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NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions)

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NCERT Exemplar for Class 11 Maths - Relations and Functions - Free PDF Download

Free PDF download of NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2 - Relations and Functions exercise questions with solutions will help you to revise the complete syllabus and score more marks in your examinations.

Competitive Exams after 12th Science

Access NCERT Exemplar Solutions for Class 11 Mathematics Chapter 2: Relations and Functions (Examples, Easy Methods and Step by Step Solutions)

Solved Examples:

1.Let $A  =  \{ 1, 2, 3, 4\}$ and  $B  =  \{ 5, 7, 9\} .$ Determine

(i). $A  \times  B$

Ans: Given: $A  =  \{ 1, 2, 3, 4\}$  and $B  =  \{ 5, 7, 9\}$ 

Use the Cartesian product.

${{A }} \times {{ B  =  \{ (1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)\} }}{{.}}$

(ii). $B  \times  A$

Ans: Given: $A  =  \{ 1, 2, 3, 4\}$  and $B  =  \{ 5, 7, 9\}$

Use the Cartesian product.

${{B }} \times {{ A  =  \{ (5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)\} }}{{.}}$

(iii). Is $ A  \times  B  =  B  \times  A ?$

Ans: Given: $A  =  \{ 1, 2, 3, 4\}$  and $B  =  \{ 5, 7, 9\}$

Use the Cartesian product.v

${{A }} \times {{ B }} \ne {{ B }} \times {{ A}}$ as the ordered pairs are different.

(iv). Is $n(A  \times  B)  =  n(B  \times  A) ?$

Ans: Given: $A  =  \{ 1, 2, 3, 4\}$  and $B  =  \{ 5, 7, 9\}$

Use the Cartesian product.

Here,

  ${{n(A }} \times {{ B)  =  n(B }} \times {{ A) }}$

  $\Rightarrow {{ n(A) }}{{. n(B)  =  4 }} \times {{ 3}}$

  $\Rightarrow {{ n(A) }}{{. n(B)  =  12}}{{.}}$ 


2.Find x and y, if

\[(i)(4x  +  3, y)  =  (3x  +  5,  -  2)\]

Ans: Given: Ordered pairs.

Elements of equal ordered pairs are equal.

Here,

  $(4{{x  +  3, y)  =  (3x  +  5,  -  2)}}$

  $\Rightarrow 4x  +  3  =  3x  +  5$ and $y  =   -  2$

  $\Rightarrow  x  =  2$ and $y  =   -  2$ 

\[(ii)(x  -  y, x  +  y)  =  (6, 10)\]

Ans: Given: Ordered pairs.

Elements of equal ordered pairs are equal.

Here,

  $({{x  -  y, x  +  y)  =  (6, 10)}}$

  $\Rightarrow {{ x  -  y  =  6 ………………...(i)}}$

  $\Rightarrow {{ x  +  y  =  10 ……………..(ii)}}$

  From (i) and (ii), we get

  $x  =  8$ and $y =  2$


3. If \[A{ }{ }\{2,4,6,9\} \text{ and } B \{4,6,18,27,54\},a\in A,b\in B,\] find the set of ordered pairs such that 'a' is factor of 'b' and a  < , b.

Ans: Given:

  ${{A  =  \{ 2, 4, 6, 9\} , a }} \in {{ A,}}$

  ${{B  =  \{ 4, 6, 18, 27, 54\} , b }} \in {{ B}}$ 

Form ordered pairs such that b' is divisible by 'a'

Required set of ordered pair will be \[\{ (2,{{ 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)\} }}{{.}}\]


4. Find the domain and range of the relation \[{{R}}\] given by $R  =  \{ (x, y) : y  =  x  +  \dfrac{6}{x};$ where $x, y \in  N \text{ and } x  <  6\} .$

Ans: Given: A relation \[{{R}}\].

Domain and range are values of x and y for which relation is defined.

  R is defined only for ${{x  =  \{ 1, 2, 3\}  , y }} \in {{ N}}$

  $\therefore$ Domain of $R  =  \{ 1, 2, 3\}$

  for, ${{x  =  1, y  =  7,}}$

  ${{x  =  2, y  =  5,}}$

  ${{x  =  3, y  =  5}}{{.}}$

  $\therefore$  Range of $R  =  \{ 7, 5\}$.


5.Is the following relation a function? Justify your answer.

\[(i).{R_1}  =  \left\{ {(2, 3), \left( {\dfrac{1}{2}, 0} \right), (2, 7), ( -  4, 6)} \right\}\]

Ans: Given: Relation \[{{{R}}_1}{{ \text{ and } }}{{{R}}_2}.\]

A relation is function only when there is only one image in Range.

Here,

  Since, (2, 3) and ${{(2, 7) }} \in {{ }}{{{R}}_1}$

  $\Rightarrow {{{R}}_1}$ is not a function 

\[(ii).{R_2}  =  \{ (x, | x |) | \text{ x is a real number}\} \]

Ans: Given: Relation \[{{{R}}_1}{{ \text{ and } }}{{{R}}_2}.\]

A relation is function only when there is only one image in Range.

Here,

for all \[x \in {{ R,}}\] There is a unique image.

Therefore, \[{{{R}}_2}\] is a function..


6.Find the domain for which the functions \[f(x)  =  2{x^2}  -  1 \text{ and } g(x)  =  1  -  3x\] are equal.

Ans: Given: \[{{f(x)  =  2}}{{{x}}^2}{{  -  1 \text{ and } g(x)  =  1  -  3x}}{{.}}\]

Find \[{{x, \text{for } f(x)  =  g(x)}}{{.}}\]

  ${{\text{for } f(x)  =  g(x)}}$

  $\Rightarrow {{ 2}}{{{x}}^2}{{  -  1  =  1  -  3x}}$

  $\Rightarrow {{ 2}}{{{x}}^2}{{  +  3x  -  2  =  0}}$ 

  $\Rightarrow {{ 2}}{{{x}}^2}{{  +  4x  -  x  -  2  =  0}}$

  $\Rightarrow {{ 2x(x  +  2)  -  1(x  +  2)  =  0}}$

  $\Rightarrow {{ (2x  -  1)(x  +  2)  =  0}}$ 

  $\Rightarrow {{ x  =  }}\dfrac{1}{2},{{  -  2}}{{.}}$

$\therefore$  Domain for $f(x)  =  g(x)$ is $\left\{ {\dfrac{1}{2},{{  -  2}}} \right\}.$ 


7.Find the domain of each of the following functions:

\[(i).f(x)  =  \dfrac{x}{{{x^2}  +  3x  +  2}}\]

Ans: Domain is the value of \[{{x}}\] for which \[{{f(x)}}\] is defined.

Here,

  ${{f(x)}}$ is defined, if

  ${{{x}}^2}{{  +  3x  +  2 }} \ne {{ 0}}$

  $\Rightarrow {{ (x  +  1)(x  +  2) }} \ne {{ 0}}$

  $\Rightarrow {{ x }} \ne {{  -  1 ,  -  2}}$

$\therefore$  Domain of $f(x)  =  R  -  \{  -  1,  -  2\}$ 

\[(ii).f(x)  =  [x]  +  x\]

Ans: Domain is the value of \[{{x}}\] for which \[{{f(x)}}\] is defined.

Here,

  $f(x)$ is defined for all $x  \in { R}$

$\therefore$ Domain of $f(x)  =  R$ 


8.Find the range of the following function:

\[(i).f(x)  =  \dfrac{{|x  -  4 |}}{{x  -  4}}\]

Ans: Write \[{{x { } \text{ in form of }{ } y}}{{.}}\] Then, find the range.

Here,

  ${{f(x)  =  }}\dfrac{{|{{x  -  4 |}}}}{{{{x  -  4}}}}$

  $\Rightarrow {{ f(x)  =  }}$

  $\dfrac{{{{x  -  4 }}}}{{{{x  -  4}}}}{{  =  1, x  >  4}}$

  $\dfrac{{ - {{ }}({{x  -  4)}}}}{{{{x  -  4}}}}{{  =   -  1, x  <  4}}$ 

$\therefore$  Range of ${{f(x)  =  \{  -  1, 1\} }}{{.}}$ 

\[(ii).f(x)  =  \sqrt {16  -  {x^2}} \]

Ans: Write \[{{x { } \text{ in form of }{ } y}}{{.}}\] Then, find the range.

Here,

  ${{Let y  =  f(x),}}$

  $\Rightarrow {{ y  =  }}\sqrt {16{{  -  }}{{{x}}^2}}$

  $\Rightarrow {{ }}{{{x}}^2}{{  =  16  -  }}{{{y}}^2}$

  ${{\text{Here}, x }} \in {{ [ -  4, 4]}}$

$\therefore$ Range of $f(x)  =  [0, 4]$ 


9.Redefine the function which is given by $ f(x)  =  | x  -  1 |  +  | 1  +  x | ,  -  2 \in x \in 2$

Ans: Given:  function \[{{f(x)  =  | x  -  1 |  +  | 1  +  x | ,  -  2 }} \leqslant {{ x }} \leqslant {{ 2}}{{.}}\]

Split the function over the given intervals.

