Answer :

Given: A = {1,2,3......9}

and R is defined as

(a,b) R (c,d) ⇒ a + b = b + c

__Reflexive:__

Consider (a,b) R (a,b) ⩝(a,b) ∈ A × A

⇒ a + b = b + a

Hence, R is reflexive

__Symmetric:__

Consider (a,b)R(c,d) given by (a,b)(c,d) ∈ A × A

(a,b) R (c,d) ⇒ a + d = b + c

⇒ b + c = a + d

⇒ c + b = d + a

⇒(c,d) R (a,b)

Hence R is symmetric

__Transitive:__

Let (a,b) R (c,d) and (c,d) R (e,f) given by (a,b),(c,d),(e,f)∈ A × A

(a,b) R (c,d) ⇒ a + b = b + c

⇒ a – c = b – d …(1)

and (c,d) R (e,f) ⇒ c + f = d + e

& c + f = d + e …(2)

adding (1) and (2), we get

a – c + c + f = b – d + d + e

⇒a + f = b + e

⇒ (a,b) R (e,f)

Hence, R is transitive

Hence, R is an equivalence relation

We need to find [(2, 5)]

So, (2, 5) will go in and (c, d) will come out

This will be possible if

a + d = b + c

⇒ 2 + d = 5 + c

⇒ d – c = 5 – 2

⇒ d – c = 3

So, in our relation [(2, 5)]

We need to find values of c and d which satisfy d – c = 3

Since (c, d) ∊ A × A

Both c and d are in set A = {1, 2, 3, …9}

[(2,5)]={(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)}is the equivalent class under relation R

**OR**

Given: f(x) = 4x^{2} + 12x + 15

To show: f: N → S is invertible and to find inverse of f

Proof:

Let f(x) = y

⇒ 4x^{2} + 12x + 15 = y

⇒ 4x^{2} + 12x + 15 – y = 0

⇒ 4x^{2} + 12x + (15 – y) = 0

Comparing equation with ax^{2} + bx + c = 0

Here, a = 4, b = 12 and c = 15 – y

Putting the values, we get

So,

As x є N, So, x is a positive real number.

So, x can’t be

Hence,

Let where g: S → N

gof = g(f(x))

= g(4x^{2} + 12x + 15)

[here, y = 4x^{2} + 12x + 15]

= x

Hence, gof = x = I_{N}

Now, we find the fog

fog = f(g(x))

= 9 + y – 6 – 3

= y

Hence, fog = y = I_{S}

Since, gof = I_{N} and fog = I_{S}

∴ f is invertible

and

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