In this article, we will be going to solve the entire Miscellaneous Exercise of Chapter 2 of the NCERT textbook. Relations and functions are fundamental concepts in mathematics, particularly in algebra and calculus. They describe how elements from one set can be associated with elements from another set.
Table of Contents
What is Relation and Function?
Relation
A relation between two sets A and B is a collection of ordered pairs where the first element is from A and the second element is from B. Formally, a relation R from set A to set B is a subset of the Cartesian product A×B.
Function
A function is a special type of relation where each element in the domain (the set from which inputs are taken) is associated with exactly one element in the codomain (the set into which outputs are mapped).
Formally, a function fff from set A to set B is a relation where for every x in A, there is a unique y in B such that (x,y) is in the relation f.
Class 11 NCERT Solutions- Chapter 2 Relation And Functions – Miscellaneous Exercise on Chapter 2
Question 1. The relation f is defined by f(x) = {x2, 0 ≤ x ≤ 3 3x, 3 ≤ x ≤ 10} The relation g is defined by g(x) = {x2, 0 ≤ x ≤ 2 3x, 2 ≤ x ≤ 10}. Show that f is a function and g is not a function.
Solution:
Given relation f is defined as:
f(x) = {x2, 0≤x≤3
3x, 3≤x≤10}
It is given that, for the condition 0 ≤ x < 3,
Solution of f(x) = x2 and
For the condition 3 < x ≤ 10, solution of f(x) = 3x.
Now for the value of x = 3, solution of f(x) by putting the value of x, f(x) = 32 = 9
or, f(x) = 3 × 3 = 9.
That means, at x = 3, f(x) = 9 [Single image]
So that, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Therefore, the given relation is a function.
Now,
In the given relation g is defined as:
g(x) = {x2, 0≤x≤2
3x, 2≤x≤10}
It is seen that, in case of both the condition, for x = 2,
The value of g(x), by putting the value of x, g(x) = 22= 4 and g(x) = 3 × 2 = 6.
So that, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6.
Therefore, this relation is not a function.
Question 2. If f(x) = x2, find f(1.1)–f(1)(1.1–1)(1.1–1)f(1.1)–f(1) .
Solution:
Given:
f(x) = x2 .
Hence, by putting the condition of f(x) in f(1.1) and f(1),
we can find the result of the given equation
((f(1.1) – f(1))/(1.1 – 1)) = (((1.1)2 – (1)2)/(1.1 – 1))
= ((1.21-1)/(0.1))
= (0.21/0.1)
= 2.1
Question 3. Find the domain of the function f(x) = x2+2x+1×2−8x+12x2−8x+12x2+2x+1((x2+2x+1)/(x2-8x+12)).
Solution:
Given function:
f(x) = ((x2+2x+1)/(x2-8x+12))
= (((x2+2x+1)/((x-6)(x-2)))
It is clearly notified that, the function f is defined for all real numbers except
at x = 6 and x = 2 as the denominator becomes zero otherwise.
Question 4. Find the domain and the range of the real function f defined by f(x) = √(x – 1).
Solution:
Given real function:
f(x) = √(x – 1).
Clearly it is notified, √(x – 1) is defined for (x – 1) ≥ 0.
Hence, the function f(x) = √(x – 1) is defined for x ≥ 1.
So that, the domain of f is the set of all real numbers greater than or equal to 1.
Domain of f = [1, ∞).
Now,
According to the condition, x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0
That’s why, the range of f is the set of all real numbers greater than or equal to 0.
Range of f = [0, ∞).
Therefore, the domain of f is R – {2, 6}.
Question 5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.
Solution:
Given real function: f(x) = |x – 1|
Clearly it is notified that, the function |x – 1| is defined for all real numbers.
Hence, Domain of f = R
Also, according to the condition , for x ∈ R, |x – 1| assumes all real numbers.
So that, the range of f is the set of all non-negative real numbers.
Question 6. Let f={(x, x21+x21+x2x2)}: x ∈ R} be a function from R into R. Determine the range of f.
Solution:
Given function:
f = {(x, x2/1+x2): x ∈ R}
Substituting values and determining the images, we have
={(0,0), (±0.5, 1/5), (±1, 1/2), (±1.5, 9/13), (±2, 4/5), (3, 9/10), (4,16/17), …..}
From the above equation, the range of f is the set of all second elements.
It can be notified that all these elements are greater than or equal to 0 but less than 1.
[As the denominator is greater than the numerator.]
Or, We know that, for x ∈ R,
x2≥ 0
Then,
x2 + 1 ≥ x2
1 ≥ (x2 / (x2 + 1))
Therefore, the range of f = [0, 1)
Question 7. Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g, and f/g.
Solution:
According to the question, let us assume, the functions f, g: R → R is defined as
given conditions f(x) = x + 1, g(x) = 2x – 3.