  ${{f(x)  =  }}$

   $- {{ x  +  1  -  1  -  x,   -  2 }} \leqslant {{ x }} \leqslant {{ 2}}$

  ${{ -  x  +  1  +  x  +  1,  -  1 }} \leqslant {{ x  <  1}}$

  ${{x  -  1  +  1  +  x, 1 }} \leqslant {{ x }} \leqslant {{ 2}}$ 

  ${{f(x)  =  }}$

   $- {{ 2x,   -  2 }} \leqslant {{ x  <   -  1}}$

  ${{2,      -  1 }} \leqslant {{ x  <  1}}$

  ${{2x,  1 }} \leqslant {{ x }} \leqslant {{ 2}}$ 


10.Find the domain of the function \[{{f}}\] given by \[f(x)  =  \dfrac{1}{{\sqrt {{{[x]}^2}  -  [x]  -  6} }}.\]

Ans: Given: Function \[{{f(x)}}{{.}}\]

find the value of \[{{x}}\] for which \[{{f(x)}}\] is defined.

  ${{f(x) \text{ is defined , if } { } [x}}{{{]}}^2}{{  -  [x]  -   6   >  0}}{{.}}$

  $\Rightarrow {{ ([x]  -  3)([x]  +  2)  >  0,}}$

  $\Rightarrow {{ [x]  <   -  2 \text{or} [x]  >  3}}$

  $\Rightarrow {{ x  <   -  2 \text{ or } x }} \geqslant {{ 4}}$

$\therefore$ Domain  ${{=  ( -  }}\infty {{,  -  2) }} \cup {{ [4, }}\infty ).$ 


Objective Type Question

11.The domain of the function \[{{f}}\] given by \[f(x)  =  \dfrac{1}{{\sqrt {x  -  | x |} }}.\]

(A).\[R\]

(B).\[{R^ + }\]

(C).\[{R^ - }\]

(D).None of these.

Ans: Given: Function \[{{f(x)}}{{.}}\]

Find the value of \[{{x}}\] for which the function \[{{f(x)}}\] is defined.

${{x  -  | x |  =  }}$

  ${{x  -  x  =  0, \text{ if x } }} \geqslant {{ 0}}$

  $2{{x ,         \text{ if } x  <  0}}$ 

  Here, f(x) is not defined for any ${{x }} \in {{ R}}{{.}}$

$\therefore$ Domain is none

Correct Answer: D


12.If \[f(x)  =  {x^3}  -  \dfrac{1}{{{x^3}}}, \text{ then } f(x)  +  f\left( {\dfrac{1}{x}} \right)\] equals to:

\[(A). 2{x^3}\]

\[(B).2\dfrac{1}{{{x^3}}}\]

\[(C).0\]

\[(D).1\]

Ans: Given: Function \[{{f(x)}}{{.}}\]

Replace the value of \[{{x \text{ in } f(x)}}{{.}}\] Then, find the sum.

  ${{\text{Here}, f(x)  =  }}{{{x}}^3}{{  -  }}\dfrac{1}{{{{{x}}^3}}}$

  $\Rightarrow {{ f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{1}{{{{{x}}^3}}}{{  -  }}\dfrac{1}{{\dfrac{1}{{{{{x}}^3}}}}}$

  $\Rightarrow {{ f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{1}{{{{{x}}^3}}}{{  -  }}{{{x}}^3}$

$\therefore {{ f(x)  +  f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}{{{x}}^3}{{  -  }}\dfrac{1}{{{{{x}}^3}}}{{  +  }}\dfrac{1}{{{{{x}}^3}}}{{  -  }}{{{x}}^3}$

  $\Rightarrow \;{{f(x)  +  f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  0}}{{.}}$ 

Correct Answer: C


13. Let A and B be any two sets such that \[n(B)  =  p, n(A)  =  q\] then the total number of functions  is equal to ________

Ans: Given: A and B be two sets and,

\[{{n(B)  =  p, n(A)  =  q}}{{.}}\]

Total number of function from \[{{A \to B \text{ is } [n(A)}}{{{]}}^{{{n(B)}}}}.\]

Every Element of the set \[{{A}}\] is related to elements of the set \[{{B}}\] in \[{{p}}\] ways. Therefore, the total number of functions  \[{{f : A }} \to {{ B}}\] is equal to \[{{{p}}^{{q}}}.\]


14.Let f and g be two functions given by

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

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

Then domain of \[{{f  +  g}}\] is _______

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 \[{{f  \text{ and } g}}{{.}}\] Then find the domain of \[{{f  +  g}}\].

  Domain of ${{f  =  \{ 2, 5, 8, 10\} ,}}$

  Domain of ${{g  =  \{ 2, 7, 8, 10, 11\} }}$

  Domain of $f  +  g$  =  (domain of $f$  $\cap$ domain of $g$ )

$\therefore$ Domain of ${{f  +  g  =  \{ 2, 8, 10\} }}{{.}}$ 


EXERCISE QUESTION:

1.If \[A  =  \{  - 1, 2, 3\}  \text{ and } B  =  \{ 1, 3\}\], then determine,

(i). \[A  \times  B\]

Ans: Given: sets \[{{A  =  \{  - 1, 2, 3\}  \text{ and } B  =  \{ 1, 3\} }}{{.}}\]

The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B

\[{{A }} \times {{ B  =  \{ ( -  1, 1), ( -  1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\} }}\]

(ii). \[B  \times  A\]

Ans: Given: sets \[{{A  =  \{  - 1, 2, 3\}  \text{ and } B  =  \{ 1, 3\} }}{{.}}\]

The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B

\[{{B }} \times {{ A  =  \{ (1,  - 1), (1, 2), (1, 3), (3,  - 1), (3, 2), (3, 3)\} }}\]

(iii). \[B  \times  B\]

Ans: Given: sets \[{{A  =  \{  - 1, 2, 3\}  \text{ and } B  =  \{ 1, 3\} }}{{.}}\]

The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B

\[{{B }} \times {{ B  =  \{ (1, 1), (1, 3), (3, 1), (3, 3)\} }}\]

(iv). \[A  \times  A\]

Ans: Given: sets \[{{A  =  \{  - 1, 2, 3\}  \text{ and } B  =  \{ 1, 3\} }}{{.}}\]

The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B

\[{{A }} \times {{ A  =  \{ ( -  1,  - 1), ( -  1, 2), ( -  1, 3), (2,  - 1), (2, 2), (2, 3), (3,  - 1), (3, 2), (3, 3)\} }}{{.}}\]


2.If \[P =  \{ x : x  <  3, x \in  N\} , Q  =  \{ x : x \le \ 2, x \in  W\} ,\] then find $(P \cup Q) \times(P \cap Q)$ where \[{{W}}\] is the set of whole numbers.

Ans: Given: 

${{P  =  \{ x : x  <  3, x }} \in {{ N\} }}$

${{Q  =  \{ x : x }} \leqslant {{ 2, x }} \in {{ W\} }}$ 

Use basic operations of sets like union and intersection of sets and then apply cartesian products.From the sets, we get

$\Rightarrow {{ P  =  \{ 1, 2\} }}$

$\Rightarrow {{ Q  =  \{ 0, 1, 2\} }}$

$\Rightarrow {{ P }} \cup {{ Q  =  \{ 0, 1, 2\}  \text{ and } P }} \cap {{ Q  =  \{ 1, 2\} }}$

$\Rightarrow {{ (P }} \cup {{ Q) }} \times {{ (P }} \cap {{ Q)  =  \{ 0, 1, 2\}  }} \times {{ \{ 1, 2\} }}$

$\Rightarrow {{ (P }} \cup {{ Q) }} \times {{ (P }} \cap {{ Q)  =  \{ (0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\}  }}$ 


3.If \[A  =  \{ x : x \in  W, x  <  2\} , B  =  \{ x : x \in  N, 1  <  x  <  5\}  \text{ and } C  =  \{ 3, 5\} \] then find

\[(i). A  \times  (B \cup C)\]

Ans: Given: The sets \[{{A  =  \{ x : x }} \in {{ W, x  <  2\} , B  =  \{ x : x }} \in {{ N, 1  <  x  <  5\}  \text{ and } C  =  \{ 3, 5\} }}\].

First convert the sets in roster form then use basic set operations and Cartesian products.

The sets in roster form will be,

\[{{A  =  \{ 0, 1\} , B  =  \{ 2, 3, 4\}  \text{ and } C  =  \{ 3, 5\} }}{{.}}\]

\[{{A }} \times {{ (B }} \cap {{ C)}}\]

  $\Rightarrow {{ A }} \times {{ (B }} \cap {{ C)  =  \{ 0, 1\}  }} \times {{ \{ 3\} }}$

  $\Rightarrow {{ A }} \times {{ (B }} \cap {{ C)  =  \{ (0, 3), (1, 3)\} }}$ 

\[(ii).A  \times  (B \cup C)\]

Ans: Given: The sets \[{{A  =  \{ x : x }} \in {{ W, x  <  2\} , B  =  \{ x : x }} \in {{ N, 1  <  x  <  5\} \text{ and } C  =  \{ 3, 5\} }}\].

First convert the sets in roster form then use basic set operations and Cartesian products.

