\(\text {(i)} (f+g)(x) =f(x)+g(x) \)
\(=2 x+1+x^{2}+1 =x^{2}+2 x+2 \)
\(\text {(ii)} (f-g)(x)=f(x)-g(x)\)
\(=(2 x+1)-\left(x^{2}+1\right) \)
\(=2 x+1-x^{2}-1=2 x-x^{2} \)
\(=-x^{2}+2 x \)
\(\text {(iii)} (f g)(x) =[f(x)][g(x)] \)
\(=(2 x+1)\left(x^{2}+1\right) \)
\(=2 x^{3}+x^{2}+2 x+1 \)
\(\text { (iv) } \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, \text { provided } g(x) \neq 0\)
\(=\frac{2 x+1}{x^{2}+1}\)

\(f \circ g =f[g(x)] \)
\(=f[-x+8] \)
\(=3(-x+8)+7 \)
\(=-3 x+24+7=-3 x+31 \)
\((f \circ g) \circ h =(f \circ g)[\mathrm{h}(\mathrm{x})] \)
\(=(f \circ g)\left[\mathrm{x}^{2}+3 \mathrm{x}-1\right] \)
\(=-3\left(\mathrm{x}^{2}+3 \mathrm{x}-1\right)+31 \)
\(=-3 \mathrm{x}^{2}-9 \mathrm{x}+3+31 \)
\(=-3 \mathrm{x}^{2}-9 \mathrm{x}+34 \)
\(g \circ h =g[h(x)] \)
\(=g\left[x^{2}+3 x-1\right] \)
\(=-\left(x^{2}+3 x-1\right)+8 \)
\(=-x^{2}-3 x+1+8=-x^{2}-3 x+9 \)
\(f \circ(g \circ h) =\mathrm{f}[\mathrm{g}(\mathrm{h}(\mathrm{x}))]=\mathrm{f}\left[\mathrm{g}\left(\mathrm{x}^{2}+3 \mathrm{x}-1\right)\right] \)
\(=\mathrm{f}\left[-\mathrm{x}^{2}-3 \mathrm{x}+9\right] \)
\(=3\left(-\mathrm{x}^{2}-3 \mathrm{x}+9\right)+7 \)
\(=-3 \mathrm{x}^{2}-9 \mathrm{x}+27+7 \)
\(=-3 \mathrm{x}^{2}-9 \mathrm{x}+34 \)
From (1) and (2)
\((f \circ g) \circ h=f \circ(g \circ h)\)
Composition of functions is Associative.

(i) Given, that the Relation R consists of elements (a, b) where 'd is a factor of 'b'.
Therefore, Relation 'R' in the roster form is
R = {(2,8), (2, 10), (3, 9),(4,8), (5, 10)}
Domain of R = {2,3,4,5}
Range of R = {8, 10,9}
(ii) Arrow diagram representing 'R'

(i) R₁ is the set of all ordered pairs whose first
component is two less than the second component.
R₁ = {(4, 6), (9, 11)}
Domain of R, - Set of all first components {4,9}
Range of R₁ = Set of all second components {6,11}
(iii) R₂ is the set of all ordered pairs whose first component is less than the second component.
R₂ = {(4,6), (9, 11), (2,3)}
Domain of R₂ = {4,9,2}
Range of R₂ = {6, 11,3)
(iii) R₃ is the set of all ordered pairs whose first component is greater than the second component.
R₃ = {(8,4), (6, 3), (3, 0)}
Domain of R₃ = {8, 6, 3}
Range of R₃ = {4,3,0}
(iv) R₄ is the set of all ordered pairs whose first component is equal to the second component.
R₄ = {(3, 3)}
Domain of R₄ = {3} and Range of R₄ = {3}.

\(Let \ \mathrm{A} \times \mathrm{B}=\mathrm{B} \times \mathrm{A} \ and \ x \in A \)
\(x \in A \Rightarrow(x, b) \in A \times B \ \forall \ b \in B \)
\(\Rightarrow(x, b) \in(B \times A) \)
\( \because A \times B=B \times A \)
\( \Rightarrow x \in B \)
\( \Rightarrow A \subseteq B \)
By the definition of Cartesian Product and let y be an arbitrary element of B
\(y \in B \Rightarrow(a, y) \in A \times B \ \forall \ a \in A \)
\( \Rightarrow(a, y) \in(B \times A) \)
\( \because \mathrm{A} \times \mathrm{B}=\mathrm{B} \times \mathrm{A} \)
By the definition of Cartesian Product
\(\Rightarrow y \in A \)
\(B \subseteq A \)
From these two results, A = B
Conversely, let A = B, then
\(A \times B =A \times A \text { and } B \times A=A \times A \)
\(\therefore A \times B =B \times A \)
\(\text { Hence } A \times B =B \times A \text { if } A=B \)

\(A =\left\{x / x^{2}-4 x+3=0\right\} \)
\(=\{x /(x-3)(x-1)=0\} \)
\(=\{x / x=1,3\} \)
\(B =\left\{x / x^{2}-x-6=0\right\} \)
\(=\{x /(x+2)(x-3)=0\} =\{x / x=-2,3\} \)
\(C =\left\{x / x^{3}-4 x=0\right\} \)
\(=\left\{x / x\left(x^{2}-4\right)=0\right\}=\{x / x=0,-2,2\} \)
\(A =\{1,3\}, B=\{-2,3\}, C=\{-2,0,2\} \)
\((i) \ \mathrm{A} \times \mathrm{B} =\{1,3\} \times\{-2,3\} =\{(1,-2),(1,3),(3,-2),(3,3)\} \)
\((ii) \mathrm{A} \times \mathrm{C}=\{1,3\} \times\{-2,0,2\} =\{(1,-2),(1,0),(1,2),(3,-2) (3,0),(3,2)\} \)
\(\text { (iii) } \mathrm{A}-\mathrm{B}=\{1,3\}-\{-2,3\}=\{1\}\)
\((\mathrm{A}-\mathrm{B}) \times \mathrm{C} =\{1\} \times\{-2,0,2\} \)
\(=\{(1,-2),(1,0),(1,2)\} \)