NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 2 Relations and Functions

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 2 - Relations and Functions aim to help students strengthen their understanding of the concepts of relations, functions, and their properties. This exercise covers various types of problems based on the key concepts learned in the chapter and serves as a great way to consolidate knowledge and apply it in different contexts.

This exercise includes problems on the domain, range, and types of functions like injective, surjective, and bijective, as well as the composition of functions and the inverse of a function. It tests the ability to work with multiple functions, their properties, and their operations in different situations.

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 2 Relations and Functions

Relations and Functions

Easy

Q.

If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

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Relations and Functions

Easy

Q.

Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

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Relations and Functions

Easy

Q.

A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

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Relations and Functions

Easy

Q.

Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

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Relations and Functions

Easy

Q.

Write the relation R = {(x, x³): x is a prime number less than 10} in roster form.

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Relations and Functions

Easy

Q.

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

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Relations and Functions

Easy

Q.

Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R

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Relations and Functions

Medium

Q.

$\begin{array}{l}\text{The relation f is defined by}\\ \text{f(x)=}\left\{\begin{array}{l}{\text{x}}^{\text{2}}\text{, 0}\le \text{x}\le \text{3}\\ \text{3x, 3}\le \text{x}\le \text{10}\end{array}\right.\\ \text{The relation g is defined by}\\ \text{g}\left(\text{x}\right)\text{=}\left\{\begin{array}{l}{\text{x}}^{\text{2}}\text{, 0}\le \text{x}\le \text{2}\\ \text{3x, 2}\le \text{x}\le \text{10}\end{array}\right.\\ \text{Show that f is a function and g is not a function}\text{.}\end{array}$

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Relations and Functions

Medium

Q.

$\text{Find the domain of the function f(x) = }\frac{{\text{x}}^{\text{2}}\text{+2x+1}}{{\text{x}}^{\text{2}}\text{– 8x+12}}\text{.}$

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Relations and Functions

Medium

Q.

$\text{Find the domain and the range of the real function f defined by (x) =}\sqrt{\left(\text{x – 1}\right)}\text{.}$

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Relations and Functions

Medium

Q.

$\text{Find the domain and the range of the real function f defined by f (x) = }\left|\text{x –\hspace{0.17em}1}\right|\text{.}$

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Relations and Functions

Medium

Q.

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.

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Relations and Functions

Medium

Q.

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.

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Question 1: Show that f is a function and g is not a function.

• The relation f is defined as x squared for x between 0 and 3, and 3x for x between 3 and 10.
• At x = 3, f(x) = 3 squared = 9, and f(x) = 3 * 3 = 9. Since the image is unique, f is a function.
• The relation g is defined as x squared for x between 0 and 2, and 3x for x between 2 and 10.
• At x = 2, g(x) = 2 squared = 4, and g(x) = 3 * 2 = 6.
• Because the element 2 has two different images (4 and 6), g is not a function.

Question 2: If f(x) = x squared, find (f(1.1) - f(1)) / (1.1 - 1).

• Substituting values: (1.1 squared - 1 squared) / (1.1 - 1).
• This equals (1.21 - 1) / 0.1 = 0.21 / 0.1.
• Answer: 2.1.

Question 3: Find the domain of f(x) = (x squared + 2x + 1) / (x squared - 8x + 12).

• The denominator factorizes to (x - 2)(x - 6).
• The function is defined for all real numbers except where the denominator is zero (x = 2 and x = 6).
• Answer: R - {2, 6}.

Question 4: Find the domain and range of f(x) = square root of (x - 1).

• The term inside the root must be non-negative, so x - 1 >= 0.
• Domain: [1, infinity).
• For x >= 1, the value of the function is always greater than or equal to zero.
• Range: [0, infinity).

Question 5: Find the domain and range of f(x) = |x - 1|.

• The function is defined for all real numbers.
• Domain: R.
• The absolute value function always gives non-negative results.
• Range: [0, infinity).

