NCERT Solutions for Class 12 Maths Chapter 1: Relation and Functions - Exercise 1.3

Exercise 1.3

1. Let \[f:\left\{ {1,{\text{ }}3,{\text{ }}4} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}5} \right\}\]and \[g:{\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}5} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}3} \right\}\]be given by\[{\text{ }}f{\text{ }} = {\text{ }}\left\{ {\left( {1,{\text{ }}2} \right),{\text{ }}\left( {3,{\text{ }}5} \right),{\text{ }}\left( {4,{\text{ }}1} \right)} \right\}\]and \[g{\text{ }} = {\text{ }}\left\{ {\left( {1,{\text{ }}3} \right),{\text{ }}\left( {2,{\text{ }}3} \right),{\text{ }}\left( {5,{\text{ }}1} \right)} \right\}.\]Write down \[gof.\]

Ans: The functions \[f:{\text{ }}\left\{ {1,{\text{ }}3,{\text{ }}4} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}5} \right\}\]and \[g:{\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}5} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}3} \right\}\]are defined as,

\[{\text{f}} = \{ (1,2),(3,5),(4,1)\} \]and \[{\text{g}} = \{ (1,3),(2,3),(5,1)\} \]

Find $\operatorname{g of} (1)$,

$\operatorname{g of} (1) = {\text{g}}[{\text{f}}(1)] = {\text{g}}(2) = 3$            ( as ${\text{f}}(1) = 2$ and ${\text{g}}(2) = 3)$

Find $\operatorname{gof} (3)$,

$\operatorname{gof} (3) = g[f(3)] = g(5) = 1$         $(\operatorname{as} \;{\text{f}}(3) = 5$ and $g(5) = 1)$

Find $\operatorname{gof} (4)$,

$\operatorname{gof} (4) = {\text{g}}[{\text{f}}(4)] = {\text{g}}(1) = 3$           (as ${\text{f}}(4) = 1$ and ${\text{g}}(1) = 3)$

Therefore,

$gof = \{ (1,3),(3,1),(4,3)\} $.

2. Let \[f,{\text{ }}g\] and \[h\] be functions from \[R\] to\[R\]. Show that

 \[\left( {f{\text{ }} + {\text{ }}g} \right)oh{\text{ }} = {\text{ }}foh{\text{ }} + {\text{ }}goh\]

\[\left( {f.g} \right)oh{\text{ }} = {\text{ }}\left( {foh} \right).\left( {goh} \right)\]

Ans:  To prove: \[\left( {f{\text{ }} + {\text{ }}g} \right)oh{\text{ }} = {\text{ }}foh{\text{ }} + {\text{ }}goh\] 

Consider,

\[LHS{\text{ }} = {\text{ }}\left[ {\left( {f{\text{ }} + {\text{ }}g} \right)oh} \right]\left( x \right)\]

            $ = (f + g)(h(x))$

            $ = f(h(x)) + g(h(x))$

            $ = (foh)(x) + (goh)(x)$ 

            $ = (foh) + (goh)(x)$

Therefore,

\[((f + g)\,o\,h)(x) = (foh) + (goh)(x)\] for all $x \in R$

Hence, $(f + g)oh = foh{\text{ }} + {\text{ }}goh$ 

To prove,

\[\left( {f.g} \right)oh{\text{ }} = {\text{ }}\left( {foh} \right).\left( {goh} \right)\]

Consider,

\[LHS{\text{ }} = {\text{ }}\left[ {\left( {f.g} \right)oh} \right]\left( x \right)\]

 $= (f.g)[h(x)] $

 $= f[h(x)] \cdot g[h(x)]  $
\[\; = (foh)(x).\;(goh)(x)\]

\[ = {\text{ }}\left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right){\text{ }} = {\text{ }}RHS\]

Thus, 

\[\left[ {\left( {f.g} \right)oh} \right]\left( x \right){\text{ }} = {\text{ }}\left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right){\text{ }}for{\text{ }}all{\text{ }}x{\text{ }} \in R\]

Hence, \[\left( {f.g} \right)oh{\text{ }} = {\text{ }}\left( {foh} \right).\left( {goh} \right)\].

