Chapter 2: Relations and Functions

Ans: Given: 

  ${{f  =  \{ (0, 1), (2, 0), (3,  -  4), (4, 2), (5, 1)\} ,}}$

  ${{g  =  \{ (1, 0), (2, 2), (3,  -  1), (4, 4), (5, 3)\} }}{{.}}$ 

First, find the domain of \[{{f}}\] and domain of \[{{g}}{{.}}\]

Then, domain of \[{{f }}{{. g  =  \text{ Domain of  }f }} \cap {{ \text{ Domain of  }g}}{{.}}\]

  ${{\text{ Domain of  }f  =  \{ 0, 2, 3, 4, 5\} ,}}$

  ${{\text{ Domain of  }g  =  \{ 1, 2, 3, 4, 5\} }}$

$\therefore {{ \text{ Domain of  }(f }}{{. g)  =  \text{ Domain of  }f }} \cap {{ \text{ Domain of  }g}}$

  $\Rightarrow {{ \text{ Domain of  }(f }}{{. g)  =  \{ 2, 3, 4, 4\} }}{{.}}$ 

Ans: Given: 

  ${{f  =  \{ (2, 4), (5, 6), (8,  -  1), (10,  -  3)\} ,}}$

  ${{g  =  \{ (2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\} }}$ 

First, find the domain of \[{{f  -  g, f  +  g, f }}{{. g, }}\dfrac{{{f}}}{{{g}}}.\] Then, find their images by using a domain.

  ${{\text{ Domain of } f  -  g, f  +  g, f }}{{. g, }}\dfrac{{{f}}}{{{g}}}{{ (\text{ is domain of } f }} \cap {\text{ domain of g)}}{{.}}$

  $\Rightarrow {{ \{ 2, 5, 8, 10\}  }} \cap {{ \{ 2, 7, 8, 10, 11\}   =  \{ 2, 8, 10\} }}{{.}}$ 

  ${{(a)  (f  -  g)(2)  =  f(2)  -  g(2)}}$

  ${{(f  -  g)(2)  =  4  -  5}}$

  $\Rightarrow {{ (f  -  g)(2)  =   -  1}}$ 

  ${{(f  -  g)(8)  =  f(8)  -  g(8)}}$

  ${{(f  -  g)(8)  =   -  1  -  4}}$

  $\Rightarrow {{ (f  -  g)(8)  =   -  5}}$

  ${{(f  -  g)(10)  =  f(10)  -  g(10)}}$

  ${{(f  -  g)(10)  =   -  3  -  13}}$

  $\Rightarrow {{ (f  -  g)(10)  =   -  16}}$

$\therefore {{ f  -  g  =  \{ (2,  -  1), (8,  -  5), (10,  -  16)\} }}{{.}}$ 

 ${{(b) (f  +  g)(2)  =  f(2)  +  g(2)}}$

  ${{(f  +  g)(2)  =  4  +  5}}$

  $\Rightarrow {{ (f  +  g)(2)  =  9}}$

  ${{(f  +  g)(8)  =  f(8)  +  g(8)}}$

  ${{(f  +  g)(8)  =   -  1  +  4}}$

  $\Rightarrow {{ (f  +  g)(8)  =  3}}$

  ${{(f  +  g)(10)  =  f(10)  +  g(10)}}$

  ${{(f  +  g)(10)  =   -  3  +  13}}$

  $\Rightarrow {{ (f  +  g)(10)  =  10}}$

$\therefore {{ f  +  g  =  \{ (2, 9), (8, 3), (10, 10)\} }}{{.}}$ 

\[{{(c) (f }}{{. g)(2)  =  f(2) }}{{. g(2)}}\]

  ${{(f }}{{. g)(2)  =  4 }} \times {{ 5}}$

  $\Rightarrow {{ (f }}{{. g)(2)  =  20}}$

  ${{(f }}{{. g)(8)  =  f(8) }}{{. g(8)}}$

  ${{(f }}{{. g)(8)  =   -  1 }} \times {{ 4}}$

  $\Rightarrow {{ (f }}{{. g)(8)  =   -  4}}$

  ${{(f }}{{. g)(10)  =  f(10) }}{{. g(10)}}$

  ${{(f }}{{. g)(10)  =   -  3 }} \times {{ 13}}$

  $\Rightarrow {{ (f }}{{. g)(10)  =   -  39}}$

$\therefore {{ f }}{{. g  =  \{ (2, 20), (8,  -  4), (10,  -  39)\} }}{{.}}$ 

\[(d)\left( {\dfrac{{{f}}}{{{g}}}} \right){{(2)  =  }}\dfrac{{{{f(2)}}}}{{{{g(2)}}}}\]

