RD Chapter 3 Functions Ex 3.3 Solutions
Question - 1 : - Find the domain of each of the following real valued functions of real variable:
(i) f (x) = 1/x
(ii) f (x) = 1/(x-7)
(iii) f (x) = (3x-2)/(x+1)
(iv) f (x) = (2x+1)/(x2-9)
(v) f (x) = (x2+2x+1)/(x2-8x+12)
Answer - 1 : -
(i) f(x) = 1/x
We know, f (x) is defined for all real values of x, except for the casewhen x = 0.
∴Domain of f = R – {0}
(ii) f(x) = 1/(x-7)
We know, f (x) is defined for all real values of x, except for the casewhen x – 7 = 0 or x = 7.
∴Domain of f = R – {7}
(iii) f(x) = (3x-2)/(x+1)
We know, f(x) is defined for all real values of x, except for the casewhen x + 1 = 0 or x = –1.
∴Domain of f = R – {–1}
(iv) f(x) = (2x+1)/(x2-9)
We know, f (x) is defined for all real values of x, except for the casewhen x2 – 9 = 0.
x2 – 9 = 0
x2 – 32 = 0
(x + 3)(x – 3) = 0
x + 3 = 0 or x – 3 = 0
x = ± 3
∴Domain of f = R – {–3, 3}
(v) f(x) = (x2+2x+1)/(x2-8x+12)
We know, f(x) is defined for all real values of x, except for the casewhen x2 – 8x + 12 = 0.
x2 – 8x + 12 = 0
x2 – 2x – 6x + 12 =0
x(x – 2) – 6(x – 2) = 0
(x – 2)(x – 6) = 0
x – 2 = 0 or x – 6 = 0
x = 2 or 6
∴Domain of f = R – {2, 6}
Question - 2 : - Find the domain of each of the following real valued functions of real variable:
(i) f (x) = √(x-2)
(ii) f (x) = 1/(√(x2-1))
(iii) f (x) = √(9-x2)
(iv) f (x) = √(x-2)/(3-x)
Answer - 2 : -
(i) f(x) = √(x-2)
We know the square of a real number is never negative.
f (x) takes real values only when x – 2 ≥ 0
x ≥ 2
∴ x ∈ [2,∞)
∴Domain (f) = [2, ∞)
(ii) f(x) = 1/(√(x2-1))
We know the square of a real number is never negative.
f (x) takes real values only when x2 –1 ≥ 0
x2 – 12 ≥ 0
(x + 1) (x – 1) ≥ 0
x ≤ –1 or x ≥ 1
∴ x ∈ (–∞,–1] ∪ [1, ∞)
In addition, f (x) is also undefined when x2 – 1 = 0 because denominator will be zero and theresult will be indeterminate.
x2 – 1 = 0 ⇒ x= ± 1
So, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}
x ∈ (–∞, –1) ∪ (1, ∞)
∴Domain (f) = (–∞, –1) ∪ (1, ∞)
(iii) f(x) = √(9-x2)
We know the square of a real number is never negative.
f (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 (f) = [–3, 3]
(iv) f(x) = √(x-2)/(3-x)
We know the square root of a real number is never negative.
f (x) takes real values only when x – 2 and 3 – x are both positive andnegative.
(a) Bothx – 2 and 3 – x are positive
x – 2 ≥ 0
x ≥ 2
3 – x ≥ 0
x ≤ 3
Hence, x ≥ 2 and x ≤ 3
∴ x ∈ [2,3]
(b) Bothx – 2 and 3 – x are negative
x – 2 ≤ 0
x ≤ 2
3 – x ≤ 0
x ≥ 3
Hence, x ≤ 2 and x ≥ 3
However, the intersection of these sets is null set. Thus, this case isnot possible.
