If f(x) = loge(1 - x) and f(x) = (x) then find: (g/f)(x). Also find (f + g)(-1), (fg)(0), (g/f)(-1), (g/f)(1/2)Please solve the problem fast.... Please....You will be marked ​brainliest

Given as

f(x) = loge (1 – x) and g(x) = [x]

As we know, f(x) takes real values only when 1 – x > 0

1 > x

x < 1, ∴ x ∈ (–∞, 1)

The domain of f = (–∞, 1)

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

The domain of g = [x], x ∈ R

= R

(i) f + g As we know,

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

The domain of f + g = Domain of f ∩ Domain of g

The domain of f + g = (–∞, 1) ∩ R

= (–∞, 1)

∴ f + g: (–∞, 1) → R is given by (f + g) (x) = loge (1 – x) + [x]

(ii) fg

As we know, (fg) (x) = f(x) g(x) (fg) (x)

= loge (1 – x) × [x]

= [x] loge (1 – x)

The domain of fg = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

∴ fg: (–∞, 1) → R is given by (fg) (x) = [x] loge (1 – x)

(iii) f/g

As we know, (f/g) (x) = f(x)/g(x) (f/g) (x)

= loge (1 – x)/[x]

The domain of f/g = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

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

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

If 0 ≤ x < 1, (f/g) (x) will be undefined as the division result will be indeterminate. T

he domain of f/g

= (–∞, 1) – [0, 1)

= (–∞, 0)

∴ f/g: (–∞, 0) → R is given by (f/g) (x) = loge (1 – x)/[x]

(iv) g/f As we know, (g/f) (x)

= g(x)/f(x) (g/f) (x)

= [x]/loge (1 – x) However,

(g/f) (x) 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 If x = 0,

(g/f) (x) will be undefined as the division result will be indeterminate.

The domain of g/f = (–∞, 1) – {0} = (–∞, 0) ∪ (0, 1)

∴ g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f) (x) = [x]/loge (1 – x)

(a) Here, we need to find (f + g) (–1). We have, (f + g) (x)

= loge (1 – x) + [x], x ∈ (–∞, 1)

By substituting x

= –1 in the above equation,

we get (f + g)(–1) = loge (1 – (–1)) + [–1]

= loge (1 + 1) + (–1)

= loge2 – 1

∴ (f + g) (–1) = loge2 – 1 (b) Here, we need to find (fg) (0).

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

= 0 in the above equation, we get (fg) (0) = [0] loge (1 – 0)

= 0 × loge1

∴ (fg) (0) = 0 (c) Here, we need to find (f/g) (1/2) We have, (f/g) (x)

= loge (1 – x)/[x],

x ∈ (–∞, 0) However,

1/2 is not in the domain of f/g.

∴ (f/g) (1/2) does not exist.

(d) Here, we need to find (g/f) (1/2)

We have, (g/f) (x)

= [x]/loge (1 – x), x ∈ (–∞, 0) ∪ (0, ∞) By substituting x

= 1/2 in the above equation, we get (g/f) (1/2)

= [x]/loge (1 – x)

= (1/2)/loge (1 – 1/2)

= 0.5/loge (1/2) = 0/loge (1/2)

= 0

Thus, (g/f) (1/2) = 0