The sets in roster form will be,

\[{{A  =  \{ 0, 1\} , B  =  \{ 2, 3, 4\} \text{ and } C  =  \{ 3, 5\} }}{{.}}\]

\[{{A }} \times {{ (B }} \cup {{ C)}}\]

  $\Rightarrow {{ A }} \times {{ (B }} \cup {{ C)  =  \{ 0, 1\}  }} \times {{ \{ 2, 3, 4, 5\} }}$

  $\Rightarrow {{ A }} \times {{ (B }} \cup {{ C)  =  \{ (0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}}{{.}}$ 


4.In each of the following cases, find a and b.

\[(i).(2a  +  b, a  -  b)  =  (8, 3)\]

Ans: Since, two ordered pairs are equal, if their corresponding first and second elements are equal.

\[(2{{a  +  b, a  -  b)  =  (8, 3)}}\]

\[\Rightarrow \;2{{a  +  b  =  8 \text{ and } a  -  b  =  3}}\]

On substituting \[{{b  =  a  -  3 \text{ in } 2a  +  b  =  8, we get}}\]

  $\Rightarrow 2{{a  +  a  -  3  =  8}}$

  $\Rightarrow \;3{{a  -  3  =  8}}$

  $\Rightarrow {{ a  =  }}\dfrac{{11}}{3}{{ }}$ 

Substituting \[{{a  =  }}\dfrac{{11}}{3}{{ \text{ in } b  =  a  -  3, we get}}\]

  $\Rightarrow {{ b  =  }}\dfrac{{11}}{3}{{  -  3}}$

  $\Rightarrow {{ b  =  }}\dfrac{2}{3}$

$\therefore {{ a  =  }}\dfrac{{11}}{3}{{ \text{ and } b  =  }}\dfrac{2}{3}$ 

\[(ii).\left( {\dfrac{a}{4}, a  -  2b} \right)  =  (0, 6  +  b)\]

Ans: Since, two ordered pairs are equal, if their corresponding first and second elements are equal.

\[\left( {\dfrac{{{a}}}{4}{{, a  -  2b}}} \right){{  =  (0, 6  +  b)}}\]

  $\Rightarrow \;\dfrac{{{a}}}{4}{{  =  0 \text {and } a  -  2b  =  6  +  b}}$

  $\Rightarrow {{ a  =  0 \text{ and } 0  -  2b  =  6  +  b}}$

  $\Rightarrow {{ a  =  0 \text{ and } b  =   -  2}}$ 


5. \[A  =  \{ 1, 2, 3, 4, 5\} , S  =  \{ (x, y) : x \in A, y \in  A\} ,\] then find the orders which satisfy the conditions given below.

\[(i). x  +  y  =  5\]

Ans: Given: The sets \[{{A  =  \{ 1, 2, 3, 4, 5\}  \text{ and } S  =  \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]

Use the condition for all the elements in sets till the condition is satisfied.

The set of ordered pairs satisfying \[{{x  +  y  =  5 \text{ is }, \{ (1, 4), (2, 3), (3, 2), (4, 1)\} }}{{.}}\]

\[(ii). x  +  y  <  5\]

Ans: Given: The sets \[{{A  =  \{ 1, 2, 3, 4, 5\}  \text{ and } S  =  \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]

Use the condition for all the elements in sets till the condition is satisfied.

The set of ordered pairs satisfying \[{{x  +  y  <  5 \text{ is }, \{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\} }}{{.}}\]

\[(iii). x  +  y  >  8\]

Ans: Given: The sets \[{{A  =  \{ 1, 2, 3, 4, 5\}  \text{ and } S  =  \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]

Use the condition for all the elements in sets till the condition is satisfied.

The set of ordered pairs satisfying \[{{x  +  y  >  8 \text{ is }, \{ (4, 5), (5, 4), (5, 5)\} }}{{.}}\]


6.If \[R  =  \{ (x, y) : x, y \in  W, {x^2}  +  {y^2}  =  25\} ,\]then find the domain and  range of \[{{R}}{{.}}\]

Ans: Given: Relation \[{{R  =  \{ (x, y) : x, y }} \in {{ W, }}{{{x}}^2}{{  +  }}{{{y}}^2}{{  =  25\} }}{{.}}\]

First, write the relation in roster form, then find the domain and range of \[{{R}}{{.}}\]\[{{R  =  \{ (0, 5), (3, 4), (4, 3), (5, 0)\} }}{{.}}\]

Domain of \[{{R}}\] is set of first element of ordered pairs in \[{{R  =  \{ 0, 3, 4, 5\} }}{{.}}\]

Range of \[{{R}}\] is set of second element of ordered pairs in \[{{R  =  \{ 5, 4, 3, 0\} }}{{.}}\]


7.If \[{R_1}  =  \{ (x, y) | y  =  2x  +  7, \text{ where } x \in  R \text{ and }  -  5 \le x \le 5\} \] is a relation. Then find the domain and range of \[{R_1}.\]

Ans: Given: A relation \[{{{R}}_1}{{  =  \{ (x, y) | y  =  2x  +  7, \text{ where } x }} \in {{ R \text{ and }  -  5 }} \leqslant {{ x }} \leqslant {{ 5\} }}{{.}}\]

The value of \[{{x}}\] represents the domain and the values of \[{{y for all x}}{{.}}\]

Domain of \[{{{R}}_1}{{  =  \{  -  5 }} \leqslant {{ x }} \leqslant {{ 5, x }} \in {{ R\} ,}}\]

\[\Rightarrow {{ }}{{{R}}_1}{{  =  [ -  5, 5]}}\].

Range of \[{{{R}}_1},\]

  ${{\text{when } x  =   - 5, \text{ then } y  =  2( -  5)  +  7  =   -  3}}$

  ${{\text{when } x  =  5, \text{ then } y  =  2(5)  +  7  =  17}}$ 

Range of \[{{{R}}_1},\]

  $\Rightarrow {{ }}{{{R}}_1} = {{ \{  -  3 }} \leqslant {{ y }} \leqslant {{ 17, y }} \in {{ R\} }}$

  $\Rightarrow {{ }}{{{R}}_1}{{  =  [ -  3, 17]}}$ 


8.If \[{R_2}  =  \{ (x, y) | x \text{ and } y \text{ are integers and } {x^2}  +  {y^2}  =  64\} \] is a relation, then find the value of \[{R_{2.}}\]

Ans: Given: A relation \[{{{R}}_2}{{  =  \{ (x, y) | |\text{ x and y are integers and }}}{{{x}}^2}{{  +  }}{{{y}}^2}{{  =  64\} }}\].

Use the given condition in a relation and then write the set in roster form.

Since, \[64\] is the sum of square of \[0{{ \text{ and } }} \pm {{ 8}}{{.}}\]

  $\Rightarrow {{ \text{ when } x  =  0, \text{ then } }}{{{y}}^2}{{  =  64,}}$

  $\Rightarrow {{ y  =  }} \pm {{ 8}}$

  $\Rightarrow {{ x  =  8, \text{ then } }}{{{y}}^2}{{  =  64  -  (8}}{{{)}}^2}{{  =  0}}$

  $\Rightarrow {{ x  =   -  8, \text{ then } }}{{{y}}^2}{{  =  64  -  ( -  8}}{{{)}}^2}{{  =  0}}$

$\therefore {{ }}{{{R}}_2}{{  =  \{ (0, 8), (0,  -  8), (8, 0), ( -  8, 0)\} }}$ 


9.If \[{{{R}}_3}{{  =  \{ (x, | x |) | x\} }}\] is a real number is a relation, then find domain and range of \[{R_3}.\]

Ans: Given: A relation \[{{{R}}_3}{{  =  \{ (x, | x |) | x\} }}\] is a real number.

The value of \[{{x}}\] represents the domain and the values of \[{\text{y for all x}}{{.}}\]

Domain of \[{{{R}}_3}{{  =  \text{ real number }}}{{.}}\]

Since, the image of any real number under \[{{{R}}_3}\] is a positive real number or zero.

Range of \[{{{R}}_3}{{  =  }}{{{R}}^ + }\; \cup {{ \{ 0\}  or (0, }}\infty {{)}}{{.}}\]


10.Is the given relation a function? Give reason for your answer.

\[(i). h  =  \{ (4, 6), (3, 9), ( -  11, 6), (3, 11)\} \]

Ans: Given: Relations are given in set-builder and roster form.

A function is said to be a relation, only and only if every element in the domain has a unique image.

\[{{h  =  \{ (4, 6), (3, 9), ( -  11, 6), (3, 11)\} }}\]

Since, \[3\] has two images \[9{{ \text{ and } 11}}{{.}}\] So, it is not a function.

\[(ii). f  =  \{ (x, x) | x\} \] is a real number

Ans: Given: Relations are given in set-builder and roster form.

A function is said to be a relation, only and only if every element in the domain has a unique image.

\[{{f  =  \{ (x, x) | x\} }}\] is a real number

We observe that every element in the domain has a unique image. So, it is a function.

\[(iii). g  =  \left\{ {\left( {x, \dfrac{1}{x}} \right) | x} \right\} \] is a positive integer

Ans: Given: Relations are given in set-builder and roster form.

A function is said to be a relation, only and only if every element in the domain has a unique image.

\[{{g  =  }}\left\{ {\left( {{{x, }}\dfrac{1}{{{x}}}} \right){{ | \text{ x is a positive integer }}}} \right\}{{ }}\]

For every \[{{x}}\], it is a positive integer and \[\dfrac{1}{{{x}}}\] is unique and distinct. Therefore, every element in the domain has a unique image. So, it is a function.

\[(iv). s  =  \{ (x, {x^2})| {x}\} \] is a positive integer

Ans: Given: Relations are given in set-builder and roster form.

