Chapter 2: Relations and Functions

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, 

${g(x)  =  \alpha x  +  \beta }$

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

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

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

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

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

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

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 ${{28  -  x }} \ne {{ 0}}$

  $\Rightarrow {{ x }} \ne {{ 28}}$

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

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, \[{{f(x)  =  1  +  3cos2x}}\]

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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,

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

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

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

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

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

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

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

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 ${{x  -  5  >  0}}$

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

  ${{Let, f(x)  =  y}}$

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

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

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

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:

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

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