OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Ex 2(f)

Continuous practice using OP Malhotra Class 11 Solutions Chapter 2 Relations and Functions Ex 2(f) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(f)

Question 1.
The range and domain of function f(x) = \(\frac { 3 }{ x }\) + 1 are subsets of A and B respectively,
where A = \(\left\{-\frac{1}{2}, 0, \frac{2}{3}, \frac{6}{7}, 1\right\}\)
and B = \(\left\{-5,0,4 \frac{1}{2}, 5,5 \frac{1}{2}\right\}\).
List the elements of the function as ordered pairs.
Solution:

Question 2.
A = {- 2, – 1, 1, 2} and f = \(\left\{\left(x, \frac{1}{x}\right), x \in A\right\}\).
(i) List the domain of f
(ii) List the range off
(iii) Is f a function ?
Solution:
Given f be a function defined by
f(x) = \(\frac { 1 }{ x }\) ∀ x ∈ A
where A = {- 2, – 1, 1, 2}
∴ f(- 2) = – \(\frac { 1 }{ 2 }\) ; f(- 1) = – 1 ∈ A
f(1) = 1 ∈ A; f(2) = \(\frac { 1 }{ 2 }\)
Thus f = \(\left\{(-1,-1),(1,1),\left(-2,-\frac{1}{2}\right),\left(2, \frac{1}{2}\right)\right\}\)
Domain (f) = {- 1, 1, – 2, 2}
Range (f) = {- 1, 1, – \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\)}
Since each element in domain of f has unique image, i.e. no two ordered pairs have same first component.
∴ f represents a function.

Question 3.
f : A → highest prime factor of x.
(i) Find the range off when the domain is {12, 13, 14, 15, 16, 17}.
(ii) State a domain of five integers for which the range is (3).
(iii) A set of positive integers is called S. What can be said about these integers iff(S) = S?
Solution:
Given f : x → highest prime factor of x
(i) Here factors of 12 are 1, 2, 3, 4, 6, 12
∴ 3 be the highest prime factor of 12.
∴ 3 ∈ R_f
Also, factors of 13 are 1 and 13
∴ 13 be the highest prime factor of 13
∴ 13 ∈ R_f
Since 1, 2, 7, 14 are the factors of 14
∴ 7 be the highest prime factor of 14.
∴ 7 ∈ R_f
Since 1, 3, 5, 15 are the factors of 15
Here 5 be the highest prime factor of 15
∴ 5 ∈ R_f
Since 1, 2, 4, 8 and 16 are the factors of 16
Here 2 be the highest prime factors of 16
∴ 2 ∈ R_f
Since 1, 17 are factors of 17. Here 17 be the highest prime factor of 17.
∴ 17 ∈ R_f
∴ range (f) = {2, 3, 5, 7, 13, 17}

(ii) given R_f = 3 i.e. highest prime factor of each element of domain of f be 3.
Since 3 be the highest prime factor of 3, 6, 9, 12, 18 [since highest prime factor of 15 be 5]
∴ Required D_f = {3, 6, 9, 12, 18}

(iii) Given f(S) = S i.e. highest prime factors of S is equal to S which is only possible if S be the set of all prime numbers. Since factors of any prime number p are 1 and p. So p be the highest prime factor of p.

Question 4.
A function f is defined on the set of integers as follows :
f(x) = \(\left\{\begin{aligned}
1+x ; & 1 \leq x<2 \\
2 x-1 ; & 2 \leq x<4 \\
3 x-10 ; & 4 \leq x<6
\end{aligned}\right\}\)
(i) Find the domain of the function
(ii) Find the range of the function
(iii) Find the value of f(4).
Solution:
(i) Given f(x) = \(\left\{\begin{aligned}
1+x ; & 1 \leq x<2 \\
2 x-1 ; & 2 \leq x<4 \\
3 x-10 ; & 4 \leq x<6
\end{aligned}\right\}\)
since f is defined on set of integers.
Clearly f(x) has different values for all 1 ≤ x ≤ 6
∴ x = 1, 2, 3, 4, 5
Thus D_f = {1, 2, 3, 4, 5}

(ii) When x = 1, f(1) = 1 + 1 = 2
When x = 2, f(2) = 2 x 2 – 1 = 3
When x = 3, f(x) = 2x – 1 ⇒ f(3) = 6 – 1 = 5
When x = 4, f (x) = 3x – 10 ⇒ f(4) = 3x – 10 = 2
When x = 5,f (x) = 3x – 10 ⇒ f(5) = 15 – 10 = 5
∴ R_f = {2, 3, 5}

