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 or प्रश्नावली 2.1 or प्रश्नावली 2.2

2 or प्रश्नावली 2.3 or विविध प्रश्नावली 2 for session 2020-21. 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 form along with Offline Apps.




NCERT Solutions for Class 11 Maths Chapter 2

Class:11
Subject:Maths – गणित
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.






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.



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.




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Important Questions on 11th Maths Chapter 2

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Number of elements in A = 3 and set B = {3,4,5},
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.
यदि G = {7, 8} और H = {5, 4, 2} तो G × H और H × G ज्ञात कीजिए।
G = {7, 8} और
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)}

If A = {-1, 1}, find A × A × A.
A = {-1,1},
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)}
If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
A × B = {(a,x),(a,y),(b,x),(b,y)}

⇒A = {a,b} and B={x,y}

मान लीजिए कि A={1, 2} और B={3, 4}. A×B लिखिए। A×B के कितने उपसमुच्चय होंगे? उनकी सूची बनाइए।
यहाँ A={1,2} और B={3,4} हैं।
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)}
मान लीजिए कि A और B दो समुच्चय हैं, जहाँ n(A)=3 और n(B)=2. यदि (x, 1), (y, 2), (z, 1), A × B में हैं, तो A और B को ज्ञात कीजिए, जहाँ x, y और z भिन्न –भिन्न अवयव हैं।
यदि (x, 1), (y, 2), (z,1), A×B में हैं,

जहाँ n(A)=3 और n(B)=2 है,
तो A = {x, y, z} और B = {1, 2} हैं।

Let A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, co-domain and range.
The relation R from A to A is given as R = {(x, y): 3x – y = 0, where x, y ∈ A}
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}
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.
R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}
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}
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.
A = {1, 2, 3, 5} and B = {4, 6, 9}
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) = 2 – 3x, x ∈ R, x > 0 का परिसर ज्ञात कीजिए
दिया है: f(x) = 2 – 3x, x ∈ R, x > 0

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।

f(x)= √((x -1) ) द्वारा परिभाषित वास्तविक फलन f का प्रांत तथा परिसर ज्ञात।
दिया है: f(x)= √((x-1) ),
√((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।
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
∴ 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}
Write the relation R = {(x, x^3): x is a prime number less than 10} in roster form.
R = {(x, x^3): x is a prime number less than 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)}
f(x)= |x-1| द्वारा परिभाषित वास्तविक फलन f का प्रांत तथा परिसर ज्ञात कीजिए।
दिया है: f(x)= |x-1|,
|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।
मान लीजिए कि f= {(1, 1), (2, 3), (0, -1), (-1, -3)} Z से Z में f(x) = ax + b, द्वारा परिभाषित एक फलन है,जहाँ a, b कोई पूर्णांक हैं। a, b को निर्धारित कीजिए।
दिया है: f= {(1,1),(2,3),(0,-1),(-1,-3)} Z से Z में f(x)=ax+b,द्वारा परिभाषित एक फलन है, जहाँ a,b कोई पूर्णांक हैं।
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।

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
It is given that A = {x, y, z} and B = {1, 2}.
∴ 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.
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.
R = {(a, b): a, b ∈ Z, a – b is an integer}
It is known that the difference between any two integers is always an integer.
∴ Domain of R = Z
Range of R = Z
मान लीजिए कि f , f = {(ab, a + b) : a, b ∈ Z} } द्वारा परिभाषित Z × Z का एक उपसमुच्चय है। क्या f, Z से Z में एक फलन है? अपने उत्तर का औचित्य भी स्पष्ट कीजिए।
हम जानते हैं कि, यदि संबंध के प्रत्येक अवयव के प्रतिबिंब अद्वितीय हों तो संबंध एक फलन होता है।
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।

मान लीजिए कि A = {9, 10, 11, 12, 13} तथा f : A→N, f (n) = n का महत्तम अभाज्य गुणक द्वारा, परिभाषित है। f का परिसर ज्ञात कीजिए।
दिया है: A = {9,10,11,12,13} तथा f : A→N, f (n) = n का महत्तम अभाज्य गुणक द्वारा, परिभाषित 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}

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.1
11 maths ex. 2.1
Class 11 Maths Chapter 2 Exercise 2.2
11 maths ex. 2.2
11 Maths Chapter 2 Exercise 2.3 solutions
11 maths ex. 2.3
NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercises
11 maths ch. 2 miscellaneous ex in english
NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercises in english medium pdf for up board
NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercises for intermediate first year
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions प्रश्नावली 2.1 in Hindi
11 maths chpater 2 exercise 2.1 in hindi
Class 11 Maths Chapter 2 प्रश्नावली 2.2 in Hindi
ex. 2.2 class 11 maths
NCERT Solutions for Class 11 Maths Chapter 2 प्रश्नावली 2.3 in Hindi
ex.2.3 maths class 11
Solutions for Class 11 Maths Chapter 2 Relations and Functions विविध प्रश्नावली
11 maths misc. ex. 2
chapter 2 class 11 maths
chapter 2 intermediate maths