# NCERT Solutions for Class 11 Maths Chapter 2

NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1 or Exercise 2.2 or Exercise 2.3 or Miscellaneous of Relations and Functions or  प्रश्नावली 2.1 or प्रश्नावली 2.2 or प्रश्नावली 2.3 or विविध प्रश्नावली 2 or संबंध एवं फलन to study online or download free in PDF form along with Offline Apps.

 Class: 11 Subject: Maths – गणित Chapter 2: Relations and Functions

## NCERT Solutions for Class 11 Maths Chapter 2

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### Class 11 Maths Chapter 2 Relations and Functions Solutions

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##### 11 Maths Chapter 2 Exercise 2.1 Sols

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.1 is given below. For other questions, please visit to Exercise 2.2 or Exercise 2.3 or Miscellaneous or go for हिंदी मीडियम Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 2 Exercise 2.2 Sols

Class 11 Maths Chapter 2  Exercise 2.2 is given below. For other questions, please visit to Exercise 2.1 or Exercise 2.3 or Miscellaneous or go for हिंदी मीडियम Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 2 Exercise 2.3 Sols

11 Maths Chapter 2 Exercise 2.3 solutions are given below. For other questions, please visit to Exercise 2.1 or Exercise 2.2 or Miscellaneous or go for हिंदी मीडियम Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 2 Miscellaneous Exercise 2 Sols

NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercises are given below. For other questions, please visit to Exercise 2.1 or Exercise 2.2 or Exercise 2.3 or go for हिंदी मीडियम Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित अध्याय 2 प्रश्नावली 2.1 के हल

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions प्रश्नावली 2.1 in Hindi is given below. For other questions, please visit to प्रश्नावली 2.2 or प्रश्नावली 2.3 or विविध प्रश्नावली 2 or go for English Medium Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित अध्याय 2 प्रश्नावली 2.2 के हल

Class 11 Maths Chapter 2 प्रश्नावली 2.2 in Hindi is given below. For other questions, please visit to प्रश्नावली 2.1 or प्रश्नावली 2.3 or विविध प्रश्नावली 2 or go for English Medium Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित अध्याय 2 प्रश्नावली 2.3 के हल

NCERT Solutions for Class 11 Maths Chapter 2 प्रश्नावली 2.3 in Hindi is given below. For other questions, please visit to प्रश्नावली 2.1 or प्रश्नावली 2.2 or विविध प्रश्नावली 2 or go for English Medium Solutions.  Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित अध्याय 2 विविध प्रश्नावली 2 के हल

Solutions for Class 11 Maths Chapter 2 Relations and Functions विविध प्रश्नावली in Hindi is given below. For other questions, please visit to प्रश्नावली 2.1 or प्रश्नावली 2.2 or प्रश्नावली 2.3 or go for English Medium Solutions.  Visit to Class 11 Maths main page or Top of the page.

Visit to Class 11 Maths main page or Top of the page

#### Important Terms on Relations and Functions

• Relation R from a non-empty set A to a non-empty set B is a subset of A × B.
• If n(A) = p, n(B) = q then n(A × B) = pq and number of relations = 2^pq.
• 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.
• 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.

##### Extra Important Questions on Relations and Functions
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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#### 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}

## 4 thoughts on “NCERT Solutions for Class 11 Maths Chapter 2”

1. SUNIL KUMAR SINGH says:

PLEASE PROVIDE HINDI MEDIUM SOLUTIONS FOR CLASS 11 AND 12……FOR VILLAGE BOYS…….THANKS SIR/MADAM

2. PLEASE PROVIDE HINDI MEDIUM SOLUTIONS FOR CLASS 11 AND 12……FOR VILLAGE BOYS

3. Krishna Tiwari says:

Sir,
Aap hindi meadium ka 11th ncert solution jald se jald upload kar dijiye ..
Is google par kewal aap hi hai jo savi solution bhej sakte hai

Thank you sir
Aap ka krishna tiwari

4. vineet kumar says:

sir hindi me samjhao please