(i) [-2, 0],
(ii) (0, 8),
(iii) (-8, -2],
(iv) [-5, -9]

A = { x | x is a whole number less than or equal to 7 } is the set-bilder form of A.

\(f\left( \frac { -1 }{ x } \right) =\frac { \frac { 1 }{ x } -1 }{ \frac { 1 }{ x } +1 } =\frac { -\left( \frac { 1 }{ x } +1 \right) }{ -\left( \frac { 1 }{ x } -1 \right) } =\frac { 1+x }{ 1-x } =\frac { 1 }{ \frac { 1-x }{ 1+x } } =\frac { 1 }{ \left( \frac { x-1 }{ x+1 } \right) } =\frac { -1 }{ f(x) } \)
Hence, \(f\left( \frac { -1 }{ x } \right) =\frac { -1 }{ f(x) } .\)

Given that \(f(x)=\frac { x-1 }{ x+1 } \)
\(f\left( \frac { 1 }{ x } \right) =\frac { \frac { 1 }{ x } -1 }{ \frac { 1 }{ x } +1 } =\frac { 1-x }{ 1+x } =\frac { -(x+1) }{ x+1 } =-f(x)\)
Hence, \(f\left( \frac { 1 }{ x } \right) =-f(x)\)

(1,7) (3, 6) (5, 5) (7,4) (9,3) (11, 2) (13, 1) not reflexive, not symmetric, not transitive, not equivalence.

4x² + 7, 16x² - 24x + 10

Given A⊂B

(i) From the diagram A ∩ B =A
(ii) A\B = Ø

Given U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
B = {2, 3, 5, 9, 10}
A' = {2, 4, 6, 8, 10}
B' = {1, 4, 6, 7, 8}
A'UB' = {1, 2, 4, 6, 7, 8, 10}

Let C = {x \(\in \) Z : x is even and < 10}
\(\Rightarrow\) C = {2, 4, 6, 8} 
\(\Rightarrow\) C is a finite set.

Let B = {x\(\in \)N : x is an odd prime number}
\(\Rightarrow\) B = {3, 5, 7, 11, .......}
\(\Rightarrow\) B is an infinite set.

Clearly A = {x: x = 4n + 1, n = 2, 3, 4, 5} = {9, 13, 17, 21}.
Hence n(A) = 4. This implies that \(n(\mathscr{P}(A))=2^4=16.\)

Given A\(\times\)B = {(x, 1)(y, 2) (z, 1)}
Since n(A) = 3 and n(B) = 2,
A \(\times\) B will have 6 elements.
The remaining elements of A \(\times\) B will be (x, 2) (y, 1) (z, 2)
\(\therefore\) A\(\times\) B = {(x,1)(y,2)(z,1)(x,2)(y,1)(z,2)}
\(\therefore\) A = {x, y, z} and B = {1, 2}

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5,1) (4, 2) (3, 3) (2, 4) and (1, 5).
Symmetric : (5, 1) \(\in \) R \(\Rightarrow\) (1, 5) \(\in \) R
(4, 2) \(\in \) R \(\Rightarrow\) (2, 4) \(\in \) R
\(\therefore\) R is symmetric

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5, 1) (4, 2) (3, 3) (2, 4) and (1, 5).
Reflexivity: R is not reflexive since (5, 5) \(\notin \) R. 

X = {a, b, c, d} 
R = {(a, a), (b, b), (a, c)} 
(i) To make R reflexive we need to include (c, c) and (d, d) 
(ii) To make R symmetric we need to include (c, a) 
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c) 
To make R symmetric we need to include (c, c) and (c, a) for transitive 
∴ The relation now becomes 
R = {(a, a), (b, b), (a, c), (c, c), (c, a)} 
∴ R is equivalence relation.