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


and


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