(iii) When x = 4, f(x) = 3x – 10
∴ f(4) = 3 x 4 – 10 = 2

Question 5.
Let f be a function whose domain is the set of all real number. If f(x) = | x | – x, what is range off?
Solution:
Given f(x) = | x | – x since | x | ≥ x ∀ x ∈ D_f
∴ | x | – x ≥ 0 ⇒ f(x) ≥ 0
∴ R_f = [0, ∞)

Question 6.
Write the domain of the following real functions
(i) \(\sqrt{9-x^2}\)
(ii) \(\sqrt{1-2 x-3 x^2}\)
(iii) 10^x
(iv) \(\frac{1}{\sqrt{x^2-7}}\)
(v) log (2 – 3x)
(vi) log\((\sqrt{x-4}+\sqrt{6-x})\)
(vii) \(\left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}}\)
(viii) \(\sin ^{-1}\left[\log _2\left(\frac{x}{2}\right)\right]\)
Solution:
(i) Given f(x) = \(\sqrt{9-x^2}\)
For D_f : f (x) must be a real number.
⇒ \(\sqrt{9-x^2}\) must be a real number.
⇒ 9 – x² > 0 ⇒ x² ≤ 9
⇒ | x | < 3 ⇒ – 3 ≤ x ≤ 3
∴ D_f = [- 3, 3]

(ii) Let f(x) = \(\sqrt{1-2 x-3 x^2}\)
For D_f : f(x) must be a real number.
⇒ \(\sqrt{1-2 x-3 x^2}\) must be a real number.
⇒ 1 – 2x – 3x² ≥ 0
⇒ 3x² + 2x – 1 ≤ 0

(iii) Let f(x) = 10^x
For D_f : f(x) must be a real number.
⇒ 10^x must be a real number,
which is a real number for all x ∈ R
∴ D_f = R

(iv) Let f(x) = \(\frac{1}{\sqrt{x^2-7}}\)
For D_f : f(x) must be a real number.
⇒ \(\frac{1}{\sqrt{x^2-7}}\) must be a real number.
⇒ x² – 7 > 0 ⇒ |x| > \(\sqrt{7}\)
⇒ x > \(\sqrt{7}\) or x < – \(\sqrt{7}\)
⇒ x ∈ (- ∞, – \(\sqrt{7}\)) ∪ (\(\sqrt{7}\), ∞)
∴ D_f = (- ∞, – \(\sqrt{7}\)) ∪ (\(\sqrt{7}\), ∞)

(v) Let f (x) = log (2 – 3x)
For D_f : f(x) must be a real number.
⇒ log (2 – 3x) must be a real number.
⇒ 2 – 3x > 0 ⇒ 2 > 3x ⇒ 3x < 2
⇒ x < \(\frac { 2 }{ 3 }\)
∴ D_f = (- ∞, \(\frac { 2 }{ 3 }\))

(vi) Let f = f(x) = log \((\sqrt{x-4}+\sqrt{6-x})\)
For D_f : f(x) must be a real number
⇒ log \((\sqrt{x-4}+\sqrt{6-x})\) must be a real number.
⇒ \(\sqrt{x-4}+\sqrt{6-x}\) > 0
⇒ x – 4 ≥ 0 and 6 – x ≥ 0
⇒ x ≥ 4 and 6 ≥ x
⇒ 4 ≤ x < 6 ⇒ x ∈ [4, 6]
∴ D_f = [4, 6]

(vii) Let y = f (x) = \(\left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{\frac{1}{2}}\)
For D_f : f(x) must be a real number.

(viii) Let f(x) = sin^-1\(\left[\log _2\left(\frac{x}{y}\right)\right]\)
We know that sin^-1 y is defined for – 1 ≤ y ≤ 1
∴ f(x) is defined only
if – 1 ≤ log₂ \(\frac { x }{ 2 }\) ≤ 1 and \(\frac { x }{ 2 }\) > 0
⇒ 2^-1 ≤ \(\frac { x }{ 2 }\) ≤ 2¹ and x > 0
⇒ \(\frac { 1 }{ 2 }\) x 2 ≤ x ≤ 4 and x > 0
⇒ 1 ≤ x ≤ 4 and x > 0
⇒ 1 ≤ x ≤ 4 ⇒ x ∈[1, 4]
∴ D_f = [1, 4]