  $\Rightarrow {{ }}\left( {\dfrac{{{f}}}{{{g}}}} \right){{(2)  =  }}\dfrac{4}{5}$

  $\left( {\dfrac{{{f}}}{{{g}}}} \right){{(8)  =  }}\dfrac{{{{f(8)}}}}{{{{g(8)}}}}$

  $\Rightarrow {{ }}\left( {\dfrac{{{f}}}{{{g}}}} \right){{(8)  =  }}\dfrac{{ - {{ 1}}}}{4}$

  $\left( {\dfrac{{{f}}}{{{g}}}} \right){{(10)  =  }}\dfrac{{{{f(10)}}}}{{{{g(10)}}}}$

  $\Rightarrow {{ }}\left( {\dfrac{{{f}}}{{{g}}}} \right){{(10)  =  }}\dfrac{{ - {{ 3}}}}{{13}}$

$\therefore {{ }}\dfrac{{{f}}}{{{g}}}{{  =  \{ (2, }}\dfrac{4}{5}{{), (8, }}\dfrac{{ - {{ 1}}}}{4}{{), (10, }}\dfrac{{ - {{ 3}}}}{{13}}{{)\} }}{{.}}$ 

Hence, the correct matches are \[{{(a) }} \to {{ (iii), (b) }} \to {{ (iv), (c) }} \to {{ (ii), (d) }} \to {{ (i)}}{{.}}\]

State True or False for the following statements given in Exercises 38 to 42:

Ans: Given:  Ordered pair \[(5,{{ 2)}}{{.}}\]

 Relation \[{{R  =  \{ (x, y) : y  =  x  -  5, x, y }} \in {{ Z\} }}{{.}}\]

The ordered pair must satisfy the relation.

${{R  =  \{ (x, y) : y  =  x  -  5, x, y }} \in {{ Z\} }}$

If ${{ x  =  5, \text{ then }}}$

${{y  =  5  -  5  =  0}}{{.}}$ 

Hence, \[(5,{{ 2)}}\] does not belong to \[{{R}}{{.}}\]

Ans: Given:  \[{{P  =  \{ 1, 2\} }}{{.}}\]

First, find the total number of elements \[{{n(P }} \times {{ P }} \times {{ P)}}{{.}}\] Then, compare.

${{P  =  \{ 1, 2\}  \text{ and } n(P)  =  2}}$

$\therefore {{ n(P }} \times {{ P }} \times {{ P)  =  8}}$

But there are 4 elements

  Therefore, False 

Ans: Given:  

  ${{A  =  \{ 1, 2, 3\} ,}}$

  ${{B  =  \{ 3, 4\} ,}}$

  ${{C  =  \{ 4, 5, 6\} }}{{.}}$ 

First, find \[{{A }} \times {{ B \text{ and } A }} \times {{ C,}}\] then find \[({{A }} \times {{ B) }} \cup {{ (A }} \times {{ C)}}{{.}}\]

  ${{A }} \times {{ B  =  \{ (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\} ,}}$

  ${{A }} \times {{ C  =  \{ (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} ,}}$

  ${{(A }} \times {{ B) }} \cup {{ (A }} \times {{ C)  =  \{ (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), }}$

  ${{(3,4), (3, 5), (3, 6)\} }}{{.}}$ 

Therefore, True.

Ans: Given: Ordered Pair \[({{x  -  2, y  +  5)  =  }}\left( { - {{ 2, }}\dfrac{1}{3}} \right).\]

Compare the elements of two ordered pairs. Then , find \[{{x \text{ and } y}}{{.}}\]  $({{x  -  2, y  +  5)  =  }}\left( { - {{ 2, }}\dfrac{1}{3}} \right)$

  $\Rightarrow {{ x  -  2  =   -  2 \text{ and } y  +  5  =  }}\dfrac{1}{3}$

  $\Rightarrow {{ x  =  0 \text{ and } y  =  }}\dfrac{{ - {{ 14}}}}{3}.$ 

Therefore, False.