A function is said to be a relation, only and only if every element in the domain has a unique image.

\[{{s  =  \{ (x, }}{{{x}}^2})|{{ x\} }}\]

 is a positive integer

Since, the square of any positive integer is unique. So, every element in the domain has a unique image. Hence, it is a function.

\[(v). t  =  \{ (x, 3) | x\} \] is a real number

Ans: Given: Relations are given in set-builder and roster form.

A function is said to be a relation, only and only if every element in the domain has a unique image.

\[{{t  =  \{ (x, 3) | x\} }}\] is a real number

Since, every element in the domain has the image \[3\]. So, it is a constant function..


11.If f and g are real functions defined by \[f(x)  =  {x^2}  +  7 \text{ and } g(x)  =  3x  +  5.\] Then find each of the following.

\[(i). f(3)  +  g( -  5)\]

Ans: Given: \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

Substitute the value of \[{{x}}\] in the functions  \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

We have,

${{f(3)  =  (3}}{{{)}}^2}{{  +  7  =  16 \text{ and } g( -  5)  =  3( -  5)  +  5  =   -  10}}$

  $\Rightarrow {{ f(3)  +  g( -  5)  =  16  -  10}}$

$\therefore {{ f(3)  +  g( -  5)  =  6}}$ 

\[(ii). f\left( {\dfrac{1}{2}} \right)  \times  g(14)\]

Ans: Given: \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

Substitute the value of \[{{x}}\] in the functions  \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

We have,

  ${{f}}\left( {\dfrac{1}{2}} \right){{  =  }}{\left( {\dfrac{1}{2}} \right)^2}{{  +  7  =  }}\dfrac{{29}}{4}$

  ${{g(14)  =  3(14)  +  5  =  47}}$

$\therefore {{ f}}\left( {\dfrac{1}{2}} \right){{ }} \times {{ g(14)  =  }}\dfrac{{29}}{{14}}{{ }} \times {{ 47  =  }}\dfrac{{1363}}{4}$ 

\[(iii). f( -  2)  +  g( -  1)\]

Ans: Given: \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

Substitute the value of \[{{x}}\] in the functions  \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

We have,

  ${{f( -  2)  =  ( -  2}}{{{)}}^2}{{  +  7  =  11,}}$

  ${{g( -  1)  =  3( -  1)  +  5  =  2}}$

$\therefore {{ f( -  2)  +  g( -  1)  =  11  +  2  =  13}}$ 

\[(iv). f(t)  -  f( -  2)\]

Ans: Given: \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

Substitute the value of \[{{x}}\] in the functions \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

We have,

  ${{f(t)  =  }}{{{t}}^2}{{  +  7,}}$

  ${{f( -  2)  =  ( -  2}}{{{)}}^2}{{  +  7  =  11}}$

$\therefore {{ f(t)  -  f( -  2)  =  }}{{{t}}^2}{{  -  4}}$ 

(v).${{ }}\dfrac{{{{f(t)  -  f(5)}}}}{{{{t  -  5}}}},{{ if t }} \ne {{ 5}}$

Ans: Given: \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

Substitute the value of \[{{x}}\] in the functions \[{{f(x)  =  }}{{{x}}^2}{{  +  7 \text{ and } g(x)  =  3x  +  5}}{{.}}\]

We have,

  ${{f(t)  =  }}{{{t}}^2}{{  +  7}}$

  ${{f(5)  =  (5}}{{{)}}^2}{{  +  7  =  32}}$

$\therefore {{ }}\dfrac{{{{f(t)  -  f(5)}}}}{{{{t  -  5}}}},{{ if t }} \ne {{ 5}}$

  ${{ =  }}\dfrac{{{{{t}}^2}{{  +  7  - 32}}}}{{{{t  -  5}}}}$

   $= {{ }}\dfrac{{{{{t}}^2}{{  -  25}}}}{{{{t  -  5}}}}$

   $= {{ }}\dfrac{{({{t  -  5)(t  +  5)}}}}{{({{t  -  5)}}}}$

   $= {{ t  +  5}}$ 


12.Let \[{{f \text{ and } g}}\] be real functions defined by \[f(x)  =  2x  +  1 \text{ and } g(x)  =  4x  -  7.\]

(i).For what real numbers \[x, f(x)  =  g(x)?\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  4x  -  7}}{{.}}\]

Substitute the real functions according to given condition and obtain the real value of \[{{x}}{{.}}\]

We have,

  $\Rightarrow {{ f(x)  =  g(x)}}$

  $\Rightarrow {{ 2x  +  1  =  4x  -  7}}$

  $\Rightarrow {{ x  =  4}}$ 

(ii). For what real numbers \[x, f(x)  <  g(x)?\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  4x  -  7}}{{.}}\]

Substitute the real functions according to given condition and obtain the real value of \[{{x}}{{.}}\]

We have,

  $\Rightarrow {{ f(x)  <  g(x)}}$

  $\Rightarrow {{ 2x  +  1  <  4x  -  7}}$

  $\Rightarrow {{  -  2x  +  1  <   - 7}}$ 

  $\Rightarrow {{  -  2x  <   -  8}}$

  $\Rightarrow {{ }}\dfrac{{{{ -  2 x}}}}{{{{  -  2}}}}{{  >  }}\dfrac{{{{ -  8}}}}{{{{ -  2}}}}$ 

\[\Rightarrow {{ x  >  4}}\]


13.If \[{{f \text{ and } g}}\] are two real valued functions defined as \[f(x)  =  2x  +  1 \text{ and } g(x)  =  2{x^2}  +  1,\] then find

\[(i). f  +  g\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  2}}{{{x}}^2}{{  +  1}}{{.}}\]

Directly perform the operation according to questions like addition, subtraction, multiplication and division.

We have,

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

  $\Rightarrow {{ (f  +  g)(x)  =  2x  +  1  +  }}{{{x}}^2}{{  +  1}}$

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

\[(ii). f  -  g\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  2}}{{{x}}^2}{{  +  1}}{{.}}\]

Directly perform the operation according to questions like addition, subtraction, multiplication and division.

We have,

  $\Rightarrow {{ (f  -  g)(x)  =  f(x)  -  g(x)}}$

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

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

\[(iii). fg\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  2}}{{{x}}^2}{{  +  1}}{{.}}\]

Directly perform the operation according to questions like addition, subtraction, multiplication and division.

We have,

  $\Rightarrow {{ (fg)(x)  =  f(x) }}{{. g(x)}}$

  $\Rightarrow {{ (fg)(x)  =  (2x  +  1)(}}{{{x}}^2}{{  +  1)}}$

  $\Rightarrow {{ (fg)(x)  =  2}}{{{x}}^3}{{  +  }}{{{x}}^2}{{  +  2x  +  1}}$ 

\[(iv).\dfrac{{{f}}}{{{g}}}\]

Ans: Given: \[{{f(x)  =  2x  +  1 \text{ and } g(x)  =  2}}{{{x}}^2}{{  +  1}}{{.}}\]

Directly perform the operation according to questions like addition, subtraction, multiplication and division.

We have,

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

  $\Rightarrow {{ }}\left( {\dfrac{{{f}}}{g}} \right){{(x)  =  }}\dfrac{{2{{x  +  1}}}}{{{{{x}}^2}{{  +  1}}}}$ 


14.Express the following functions as a set of ordered pairs and determine their range.

Ans: Given: A function \[{{f : x }} \to {{ R, f(x)  =  }}{{{x}}^3}{{  +  1, \text{ where } x  =  \{  -  1, 0, 3, 9, 7\} }}\].

Substitute the value of \[{{x}}\] in the function to obtain values of \[{{y}}\] as range.

We have,

  ${{f : X }} \to {{ R, f(x)  =  }}{{{x}}^3}{{  +  1, \text{ where } x  =  \{  -  1, 0, 3, 9, 7\} }}$

  ${{x  =   -  1, \text{ then } f( -  1)  =  ( -  1}}{{{)}}^3}{{  +  1}}$

  $\Rightarrow \;{{f( -  1)  =  0}}$

  ${{x  =  0, \text{ then } f(0)  =  (0}}{{{)}}^3}{{  +  1}}$

  $\Rightarrow \;{{f(0)  =  1}}$ 

  $\Rightarrow \;{{f(3)  =  28}}$

  ${{x  =  9, \text{ then } f(9)  =  (9}}{{{)}}^3}{{  +  1}}$

  $\Rightarrow {{ f(9)  =  730}}$ 

  ${{x  =  3, \text{ then } f(3)  =  (3}}{{{)}}^3}{{  +  1}}$

  ${{x  =  7, \text{ then } f(7)  =  (7}}{{{)}}^3}{{  +  1}}$

  $\Rightarrow {{ f(7)  =  344}}$

  $\Rightarrow {{ f  =  \{ ( -  1, 0), (0, 1), (3, 28), (9, 730), (7, 344)\} }}$

$\therefore \;{{\text{ Range of }f  =  \{ 0, 1, 28, 730, 344\} }}$ 


15.Find the value of \[{{x}}\] for which the functions \[f(x)  =  3{x^2}  -  1 \text{ and } g(x)  =  3  +  x\] are equal.

