\({ \{ }(X,Y):x\in P,y\in Q,y+2\} \)

\(\because \) (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}

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

Here, first element of each ordered pair of A x B gives the elements of set A and corresponding second element gives the elements of set B
\(\therefore\) A = {a, b} and B = {l, 3, 2}

Let A = \(\left\{ 1,2,3,4 \right\} \), then \({ f }_{ 1 },{ f }_{ 2 }\) and \({ f }_{ 3 }\) are defined from A to A.
First express f₁, f₂ and f₃ as sets of ordered pairs, write
\({ f }_{ 1 }=\left\{ (1,2),(2,3),(3,4) \right\} \)
\({ f }_{ 2 }=\left\{ (1,4),(2,3),(3,2),(4,1) \right\} \)
\({ f }_{ 3 }=\left\{ (1,4),(2,3),(2,4),(3,3),(3,2),(3,4),(4,1),(4,2),(4,3),(4,4) \right\} \)
Now in f₁ we observe that an element 4 \(\epsilon\) A is not appeared at first place-of any ordered pair of f₁.
So, f₁ is not a function from A to A. In f₂ we observe that each element of set A is appeared at first place in one and only one ordered pair of f₂.
So, f₂ is a function from A to A and Range of f₂ = {1, 2, 3, 4}.
In f₃ we observe that 2, 3, 4 \(\in\) A have appeared at first place of ordered pair more than one time. So, f₃ is not a function.

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

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

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.