Answer :

As we have to find g: RR such that-


fog = gof = IR


We know that IR represents an identity function.


By identity function we mean that –


I(x) = x


fog(x) = I(x)


f(g(x)) = x


f(x) = 10x + 7


fog(x) = 10{g(x)} + 7 = x


10g(x) = x – 7


g(x) =


G (x) is defined everywhere on a set of real numbers, and its range is also R.


g: RR can be defined by g(x) =


OR


Given,


‘*’ is a binary operation defined on the set {0, 1, 2, 3, 4, 5} as –


…(1)


To prove: 0 is an identity for ‘*.’


Let a be an arbitrary element of set {0, 1, 2, 3, 4, 5}


0≤ a ≤ 5


As (a + 0)< 6


a*0 = a + 0 = a {from 1}


Also 0*a = 0 + a = a {from 1}


a*0 = 0*a = a a {0, 1, 2, 3, 4, 5}


0 is an identity for the operation ‘*’


To prove that (6-a) is the inverse –


Let a be an arbitrary element of set {0, 1, 2, 3, 4, 5}


0≤ a ≤ 5


a + (6-a) = 6


a*(6-a) = a + 6 – a – 6 = 0 {from 1}


And, (6-a)*a = 6 – a + a – 6 = 0 {from 1}


Clearly we observe that –


a*(6-a) = (6-a)*a = 0 (identity)


For all ‘a’ {0, 1, 2, 3, 4, 5}, when it is operated with (6 – a), we get the identity element.


All elements of the given set are invertible and 6 – a is the inverse where ‘a’ is any arbitrary element of the given set.


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