Given,   \(f\left( x \right) =\frac { x-2 }{ 3-x } =y\)  (say)
Here, denominator = 3 - x
Put denominator = 0, we get 3 - x = 0 \(\Rightarrow \) x = 3
Therefore, Domain of y = f(x) is R - {3}
Now, consider \(y=\frac { x-2 }{ 3-x } \)
\(\Rightarrow y(3-x)=(x-2)\)
\(\Rightarrow 3y-xy=x-2\)
\(\Rightarrow x+xy=3y+2\)
\(\Rightarrow x(1+y)=3y+2\)
\(\Rightarrow x=\frac { 3y+2 }{ 1+y } \)
It will define, if \(1+y\neq 0\ \Rightarrow y\neq -1\)
So, x=g(y) takes real values, if \(y\in R-\{ -1\} \)
Also, \(\frac { 3y+2 }{ 1+y } \neq 3\) for all \(y\in R-\{ -1\} \)
[ \(\therefore \) if \(\frac { 3y+2 }{ 1+y } =3,\)then 3y+2 = 3+3y \(\Rightarrow \) 2 = 3, which is absurd]
\(\Rightarrow \frac { 3y+2 }{ 1+y } \neq 3\quad y\in R-\{ -1\} \)
Thus, \(x=\frac { 3y+2 }{ 1+y } \in \) Domain (f), \(\forall \ y\in R-\{ 1\} \)
Hence, the range of f(x) is R-{-1}.

We have, \(y=f(x)=\frac { 1-x }{ 1+x } \)
Then, \(f(y)=\frac { 1-y }{ 1+y } =\frac { 1-\left( \frac { 1-x }{ 1+x } \right) }{ 1+\left( \frac { 1-x }{ 1+x } \right) } \)  [from Eq. (i)]
\(\frac { \frac { 1+x-(1-x) }{ 1+x } }{ \frac { 1+x+1-x }{ 1+x } } =\frac { 2x }{ 2 } \)
\(\Rightarrow \) f(y) = x
\(\therefore \)  x = f(y), when \(y=f(x)=\frac { 1-x }{ 1+x } \)

\(f+g=3 x-2\)
\(f-g=-x+4\)
\(\frac { f }{ g } :R-\left\{ \frac { 3 }{ 2 } \right\} \rightarrow R,\)  \(\frac { f }{ g } =\frac { x+1 }{ 2x+1 } \)

\(\because \) (a,2),(b,3),(c,2),(d,3),(e,2) \(\in \) A x B
\(\therefore \) a,b,c,d,e \(\in \) A and 2,3 \(\in \) A x B
Also, it is given that n(A) = 5, n(B) = 2
\(\therefore \) A = {a,b,c,d,e}, B = {2,3}

\(A\times C\) = {(1,5),(1,6),(2,5),(2,6),(3,5),(3,6)}
\(B\times D\) = {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}
Here, elements (1,5),(1,6),(2,5),(2,6),(3,5),(3,6) \(\in \) \(A\times C\) are also belongs to \(B\times D\), therefore \((A\times C)\subset (B\times D)\)

\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1,\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

We have,
\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

We have,
\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

R = {(x,y) : x, y \(\in\) N, x + 2y = 8}
\(\therefore \)  x + 2y = 8 
When x=2, y=3
When x=4, y=2
When x=6, y=1
\(\Rightarrow \) R= {(2,3),(4,2),(6,1)}
\(\Rightarrow \)\({ R }^{ -1 }\)= {(3,2),(2,4),(1,6)}

Let\(f\left( x \right) =x,g\left( x \right) =\frac { 1 }{ x }\) ;
Domain\((f)=R={ D }_{ 1 }\)
Domain\((g)=R-\left\{ 0 \right\} ={ D }_{ 2 };{ D }_{ 1 }\cap { D }_{ 2 }=R-\{ 0\} \) 
Ans Sum= \(x+\frac { 1 }{ x } \)and difference \(x-\frac { 1 }{ x } \)

Domain  \(f=\left[ 2,\infty \right] ={ D }_{ 1 }\) [say]   
Domain g= \(\left[ -2,\infty \right] ={ D }_{ 2 }\) [say];   
Ans \(\left( \sqrt { x-2 } +\sqrt { x+2 } \right) \)and\(\left( \sqrt { x-2 } -\sqrt { x+2 } \right) \)

