RD Chapter 3 Functions Ex 3.4 Solutions
Question - 1 : - Find f + g, f – g, cf (c ∈ R, c ≠ 0), fg, 1/f and f/g in each of the following:
(i) f (x) = x3 + 1 and g (x) = x + 1
(ii) f (x) = √(x-1) and g (x) = √(x+1)
Answer - 1 : -
(i) f(x) = x3 + 1 and g(x) = x+ 1
We have f(x): R → R and g(x): R → R
(a) f + g
We know, (f + g) (x) = f(x) + g(x)
(f + g) (x) = x3 + 1+ x + 1
= x3 + x + 2
So, (f + g) (x): R → R
∴f + g: R → R is given by (f + g) (x) = x3 + x + 2
(b) f – g
We know, (f – g) (x) = f(x) – g(x)
(f – g) (x) = x3 + 1– (x + 1)
= x3 + 1 – x – 1
= x3 – x
So, (f – g) (x): R → R
∴f – g: R → R is given by (f – g) (x) = x3 – x
(c) cf (c ∈ R, c ≠ 0)
We know, (cf) (x) = c × f(x)
(cf)(x) = c(x3 + 1)
= cx3 + c
So, (cf) (x) : R → R
∴cf: R → R is given by (cf) (x) = cx3 +c
(d) fg
We know, (fg) (x) = f(x) g(x)
(fg) (x) = (x3 + 1)(x + 1)
= (x + 1) (x2 – x +1) (x + 1)
= (x + 1)2 (x2 – x + 1)
So, (fg) (x): R → R
∴fg: R → R is given by (fg) (x) = (x + 1)2(x2 –x + 1)
(e) 1/f
We know, (1/f) (x) = 1/f (x)
1/f (x) = 1 / (x3 +1)
Observe that 1/f(x) is undefined when f(x) = 0 or when x = – 1.
So, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x3 + 1)
(f) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = (x3 + 1)/ (x + 1)
Observe that (x3 +1) / (x + 1) is undefined when g(x) = 0 or when x = –1.
Using x3 + 1 = (x +1) (x2 – x + 1), we have
(f/g) (x) = [(x+1) (x2–x+1)/(x+1)]
= x2 – x + 1
∴f/g: R – {–1} → R is given by (f/g) (x) = x2 – x + 1
(ii) f(x) = √(x-1) and g (x) = √(x+1)
We have f(x): [1, ∞) → R+ andg(x): [–1, ∞) → R+ asreal square root is defined only for non-negative numbers.
(a) f + g
We know, (f + g) (x) = f(x) + g(x)
(f+g) (x) = √(x-1) + √(x+1)
Domain of (f + g) = Domain of f ∩ Domain of g
Domain of (f + g) = [1, ∞) ∩ [–1, ∞)
Domain of (f + g) = [1, ∞)
∴f + g: [1, ∞) → R is given by (f+g) (x) = √(x-1) + √(x+1)
(b) f – g
We know, (f – g) (x) = f(x) – g(x)
(f-g) (x) = √(x-1) – √(x+1)
Domain of (f – g) = Domain of f ∩ Domain of g
Domain of (f – g) = [1, ∞) ∩ [–1, ∞)
Domain of (f – g) = [1, ∞)
∴f – g: [1, ∞) → R is given by (f-g) (x) = √(x-1) – √(x+1)
(c) cf (c ∈ R, c ≠ 0)
We know, (cf) (x) = c × f(x)
(cf) (x) = c√(x-1)
Domain of (cf) = Domain of f
Domain of (cf) = [1, ∞)
∴cf: [1, ∞) → R is given by (cf) (x) = c√(x-1)
(d) fg
We know, (fg) (x) = f(x) g(x)
(fg) (x) = √(x-1) √(x+1)
= √(x2 -1)
Domain of (fg) = Domain of f ∩ Domain of g
Domain of (fg) = [1, ∞) ∩ [–1, ∞)
Domain of (fg) = [1, ∞)
∴fg: [1, ∞) → R is given by (fg) (x) = √(x2 -1)
(e) 1/f
We know, (1/f) (x) = 1/f(x)
(1/f) (x) = 1/√(x-1)
Domain of (1/f) = Domain of f
Domain of (1/f) = [1, ∞)
Observe that 1/√(x-1) is also undefined when x – 1 = 0 or x = 1.
∴1/f: (1, ∞) → R is given by (1/f) (x) = 1/√(x-1)
(f) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = √(x-1)/√(x+1)
(f/g) (x) = √[(x-1)/(x+1)]
Domain of (f/g) = Domain of f ∩ Domain of g
Domain of (f/g) = [1, ∞) ∩ [–1, ∞)
Domain of (f/g) = [1, ∞)
∴ f/g:[1, ∞) → R is given by (f/g) (x) = √[(x-1)/(x+1)]
Question - 2 : - Let f(x) = 2x + 5 and g(x) = x2 + x. Describe
(i) f + g
(ii) f – g
(iii) fg
(iv) f/g
Find the domain in each case.
Answer - 2 : -
Given:
f(x) = 2x + 5 and g(x) = x2 +x
Both f(x) and g(x) are defined for all x ∈ R.
So, domain of f = domain of g = R
(i) f+ g
We know, (f + g)(x) = f(x) + g(x)
(f + g)(x) = 2x + 5 + x2 +x
= x2 + 3x + 5
(f + g)(x) Is defined for all real numbers x.
∴ Thedomain of (f + g) is R
(ii) f– g
We know, (f – g)(x) = f(x) – g(x)
(f – g)(x) = 2x + 5 – (x2 +x)
= 2x + 5 – x2 – x
= 5 + x – x2
(f – g)(x) is defined for all real numbers x.
