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# RD Chapter 3- Functions Ex-3.4 Interview Questions Answers

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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)

(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)(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.

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).

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

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).

Given:

f(x) = log(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) = log(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) = log(1 – x) + [x]

(ii) fg

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

(fg) (x) = log(1 –x) × [x]

= [x] log(1 – x)

Domain of fg = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

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

(iii) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = log(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) = log(1 – x) / [x]

(iv) g/f

We know, (g/f) (x) = g(x)/f(x)

(g/f) (x) = [x] / log(1– x)

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

log(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]/ log(1 – x)

(a) We need to find (f + g) (–1).

We have, (f + g) (x) = log(1– x) + [x], x  (–∞, 1)

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

(f + g)(–1) = log(1– (–1)) + [–1]

= log(1 + 1) +(–1)

= loge2 – 1

(f + g) (–1) = loge2 – 1

(b) We need to find (fg) (0).

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

Substituting x = 0 in the above equation, we get

(fg) (0) = [0] log(1– 0)

= 0 × loge1

(fg)(0) = 0

(c) We need to find (f/g) (1/2)

We have, (f/g) (x) = log(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] / log(1– x), x  (–∞, 0)  (0, ∞)

Substituting x=1/2 in the above equation, we get

(g/f) (1/2) = [x] / log(1– x)

= (1/2)/ log(1 –1/2)

= 0.5/ log(1/2)

= 0 / log(1/2)

= 0

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

krishan