Ans: Given: \[{{A }} \times {{ B  =  \{ (a, x), (a, y), (b, x), (b, y)\} }}{{.}}\] First element and second element in \[{{A }} \times {{ B}}\]  forms set \[{{A \text{ and } B}}{{.}}\]

\[{{A  = }}\] Set of first elements of ordered pair in \[{{A }} \times {{ B  =  \{ a, b\} }}{{.}}\]

\[{{B  = }}\] Set of second elements of ordered pair in \[{{A }} \times {{ B  =  \{ x, y\} }}{{.}}\]

Therefore, True.

Easy solutions for Class 11 Maths Chapter 2 Relations and Functions

NCERT Exemplar for Class 11 Maths - relations and functions free PDF is available at Vedantu. The pdf of chapter 2 - relations and functions is designed by the expert faculty of Vedantu on the latest syllabus pattern advised by the Central Board of Education. The pdf consists of solutions to the textbook and additional questions, these solutions of the Exemplar help students to solve the problems and to find the correct solution. 

The practice pattern mentioned in the Exemplar increases the efficiency of the students. The solutions in the Exemplar are designed in such a way that students understand the step-by-step representation of the answer which helps them to understand the concepts easily. The Exemplar has every question related to every topic in the chapter, students can easily learn and revise the concepts. 

About Vedantu

Class 11 Maths Chapter - 2 Relations and Functions NCERT Exemplar solutions available at Vedantu explain the concepts that are related to relations and functions. The concept of relation is defined as the set of output as well as inputs. The mapping diagram and relation of a graph both are used for the representation of the concept of relation. The best way of visual representation of the relationship between the inputs and the outputs is the graph of the relationship. Whereas the relationship between the single input and output is shown by Functions. The topics that are mentioned below are used as a reference tool for the students while preparing for the exams. 

Relationship between sets, Identify function, Cartesian product of sets, Functions on sets, Rational function, Polynomial function, Signum function, The modulus function, Greatest integer function, Algebra of Functions, Subtraction of other function from a real function, the addition of two real functions, multiplication of the two real functions, multiplication by a scalar, the quotient of two real functions, graphical representation of functions, real function algebraic real function.  

Vedantu also offers online study material for the students of chapter 2 - relations and functions. Students can get notes, additional questions, sample papers, and last year’s paper on the respective topic. These notes and additional study material will give an additional point of view to the students about those challenging questions that they have not prepared for. Students can also ask their doubts through live online sessions. This will give an upper hand to the student so that they can score enhanced marks. 

1. Let A = {-1, 2, 3} and B = (1, 3}. Determine

(i) A×B

(ii) B×A

(iii) B×B

(iv) A×A

Given that: A = {-1, 2, 3} and B = {1, 3}

(i) A × B = {-1, 2, 3} × {1, 3} = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)}

(ii) B × A = {1, 3} × {-1, 2, 3} = {(1, -1), (3, -1), (1, 2), (3, 2), (1, 3), (3, 3)}

(iii) B × B = {1, 3} × {1, 3} = {(1, 1), (1, 3), (3, 1), (3, 3)}

(iv) A × A = {-1, 2, 3} × {-1, 2, 3} = {{-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)}

2. Given R = {(x, y) : x, y ∈ W, x² + y² = 25}, find the domain and range of R.

Given that: R = {(x, y) : x, y ∈ W, x² + y² = 25}

So, the ordered pairs satisfying the given condition x² + y² = 25 are (0, 5), (3, 4), (5, 0), (4, 3)

Hence, the domain = {0, 3, 4, 5} and the range = {0, 3, 4, 5}.

As you can see, how easily the subject matter experts have solved the exercise problems included in NCERT Class 11 Exemplar. Download this solution file today from Vedantu for this chapter and focus on learning how to approach these problems without any hassle. Complete learning all the concepts of this chapter at ease and utilize your foundation to score more in the board exams.