Domain = R – = R –

Range = {0} ∪ = {0} ∪

Sol. f(x) =

case(i) if sin x ∈ (0, 1)

(sin n x) → 0

f(x) = 0 and sin x ∈ (0, 1) ⇒ x ∈ (2n π , 2n π + π )

case(ii) if sin x = 0 ⇒ x ∈ k π , k ∈ z

f(x) = 0

case(iii) if sin x = 1 ⇒ x ∈ 2k π + , k ∈ z

f(x) =

case(iv) if sin x ∈ (–1, 0)

(sin n x) → 0, x ∈ (2k π + π , 2k π + 2 π ) ; k ∈ z

f(x) = 0

case(v) if sin x = –1

(sin n x) = – 1

then f(x) will not take any definite value

so if sinx = –1

x = these values not lie in the domain

For Domain ; x ∈ R –

For Range ; f(x) = 0, for x ∈ (2k π + π , 2k π + 2 π ) ∪ {k π } ∪ (2k π , 2k π + π ) ; k ∈ z

f(x) = , for x ∈ 2k π + , k ∈ z

Range = {0} ∪

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