Let f : R →

As we have to find g: R→R such that-

fog = gof = I_R

We know that I_R 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: R→R 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.