1. The problem asks to identify properties of relation $$R = \{(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)\}$$ on set $$A=\{1,2,3,4\}$$. \nCheck reflexivity: Elements 1, 2, 3, 4 must relate to themselves. Here, $(3,3)$, $(2,2)$, $(1,1)$, and $(4,4)$ are present, so reflexive. \nCheck symmetry: Since $(1,2)\in R$ but $(2,1)\notin R$, relation is not symmetric. \nCheck transitivity: $(1,2)$ and $(2,2)$ in R implies $(1,2)$ in R (true), but $(1,3)$ and $(3,2)$ in R implies $(1,2)$ in R (true), and others confirmed. Transitive. \nHence, answer is (b) reflexive and transitive but not symmetric. \n\n2. Relation $$R=\{(a,b),(b,a),(a,a)\}$$ on $$A=\{a,b,c,d\}$$. \nReflexive requires $(b,b)$, $(c,c)$, $(d,d)$ which are missing, so not reflexive. \nSymmetric: Since $(a,b)$ and $(b,a)$ present, symmetric. \nTransitive: $(a,b)$ and $(b,a)$ requires $(a,a)$ (present), but $(b,b)$ missing so not transitive. \nAnswer: (a) symmetric only. \n\n3. Relation defined by $xRy$ if $x - y + \sqrt{2}$ is irrational. \nCheck reflexive: $x - x + \sqrt{2} = \sqrt{2}$ irrational, so reflexive. \nSymmetric: If $xRy$, then $x - y + \sqrt{2}$ irrational, so $y - x + \sqrt{2} = -(x-y) + \sqrt{2}$ also irrational; symmetric. \nTransitive: two irrational sums may not imply a third irrational; not transitive. \nAnswer: (b) symmetric. \n\n4. Relation R is "a is brother of b" on children set. \nBrother relation is not symmetric (a brother of b doesn’t imply b brother of a if gender differs), not transitive (a brother to b and b brother to c doesn’t imply a brother to c). \nAnswer: (c) neither symmetric nor transitive. \n\n5. Number of equivalence relations on 3-element set equals number of partitions (Bell number) for 3 is 5. \nAnswer: (d) 5. \n\n6. Relation R on lines where $lRm$ iff $l$ perpendicular to $m$. \nReflexive: No line is perpendicular to itself, so not reflexive. \nSymmetric: If $l$ perpendicular to $m$, then $m$ perpendicular to $l$, symmetric. \nTransitive: If $l$ perpendicular to $m$ and $m$ perpendicular to $n$, $l$ and $n$ may not be perpendicular, so not transitive. \nAnswer: (b) symmetric. \n\n7. Number of relations on $A=\{1,2,3\}$ containing $(1,2)$ and $(1,3)$ that are reflexive and symmetric but not transitive is 2. \nAnswer: (b) 2. \n\n8. Number of equivalence relations on $A=\{1,2,3\}$ containing $(1,2)$ is 2. \nAnswer: (b) 2. \n\n9. Number of relations from $A$ (m elements) to $B$ (n elements) is $2^{mn}$. \nAnswer: (a) $2^{mn}$. \n\n10. Number of injective mappings from $|A|=3$ to $|B|=4$ is permutations of 4 elements taken 3 at a time: $P(4,3)=4\times3\times2=24$. \nAnswer: (c) 24. \n\n11. $f(x)=2^x + 2|x|$ is many-one because absolute value causes $f(-x)\neq f(x)$ strictly, and not onto since output is always positive. \nAnswer: (d) many-one and into. \n\n12. Number of one-one onto mappings from set with 5 to 6 elements is 0 (onto impossible if codomain larger). \nAnswer: (c) 0. \n\n13. Bijection from integers to integers: $f(x)=x^3$ and $f(x)=x+2$ are bijections, $f(x)=2x+1$ is bijection, $f(x)=x^2+1$ not (not one-one). Among options, only $f(x)=x^3$ fits well. \nAnswer: (a) $f(x)=x^3$. \n\n14. $f(x)=x^2 - 4x + 5 = (x-2)^2 + 1$ with domain $[2, \infty)$ outputs values $\ge 1$. \nAnswer: (b) $[1, \infty)$. \n\n15. $f(x)=x^2+1$. Solve $x^2+1=17$ gives $x=\pm4$. Solve $x^2+1=-3$ no real solution. \nPreimages: 17 is $\{4,-4\}$, -3 is $\emptyset$. \nAnswer: (c). \n\n16. $f(x)=2x + \sin x$ is strictly increasing (sum of strictly increasing $2x$ and bounded $\sin x$ change). Hence, one-one and onto. \nAnswer: (d). \n\n17. Relation $R=\{(a,b): a=b-2, b>6 \}$. \nCheck options: (2,4): $2=4-2$, $4>6$ false. \n(3,8): $3=8-2=6$ false. \n(6,8): $6=8-2=6$ and $8>6$ true. \n(8,7): $8=7-2=5$ false. \nAnswer: (c). \n\n18. $f(x)=x^4$ on $R$, not one-one (even powers), not onto (negative numbers missed). \nAnswer: (d). \n\n19. $f(x)=3x$ on $R$ is one-one and onto linear function. \nAnswer: (a). \n\n20. $f(x)=2x^3+2x^2+300x+5\sin x$ is polynomial dominant cubic function plus small term, strictly monotone, hence one-one and onto. \nAnswer: (a). \n\n21. $f(x) = x^2+1$. Preimage of 5: $x^2+1=5 \Rightarrow x^2=4 \Rightarrow x=\pm2$. Preimage of -5 no solution. \nAnswer: (c).