Now,
We find that (f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
So that, (f + g) (x) = 3x – 2
Now, we find that, (f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
So that, (f – g) (x) = -x + 4
(f/g(x)) = f(x)/g(x), g(x) ≠ 0, x ∈ R
(f/g(x)) = x + 1/ 2x – 3, 2x – 3 ≠ 0
So that, (f/g(x)) = x + 1/ 2x – 3, x ≠ 3/2.
Question 8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution:
Given the values, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}
And according to the question, the function is defined as, f(x) = ax + b
For (1, 1) ∈ f
We have, f(1) = 1
So, a × 1 + b = 1
a + b = 1 …. (i)
And for (0, -1) ∈ f
We have f(0) = -1
a × 0 + b = -1
So, b = -1
Now, On substituting b = –1 in (i), we get
Putting the value here, a + (–1) = 1 ⇒ a = 1 + 1 = 2.
Hereby, the values of a and b are 2 and –1 respectively.
Question 9. Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
Justify your answer in each case.
Solution:
According to the question, Given relation R = {(a, b): a, b ∈ N and a = b2}
(i) It can be notified that 2 ∈ N; however, 2 ≠ 22 = 4.
Hence, it is notified that the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) Its clearly notified that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.
Now, it’s clear that 3 ≠ 92 = 81; therefore, (3, 9) ∉ N
Hence, it is notified that the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) Its clearly notified that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.
Now, it is clear that 16 ≠ 22 = 4; therefore, (16, 2) ∉ N
Hence, it is notified that the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.
Question 10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
(i) f is a relation from A to B
(ii) f is a function from A to B.
Justify your answer in each case.
Solution:
Given, A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
So,
A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5),
(2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11),
(3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
Also given in the question that, f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(i) The verified statement is : A relation from a non-empty set A to a non-empty
set B is a subset of the Cartesian product A × B.
It’s clearly notified that f is a subset of A × B.
Hence, it’s clear that f is a relation from A to B.
(ii) From the given condition, as the same first element i.e., 2 corresponds to two
different images (9 and 11), relation f is not a function.
Question 11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer?
Solution:
According to the question, Given relation f is defined as f = {(ab, a + b): a, b ∈ Z}.
Now, we know that a relation f from a set A to a set B is said to be a function
if every element of set A has unique images in set B.
As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
i.e., (12, 8), (12, –8) ∈ f
It’s clearly notified that, the same first element, 12 corresponds
to two different images (8 and –8).
Therefore, the relation f is not a function.
Question 12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Solution:
Given, A = {9, 10, 11, 12, 13}
Now, f: A → N is defined as
f(n) = The highest prime factor of n.
So, Prime factor of 9 = 3
Prime factors of 10 = 2, 5
Prime factor of 11 = 11
Prime factors of 12 = 2, 3
Prime factor of 13 = 13
Hence, it can be expressed as:
Here,
f(9) means the highest prime factor of 9 = 3
f(10) means the highest prime factor of 10 = 5
f(11) means the highest prime factor of 11 = 11
f(12) means the highest prime factor of 12 = 3
f(13) means the highest prime factor of 13 = 13
So that, the range of f is the set of all f(n), where n ∈ A.
Therefore,
Range of f = {3, 5, 11, 13}.
Summary
Chapter 2 of the NCERT textbook covers the fundamental concepts of Relations and Functions. A relation is a collection of ordered pairs derived from two sets, where the first element belongs to one set and the second to another. A function is a special type of relation that associates each element in the domain with exactly one element in the codomain. This chapter’s miscellaneous exercise helps solidify these concepts by applying them to various problems, including finding domains and ranges, determining whether a given relation is a function, and working with composite functions.
Relation And Function – FAQs
What is the difference between a relation and a function?
A relation is a set of ordered pairs from two sets, while a function is a specific type of relation where each element in the domain is associated with exactly one element in the codomain.
How can you determine if a relation is a function?
A relation is a function if for every element in the domain, there is a unique corresponding element in the codomain. If any element in the domain is associated with more than one element in the codomain, the relation is not a function.
What is the domain of a function?
The domain of a function is the set of all possible input values (x-values) for which the function is defined.
Relations and Functions NCERT Solutions for Class 11 Maths Chapter 2 Free PDF Download
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions are solved in detail in the PDF given below. All the solutions to the problems in the exercises are created in such a way that it enables the students to prepare for the exam and ace it. The NCERT Solutions are prepared by the most experienced teachers in the education space, making the explanation of each solution simple, understandable, and according to the latest CBSE Syllabus. The solution helps Class 11 students to master the concept of Relations and Functions.
The solutions provide a good understanding of the fundamental concepts before they solve the equations. Through regular practice, students will know the difference between relations and functions, which are included under the syllabus, and become well-versed in its concepts. Numerous examples are present in the textbook before the exercise questions to help them understand the methodologies to be followed while solving the problems. Referring to the NCERT Class 11 Solutions PDF, students can get a glimpse of the important concepts before facing their final exams.