Hence, x ∈ [2, 3] – {3}
x ∈ [2, 3]
∴Domain (f) = [2, 3]
Question - 3 : - Find the domain and range of each of the following real valued functions:
(i) f (x) = (ax+b)/(bx-a)
(ii) f (x) = (ax-b)/(cx-d)
(iii) f (x) = √(x-1)
(iv) f (x) = √(x-3)
(v) f (x) = (x-2)/(2-x)
(vi) f (x) = |x-1|
(vii) f (x) = -|x|
(viii) f (x) = √(9-x2)
Answer - 3 : -
(i) f(x) = (ax+b)/(bx-a)
f(x) is defined for all real values of x, except for the case when bx – a= 0 or x = a/b.
Domain (f) = R – (a/b)
Let f (x) = y
(ax+b)/(bx-a) = y
ax + b = y(bx – a)
ax + b = bxy – ay
ax – bxy = –ay – b
x(a – by) = –(ay + b)
∴x = – (ay+b)/(a-by)
When a – by = 0 or y = a/b
Hence, f(x) cannot take the value a/b.
∴Range (f) = R – (a/b)
(ii) f(x) = (ax-b)/(cx-d)
f(x) is defined for all real values of x, except for the case when cx – d= 0 or x = d/c. Domain (f) = R – (d/c)
Let f (x) = y
(ax-b)/(cx-d) = y
ax – b = y(cx – d)
ax – b = cxy – dy
ax – cxy = b – dy
x(a – cy) = b – dy
∴x = (b-dy)/(a-cy)
When a – cy = 0 or y = a/c,
Hence, f(x) cannot take the value a/c.
∴Range (f) = R – (a/c)
(iii) f(x) = √(x-1)
We know the square of a real number is never negative.
f(x) takes real values only when x – 1 ≥ 0
x ≥ 1
∴ x ∈ [1,∞)
Thus, domain (f) = [1, ∞)
When x ≥ 1, we have x – 1 ≥ 0
Hence, √(x-1) ≥ 0 ⇒ f (x) ≥ 0
f(x) ∈ [0, ∞)
∴Range (f) = [0, ∞)
(iv) f(x) = √(x-3)
We know the square of a real number is never negative.
f (x) takes real values only when x – 3 ≥ 0
x ≥ 3
∴ x ∈ [3,∞)
Domain (f) = [3, ∞)
When x ≥ 3, we have x – 3 ≥ 0
Hence, √(x-3) ≥ 0 ⇒ f (x) ≥ 0
f(x) ∈ [0, ∞)
∴Range (f) = [0, ∞)
(v) f(x) = (x-2)/(2-x)
f(x) is defined for all real values of x, except for the case when 2 – x =0 or x = 2.
Domain (f) = R – {2}
We have, f (x) = (x-2)/(2-x)
f (x) = -(2-x)/(2-x)
= –1
When x ≠ 2, f(x) = –1
∴Range (f) = {–1}
(vi) f(x) = |x-1|
Now we have,
Hence, f(x) is defined for all real numbers x.
Domain (f) = R
When, x < 1, we have x – 1 < 0 or 1 – x > 0.
|x – 1| > 0 ⇒ f(x) > 0
When, x ≥ 1, we have x – 1 ≥ 0.
|x – 1| ≥ 0 ⇒ f(x) ≥ 0
∴ f(x)≥ 0 or f(x) ∈ [0, ∞)
Range (f) = [0, ∞)
(vii) f(x) = -|x|
Now we have,
Hence, f(x) is defined for all real numbers x.
Domain (f) = R
When, x < 0, we have –|x| < 0
f (x) < 0
When, x ≥ 0, we have –x ≤ 0.
–|x| ≤ 0 ⇒ f (x) ≤ 0
∴ f(x) ≤ 0 or f (x) ∈ (–∞, 0]
Range (f) = (–∞, 0]
(viii) f(x) = √(9-x2)
We know the square of a real number is never negative.
f(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 (f) = [–3, 3]
When, x ∈ [–3, 3], we have 0 ≤ 9 – x2 ≤ 9
0 ≤ √(9-x2) ≤ 3 ⇒0 ≤ f (x) ≤ 3
∴ f(x) ∈ [0,3]
Range (f) = [0, 3]