Ans: Given: \[{{f(x)  =  g(x)}}\]

Use the given condition, substitute the functions equal and then find \[{{x}}\]

  $\Rightarrow {{ f(x)  =  g(x)}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  1  =  3  +  x}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  4x  +  3x  -  4  =  0}}$ 

  $\Rightarrow {{ x(3x  -  4)  +  1(3x  -  4)}}$

  $\Rightarrow {{ (3x  -  4)(x  +  1)}}$

  $\Rightarrow {{ x  =   - 1, }}\dfrac{4}{3}$ 


Long Answer Type Questions:

16.Is \[g  =  \{ (1, 1), (2, 3), (3, 5), (4, 7)\} \] a function, justify. If this is described by the relation, $g(x)  =  \alpha x  +  \beta$  then what values of should be assigned to $\alpha$ and  $\beta$

Ans: Given: A relation \[{{g  =  \{ (1, 1), (2, 3), (3, 5), (4, 7)\} }}\] and $g(x)  =  \alpha x  +  \beta$

First, find the two equation by substitutions of different values of \[{{x \text{ and } g(x)}}{{.}}\]

We have,

\[{{g  =  \{ (1, 1), (2, 3), (3, 5), (4, 7)\} }}\]

Since, every element has a unique image under \[{{g}}\]. So, \[{{g}}\] is a function.

Now, 

${g(x)  =  \alpha x  +  \beta }$

  ${x  = 1, \text{ then } g(1)  =  \alpha (1)  +  \beta }$

  $\Rightarrow { 1  =  \alpha   +  \beta          - (i)}$

  ${x  =  2, \text{ then } g(2)  =  \alpha (2)  +  \beta }$

  $\Rightarrow { 3  =  2\alpha   +  \beta         - (ii)}$

  ${\text{From (i) and (ii), we get}}$

  ${\alpha   =  2, \beta   =   -  1}$ 


17.Find the domain of each of the following functions given by

\[(i). f(x)  =  \dfrac{1}{{\sqrt {1  -  cosx} }}\]

Ans: The value of \[{{x}}\] in the function represents the domain of functions.

We have, \[{{f(x)  =  }}\dfrac{1}{{\sqrt {1{{  -  cosx}}} }}\]

  $\Rightarrow {{  -  1 }} \leqslant {{ cosx }} \leqslant {{ 1}}$

  $\Rightarrow {{  -  1 }} \leqslant {{  -  cosx }} \leqslant {{ 1}}$

  $\Rightarrow {{ 0 }} \leqslant {{ 1  -  cosx }} \leqslant {{ 2}}$ 

  $f(x)$ is defined, if ${{1  -  cosx }} \ne {{ 0}}$

  $\Rightarrow {{ cosx }} \ne {{ 1}}$

  $\Rightarrow {{ cosx }} \ne {{ 2n\pi }}$

  Domain of $f  =  R - { 2n \pi  : n} \in {{ Z}}$ 

\[(ii). f(x)  =  \dfrac{1}{{\sqrt {x  +  | x |} }}\]

Ans: The value of \[{{x}}\] in the function represents the domain of functions.

We have, \[{{f(x)  =  }}\dfrac{1}{{\sqrt {{{x  +  | x |}}} }}\]

  $\Rightarrow {{ x  +  | x |  =  x  -  x}}$

  $\Rightarrow {{ 0, x  <  0}}$

  $\Rightarrow {{ x  +  x  =  2x, x }} \geqslant {{ 0}}$ 

Hence, domain of \[{{f  =  }}{{{R}}^ + }.\]

\[(iii). f(x)  =  x | x |\]

Ans: The value of \[{{x}}\] in the function represents the domain of functions.

We have, \[{{f(x)  =  x | x |}}\]

Clearly, function is defined for all real numbers.

Domain of \[{{f  =  R}}{{.}}\]

\[(iv). f(x)  =  \dfrac{{{x^3}  -  x  +  3}}{{{x^2}  -  1}}\]

Ans: The value of \[{{x}}\] in the function represents the domain of functions.

We have, \[{{f(x)  =  }}\dfrac{{{{{x}}^3}{{  -  x  +  3}}}}{{{{{x}}^2}{{  -  1}}}}\]

  ${\text{f(x) is not defined, if }}{{{x}}^2}{{  -  1  =  0}}$

  $\Rightarrow {{ (x  -  1)(x  + 1)  =  0}}$

  $\Rightarrow {{ x  =   -  1, 1}}$

  ${{\text{ Domain of  }f  =  R  -  \{  -  1, 1\} }}{{.}}$ 

\[(v). f(x)  =  \dfrac{{3x}}{{28  -  x}}\]

Ans: The value of \[{{x}}\] in the function represents the domain of functions.

We have, \[{{f(x)  =  }}\dfrac{{3{{x}}}}{{28{{  -  x}}}}\]

  $f(x)$ is defined, if ${{28  -  x }} \ne {{ 0}}$

  $\Rightarrow {{ x }} \ne {{ 28}}$

  ${{\text{ Domain of  }f  =  R  -  \{ 28\} }}{{.}}$ 


18.Find the range of the following functions is given by

\[(i). f(x)  =  \dfrac{3}{{2  -  {x^2}}}\]

Ans: First, find the value of \[{{x}}\] in terms of \[{{y, \text{ where } y  =  f(x)}}{{.}}\] Then, find the values of \[{{y}}\] for which \[{{x}}\] attain real values.

We have, \[{{f(x)  =  }}\dfrac{3}{{2{{  -  }}{{{x}}^2}}}\]

Let  ${{ y  =  f(x)}}$

  ${{\text{ then }, y  =  }}\dfrac{3}{{2{{  -  }}{{{x}}^2}}}$

  $\Rightarrow {{ x  =  }}\sqrt {\dfrac{{2{{y  -  3}}}}{{{y}}}}$

  $\Rightarrow {{ 2y  -  3 }} \geqslant {{ 0 \text{ and } y  >  0}}$

  $\Rightarrow {{ y }} \geqslant {{ }}\dfrac{3}{2}$

  ${{\text{ Range of }f  =  }}\left[ {\dfrac{3}{2},{{ }}\infty } \right).$ 

\[(ii). f(x)  =  1  -  | x  -  2 |\]

Ans: First, find the value of \[{{x}}\] in terms of \[{{y, \text{ where } y  =  f(x)}}{{.}}\] Then, find the values of \[{{y}}\] for which \[{{x}}\] attain real values.

We have, \[{{f(x)  =  1  -  | x  -  2 |}}\]

  $\Rightarrow {{ | x  -  2 | }} \geqslant {{ 0}}$

  $\Rightarrow {{  -  | x  -  2 | }} \leqslant {{ 0}}$

  $\Rightarrow {{ 1  -  | x  -  2 | }} \leqslant {{ 1}}$

  $\Rightarrow {{ f(x) }} \leqslant {{ 1}}$

  ${{\text{ Range of }f  =  (  -  }}\infty {{, 1]}}{{.}}$ 

\[(iii). f(x)  =  | x  -  3 |\]

Ans: First, find the value of \[{{x}}\] in terms of \[{{y, \text{ where } y  =  f(x)}}{{.}}\] Then, find the values of \[{{y}}\] for which \[{{x}}\] attain real values.

We have, \[{{f(x)  =  | x  -  3 |}}\]

  $\Rightarrow {{ | x  -  3 | }} \geqslant {{ 0}}$

  $\Rightarrow {{ f(x) }} \geqslant {{ 0}}$

  ${{\text{ Range of }f  =  [0, }}\infty {{)}}{{.}}$ 

\[(iv). f(x)  =  1  +  3cos2x\]

Ans: First, find the value of \[{{x}}\] in terms of \[{{y, \text{ where } y  =  f(x)}}{{.}}\] Then, find the values of \[{{y}}\] for which \[{{x}}\] attain real values.

We have, \[{{f(x)  =  1  +  3cos2x}}\]

  $\Rightarrow {{  -  1 }} \leqslant {{ cos2x }} \leqslant {{ 1}}$

  $\Rightarrow {{  -  3 }} \leqslant {{ 3cos2x }} \leqslant {{ 3}}$

  $\Rightarrow {{ 1  -  3 }} \leqslant {{ 1  +  3cos2x }} \leqslant {{ 1  +  3}}$

  $\Rightarrow {{  -  2 }} \leqslant {{ f(x) }} \leqslant {{ 4}}$

  ${{\text{ Range of }f  =  [ -  2, 4]}}{{.}}$ 


19.Redefine the function, \[f(x)  =  | x  -  2 |  +  | 2  +  x |,  -  3 \leqslant x \leqslant 3.\]

Ans: Given: Function, \[{{f(x)  =  | x  -  2 |  +  | 2  +  x |,  -  3 }} \leqslant {{ x }} \leqslant {{ 3}}{{.}}\]

First, find the interval in which \[{{| x  -  2 | \text{ and } | 2  +  x |}}\] is defined, then find the value of \[{{f(x)}}\] in that interval.