3. Find \[gof\] and\[fog\], if

i. $f(x) = |x|$ and $g(x) = |5x - 2|$

Ans: To find \[gof\]and \[fog\]if,

$f(x) = |x|$ and $g(x) = |5x - 2|$

$\therefore {\text{ }}go{\text{ }}f\left( x \right){\text{ }} = {\text{ }}g\left( {f\left( x \right)} \right){\text{ }} $

 $ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ }}g\left( {|x|} \right){\text{ }} $

 $ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ }}\left| 5 \right|x\left| { - 2} \right|$

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

  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ }}f\left( {|5x - 2|} \right){\text{ }}$

  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ }}\left| {\left| {5x - 2} \right|} \right|{\text{ }}$

  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ }}\left| {5x - 2} \right|$

(ii) $f(x) = 8{x^3}$ and $g(x) = {x^{\dfrac{1}{3}}}$

Ans: To find \[gof\]and\[fog\] if,

$f(x) = 8{x^3}$ and $g(x) = {x^{\dfrac{1}{3}}}$

$ \therefore {\text{\;gof\;}}(x) = g(f(x)) $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = g\left( {8{x^3}} \right) $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {8{X^3}} \right)^{\dfrac{1}{3}}} $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x $ 

$ fog(x) = f(g(x)) $

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

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8{\left( {{x^{\dfrac{1}{3}}}} \right)^3} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8x $ 

4. If $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}},x \ne \dfrac{2}{3}$, show that $fof(x) = x$, for all $x \ne \dfrac{2}{3}.$ What is the inverse of \[f?\]

Ans: Given function,

\[{\text{f}}({\text{x}}) = \dfrac{{(4x + 3)}}{{(6x - 4)}},x \ne \dfrac{2}{3}\] 

\[(fof)(x) = f(f(x)) = f\left( {\dfrac{{(4x + 3)}}{{(6x - 4)}} = \dfrac{{4\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) - 4}}} \right.\] 

$ = \dfrac{{16x + 12 + 18x - 12}}{{24x + 18 - 24s + 16}} = \dfrac{{34x}}{{34}}$

Therefore,

$fof(x) = x$, for all $x \ne \dfrac{2}{3}$

$ \Rightarrow fof = {{\text{I}}_{\text{x}}}$

Hence, the given function ${\text{f}}$ is invertible and,

The inverse of the given function ${\text{f}}$ is \[f\] itself.

5. State with reason whether following functions have inverse

\[\left( i \right){\text{ }}f:{\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4} \right\}{\text{ }} \to {\text{ }}\left\{ {10} \right\}\]with

\[f{\text{ }} = {\text{ }}\left\{ {\left( {1,{\text{ }}10} \right),{\text{ }}\left( {2,{\text{ }}10} \right),{\text{ }}\left( {3,{\text{ }}10} \right),{\text{ }}\left( {4,{\text{ }}10} \right)} \right\}\] 

\[\left( {ii} \right){\text{ }}g:{\text{ }}\left\{ {5,{\text{ }}6,{\text{ }}7,{\text{ }}8} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4} \right\}\]with

\[g{\text{ }} = {\text{ }}\left\{ {\left( {5,{\text{ }}4} \right),{\text{ }}\left( {6,{\text{ }}3} \right),{\text{ }}\left( {7,{\text{ }}4} \right),{\text{ }}\left( {8,{\text{ }}2} \right)} \right\}\]

\[\left( {iii} \right){\text{ }}h:{\text{ }}\left\{ {2,{\text{ }}3,{\text{ }}4,{\text{ }}5} \right\}{\text{ }} \to {\text{ }}\left\{ {7,{\text{ }}9,{\text{ }}11,{\text{ }}13} \right\}\]with

\[h{\text{ }} = {\text{ }}\left\{ {\left( {2,{\text{ }}7} \right),{\text{ }}\left( {3,{\text{ }}9} \right),{\text{ }}\left( {4,{\text{ }}11} \right),{\text{ }}\left( {5,{\text{ }}13} \right)} \right\}\]

Ans: Consider,

\[\left( i \right){\text{ }}f:{\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4} \right\}{\text{ }} \to {\text{ }}\left\{ {10} \right\}\] defined as \[f{\text{ }} = {\text{ }}\left\{ {\left( {1,{\text{ }}10} \right),{\text{ }}\left( {2,{\text{ }}10} \right),{\text{ }}\left( {3,{\text{ }}10} \right),{\text{ }}\left( {4,{\text{ }}10} \right)} \right\}\]