Question 7.
Find the range of each of the following function
(i) |x – 3|
(ii) \(\sqrt{x-5}\)
(iii) cos(\(\frac { x }{ 3 }\))
(iv) \(\frac{x+2}{|x+2|}\)
(v) \(\sec \left(\frac{\pi}{4} \cos ^2 x\right)\), – ∞ < x ∞
(vi) \(\frac{x^2+x+2}{x^2+x+1}\)
(vii) \(\frac{x}{1+x^2}\)
(viii) f(x) = \(\frac{3}{2-x^2}\)
Solution:
(i) Let y = f(x) = | x – 3 | if defined for all x ∈ R
∴ D_f = R = (- ∞, ∞)
For R_f since | x | ≥ 0 ∀ x ∈ R
⇒ |x – 3| ≥ 0 ∀ x ∈ R
⇒ f(x) ≥ 0 ∀ X ∈ R
∴ R_f = [0, ∞)

(ii) y = f(x) = \(\sqrt{x-5}\)
For D_f : f(x) must be a real number.
⇒ \(\sqrt{x-5}\) must be a real number.
⇒ x – 5 ≥ 0
⇒ x ≥ 5
∴ D_f = [5, ∞)
For R_f : Let y = f(x) = \(\sqrt{x-5}\) ∀ x ∈ D_f
∴ y ≥ 0 [since square root of a real no. is always non-negative]
⇒ y² = x – 5 ⇒ x = y² + 5
since x ≥ 5 ⇒ y² + 5 ≥ 5
⇒ y² ≥ 0 ∀ y ∈ R
∴ R_f = [0, ∞)

(iii) Let y = f(x) = cos\(\frac { x }{ 3 }\)
Here f(x) is defined for all x ∈ R
∴ D_f = R
For R_f : since – 1 ≤ cos z ≤ 1 ∀ z ∈ R
⇒ -1 ≤ cos \(\frac { x }{ 3 }\) ≤ 1
⇒ – 1 ≤ y ≤ 1
∴ R_f = [- 1, 1]

(iv) Given f(x) = \(\frac{x+2}{|x+2|}\)
For D_f : f(x) must be a real number.
⇒ \(\frac{x+2}{|x+2|}\) must be a real number
⇒ x + 2 ≠ 0 ⇒ x ≠ – 2
∴ D_f = R – {- 2}
For R_f : Let y = \(\frac{x+2}{|x+2|}\) ∀ x ∈ D_f
Case-I : When x < – 2 ⇒ x + 2 < 0 ∴ | x + 2 | = – (x + 2) ∴ y = \(\frac{x+2}{-(x+2)}\) = 1 [ ∵ x ≠ – 2] Case-II : When x > – 2 ⇒ x + 2 > 0
⇒ | x + 2 | = x + 2
∴ y = \(\frac{x+2}{x+2}\) = 1 [ ∵ x ≠ – 2]
Hence, R_f = {- 1, 1}

(v) Let y = f (x) = sec \(\left(\frac{\pi}{4} \cos ^2 x\right)\)

(vi) Let y = f (x) = \(\frac{x^2+x+2}{x^2+x+1}\), x ∈ R

(vii) Let y = \(\frac{x}{1+x^2}\) = f(x)
For D_f : f(x) must be a real number
⇒ \(\frac{x}{1+x^2}\) must be a real number
which is a real number [∵ 1 + x² > 0 ∀ x ∈ R]

(viii) Let y = f(x) = \(\frac{3}{2-x^2}\)
For D_f : f (x) must be a real number.

Question 8.
Find the domain and range of the function \(\frac{x^2-4}{x-2}\).
Solution:
Let y = f(x) = \(\frac{x^2-4}{x-2}\)
For D_f : f(x) must be a real number.
⇒ \(\frac{x^2-4}{x-2}\) must be a real number.
⇒ x – 2 ≠ 0 ⇒ x = ≠ 2
∴ D_f = R – {2}
For R_f : \(\frac{x^2-4}{x-2}\) ∀ x ∈ D_f i.e. x ≠ 2
⇒ y = \(\frac{(x-2)(x+2)}{x-2}\) = x + 2 [∵ x – 2 ≠ 0]
⇒ x = y – 2
Since x ∈ D_f ⇒ x ≠ 2 ⇒ y – 2 ≠ 2 ⇒ y ≠ 4
∴ R_f = R – {4}

Question 9.
If the domain of the function \(\frac{|x|}{x}\) be [3, 7], then its range is
(a) [- 1, 1]
(b) [- 1, 1]
(c) {1}
(d) {- 1}
Solution:
Given domain of function f (x) = \(\frac{|x|}{x}\) be [3, 7].
∴ 3 ≤ x ≤ 7 ⇒ | x | = x
Thus, f(x) = \(\frac{|x|}{x}=\frac{x}{x}\) = 1 [∵ x ≠ 0 and x > 0]
∴ R_f = {1}