${{Since, | x  -  2 |  =   -  (x  -  2), x  <  2}}$

$\Rightarrow {{ x  -  2, x }} \geqslant {{ 2}}$

${{\text{ and }, | 2  +  x |  =   -  (2  +  x), x  <   -  2}}$

$\Rightarrow {{ (2  +  x), x }} \geqslant {{  -  2}}$

$\therefore {{ f(x)  =  | x  -  2 |  +  | 2  +  x |,  -  3 }} \leqslant {{ x }} \leqslant {{ 3}}$

  ${{ -  (x  -  2)  -  (2  +  x),  -  3 }} \leqslant {{ x  <   -  2}}$

  ${{ -  (x  -  2)  +  2  +  x,  -  2 }} \leqslant {{ x  <  2}}$

  ${{x  -  2  +  2  +  x, 2 }} \leqslant {{x }} \leqslant {{ 3}}$ 

  ${{ -  2x,  -  3 }} \leqslant {{ x  <   -  2}}$

  ${{4,      -  2 }} \leqslant {{ x  <  2}}$

  ${{2x,    2 }} \leqslant {{x }} \leqslant {{ 3}}$ 


20. If \[f(x)  =  \dfrac{{x  -  1}}{{x  +  1}},\] then show that 

\[(i). f\left( {\dfrac{1}{x}} \right)  =   -  f(x)\]

Ans: Given: \[{{f(x)  =  }}\dfrac{{{{x  -  1}}}}{{{{x  +  1}}}}.\]

Replace the value of \[{{x}}\] according to functions and then substitute them equal

Taking L.H.S,

  ${{f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{\dfrac{1}{{{x}}}{{  -  1}}}}{{\dfrac{1}{{{x}}}{{  +  1}}}}{{ }}$

  $\Rightarrow {{ f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{1{{  -  x}}}}{{1{{  +  x}}}}$

  $\Rightarrow {{ f}}\left( {\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{{{ -  (x  -  1)}}}}{{1{{  +  x}}}}$

  $\Rightarrow {{ f}}\left( {\dfrac{1}{{{x}}}} \right){{  =   -  f(x)}}$ 

\[(ii). f\left( { -  \dfrac{1}{x}} \right)  =  \dfrac{{ -  1}}{{f(x)}}\]

Ans: Given: \[{{f(x)  =  }}\dfrac{{{{x  -  1}}}}{{{{x  +  1}}}}.\]

Replace the value of \[{{x}}\] according to functions and then substitute them equal.

Taking L.H.S

  ${{f}}\left( {{{ -  }}\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{{{ -  }}\dfrac{1}{{{x}}}{{  -  1}}}}{{{{ -  }}\dfrac{1}{{{x}}}{{  +  1}}}}$

  $\Rightarrow {{ f}}\left( {{{ -  }}\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{{{ -  (x  +  1)}}}}{{{{x  -  1}}}}$

Now,

  $\dfrac{{{{ -  1}}}}{{{{f(x)}}}}{{  =  }}\dfrac{{{{ -  1}}}}{{\dfrac{{{{x  -  1}}}}{{{{x  +  1}}}}}}$

  $\Rightarrow {{ }}\dfrac{{{{ -  1}}}}{{{{f(x)}}}}{{  =  }}\dfrac{{{{ -  (x  +  1)}}}}{{{{x  -  1}}}}$

$\therefore {{ f}}\left( {{{ -  }}\dfrac{1}{{{x}}}} \right){{  =  }}\dfrac{{{{ -  1}}}}{{{{f(x)}}}}$ 


21.If \[f(x)  =   \sqrt x  \text{ and } g(x) =  x\] be two functions defined in the domain \[{{R}^{+}}\cup { }\!\!~\!\!{ }\{0\},\] then find the value of

\[(i).(f  +  g)(x)\]

Ans: Given: \[{{f(x)  =   }}\sqrt {{x}} {{ \text{ and } g(x)  =  x}}\]

Use the two functions according to operations like addition, subtraction, multiplication and division.

Value of \[({{f  +  g)(x),}}\]

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

  $\Rightarrow {{ }}({{f  +  g)(x)  =  }}\sqrt {{x}} {{  +  x}}$ 

\[(ii).(f  -  g)(x)\]

Ans: Given: \[{{f(x)  =   }}\sqrt {{x}} {{ \text{ and } g(x)  =  x}}\]

Use the two functions according to operations like addition, subtraction, multiplication and division.

Value of \[({{f  -  g)(x),}}\]

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

  $\Rightarrow {{ }}({{f  -  g)(x)  =  }}\sqrt {{x}} {{  -  x}}$ 

\[(iii).(fg)(x)\]

Ans: Given: \[{{f(x)  =   }}\sqrt {{x}} {{ \text{ and } g(x)  =  x}}\]

Use the two functions according to operations like addition, subtraction, multiplication and division.

Value of \[({{fg)(x),}}\]

  $({{fg)(x)  =  f(x)}}{{.g(x)}}$

  $\Rightarrow {{ }}({{fg)(x)  =  }}\sqrt {{x}} .{{x}}$

  $\Rightarrow {{ }}({{fg)(x)  =  }}{{{x}}^{\dfrac{3}{2}}}$ 

\[(iv).\left( {\dfrac{f}{g}} \right)(x)\]

Ans: Given: \[{{f(x)  =   }}\sqrt {{x}} {{ \text{ and } g(x)  =  x}}\]

Use the two functions according to operations like addition, subtraction, multiplication and division.

Value of \[\left( {\dfrac{{{f}}}{{{g}}}} \right){{(x),}}\]

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

  $\Rightarrow {{ }}\dfrac{{{{f(x)}}}}{{{{g(x)}}}}{{  =  }}\dfrac{{\sqrt {{x}} }}{{{x}}}$

  $\Rightarrow {{  }}\left( {\dfrac{{{f}}}{{{g}}}} \right){{(x)  =  }}\dfrac{1}{{\sqrt {{x}} }}$ 


22.Find the domain and range of the function \[f(x)  =  \dfrac{1}{{\sqrt {x  -  5} }}.\]

Ans: Given: \[{{f(x)  =  }}\dfrac{1}{{\sqrt {{{x  -  5}}} }}.\]

The value of \[{{x}}\] represents the domain of function and the value of \[{{y at x}}{{.}}\]

  $f(x)$ is defined, if ${{x  -  5  >  0}}$

  ${{\text{ Domain of  }f  =  (5, }}\infty {{)}}{{.}}$

  ${{Let, f(x)  =  y}}$

  $\Rightarrow {{ y  =  }}\dfrac{1}{{\sqrt {{{x  -  5}}} }}$

  $\Rightarrow \,{{x  =  }}\dfrac{1}{{{{{y}}^2}}}{{  +  5}}$

  ${{\text{ Range of }f  =  }}{{{R}}^ + }.$ 


23. If \[f(x)  =  y  =  \dfrac{{ax  -  b}}{{cx  -  a}},\] then prove that \[f(y)  =  x.\]

Ans: Given: \[{{f(x)  =  y  =  }}\dfrac{{{{ax  -  b}}}}{{{{cx  -  a}}}}.\]

Use the given condition in question and substitute the value of \[{{y}}\] in the function.

We have, 

  ${{f(x)  =  y  =  }}\dfrac{{{{ax  -  b}}}}{{{{cx  -  a}}}}$

  ${{f(y)  =  }}\dfrac{{{{ay  -  b}}}}{{{{cy  -  a}}}}$ 

  $\Rightarrow {{ f(y)  =  }}\dfrac{{{{a}}\left( {\dfrac{{{{ax  -  b}}}}{{{{cx  -  a}}}}} \right){{  -  b}}}}{{{{c}}\left( {\dfrac{{{{ax  -  b}}}}{{{{cx  -  a}}}}} \right){{  -  a}}}}$

  $\Rightarrow {{ f(y)  =  }}\dfrac{{{{{a}}^2}{{x  -  ab  -  bcx  +  ab}}}}{{{{acx  -  bc  -  acx  +  }}{{{a}}^2}}}$

  $\Rightarrow {{ f(y)  =  }}\dfrac{{{{{a}}^2}{{x  -  bcx}}}}{{{{{a}}^2}{{  -  bc}}}}$

  $\Rightarrow {{ f(y)  =  }}\dfrac{{{{x(}}{{{a}}^2}{{  -  bc)}}}}{{{{(}}{{{a}}^2}{{  -  bc)}}}}$

  $\Rightarrow {{ f(y)  =  x}}{{.}}$ 

Hence proved.


Objective Type Questions:

24.Let \[n(A)  =  m \text{ and } n(B)  =  n.\] Then, total number of non-empty relations that can be defined from \[A \text{ to } B\] is

(A). \[{m^n}\]

(B). \[{n^m}  -  1\]

(C). \[mn  -  1\]

(D). \[{2^{mn}}  -  1\]

Ans: Given: \[{{n(A)  =  m \text{ and } n(B)  =  n}}{{.}}\]

First, find the number of elements in \[{{A }} \times {{ B}}\] and then find the number of relation by using \[{{{2}}^{{{m(A }} \times {{ B)}}}}{{  -  1}}{{.}}\]

We have,

  ${{n(A) }}{{ =  m \text{ and } n(B)  =  n}}$

  ${{n(A }} \times {{ B)  =  n(A)}}{{.n(B)}}$

  ${{n(A }} \times {{ B)  =  mn}}$ 

Total number of relations from \[{{A \text{ to } B  =  }}{{{2}}^{{{mn}}}}{{  -  1}}{{.}}\]

Correct Answer: D


25.If \[{[x]^2}  -  5[x]  +  6  =  0, \text{ where } [ . ]\] represents the greatest integer function, then

\[(A) x \in [3, 4] \]

\[(B) x \in  (2, 3] \]

\[(C) x \in  [2, 3] \]

\[(D) x \in  [2, 4) \]

Ans: Given:  \[{{{[x]}}^2}{{  -  5[x]  +  6  =  0, \text{ where } [ }}{{. ]}}\] represents Greatest integer function.

