Q. 20

# Let A = {1,

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) = 4x2 + 12x + 15

To show: f: NS is invertible and to find inverse of f

Proof:

Let f(x) = y

4x2 + 12x + 15 = y

4x2 + 12x + 15 – y = 0

4x2 + 12x + (15 – y) = 0

Comparing equation with ax2 + 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: SN

gof = g(f(x))

= g(4x2 + 12x + 15) [here, y = 4x2 + 12x + 15]      = x

Hence, gof = x = IN

Now, we find the fog

fog = f(g(x))     = 9 + y – 6 – 3

= y

Hence, fog = y = IS

Since, gof = IN and fog = IS

f is invertible

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