Q. 115.0( 1 Vote )

# Let f : R →

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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