Domain (R) = {1, 2, 3, 4, 5, 6, 7}
Range (R) = {7, 6, 5, 4, 3, 2, 1}

\(f(x)=\sqrt { x+2 } ,g(x)=\sqrt { 4-{ x }^{ 2 } } \)
\(f\)(x) is defined for \(x+2\ge 0\Rightarrow \ge -2\)
Domain(\(f\)) = \([-2,\infty )\)
\(g(x)\)is defined for \(4-{ x }^{ 2 }\ge =0\Rightarrow { x }^{ 2 }-4\le 0\)
\(\Rightarrow (x-2)(x+2)\le 0\Rightarrow x\in \left[ -2,2 \right] ;Domain(g)=\left[ -2,2 \right] \)
Domain(\(f\)) \(\cap \) domain(\(g\)) = [-2,2]
\(1/\sqrt { 2-x } \)

\(f(x)=\sqrt { x+2 } ,g(x)=\sqrt { 4-{ x }^{ 2 } } \)
\(f\)(x) is defined for \(x+2\ge 0\Rightarrow \ge -2\)
Domain(\(f\))=\([-2,\infty )\)
\(g(x)\)is defined for \(4-{ x }^{ 2 }\ge =0\Rightarrow { x }^{ 2 }-4\le 0\)
\(\Rightarrow (x-2)(x+2)\le 0\Rightarrow x\in \left[ -2,2 \right] ;Domain(g)=\left[ -2,2 \right] \)
Domain(\(f\)) \(\cap \) domain(\(g\)) = [-2,2]
\(\left( x+2 \right) \sqrt { 2-x } \)

\(f(x)=\sqrt { x+2 } ,g(x)=\sqrt { 4-{ x }^{ 2 } } \)
\(f\)(x) is defined for \(x+2\ge 0\Rightarrow \ge -2\)
Domain(\(f\))=\([-2,\infty )\)
\(g(x)\)is defined for \(4-{ x }^{ 2 }\ge =0\Rightarrow { x }^{ 2 }-4\le 0\)
\(\Rightarrow (x-2)(x+2)\le 0\Rightarrow x\epsilon \left[ -2,2 \right] ;Domain(g)=\left[ -2,2 \right] \)
Domain (\(f\))  \(\cap \) domain (\(g\)) = [-2,2]
\(\left( \sqrt { x+2 } \right) \left[ 1+\sqrt { 2-x } \right] \)

Given functions are
\(f(x)=\frac { 1 }{ (x-2) } ,x\neq 2\) and \(g(x)=(x-2)^{ 2 }\)
\((f+g)(x)=f(x)+g(x)=\frac { 1 }{ (x-2) } +(x-2)^{ 2 },\quad x\neq 2\)
\(=\frac { 1+(x-2)^{ 3 } }{ x-2 } ,x\neq 2\)

We have, A = {a,b} and B = {2,3},
\(\therefore \ n\left( A\times B \right) =n\left( A \right) \times n\left( B \right) =2\times 2=4\)
Now, number of subsets of A x B
\(={ 2 }^{ n\left( A\times B \right) }={ 2 }^{ 4 }=16\)  
Thus, the number of relations from A to B is 16.

Given sets are A={a,b,c} and B={x: x \( \in \)N, x is a prime number leaa than 5}={2,3}
For element a of set A, All ordered pairs are (a,2), (a,3).
For element b of set A, All ordered pairs are (b,2), (b,3).
For element c of set A, All ordered pairs are (c,2), (c,3).
\( \therefore \)  A X B = {(a,2)(a,3),(b,2)(b,3)(c,2)(c,3)}
For element 2 of set B, all ordered pairs are (2,a), (2,b),(2,c).
For element 3 of set B, all ordered pairs are (3,a), (3,b), (3,C).
\( \therefore \)B X A = {(2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}
Since,  (a,2) \( \neq \) (2,a)
 A X B​​​ \( \neq \) B X A

\(\left\{ (2,0),(4,8),(2,1),(3,6) \right\} \)
It is not a function because first elements of (2,0) and (2,1) are same.

\( \left\{ (3,3),(4,2),(5,1),(6,0),(7,7) \right\} \)
It is a function because the first element of each ordered pair is different.

We know that, two ordered pairs are equal, if their corresponding elements are equal.
(a-3,b+7)=(3,7)\(\Rightarrow \) a-3=3 and b+7=7
[equating corresponding elements]
\(\Rightarrow \)   a=3+3 and b=7-7
\(\Rightarrow \)   a=6   and b=0

We know that, two ordered pairs are equal, if their corresponding elements are equal.
(2a-5,4)=(5,b+6)\(\Rightarrow \)2a-5=5 and 4=b+6
[equating corresponding elements]
\(\Rightarrow \)  2a=5+5 and and 4-6=b
\(\Rightarrow \)  2a=10 and -2=b
\(\Rightarrow \) a=5 and b=-2

The relation R={(1,1),(2,4),(3,9),(4,16)}

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