From the given definition of\[f\], we can see that \[\;f\] is a many one function as,

\[f\left( 1 \right){\text{ }} = {\text{ }}f\left( 2 \right){\text{ }} = {\text{ }}f\left( 3 \right){\text{ }} = {\text{ }}f\left( 4 \right){\text{ }} = {\text{ }}10\]

They do not have unique image,

Therefore, \[f\] is not one – one

Hence, function \[f\] does not have an inverse.

\[\left( {ii} \right){\text{ }}g:{\text{ }}\left\{ {5,{\text{ }}6,{\text{ }}7,{\text{ }}8} \right\}{\text{ }} \to {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4} \right\}\]with

\[g{\text{ }} = {\text{ }}\left\{ {\left( {5,{\text{ }}4} \right),{\text{ }}\left( {6,{\text{ }}3} \right),{\text{ }}\left( {7,{\text{ }}4} \right),{\text{ }}\left( {8,{\text{ }}2} \right)} \right\}\]

From the given definition of\[g\], it is seen that \[g\]is a many one function as \[g\left( 5 \right) = g\left( 7 \right) = 4.\]

\[\therefore g\]is not one – one

Therefore, \[g\]has no inverse.

Hence, function\[g\]does not have an inverse.

\[\left( {iii} \right){\text{ }}h:{\text{ }}\left\{ {2,{\text{ }}3,{\text{ }}4,{\text{ }}5} \right\}{\text{ }} \to {\text{ }}\left\{ {7,{\text{ }}9,{\text{ }}11,{\text{ }}13} \right\}\] defined as,

 \[h{\text{ }} = {\text{ }}\left\{ {\left( {2,{\text{ }}7} \right),{\text{ }}\left( {3,{\text{ }}9} \right),{\text{ }}\left( {4,{\text{ }}11} \right),{\text{ }}\left( {5,{\text{ }}13} \right)} \right\}\]

It is seen that all distinct elements of the set \[\left\{ {2,{\text{ }}3,{\text{ }}4,{\text{ }}5} \right\}\]have distinct images under \[h.\]

Therefore, Function \[h\]is one – one

Also, \[h\] is onto since for every element \[y\] of the set \[\left\{ {7,{\text{ }}9,{\text{ }}11,{\text{ }}13} \right\},\]there exists an element \[x\] in the set \[\left\{ {2,{\text{ }}3,{\text{ }}4,{\text{ }}5} \right\},\]such that \[h\left( x \right){\text{ }} = {\text{ }}y.\]

All elements have different images under\[h\]. So \[h\]is one-one onto function, 

Therefore, \[h\]has an inverse.

6. Show that f: $[ - 1,1] \to {\text{R}}$, given by $(x) = X(X + 2)$ is one- one. Find the inverse of the function ${\text{f}}:[ - 1,1] \to $ Range ${\text{f}}$.

(Hint: For ${\text{y}} \in $ Range \[f\], ${\text{y}} = (x) = (X + 2))$, for some ${\text{x}}$ in $[ - 1,1]$, i.e., $x = 2y(1 - y))$

Ans: Given function,

\[f:{\text{ }}\left[ { - 1,{\text{ }}1} \right]{\text{ }} \to {\text{ }}R\] is given as \[\left( x \right) = X\left( {X + 2} \right)\]

For one - one

Let\[f\left( x \right) = f\left( y \right)\]

$ \Rightarrow (X + 2) = (Y + 2)$

$ \Rightarrow xy + 2x = xy + 2y$

$ \Rightarrow 2x = 2y$

$ \Rightarrow {\text{x}} = {\text{y}}$

$\therefore $ \[f\] is a one -one function.

It is clear that \[f:{\text{ }}\left[ { - 1,{\text{ }}1} \right]{\text{ }} \to \]Range \[\;f\] is onto.

Therefore,

 \[f:{\text{ }}\left[ { - 1,{\text{ }}1} \right]{\text{ }} \to \]Range\[\;f\] is one – one and onto and therefore, the inverse of the function \[f:{\text{ }}\left[ { - 1,{\text{ }}1} \right]{\text{ }} \to \]Range\[\;f\]exists.