Question 6: Determine the range of the function f = {(x, x squared / (1 + x squared)) : x is in R}.

• By calculating values like (0,0), (1, 0.5), and (2, 0.8), we see the second elements are always between 0 and 1.
• Answer: Range = [0, 1).

Question 7: Let f(x) = x + 1 and g(x) = 2x - 3. Find f+g, f-g, and f/g.

• f + g: (x + 1) + (2x - 3) = 3x - 2.
• f - g: (x + 1) - (2x - 3) = -x + 4.
• f / g: (x + 1) / (2x - 3), where x is not equal to 3/2.

Question 8: Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be defined by f(x) = ax + b. Find a and b.

• Using (0, -1): a(0) + b = -1, so b = -1.
• Using (1, 1): a(1) + (-1) = 1, so a = 2.
• Answer: a = 2, b = -1.

Question 9: Let R be a relation from N to N defined by a = b squared. Are the following true?

• (i) (a, a) is in R for all a: False. For example, 2 is not equal to 2 squared.
• (ii) (a, b) in R implies (b, a) in R: False. (9, 3) is in R because 9 = 3 squared, but (3, 9) is not.
• (iii) (a, b) and (b, c) in R implies (a, c) in R: False. For example, (16, 4) and (4, 2) are in R, but (16, 2) is not.

Question 10: Let A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}. Is f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} a relation and a function?

• Relation: True, because f is a subset of the Cartesian product A x B.
• Function: False, because the element 2 has two different images (9 and 11).

Question 11: Let f = {(ab, a+b) : a, b are integers}. Is f a function from Z to Z?

• Answer: No.
• Reason: If we take a=2, b=6, then (12, 8) is in f. If we take a=-2, b=-6, then (12, -8) is in f. The element 12 has two different images.

Question 12: Let A = {9, 10, 11, 12, 13} and f(n) = the highest prime factor of n. Find the range.

• f(9) = 3.
• f(10) = 5.
• f(11) = 11.
• f(12) = 3.
• f(13) = 13.
• Answer: Range = {3, 5, 11, 13}.

FAQs – Class 11 Maths Miscellaneous Exercise Chapter 2 Relations and Functions

Q1. What is the focus of the Miscellaneous Exercise in Chapter 2?
The Miscellaneous Exercise in Chapter 2 focuses on reinforcing the concepts of relations, functions, and their operations, such as composition and inverse functions. It also includes a variety of problems to test your understanding of different types of relations and functions.

Q2. Why is this exercise important for exams?
This exercise is important because it combines all the concepts from the chapter in a variety of problem types. By practicing this exercise, students can improve their problem-solving skills and enhance their understanding of functions, relations, and their applications, which are crucial for future chapters like calculus.

Q3. What are the different types of functions covered in the chapter?
The chapter covers the following types of functions:

• Injective (One-to-One): Each element of the domain is mapped to a unique element in the range.

• Surjective (Onto): Every element of the range has at least one element from the domain mapping to it.

• Bijective (One-to-One Correspondence): A function that is both injective and surjective.

Q4. What is the composition of functions?
The composition of two functions

$f$

f and

$g$

g, written as

$f \circ g$

f∘g, means applying

$g$

g first and then applying

$f$

f to the result:

$$(f \circ g)(x) = f(g(x))$$

(f∘g)(x)=f(g(x))

Q5. How is the inverse of a function found?
To find the inverse of a function

$f(x)$

f(x), follow these steps:

1. Replace

   $f(x)$f(x) with

   $y$y.

2. Swap

   $x$x and

   $y$y in the equation.

3. Solve for

   $y$y.

4. Replace

   $y$y with

   $f^{-1}(x)$f−1(x), which gives the inverse function.

Q6. How can students prepare effectively for this Miscellaneous Exercise?

• Review the definitions of domain, range, and the types of functions.

• Practice composition and inversion of functions.

• Solve all the problems in the Miscellaneous Exercise, as they include a variety of concepts from the chapter.