It is clear from the arrow diagram,that domain={2,3,4,5} ,codomain={1,2,3,4,5} and range={1,2,3,4}

f (x ) = 1 - x² ,
\(\forall x\in { R }^{ + }\)
Domain of f = (0 , \(\infty \) ) and range of f = ( - \(\infty \)  , 1)

 \(\frac { f }{ g } :(-\infty ,0)\rightarrow R,\frac { f }{ g } (x)=\frac { \log { (1-x) } }{ [x] } \)

\(B\cap C\) = {b,c};
\(A\times (B\cap C)\) = {(a,b),(a.c),(d,b),(d,c)}
and \((A\times B)\cap (A\times C)\) = {(a,b),(a,c),(d,b),(d,c)}
\(\Rightarrow\)\(A\times (B\cap C)=(A\times B)\cap (A\times C)\)

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

Here A = {1, 2, 3, 4} and B = {5, 6, 7}, a \(\in\) A, b \(\in\) B.
\(\therefore\) a - b = 1 - 5, 1 - 6, 1 - 7, 2 - 5, 2 - 6, 2 - 7, 3 - 5, 3 - 6, 3 - 7, 4 - 5, 4-6,4-7
= -4, -5, -6, -3, -4, -5, -2, -3, -4, -1, -2,-3
R = {(1, 5), (1, 7), (2, 6), (3,5), (3, 7), (4,6)}.

Here f(x) = x².
At x=1.1
f(1.1) = (1.1)² = 1.21
f(1) = (1)² =1
\(\therefore {f(1.1)-f(1)\over (1.1-1)}={1.21-1\over 0.1}={0.21\over 0.1}=2.1\)

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].

Here (-1,0) \(\in\) A x A \(\Rightarrow\)-1, 0 \(\in\) A and (0, 1) \(\in\)  A  x  A \(\Rightarrow\)0, 1\(\in\) A
\(\therefore -1,0,1\)\(\in\) A.
It is given that n(A x A) = 9 which implies that n(A) = 3
\(\therefore\)A = {-1,0,1}
\(\therefore\)A x A = {(-1, -1), (-1, 0), (-1, 1),(0, -1), (0, 0), (0, 1), (1, -1),(1, 0), (1, 1)}
So the remaining elements of A x A are
(-1, 1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0) and (1,1).

 Here A = {1, 2} and B = {3, 4}
\(\therefore\)A x B = {1, 2} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
Number of elements in A x B = 4
\(\therefore\) Number of subsets of A x B = 2⁴ = 16
The subsets are:
\(\phi\), {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)},
{(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)},
{1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)},
{(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)} {(1, 4),
(2,3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}