Let \[g:\]Range \[f \to \left[ { - 1,{\text{ }}1} \right]\]be the inverse of \[f.\]

Let \[y\] be an arbitrary element of range \[f.\]

Since \[f:{\text{ }}\left[ { - 1,{\text{ }}1} \right]{\text{ }} \to \]Range\[\;f\] is onto, we have,

\[y = f\left( x \right)\]for some \[x \in \left[ { - 1,{\text{ }}1} \right]\]

$ \Rightarrow y = \dfrac{x}{{(x + 2)}}$

$ \Rightarrow xy + 2y = x$

$ \Rightarrow x(1 - y) = 2y$

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

Now, let us define g: Range ${\text{f}} \to [ - 1,1]$ as,

$g(y) = \dfrac{{2y}}{{1 - y}},y \ne 1$

Now find \[(gof)(x)\],

$ (gof)(x) = g(f(x)) $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = g\left( {\dfrac{x}{{(x + 2)}}} \right) $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {\dfrac{{2\left( {\dfrac{x}{{x + 2}}} \right)}}{{1 - \left( {\dfrac{x}{{x + 2}}} \right)}}} \right) $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2x}}{{x + 2 - x}} $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2x}}{2} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{x}} $ 

Find $(fog)(y)$,

$ (fog)(y) = f(g(y)) $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {\dfrac{{2y}}{{1 - y}}} \right) $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\dfrac{{2y}}{{1 - y}}}}{{\dfrac{{2y}}{{1 - y}} + 2}} $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2y}}{{2y + 2 - 2y}} $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2y}}{2} $

$  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = y $ 

Therefore,

$gof = x = {I_{[ - 1,1]}}$ and $fog = y = I$Range $f$

$\therefore {f^{ - 1}} = g$

$ \Rightarrow {f^{ - 1}}(y) = \dfrac{{2y}}{{1 - y}},y \ne 1$

7. Consider \[f:{\text{ }}R{\text{ }} \to {\text{ }}R\] given by\[f\left( x \right) = 4x + 3\]. Show that $f$ is invertible. Find the inverse of $f$.

Ans: Consider\[f:{\text{ }}R{\text{ }} \to {\text{ }}R\]given by \[f\left( x \right) = 4x + 3\]

Let \[f\left( x \right){\text{ }} = {\text{ }}f\left( y \right)\]

$\Rightarrow 4 x+3=4 y+3$

$\Rightarrow 4 x=4 y$

$\Rightarrow x=y$

Therefore\[,{\text{ }}f\] is a one – one function.

Let $y \in $ Range of $f$

$y = 4x + 3$

or $x = \dfrac{{(y - 3)}}{4}$

Here, $f\left( {\dfrac{{(y - 3)}}{4}} \right) = 4\left( {\dfrac{{(y - 3)}}{4}} \right) + 3 = y$

This implies $f(x) = y$

So $f$ is onto

Therefore, $f$ is invertible.

The inverse of $f$ is $x = {f^{ - 1}}(y) = \dfrac{{(y - 3)}}{4}$.

8. Consider $f:{\mathbf{R}} +  \to [4,\infty )$ given by ${\text{f}}({\text{x}}) = {{\text{x}}^2} + 4$. Show that ${\text{f}}$is invertible with the inverse $f - 1$ of given ${\text{f}}$ by ${f^{ - 1}}(y) = \sqrt {Y - 4} $, where $R + $ is the set of all non-negative real numbers.

Ans:

Consider $f:{\mathbf{R}} +  \to [4,\infty )$ given by ${\text{f}}({\text{x}}) = {{\text{x}}^2} + 4$.

Let $x,y \in R \to [4,\infty )$ then,

$f(x) = {x^2} + 4$ and

$f(y) = {y^2} + 4$

if $f(x) = f(y)$ then,

 ${x^2} + 4 = {y^2} + 4$

or $x = y$

${\text{f}}$ is one-one.

Now $y = f(x) = {x^2} + 4$ or $x = \sqrt {y - 4} $ as $x > 0$

$f(\sqrt {y - 4} ) = {(\sqrt {y - 4} )^ \wedge }2 + 4 = y$

$f(x) = y$

$f$ is onto function. 

Therefore, $f$ is invertible and Inverse of $f$ is${f^{ - 1}}(y) = \sqrt {y - 4} $.

