Question -
Answer -
Given:
f(x) = loge (1 – x)and g(x) = [x]
We know, f(x) takes real values only when 1 – x > 0
1 > x
x < 1, ∴ x ∈ (–∞, 1)
Domain of f = (–∞, 1)
Similarly, g(x) is defined for all real numbers x.
Domain of g = [x], x ∈ R
= 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
= (–∞, 1)
∴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]
= [x] loge (1 – x)
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
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
= (–∞, 1) ∩ R
= (–∞, 1)
However, (f/g) (x) 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, (f/g) (x) will be undefined as the divisionresult will be indeterminate.
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
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
When x = 0, (g/f) (x) will be undefined as the division resultwill be indeterminate.
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) We need to find (f + g) (–1).
We have, (f + g) (x) = loge (1– x) + [x], x ∈ (–∞, 1)
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) We need to find (fg) (0).
We have, (fg) (x) = [x] loge (1– x), x ∈ (–∞, 1)
Substituting x = 0 in the above equation, we get
(fg) (0) = [0] loge (1– 0)
= 0 × loge1
∴ (fg)(0) = 0
(c) 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) We need to find (g/f) (1/2)
We have, (g/f) (x) = [x] / loge (1– x), x ∈ (–∞, 0) ∪ (0, ∞)
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
∴(g/f) (1/2) = 0