }R\text{ be defined by }R=\{(L_1,L_2):L_1\parallel L_2,\text{ where }
\displaystyle L_1,L_2\in L\}. \ \text{Then }R\text{ is a/an}
\displaystyle \text{(a) equivalence relation}\qquad\text{(b) only reflexive}
\displaystyle \text{(c) not reflexive}\qquad\text{(d) symmetric but not transitive}
\displaystyle \text{Answer:}
\displaystyle \text{Using the convention that a line is parallel to itself, }L_1\parallel L_1.
\displaystyle \therefore R\text{ is reflexive.}
\displaystyle \text{If }L_1\parallel L_2,\text{ then }L_2\parallel L_1.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{If }L_1\parallel L_2\text{ and }L_2\parallel L_3,\text{ then }L_1\parallel L_3.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \therefore R\text{ is an equivalence relation.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }R=\{(L_1,L_2):L_1\perp L_2,\text{ where }L_1,L_2\in L\}.\text{ Which of the following is true?}
\displaystyle \text{(a) }R\text{ is symmetric but neither reflexive nor transitive}
\displaystyle \text{(b) }R\text{ is reflexive and transitive but not symmetric}
\displaystyle \text{(c) }R\text{ is reflexive but neither symmetric nor transitive}
\displaystyle \text{(d) }R\text{ is an equivalence relation}
\displaystyle \text{Answer:}
\displaystyle \text{A line cannot be perpendicular to itself. Hence, }R\text{ is not reflexive.}
\displaystyle \text{If }L_1\perp L_2,\text{ then }L_2\perp L_1.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \text{If }L_1\perp L_2\text{ and }L_2\perp L_3,\text{ then }L_1\parallel L_3.
\displaystyle \therefore L_1\not\perp L_3,\text{ in general. Hence, }R\text{ is not transitive.}
\displaystyle \therefore R\text{ is symmetric but neither reflexive nor transitive.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The function }f:\mathbb{R}\to\mathbb{R}\text{ defined by }f(x)=x-4\text{ is}
\displaystyle \text{(a) bijective}\qquad\text{(b) surjective but not injective}
\displaystyle \text{(c) injective but not surjective}\qquad\text{(d) neither surjective nor injective}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_1)=f(x_2).
\displaystyle x_1-4=x_2-4\implies x_1=x_2.
\displaystyle \therefore f\text{ is injective.}
\displaystyle \text{For any }y\in\mathbb{R},\text{ let }x=y+4\in\mathbb{R}.
\displaystyle \text{Then }f(x)=f(y+4)=y+4-4=y.
\displaystyle \therefore f\text{ is surjective.}
\displaystyle \therefore f\text{ is both injective and surjective, and hence bijective.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }f:\mathbb{R}\to\mathbb{R}\text{ be defined by }f(x)=x-4.\text{ Then the range of }f\text{ is}
\displaystyle \text{(a) }\mathbb{R}\qquad\text{(b) }\mathbb{Z}\qquad\text{(c) }\mathbb{W}\qquad\text{(d) }\mathbb{Q}
\displaystyle \text{Answer:}
\displaystyle f(x)=x-4.
\displaystyle \text{For every }y\in\mathbb{R},\text{ choosing }x=y+4\in\mathbb{R}\text{ gives }f(x)=y.
\displaystyle \therefore \text{Every real number has a pre-image under }f.
\displaystyle \therefore \text{Range of }f=\mathbb{R}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }R=\{(L_1,L_2):L_1\text{ is parallel to }L_2\text{ and }
\displaystyle L_1:y=x-4\}.\text{ Which of the} \ \text{following can be taken as }L_2\text{?}
\displaystyle \text{(a) }2x-2y+5=0\qquad\text{(b) }2x+y=5
\displaystyle \text{(c) }2x+2y+7=0\qquad\text{(d) }x+y=7
\displaystyle \text{Answer:}
\displaystyle L_1:y=x-4.
\displaystyle \therefore \text{Slope of }L_1=1.
\displaystyle \text{For option (a), }2x-2y+5=0.
\displaystyle 2y=2x+5\implies y=x+\frac{5}{2}.
\displaystyle \therefore \text{Slope of this line}=1.