If \[{{a \text{ and } b}}\] are two successive positive integer and \[{{[x]  =  a, b, \text{ then } x }} \in {{ [a, b]}}{{.}}\]

We have,

  ${{{[x]}}^2}{{  -  5[x]  +  6  =  0}}$

  $\Rightarrow {{ [x}}{{{]}}^2}{{  -  3[x]  +  2[x]  +  6  =  0}}$

  $\Rightarrow {{ [x]([x]  -  3)  -  2([x]  -  3)  =  0}}$

  $\Rightarrow {{ ([x]  -  3)}}([{{x]  -  2)  =  0}}$

  $\Rightarrow {{ [x]  =  2, 3}}$

$\therefore {{ x }} \in {{ [2, 3]}}$ 

Correct Answer: C


26.Range of \[f(x)  =  \dfrac{1}{{1  -  2cosx}}\] is

\[(A)\left[ {\dfrac{1}{3}, 1} \right]\]

\[(B)\left[ { -  1, \dfrac{1}{3}} \right]\]

$\left.(C)(-\infty,-1] \cup \frac{1}{3}, \infty\right)$

$(D)\left[-\frac{1}{3}, 1\right]$

Ans: Given: \[{{f(x)  =  }}\dfrac{1}{{1{{  -  2cosx}}}}\]

Range of function represents the value the function attains. Start with the range of functions you know.

Range of cosine function is,

  $\Rightarrow {{  -  1 }} \leqslant {{  -  cosx }} \leqslant {{ 1}}$

  $\Rightarrow {{  -  2 }} \leqslant {{  -  2cosx }} \leqslant {{ 2}}$

  $\Rightarrow {{ 1  -  2 }} \leqslant {{ 1  -  2cosx }} \leqslant {{ 1  +  2}}$ 

  $\Rightarrow {{  -  1 }} \leqslant {{ 1  -  2cosx }} \leqslant {{ 3}}$

  $\Rightarrow {{  -  1 }} \leqslant {{ }}\dfrac{1}{{1{{  -  cos2x}}}}{{ }} \leqslant {{ }}\dfrac{1}{3}$

  $\Rightarrow {{  -  1 }} \leqslant {{ f(x) }} \leqslant {{ }}\dfrac{1}{3}$

  ${{\text{ Range of }f  =  }}\left[ { - {{ 1, }}\dfrac{1}{3}} \right].$ 

Correct Answer: B


27.Let \[f(x)  =  \sqrt {1  +  {x^2}} ,\]  then

\[(A) f(xy)  =  f(x) . f(y)\]

$(B) f(x y) \geq f(x) \cdot f(y)$

\[(C) f(xy) \le f(x) . f(y)\]

 (D) None of these

Ans: Given: Function \[{{f(x)  =  }}\sqrt {1{{  +  }}{{{x}}^2}} .\]

Replace the variables of function as \[{{x }} \to {{ xy}}\] and then compare.

We have,

  ${{f(x)  =  }}\sqrt {1{{  +  }}{{{x}}^2}}$

  ${{f(xy)  =  }}\sqrt {1{{  +  }}{{{x}}^2}{{{y}}^2}}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {1{{  +  }}{{{x}}^2}} {{ }}{{. }}\sqrt {1{{  +  }}{{{y}}^2}}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {(1{{  +  }}{{{x}}^2})(1{{  +  }}{{{y}}^2})}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {1{{  +  }}{{{x}}^2}{{  +  }}{{{y}}^2}{{  +  }}{{{x}}^2}{{{y}}^2}}$

  $\sqrt {1{{  +  }}{{{x}}^2}{{{y}}^2}} {{ }} \leqslant {{ }}\sqrt {1{{  +  }}{{{x}}^2}} {{ }}{{. }}\sqrt {1{{  +  }}{{{y}}^2}}$

  $\Rightarrow {{ f(xy) }} \leqslant {{ f(x) }}{{. f(y)}}{{.}}$ 

Correct Answer: C


28.Domain of \[\sqrt {{a^2}  -  {x^2}}  (a  >  0) is\]

\[(A)( -  a, a)\]

\[(B)[ -  a, a]\]

\[(C) [0, a]\]

\[(D)( -  a, 0]\]

Ans: Given: \[{{f(x)  =  }}\sqrt {{{{a}}^2}{{  -  }}{{{x}}^2}} \]

Domain of the function is the value of \[{{x}}\]for which the function is defined.

 Let ${{ f(x)  =  }}\sqrt {{{{a}}^2}{{  -  }}{{{x}}^2}}$

  $f(x)$ is defined, if ${{}}{{{a}}^2}\; - \;{{{x}}^2}{{ }} \geqslant \;{{0}}$

  $\Rightarrow {{ }}{{{x}}^2}\; - \;{{{a}}^2}\; \leqslant \;0$

  $\Rightarrow {{ (x  -  a)(x  +  a) }} \leqslant {{ 0}}$

  $\Rightarrow {{  -  a }} \leqslant {{ x }} \leqslant {{ a}}$

$\therefore {{ \text{ Domain of  }f  =  [ -  a, a]}}{{.}}$

Correct Answer: B


29. If \[f(x)  =  ax  +  b,\] where a and  b are integers, \[f( -  1)  =   -  5 { \text{ and } } f(3)  =  3\], then a and b are equal to

\[(A) a  =   -  3, b  =   -  1\]

\[(B) a  =  2, b  =   -  3\]

\[(C) a  =  0, b  =  2\]

\[(D) a  =  2, b  =  3\]

Ans: Given:

  ${{f(x)  =  ax  +  b}}$

  ${{f( -  1)  =   -  5 }}$

  ${{f(3)  =  3}}$ 

First, substitute the values in functions. Solve equations for \[{{a \text{ and } b}}{{.}}\]

  ${{for, f(x)  =  ax  +  b}}$

  $\Rightarrow {{ f( -  1)  =  a( -  1)  +  b}}$

  $\Rightarrow {{  -  5  =   -  a  +  b       - (i)}}$

  ${{\text{ and },}}$

  ${{f(3)  =  a(3)  +  b}}$

  $\Rightarrow {{ 3  =  3a  +  b              - (ii)}}$

  ${\text{from (i) and (ii), we get}}$

  ${{a  =  2 \text{ and } b  =   -  3}}$ 

Correct answer: B


30.The domain of the function \[{{f}}\] defined by \[f(x)  =  \sqrt {4  -  x}   +  \dfrac{1}{{\sqrt {{x^2}  -  1} }}\] is equal to

$(A)(-\infty,-1) \cup(1,4]$

$(B)(-\infty,-1] \cup(1,4]$

$(C)(-\infty, 1) \cup[1,4]$

$(D)(-\infty, 1) \cup[1,4)$

Ans: Given: Function \[{{f(x)  =  }}\sqrt {4{{  -  x}}} {{  +  }}\dfrac{1}{{\sqrt {{{{x}}^2}{{  -  1}}} }}\]

Domain of the function is the value of \[{{x}}\]for which the function is defined.

  $f(x)$ is defined, if ${{}}$

  ${{4  -  x }} \geqslant {{ 0 \text{ or } }}{{{x}}^2}{{  -  1  >  0}}$

  ${{x  -  4 }} \leqslant {{ 0 \text{ or } (x  +  1)(x  -  1)  >  0}}$

  ${{x }} \leqslant {{ 4 \text{ or } x  <   -  1 \text{ and } x  >  1}}$

$\therefore {{ \text{ Domain of  }f  =  ( -  }}\infty ,{{  -  1) }} \cup {{ (1, 4]}}{{.}}$ 

Correct Answer: A


31.The domain and range  of the real functions \[{{f}}\] defined by \[f(x)  =  \dfrac{{4  -  x}}{{x  -  4}}\]is given by

\[(A) Domain  =  R, Range  =  \{  - 1, 1\} \]

\[(B) Domain  =  R  -  \{ 1\} , Range  =  R\]

\[(C) Domain  =  R  -  \{ 4\} , Range  =  \{  - 1\} \]

\[(D) Domain  =  R  -  \{  -  4\} , Range  =  \{  - 1, 1\} \]

Ans: Given: \[{{f(x)  =  }}\dfrac{{4{{  -  x}}}}{{{{x  -  4}}}}\].

A rational function  \[\dfrac{{{{f(x)}}}}{{{{g(x)}}}}\] is defined if \[{{g(x) }} \ne {{ 0}}{{.}}\]

  $f(x)$ is defined, if ${{x  -  4 }} \ne {{ 0 \text{ or } x }} \ne {{ 4}}$

$\therefore {{ \text{ Domain of  }f  =  R  -  \{ 4\} }}{{.}}$

  ${{Let f(x)  =  y}}$

  $\Rightarrow {{ y  =  }}\dfrac{{4{{  -  x}}}}{{{{x  -  4}}}}$

  $\Rightarrow {{ xy  -  4y  =  4  -  x}}$

  $\Rightarrow {{ xy  +  x  =  4  +  4y}}$

  $\Rightarrow {{ x(y  +  1)  =  4(1  +  y)}}$ 

\[\Rightarrow {{ x  =  }}\dfrac{{4(1{{  +  y)}}}}{{{{y  +  1}}}}\]

${\text{x is defined, if y  +  1 }} \ne {{ 0, y }} \ne {{  -  1}}$

$\therefore$  Range of ${{f  =  R  -  \{  -  1\} }}{{.}}$ 

Correct Answer: C


32.The domain and range of real function \[{{f}}\] defined by \[f(x)  =  \sqrt {x  -  1} \] is given by

$(A)$ Domain $=(1, \infty)$, Range $=(0, \infty)$

$(B)$ Domain $=[1, \infty)$, Range $=(0, \infty)$

$(C)$ Domain $=(1, \infty)$, Range= $[0, \infty)$

$(D)$ Domain $=[1, \infty)$, Range $=[0, \infty)$

Ans: Given: \[{{f(x)  =  }}\sqrt {{{x  -  1}}} \].

