# 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 PDF

#### Class 11 Maths Chapter 2 Exercise 2.1 Solution in Videos

Class 11 Maths Chapter 2 Exercise 2.1 Solution
Class 11 Maths Exercise 2.1 Solution in Hindi

#### Class 11 Maths Exercise 2.1 and 2.2 Solution in Hindi Videos

Class 11 Maths Chapter 2 Exercise 2.2 Solution
Class 11 Maths Exercise 2.2 Solution in Hindi

#### Class 11 Maths Exercise 2.3 and Miscellaneous Solution in Hindi

Class 11 Maths Chapter 2 Exercise 2.3 Solution
Class 11 Maths Exercise 2.3 Solution in Hindi

#### Class 11 Maths Chapter 2 Miscellaneous Solution in Videos

Class 11 Maths Chapter 2 Miscellaneous Solution
Class 11 Maths Miscellaneous 2 Solution in Hindi

#### 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.

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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}                    