\displaystyle \text{Since the two lines have the same slope, they are parallel.}
\displaystyle \therefore L_2:2x-2y+5=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 5}

\displaystyle \text{Raji visited an exhibition along with her family. The exhibition had a huge swing, which}
\displaystyle \text{attracted many children. Raji found that the swing traced the path of a parabola given by}\displaystyle y=x^2.  \ \text{Answer the following questions using the above information.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Let }f:\mathbb{R}\to\mathbb{R}\text{ be defined by }f(x)=x^2.\text{ Then }f\text{ is}
\displaystyle \text{(a) neither surjective nor injective}\qquad\text{(b) surjective}
\displaystyle \text{(c) injective}\qquad\text{(d) bijective}
\displaystyle \text{Answer:}
\displaystyle f(1)=1\quad\text{and}\quad f(-1)=1.
\displaystyle \text{Since }1\ne-1\text{ but }f(1)=f(-1),\ f\text{ is not injective.}
\displaystyle \text{Also, }f(x)=x^2\geq0\text{ for every }x\in\mathbb{R}.
\displaystyle \text{Hence, no negative real number has a pre-image under }f.
\displaystyle \therefore f\text{ is not surjective.}
\displaystyle \therefore f\text{ is neither injective nor surjective.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }f:\mathbb{N}\to\mathbb{N}\text{ be defined by }f(x)=x^2.\text{ Then }f\text{ is}
\displaystyle \text{(a) surjective but not injective}\qquad\text{(b) surjective}
\displaystyle \text{(c) injective}\qquad\text{(d) bijective}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_1)=f(x_2).
\displaystyle x_1^2=x_2^2.
\displaystyle \text{Since }x_1,x_2\in\mathbb{N},\ x_1=x_2.
\displaystyle \therefore f\text{ is injective.}
\displaystyle \text{However, natural numbers such as }2,3,5,\ldots\text{ are not squares of natural numbers.}
\displaystyle \therefore f\text{ is not surjective.}
\displaystyle \therefore f\text{ is injective but not surjective.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }f:\{1,2,3,\ldots\}\to\{1,4,9,\ldots\}\text{ be defined by }f(x)=x^2.
\displaystyle \text{Then }f\text{ is}
\displaystyle \text{(a) bijective}\qquad\text{(b) surjective but not injective}
\displaystyle \text{(c) injective but not surjective}\qquad\text{(d) neither surjective nor injective}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x_1)=f(x_2).
\displaystyle x_1^2=x_2^2.
\displaystyle \text{Since }x_1,x_2\in\{1,2,3,\ldots\},\ x_1=x_2.
\displaystyle \therefore f\text{ is injective.}
\displaystyle \text{Also, every element of }\{1,4,9,\ldots\}\text{ is the square of some positive integer.}
\displaystyle \therefore f\text{ is surjective.}
\displaystyle \therefore f\text{ is both injective and surjective, and hence bijective.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }f:\mathbb{N}\to\mathbb{R}\text{ be defined by }f(x)=x^2.\text{ The range of }f\text{ is}
\displaystyle \text{(a) }\{1,4,9,16,\ldots\}\qquad\text{(b) }\{1,4,8,9,10,\ldots\}
\displaystyle \text{(c) }\{1,4,9,15,16,\ldots\}\qquad\text{(d) }\{1,4,8,16,\ldots\}
\displaystyle \text{Answer:}
\displaystyle f(x)=x^2,\qquad x\in\mathbb{N}.
\displaystyle f(1)=1,\quad f(2)=4,\quad f(3)=9,\quad f(4)=16,\ldots
\displaystyle \therefore \text{Range of }f=\{1,4,9,16,\ldots\}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The function }f:\mathbb{Z}\to\mathbb{Z}\text{ defined by }f(x)=x^2\text{ is}
\displaystyle \text{(a) neither injective nor surjective}\qquad\text{(b) injective}
\displaystyle \text{(c) surjective}\qquad\text{(d) bijective}
\displaystyle \text{Answer:}
\displaystyle f(1)=1\quad\text{and}\quad f(-1)=1.
\displaystyle \text{Since }1\ne-1\text{ but }f(1)=f(-1),\ f\text{ is not injective.}
\displaystyle \text{Also, }f(x)=x^2\geq0\text{ for every }x\in\mathbb{Z}.
\displaystyle \text{Hence, negative integers have no pre-image under }f.
\displaystyle \therefore f\text{ is not surjective.}
\displaystyle \therefore f\text{ is neither injective nor surjective.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

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