\(f(x)=\frac{x}{x^{2}-5 x+6}=\frac{x}{(x-3)(x-2)}\)
When x = 2 or 3, f(x) becomes undefined
therefore Domain of f = R = {2,3}

The first elements of two ordered pairs (1,4) and (1, 5) in 'f' are same.
Therefore f = {(1, 4), (1, 5), (2,4),(3, 5)} is not a function from A to B

(i) The two ordered pairs (2, b) and, (2,c) have the same first co-ordinate. Therefore, (1, a), (2, b),(2,c), (3, c) does not represent a function A\(\rightarrow\) B.
(ii) Since one element 1 of A is not associated with some element of B. i.e., 1 is not the first co-ordinate of any ordered pair. So {(2, b), (3, b)} does not represent a function from A\(\rightarrow\) B.

f(x) = 2x - 5
f(0) = 2(0)-5 = -5
f(7) - 2(7)-5 = 14 - 5 = 9
f(-3) = 2(-3) -5 = -6 -5 = -11

R - {(a, b) \(\in\) A x B, 'a' month has 'b' number of days}
\(\therefore\) R = {( January, 31), (February, 28),(February, 29), (April, 30), (June, 30), (September 30), (October, 31 ),(November, 30), December, 31)}
\(\therefore\) The subset of relation R connected with Teachers Day is {(September, 30}

Set A has 'n' elements.
\(\therefore\) Number of elements in A x A = n x n = n²
The number of possible subsets of A x A = \(2^{n ^2}\)
Since, each subset of A x A is a relation on A, the total number of relations on A is \(2^{n ^2}\)

Since a \(\in\) A and b \(\in\) B,
a - b = (3 -7),(3 - 11), (5 -7),(5- 11)
= - 4,- 8, - 2, - 6 None of them is an odd number.
Therefore, R is an empty relation

A = Set of all first entries = {a, b} .
S = Set of all second entries = {x, y}

\(\mathrm{A} \times \mathrm{A}=\{-1,1\} \times\{-1,1\} \)
\(=\{(-1,-1),(-1,1),(1,-1),(1,1)\} \)
\(\mathrm{A} \times \mathrm{A} \times \mathrm{A}=\{(-1,-1),(-1,1),(1,-1),(1,1)\} \times\{-1,1\} \)
\(=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1), (1,1,-1),(1,1,1)\}\)

G x H = {7,8} x {5,4,2}
= {(7,5),(7,4),(7,2), (9,5), (9,4), (8,2)}
H x G = {5,4,2} x {7,8}
= {(5,7), (5, 8), (4,7), (4, 8), (2,7), (2, 8)}

A has 3 elements and 'B' has 3 elements, then 'A x B' will have 9 elements.

It is not a function as the vertical line PQ cuts the graph at two points

A = {1,2,3},B =  {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) =  2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(ii) An arrow diagram

(iv) A Graph f = \(\{(x, f(x) / x \in A\}\)
= {(0, 1), (1, 3), (2, 5), (3, 7)}

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

f(k) - 2k - 1
f o f(k) = 5
f(f(k)) = f(2k - 1) = 5
⇒ 2(2k-1) -1 = 5
4k - 2 -1 = 5 ⇒ 4k = 8
k = 2