Q. 54.2( 5 Votes )

# If f(x) = log<sub

Given f(x) = loge(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) = loge(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) = loge(1 – x) + [x]

ii. fg

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

(fg)(x) = loge(1 – x) × [x]

(fg)(x) = [x]loge(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]loge(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 loge(1 – x) = 0.

loge(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) = loge(1 – x) + [x], x (–∞, 1)

We need to find (f + g)(–1).

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

(f + g)(–1) = loge(1 – (–1)) + [–1]

(f + g)(–1) = loge(1 + 1) + (–1)

(f + g)(–1) = loge2 – 1

Thus, (f + g)(–1) = loge2 – 1

We have (fg)(x) = [x]loge(1 – x), x (–∞, 1)

We need to find (fg)(0).

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]loge(1 – 0)

(fg)(0) = 0 × loge1

(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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