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Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.3

  • Last Updated : 30 Apr, 2021
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Question 1. Find the domain of each of the following real valued functions of real variable:

(i) f (x) = 1/x 

Solution:

We are given, f (x) = 1/x.  

Here, f (x) is defined for all real values of x, except for the case when x = 0.

Therefore, domain of f = R – {0}

(ii) f (x) = 1/(x−7)

Solution:



We are given, f (x) = 1/(x−7).

Here, f (x) is defined for all real values of x, except for the case when x – 7 = 0 or x = 7.

Therefore, domain of f = R – {7}

(iii) f (x) = (3x−2)/(x+1)

Solution:

We are given, f (x) = (3x−2)/(x+1).

Here, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = –1.

Therefore, domain of f = R – {–1}

(iv) f (x) = (2x+1)/(x2−9)

Solution:



We are given, f (x) = (2x+1)/(x2−9).

Here, f (x) is defined for all real values of x, except for the case when x2 – 9 = 0.

=> x2 – 9 = 0

=> (x + 3)(x – 3) = 0

=> x + 3 = 0 or x – 3 = 0

=> x = ± 3

Therefore, domain of f = R – {–3, 3}

(v) f (x) = (x2+2x+1)/(x2–8x+12)

Solution:

We are given, f (x) = (x2+2x+1)/(x2–8x+12).

Here, f(x) is defined for all real values of x, except for the case when 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

Therefore, domain of f = R – {2, 6}

Question 2. Find the domain of each of the following real-valued functions of a real variable:

(i) f (x) = √(x–2)

Solution:

We are given, f (x) = √(x–2).



Here, f (x) takes real values only when x – 2 ≥ 0 as the square of a real number cannot be negative.

=> x – 2 ≥ 0

=> x ≥ 2

=> x ∈ [2, ∞)

Therefore, domain of f = [2, ∞)

(ii) f (x) = 1/(√(x2–1))

Solution:

We are given, f (x) = 1/(√(x2–1)).

Here, f (x) takes real values only when x2 – 1 > 0 as the square of a real number cannot be negative and the denominator x2 – 1 cannot be zero.

=> x2 – 1 > 0

=> (x + 1) (x – 1) > 0

=> x < –1 or x > 1

=> x ∈ (–∞, –1) ∪ (1, ∞)

Therefore, domain of f = (–∞, –1) ∪ (1, ∞)

(iii) f (x) = √(9–x2)

Solution:

We are given, f (x) = √(9–x2).

Here, f (x) takes real values only when 9 – x2 ≥ 0 as the square of a real number cannot be negative.

=> 9 – x2 ≥ 0

=> 9 ≥ x2

=> x2 ≤ 9

=> x2 – 9 ≤ 0



=> (x + 3)(x – 3) ≤ 0

=> x ≥ –3 and x ≤ 3

=> x ∈ [–3, 3]

Therefore domain of f = [–3, 3]

(iv) f (x) = √[(x–2)/(3–x)]

Solution:

We are given, f (x) = √[(x–2)/(3–x)].

Here, f (x) takes real values only when x – 2 and 3 – x are both positive and negative.

Case 1. x – 2 ≥ 0 and 3 – x ≥ 0

=> x ≥ 2 and x ≤ 3

Therefore, x ∈ [2, 3]

Case 2. x – 2 ≤ 0 and 3 – x ≤ 0.

=> x ≤ 2 and x ≥ 3

This case is not possible as the intersection of these sets is null set. 

Hence, x ∈ [2, 3] – {3}

=> x ∈ [2, 3)

Therefore, domain of 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)

Solution:

We are given, f (x) = (ax+b)/(bx–a).

Here, f(x) is defined for all real values of x, except for the case when bx – a = 0 or x = a/b.

So, domain of f = R – (a/b)

Let f (x) = y. So, (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.

Therefore, range of f = R – (a/b)

(ii) f (x) = (ax–b)/(cx–d)

Solution:

We are given, f (x) = (ax–b)/(cx–d).



Here, f(x) is defined for all real values of x, except for the case when cx – d = 0 or x = d/c. 

So, domain of f = R – (d/c)

Let f (x) = y. So, (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.

Therefore, range of f = R – (a/c)

(iii) f (x) = √(x–1)

Solution:

We are given, f (x) = √(x–1). 

Here, f(x) takes real values only when x – 1 ≥ 0.

=> x ≥ 1

=> x ∈ [1, ∞)

So, domain of f = [1, ∞)

When x ≥ 1, we have x – 1 ≥ 0. So, √(x–1) ≥ 0. 

=> f (x) ≥ 0

=> f(x) ∈ [0, ∞)

Therefore, range of f = [0, ∞)

(iv) f (x) = √(x–3)

Solution:

We are given, f (x) = √(x–3).

Here, f (x) takes real values only when x – 3 ≥ 0.

=> x ≥ 3

=> x ∈ [3, ∞)

So, domain of f = [3, ∞)

When x ≥ 3, we have x – 3 ≥ 0. Hence, √(x–3) ≥ 0 

=> f (x) ≥ 0

=> f(x) ∈ [0, ∞)

Therefore, range of f = [0, ∞)

(v) f (x) = (x–2)/(2–x)

Solution:

We are given, f (x) = (x–2)/(2–x).

Here, f(x) is defined for all real values of x, except for the case when 2 – x = 0 or x = 2.

So, domain of f = R – {2}

And also, f (x) = –(2–x)/(2–x) = –1

Hence, when x ≠ 2, f(x) = –1

Therefore, range of f = {–1}

(vi) f (x) = |x–1|

Solution:

We are given, f (x) = |x–1|.

Clearly, f(x) is defined for all real numbers x.

So, domain of f = R

Let f (x) = y. So, |x–1| = y. 

Therefore, y can take only the positive values. So, y ≥ 0.

Therefore, range of f = (0, ∞]

(vii) f (x) = –|x|

Solution:

We are given, f (x) = –|x|.

Clearly, f(x) is defined for all real numbers x.

So, domain of f = R

Let f (x) = y. So, y = –|x|.

Therefore, y can take only the negative values. So, y ≤ 0.



Therefore, range of f = (–∞, 0]

(viii) f (x) = √(9–x2)

Solution:

We are given, f (x) = √(9–x2)

Here, f(x) takes real values only when 9 – x2 ≥ 0.

=> 9 ≥ x2

=> x2 ≤ 9

=> x2 – 9 ≤ 0

=> (x + 3)(x – 3) ≤ 0

=> x ≥ –3 and x ≤ 3

=> x ∈ [–3, 3]

So, domain of 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]

Therefore, range of f = [0, 3]

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