A square root function \[{{f(x)  =  }}\sqrt {{x}} \] is defined by \[{{x }} \geqslant {{ 0}}{{.}}\]

  $f(x)$ is defined, if ${{x  -  1 }} \geqslant {{ 0}}$

  $\Rightarrow {{ x }} \geqslant {{ 1}}$

$\therefore {{ \text{ Domain of  }f  =  [1, }}\infty {{)}}{{.}}$

  ${{Let f(x)  =  y}}$

  $\Rightarrow {{ y  =  }}\sqrt {{{x  -  1}}}$

  $\Rightarrow {{ }}{{{y}}^2}{{  =  x  -  1}}$

  $\Rightarrow {{ x  =  }}{{{y}}^2}{{  +  1}}$

  $x$ is defined for ${{y }} \in {{ R}}{{.}}$

  ${{but, y }} \geqslant {{ 0}}$

$\therefore$  Range of ${{f  =  [0, }}\infty ).$ 

Correct Answer: D


33.The domain of function \[{{f}}\] given by \[f(x)  =  \dfrac{{{x^2}  +  2x  +  1}}{{{x^2}  -  x  -  6}}.\]

\[(A) R  -  \{ 3,  -  2\} \]

\[(B) R  -  \{  -  3, 2\} \]

\[(C) R  -  [3,  -  2]\]

\[(D) R  -  (3,  -  2)\]

Ans: Given: \[{{f(x)  =  }}\dfrac{{{{{x}}^2}{{  +  2x  +  1}}}}{{{{{x}}^2}{{  -  x  -  6}}}}.\]

The domain is the value of \[{{x}}\] for which function is defined.

  ${{f(x) \text{ is defined, if } }}{{{x}}^2}{{  -  x  -  6  =  0}}$

  $\Rightarrow {{ }}{{{x}}^2}{{  -  3x  +  2x  -  6  =  0}}$

  $\Rightarrow {{ x(x  -  3)  +  2(x  -  3)  =  0}}$ 

  $\Rightarrow {{ (x  -  3)(x  +  2)  =  0}}$

  $\Rightarrow {{ x  =  3,  -  2}}$

$\therefore$ Domain of ${{f  =  R  -  \{ 3,  -  2\} }}{{.}}$ 

Correct Answer: A


34.The domain and range of the function \[{{f}}\] given by \[f(x)  =  2  -  | x  -  5 |\] is

$(A)$ Domain $=R^{+}$, Range $=(-\infty, 1]$

$(B)$ Domain $=R$, Range $=(-\infty, 2]$

$(C)$ Domain $=R$, Range $=(-\infty, 2)$

$(D)$ Domain $=R^{+}$, Range $=(-\infty, 2]$

Ans: Given: \[{{f(x)  =  2  -  | x  -  5 |}}{{.}}\]

The absolute modulus function \[|{{x | }} \geqslant {{ 0}}{{.}}\]

  ${\text{f(x) is defined for all x }} \in {{ R}}$

$\therefore {{ \text{ Domain of  }f  =  R}}{{.}}$

  ${{Now, | x  -  5 | }} \geqslant {{ 0}}$ 

  $\Rightarrow {{  -  | x  -  5 | }} \leqslant {{ 0}}$

  $\Rightarrow {{ 2  -  | x  -  5 | }} \leqslant {{ 2}}$

  $\Rightarrow {{ f(x)}}{{ }} \leqslant {{ 2}}$

$\therefore$  Range of ${{f  =  ( -  }}\infty {{, 2]}}{{.}}$ 

Correct Answer: B


35.The domain for which the functions defined by \[f(x)  =  3{x^2}  -  1 { \text{ and } } g(x)  =  3  +  x\] are equal to

\[(A)\left[ { -  1, \dfrac{4}{3}} \right]\]

\[(B)\left[ {1, \dfrac{4}{3}} \right]\]

\[(C)\left[ { -  1,  -  \dfrac{4}{3}} \right]\]

\[(D)\left[ { -  2,  -  \dfrac{4}{3}} \right]\]

Ans: Given: \[{{f(x)  =  3}}{{{x}}^2}{{  -  1 \text{ and } g(x)  =  3  +  x}}{{.}}\]

Substitute the functions equal and then find \[{{x}}{{.}}\]

  ${{\text{For}, f(x)  =  g(x)}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  1  =  3  +  x}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  x  -  4  =  0}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  4x  +  3x  -  4  =  0}}$

  $\Rightarrow {{ x(3x  -  4)  +  1(x  +  1)  =  0}}$

  $\Rightarrow {{ (3x  -  4)(x  +  1)  =  0}}$

  $\Rightarrow {{ x  =   - 1, }}\dfrac{4}{3}$

$\therefore {{ \text{Domain for } f(x)  =  g(x) \text{ is } }}\left[ { - {{ 1, }}\dfrac{4}{3}} \right].$ 

Correct Answer: A


Fill in the blanks:

36.Let \[f \text{ and } g\] be two real functions given by \[f  =  \{ (0, 1), (2, 0), (3,  -  4), (4, 2), (5, 1)\}  \text{ and } g  =  \{ (1, 0), (2, 2), (3,  -  1), (4, 4), (5, 3)\} ,\] then the domain of \[f . g \] is given by ………

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

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

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

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

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


37.Let \[f  =  \{ (2, 4), (5, 6), (8,  -  1), (10,  -  3)\}  \text{ and } g  =  \{ (2, 5), (7, 1), (8, 4), (10, 13), (11, 5)\} \] be two real functions. Then match the following.

  (a) f  -  g 

(i) $\left\{ {\left( {2, \dfrac{4}{5}} \right), \left( {8, \dfrac{{ -  1}}{4}} \right), \left( {10, \dfrac{{ -  3}}{{13}}} \right)} \right\}$

(b) f  +  g 

  (ii) $\left\{ {(2, 20), (8,  -  4), (10,  -  39)} \right\}$

(c) f . g 

(iii) $\{ (2,  -  1), (8,  -  5), (10,  -  16)\}$

(d) $\dfrac{f}{g}$   

(iv) $\{ (2,9), (8, 3), (10,  -  10)\}$ 

The domain of \[{{f  -  g, f  +  g, f }}{{. g, }}\dfrac{{{f}}}{{{g}}}\] is domain of \[{{f }} \cap \] domain of \[{{g}}{{.}}\] Then, find their images.

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:

38.The ordered pair \[(5, 2)\] belongs to the relation \[R  =  \{ (x, y) : y  =  x  -  5, x, y \in  Z\} .\]

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 \[{{R}}{{.}}\]


39.If \[P  =  \{ 1, 2\} ,\] then \[P  \times  P  \times  P  =  \{ (1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)\} .\]

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 


40.If \[A  =  \{ 1, 2, 3\} , B  =  \{ 3, 4\}  \text{ and } C  =  \{ 4, 5, 6\} ,\] then 

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

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.


41.If \[(x  -  2, y  +  5)  =  \left( { -  2, \dfrac{1}{3}} \right)\] are two equal ordered pairs, then \[x  =  4, y  =  \dfrac{{ -  14}}{3}.\]

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.


42.If \[A  \times  B  =  \{ (a, x), (a, y), (b, x), (b, y)\} ,\] then \[A  =  \{ a, b\}  \text{ and } B  =  \{ x, y\} .\]

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. 

 

Some Solved Examples

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. 

FAQs on NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions)

1. Where can students get the NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions)?

Students can easily get the Exemplar Solutions for Class 11 chapter 2 relations and functions from  Vedantu’s official website i.e. vedantu. 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 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. 

2. What are the main topics that students can find in the NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions)?

The chapter consists of many important topics regarding relations and functions but the main topics which are covered in the pdf are -

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

3. What are the benefits Vedantu offers to their students?

Vedantu also offers many other benefits for chapter 2 - relations and functions such as- online study material for the students of chapter 2 - relations and functions, notes, additional questions, sample papers, and last year’s paper on the respective topic. Students can also ask their doubts through live online sessions. 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. 

4. Is NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions) important for the Class 11 students?

Class 11 is important for a student’s conceptual foundation and for future career choices. All the chapters they study in the respective classes are important. This is why students should study with the help of NCERT Exemplar for Class 11 Maths Chapter 2 - Relations and Functions (Book Solutions). Chapter 2 i.e. relation and functions are really important for Class 11 as it consists of many important topics that will be used in the future years. Students will learn some topics which will increase their reasoning sense. Students will be beneficial by learning such topics, not only for board exams but also for the upcoming professional courses they will take after completing the higher secondary level education.