∴ Thedomain of (f – g) is R
(iii) fg
We know, (fg)(x) = f(x)g(x)
(fg)(x) = (2x + 5)(x2 +x)
= 2x(x2 + x) + 5(x2 + x)
= 2x3 + 2x2 + 5x2 + 5x
= 2x3 + 7x2 + 5x
(fg)(x) is defined for all real numbers x.
∴ Thedomain of fg is R
(iv) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = (2x+5)/(x2+x)
(f/g) (x) is defined for all real values of x, except for the casewhen x2 + x = 0.
x2 + x = 0
x(x + 1) = 0
x = 0 or x + 1 = 0
x = 0 or –1
When x = 0 or –1, (f/g) (x) will be undefined as the divisionresult will be indeterminate.
∴The domain of f/g = R – {–1, 0}
Question - 3 : - If f(x) be defined on [–2, 2] and is given by
and g(x) = f(|x|) + |f(x)|. Find g(x).
Answer - 3 : -
Given:
Question - 4 : - Let f, g be two real functions defined by f(x) = √(x+1) and g(x) = √(9-x2). Then, describe each of the following functions.
(i) f + g
(ii) g – f
(iii) fg
(iv) f/g
(v) g/f
(vi) 2f – √5g
(vii) f2 + 7f
(viii) 5/g
Answer - 4 : -
Given:
f(x) = √(x+1) and g(x) = √(9-x2)
We know the square of a real number is never negative.
So, f(x) takes real values only when x + 1 ≥ 0
x ≥ –1, x ∈ [–1, ∞)
Domain of f = [–1, ∞)
Similarly, g(x) takes real values only when 9 – x2 ≥ 0
9 ≥ x2
x2 ≤ 9
x2 – 9 ≤ 0
x2 – 32 ≤ 0
(x + 3)(x – 3) ≤ 0
x ≥ –3 and x ≤ 3
∴ x ∈ [–3,3]
Domain of g = [–3, 3]
(i) f+ g
We know, (f + g)(x) = f(x) + g(x)
(f + g) (x) = √(x+1) + √(9-x2)
Domain of f + g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴f + g: [–1, 3] → R is given by (f + g) (x) = f(x) + g(x) = √(x+1) +√(9-x2)
(ii) g– f
We know, (g – f)(x) = g(x) – f(x)
(g – f) (x) = √(9-x2) –√(x+1)
Domain of g – f = Domain of g ∩ Domain of f
= [–3, 3] ∩ [–1, ∞)
= [–1, 3]
∴g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = √(9-x2) – √(x+1)
(iii) fg
We know, (fg) (x) = f(x)g(x)
(fg) (x) = √(x+1) √(9-x2)
= √[(x+1) (9-x2)]
= √[x(9-x2) + (9-x2)]
= √(9x-x3+9-x2)
= √(9+9x-x2-x3)
Domain of fg = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = √(x+1) √(9-x2) = √(9+9x-x2-x3)
(iv) f/g
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = √(x+1) / √(9-x2)
= √[(x+1) / (9-x2)]
Domain of f/g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
However, (f/g) (x) is defined for all real values of x ∈ [–1,3], except for the case when 9 – x2 =0 or x = ± 3
When x = ±3, (f/g) (x) will be undefined as the division resultwill be indeterminate.
Domain of f/g = [–1, 3] – {–3, 3}
Domain of f/g = [–1, 3)
∴f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = √(x+1)/ √(9-x2)
(v) g/f
We know, (g/f) (x) = g(x)/f(x)
(g/f) (x) = √(9-x2) /√(x+1)
= √[(9-x2) / (x+1)]
Domain of g/f = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
However, (g/f) (x) is defined for all real values of x ∈ [–1,3], except for the case when x + 1 = 0 or x = –1
When x = –1, (g/f) (x) will be undefined as the division resultwill be indeterminate.
Domain of g/f = [–1, 3] – {–1}
Domain of g/f = (–1, 3]
∴g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) = √(9-x2) / √(x+1)
(vi) 2f– √5g
We know, (2f – √5g) (x) = 2f(x) – √5g(x)
(2f – √5g) (x) = 2f (x) – √5g (x)
= 2√(x+1) – √5√(9-x2)
= 2√(x+1) – √(45- 5x2)
Domain of 2f – √5g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f (x) – √5g(x) = 2√(x+1) – √(45- 5x2)
(vii) f2 + 7f
We know, (f2 + 7f)(x) = f2(x) + (7f)(x)
(f2 + 7f) (x) = f(x)f(x) + 7f(x)
= √(x+1) √(x+1) + 7√(x+1)
= x + 1 + 7√(x+1)
Domain of f2 + 7f issame as domain of f.
Domain of f2 + 7f =[–1, ∞)
∴f2 + 7f: [–1,∞) → R is given by (f2 +7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7√(x+1)
(viii) 5/g
We know, (5/g) (x) = 5/g(x)
(5/g) (x) = 5/√(9-x2)
Domain of 5/g = Domain of g = [–3, 3]
However, (5/g) (x) is defined for all real values of x ∈ [–3,3], except for the case when 9 – x2 =0 or x = ± 3
When x = ±3, (5/g) (x) will be undefined as the division resultwill be indeterminate.
Domain of 5/g = [–3, 3] – {–3, 3}
= (–3, 3)
∴5/g: (–3, 3) → R is given by (5/g) (x) = 5/g(x) = 5/√(9-x2)
Question - 5 : - If f(x) = loge (1 – x) and g(x) = [x], then determine each of the following functions:
(i) f + g
(ii) fg
(iii) f/g
(iv) g/f
Also, find (f + g) (–1), (fg) (0), (f/g) (1/2) and (g/f) (1/2).
Answer - 5 : -
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