f(1) = 5
f(-1) = -5
f(0) = 1
f(2) = 9

Domain = \((-\infty ,-1)\cup (1,4)\)

To determine \(A\times (B\cup C)\)
\(B\cup C\) = {b,c,e} \(\cup \) {b,c,f} = {b,c,e,f}
\(\therefore \)\(A\times (B\cup C)\) = {a,d} x {b,c,e,f}
={(a,b),(a,c),(a,e),(a,f),(d,b),(d,c)(d,e),(d,f)}........(i)
To determine \((A\times B)\cup (A\times C)\)
\(A\times B\)={a,d} x {b,c,e}
={(a,b),(a,c),(a,e),(d,b),(d,c),(d,e)}
\(A\times C\)={a,d} x {b,c,f}
={(a,b),(a,c),(a,f),(d,b),(d,c),(d,f)}
\(\therefore \)\((A\times B)\cup (A\times C)\) = {(a,b),(a,c),(a,e),(a,f),(d,b),(d,c),(d,e),(d,f)}......(ii)
From Eqs.(i) and (ii) we get
\(A\times (B\cup C)=(A\times B)\cup (A\times C)\)

Domain = {1, 2, 3, 4, 5} Range = {5, 7, 9, 11, 13}

Domain \((f)\cap \)  Domain \((g)=\left[ -1,3 \right] \)
 \(\left( \frac { f }{ g } \right) (x)=\sqrt { \frac { x+1 }{ 9-{ x }^{ 2 } } } \)

Let A = B.
Now A = B\(\Rightarrow\)A x B = A x A....(i)
and A = B \(\Rightarrow\) B x A = A x A...(ii)
From (i) and (ii), we have
AxB=BxA
Let A x B = B x A and x be any element of A and Y be any element of B.
\(\Rightarrow\) (x, y) \(\in\)A x B
\(\Rightarrow\) (x, y)\(\in\) B x A (\(\because\)A x B = B x A)
\(\Rightarrow\) X\(\in\)B
\(\therefore\) x\(\in\)A\(\Rightarrow\) x\(\in\)B
\(\therefore\) A\(\subset\)B ...(iii)
Let x be any element of B and y be any element of A.
\(\Rightarrow\)(x, y) \(\in\)B x A 
\(\Rightarrow\) (x, y)\(\in\) A x B (\(\because\) A x B = B x A)
\(\Rightarrow\) x\(\in\)A
\(\therefore\) x\(\in\)B\(\Rightarrow\) xEA
\(\therefore\) B \(\subset\)A...(iv)
From (iii) and (iv), we have
A=B

Let (x, y) \(\in\) A x A
\(\Rightarrow x,y \in A\)  \((\therefore A\subset B)\)
\(\Rightarrow x,y \in B\)
\(\therefore\) \(x,y \in A,y\in B\Rightarrow (x,y)\in A\times B\)
x \(\in\) B, Y\(\in\)A \(\Rightarrow\) (x, y) \(\in\) B x A
\(\Rightarrow\) (x, y)\(\in\) (A x B) \(\cap\) (B x A)
\(\therefore\) A x A\(\subset\)(A x B) \(\cap\) (B x A) ...(i)
Let (x, y)\(\in\)(A x B) \(\cap\)(B x A)
\(\Rightarrow\) (x, y)\(\in\) A x B and (x, y) \(\in\) B x A
\(\Rightarrow\) [x\(\in\) A and x\(\in\) B] and [x\(\in\) Band Y\(\in\)A]
\(\Rightarrow\) x, Y\(\in\)A \(\Rightarrow\) x, Y\(\in\)A x A
\(\therefore\)(A x B) \(\cap\) (B x A) \(\subset\)A x A ... (ii)
From (i) and (ii), we have
\(\therefore\) A x A = (A x B)\(\cap\) (B x A).

Here
f(x) = x² 0\(\le\)x \(\le\)3
f(x) = 3x 3 \(\le\) x \(\le\)10
At x=3
f(3) = (3)² = 9 and f(3) = 3 x 3 = 9.
We observe that f(x) takes unique value at each point in its domain [0,10]. So f is a function.
Now g(x) = x² 0\(\le\)x \(\le\)2
g(x) = 3x  2\(\le\)x \(\le\)10
At x=2
g(2) = (2)² = 4 and g(2) = 3 x 2 = 6
So g(x) does not have unique value at x = 2.
Hence g(x) is not a function.

f(x) = \(\sqrt{9-x^{2}}\) 
Since \(\sqrt{9-x^{2}}\) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].
For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.
∴ The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

16; \(\phi\), {(x, a)}, {(x, b)}, {(y, a)}, {(y, b)}
{(x, a), (x, b)}, {(x, a), (y, a)}, {(x, a), (y, b)}, {(x, b), (y, a)}
{(x, b), (y, b)}, {(y, a), (y, b)}, {(x, a), (x, b), (y, a)}
{(x, a), (x, b), (y, b)}, {(x, a), (y, a), (y, b)}, {(x, b), (y, a), (y, b)}
{(x, a), (x, b), (y, a), (y, b)}

Domain of R = {1, 2, 3, 4}
Range of R = {4, 7, 10, 13}