Chapter 2: Relations and Functions

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

Use the Cartesian product.

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

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

Use the Cartesian product.

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

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.

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

Use the Cartesian product.

Here,

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

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

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

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 $y =  2$

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 \[\{ (2,{{ 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)\} }}{{.}}\]

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 $R  =  \{ 7, 5\}$.

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..

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 $\left\{ {\dfrac{1}{2},{{  -  2}}} \right\}.$ 

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 $f(x)  =  R$ 

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 $f(x)  =  [0, 4]$ 

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

Split the function over the given intervals.

  ${{f(x)  =  }}$

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

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

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

  ${{f(x)  =  }}$

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

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

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

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