Answer :

Given f(x) = log_{e}(1 – x) and g(x) = [x]

Clearly, f(x) takes real values only when 1 – x > 0

⇒ 1 > x

⇒ x < 1

∴ x ∈ (–∞, 1)

Thus, domain of f = (–∞, 1)

g(x) is defined for all real numbers x.

Thus, domain of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = log_{e}(1 – x) + [x]

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = (–∞, 1) ∩ R

∴ Domain of f + g = (–∞, 1)

Thus, f + g : (–∞, 1) → R is given by (f + g)(x) = log_{e}(1 – x) + [x]

ii. fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = log_{e}(1 – x) × [x]

∴ (fg)(x) = [x]log_{e}(1 – x)

Domain of fg = Domain of f ∩ Domain of g

⇒ Domain of fg = (–∞, 1) ∩ R

∴ Domain of fg = (–∞, 1)

Thus, f – g : (–∞, 1) → R is given by (fg)(x) = [x]log_{e}(1 – x)

iii.

We know

As earlier, domain of = (–∞, 1)

However, is defined for all real values of x ∈ (–∞, 1), except for the case when [x] = 0.

We have [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

When 0 ≤ x < 1, will be undefined as the division result will be indeterminate.

⇒ Domain of = (–∞, 1) – [0, 1)

∴ Domain of = (–∞, 0)

Thus, : (–∞, 0) → R is given by

iv.

We know

As earlier, domain of = (–∞, 1)

However, is defined for all real values of x ∈ (–∞, 1), except for the case when log_{e}(1 – x) = 0.

log_{e}(1 – x) = 0 ⇒ 1 – x = 1 or x = 0

When x = 0, will be undefined as the division result will be indeterminate.

⇒ Domain of = (–∞, 1) – {0}

∴ Domain of = (–∞, 0) ∪ (0, ∞)

Thus, : (–∞, 0) ∪ (0, ∞) → R is given by

We have (f + g)(x) = log_{e}(1 – x) + [x], x ∈ (–∞, 1)

We need to find (f + g)(–1).

Substituting x = –1 in the above equation, we get

(f + g)(–1) = log_{e}(1 – (–1)) + [–1]

⇒ (f + g)(–1) = log_{e}(1 + 1) + (–1)

∴ (f + g)(–1) = log_{e}2 – 1

Thus, (f + g)(–1) = log_{e}2 – 1

We have (fg)(x) = [x]log_{e}(1 – x), x ∈ (–∞, 1)

We need to find (fg)(0).

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]log_{e}(1 – 0)

⇒ (fg)(0) = 0 × log_{e}1

∴ (fg)(0) = 0

Thus, (fg)(0) = 0

We have, x ∈ (–∞, 0)

We need to find

However, is not in the domain of.

Thus, does not exist.

We have, x ∈ (–∞, 0) ∪ (0, ∞)

We need to find

Substituting in the above equation, we get

Thus,

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