NCERT Solutions for Class 11 Maths Chapter 2
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.1, Exercise 2.2, Exercise 2.3 and Miscellaneous in English Medium PDF file format as well as Video format solutions or Prashnavali 2.1
1 or Prashnavali 2.2 or Prashnavali 2.3 or Vividh Prashnavali 2 in Hindi Medium free for session 2020-2021. UP Board students are also using NCERT Textbooks, so they also can download UP Board Solution for Class 11 Maths Chapter 2 in Hindi Medium free. NCERT Solutions 2020-2021 and Offline Apps are updated according to NCERT Books 2020-21 following the latest CBSE Syllabus 2020-2021. All the contents are free to study online or download free in PDF format along with Offline Apps.
NCERT Solutions for Class 11 Maths Chapter 2
Class: 11 | Maths (English and Hindi Medium) |
Chapter 2: | Relations and Functions |
11th Maths Chapter 2 Solutions
NCERT Solutions for Class 11 Maths Chapter 2 are given below updated for 2020-21. All NCERT Solutions 2020-2021 are done properly by experts removing errors, if still any mistake, please inform us. We will try to remove as soon as possible. These are appropriate for CBSE Board, Uttarakhand Board, UP Board, MP Board, Gujrat Board and all other board who are following NCERT Books based on New Updated CBSE Curriculum 2020-21.
11th Maths Chapter 2 Solutions in English Medium
11th Maths Chapter 2 Solutions in Hindi Medium
11th Maths Chapter 2 Solutions in PDF
Class 11 Maths Chapter 2 Exercise 2.1 Solution in Videos
Class 11 Maths Exercise 2.1 and 2.2 Solution in Hindi Videos
Class 11 Maths Exercise 2.3 and Miscellaneous Solution in Hindi
Class 11 Maths Chapter 2 Miscellaneous Solution in Videos
Important Terms on Relations and Functions
1. Relation R from a non-empty set A to a non-empty set B is a subset of A × B.
2. If n(A) = p, n(B) = q then n(A × B) = pq and number of relations = 2^pq.
3. A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.
4. Let A and B be two non-empty finite sets such that n(A) = p and n(B) = q then number of functions from A to B = q^p.
Class 11 Maths Chapter 2 Important Questions for Practice
1. If A and B are finite sets such that n(A) = 5 and n(B) = 7, then find the number of functions from A to B.
2. If A = {2, 4, 6, 9} B = {4, 6, 18, 27, 54} and a relation R from A to B is defined by R = {(a, b): a belongs to A, b belongs to B, a is a factor of b and a < b}, then find in Roster form. Also find its domain and range. 3. Determine a quadratic function (f) is defined by f(x) = ax² + bx + c. If f(0) = 6; f(2) = 11, f(–3) = 6.
Feedback & Suggestions
If Users are facing any problem regarding any thing based on education, please feel free to contact us. You may opt ENGLISH or HINDI language for communication. We are here to help you, if you don’t want to call us, just Contact Us by sending a Text Message or Whats App or Mail for communication. Always give feedback or suggestions to improve the contents of this website for academic session 2020-21.
Important Questions on 11th Maths Chapter 2
Number of elements in B = 3
Number of elements in A×B=(Number of elements in A)×(Number of elements in B )
= 3 × 3
= 9
Hence,the number of elements in A × B is 9.
H = {5, 4, 2},
इसलिए,
G × H
= {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
तथा
H × G
= {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}
Therefore,
A×A
={-1,1}×{-1,1}={(-1,-1),(-1,1),(1,-1),(1,1)}
and
A×A×A
={(-1,-1),(-1,1),(1,-1),(1,1)} × {-1,1}
={(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}
⇒A = {a,b} and B={x,y}
इसलिए,
A × B={(1,3),(1,4),(2,3),(2,4)}
A×B में अवयवों की संख्या = 4
इसलिए,
उपसमुच्चयों की संख्या = 2^4=16
A×B के उपसमुच्चय
= ϕ,{(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},{(1,3),(2,3)},
{(1,3),(2,4)},{(1,4),(2,3)},{(1,4),(2,4)},{(2,3),(2,4)},{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},
{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)},{(1,3),(1,4),(2,3),(2,4)}
जहाँ n(A)=3 और n(B)=2 है,
तो A = {x, y, z} और B = {1, 2} हैं।
i.e., R = {(x, y): 3x = y, where x, y ∈ A}
∴ The roster form is given by R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3, 4}
The whole set A is the co-domain of the relation R.
∴ Co-domain of R = A = {1, 2, 3… 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {3, 6, 9, 12}
The natural numbers less than 4 are 1, 2, and 3.
∴ R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {6, 7, 8}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
∴ The roster form of R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
f(x) का मान अधिकतम होगा यदि 2-3x का मान अधिकतम (2) हो अर्थात x का मान न्यूनतम (x=0) हो।
परन्तु , दिया है x>0, अतः फलन f का परिसर (-∞, 2) है।
√((x-1) ) सभी x≥1 वास्तविक संख्याओं के लिए परिभाषित है।
अतः, f, का प्रांत [1,∞) है।
यदि x≥1 तो x-1≥0, इसलिए, f(x)= √((x-1) ) ≥ 0 है।
अतः, f, का परिसर [0, ∞) है।
∴ The roster form R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
∴ Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}
|x-1| सभी वास्तविक संख्याओं के लिए परिभाषित है। अतः, f, का प्रांत वास्तविक संख्याओं का समुच्चय R है।
यदि x एक वास्तविक संख्या है तो |x-1|, सदैव वास्तविक तथा धनात्मक होगा।
अतः, f, का परिसर सभी धनात्मक वास्तविक संख्याओं का समुच्चय R+ है।
यहाँ, (1,1)∈f
⇒f(1)=1
⇒1=a(1)+b
⇒b=1-a … (1)
इसीप्रकार,
(2,3)∈f
⇒f(2)=3
⇒3=a(2)+b
⇒2a+b=3
समीकरण (1) से b का मान रखने पर
2a+(1-a)=3
⇒2a+1-a=3
⇒a=2
समीकरण (1) में a का मान रखने पर
b=1-a=1-2=-1
अतः, a=2 तथा b=-1 है।
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A × B) = 6, the number of subsets of A × B is 26.
Therefore, the number of relations from A to B is 26.
It is known that the difference between any two integers is always an integer.
∴ Domain of R = Z
Range of R = Z
दिया है: f = {(ab, a + b) : a, b ∈ Z} + b) : a, b ∈ Z}
यदि a = 2 ∈ Z और b = 1 ∈ Z हो तो
(2×1,2+1) ∈ f
⇒(2,3) ∈ f
यदि a = -2 ∈ Z और b = -1 ∈ Z हो तो
((-2)×(-1),-2-1) ∈ f
⇒(2,-3) ∈ f
यहाँ, एक ही अवयव 2, दो भिन्न-भिन्न प्रतिबिंबों 3 और -3 से संबंधित है, इसलिए यह संबंध फलन नहीं है।
9 का अभाज्य गुणक = 3
10 का महत्तम अभाज्य गुणक = 2, 5
11 का अभाज्य गुणक = 11
12 का अभाज्य गुणक = 2, 3
13 का अभाज्य गुणक = 13
∴ f(9) = 9 का महत्तम अभाज्य गुणक = 3
f(10) = 10 का महत्तम अभाज्य गुणक = 5
f(11) = 11 का महत्तम अभाज्य गुणक = 11
f(12) = 12 का महत्तम अभाज्य गुणक = 3
f(13) = 13 का महत्तम अभाज्य गुणक = 13
∴ f का परिसर = {3, 5, 11, 13}