9. Consider f: ${{\mathbf{R}}_ + } \to [ - 5,\infty )$ given by ${\text{f}}({\text{x}}) = 9{{\text{x}}^2} + 6{\text{x}} - 5$. Show that ${\text{f}}$ is invertible with ${f^{ - 1}}(y) = \left( {\dfrac{{(\sqrt {y + 6} ) - 1}}{3}} \right)$.

Ans: Consider $f:{R_ + } \to [ - 5,\infty )$ given by $f(x) = 9{x^2} + 6x - 5$

Consider $f:{R_ + } \to [4,\infty )$ given by $f(x) = {x^2} + 4$

Let $x,y \in R \to [ - 5,\infty )$ then,

Let $y = 9{x^2} + 6x - 5$

Solve the quadratic equation,

$ \Rightarrow {\text{y}} = {(3{\text{x}} + 1)^2} - 1 - 5 = {(3{\text{x}} + 1)^2} - 6$

$ \Rightarrow y + 6 = {(3x + 1)^2}$

$ \Rightarrow 3x + 1 = \sqrt {{\text{Y}} + 6} \quad [$ as $y \geqslant  - 5 \Rightarrow y + 6 > 0]$

$ \Rightarrow x = \left( {\dfrac{{(\sqrt {y + 6} ) - 1}}{3}} \right)$

Thus,${\text{f}}$ is onto, thereby range ${\text{f}} = [ - 5,\infty )$.

Let us define $g:[ - 5,\infty ) \to {\text{R}} + $ as ${\text{g}}({\text{y}}) = \left( {\dfrac{{(\sqrt {{\text{y}} + 6} ) - 1}}{3}} \right)$

Now, $\left( {gof} \right)({\text{x}}) = {\text{g}}({\text{f}}({\text{x}})) = {\text{g}}\left( {9{{\text{x}}^2} + 6{\text{x}} - 5} \right) = {\text{g}}\left( {{{(3{\text{x}} + 1)}^2} - 6} \right)$

$ = \sqrt {{{(3x + 1)}^2} - 6 + 6}  - 1$

$ = \dfrac{{3x + 1 - 1}}{3} = \dfrac{{3x}}{3} = X$

and

$(fog)(y) = f(g(y)) = \left( {\dfrac{{\sqrt {Y + 6}  - 1}}{3}} \right) = \left[ {3\left( {\dfrac{{\sqrt {Y + 6}  - 1}}{3}} \right) + {1^2}} \right] - 6$

$ = {(\sqrt {Y + 6} )^2} - 6 =  + 6 - 6 = y$

Therefore,

$\;gof = {\text{x}} = {{\text{I}}_{\text{R}}}$ and fog $ = {\text{y}} = {l_{{\text{Range }}f}}$

Hence, ${\text{f}}$is invertible and the inverse of ${\text{f}}$is given by,

${f^{ - 1}}(y) = g(y) = \left( {\dfrac{{\sqrt {Y + 6}  - 1}}{3}} \right)$

10. Let ${\text{f}}:{\text{X}} \to {\text{Y}}$ be an invertible function. Show that ${\text{f}}$ has a unique inverse.

(Hint: suppose ${g_1}$ and ${g_2}$ are two inverses of ${\text{f}}$. Then for all ${\text{y}} \in {\text{Y}},{\operatorname{fog} _1}(y) = \operatorname{Ir} (y) = {\operatorname{fog} _2}(y)$. Use one - one ness of ${\text{f}}$).

Ans: Given, $f:X \to Y$ be an invertible function. And ${g_1}$ and${g_2}$ are two inverses of $f$.

For all $y \in Y$, we get

${\operatorname{fog} _1}(y) = {1_Y}(y) = f{\operatorname{fog} _2}(y)$

$f\left( {{g_1}(y)} \right) = f(g < (y))$

${g_1}(y) = {g_2}(y)$

${g_1} = {g_2}$

Therefore, $f$ has a unique inverse.

11. Consider ${\text{f}}:\{ 1,2,3\}  \to \{ {\text{a}},{\text{b}},{\text{c}}\} $ given by ${\text{f}}(1) = {\text{a}},{\text{f}}(2) = {\text{b}}$ and ${\text{f}}(3) = {\text{c}}$. Find ${{\text{f}}^{ - 1}}$

and show that ${\left( {{{\text{f}}^{ - 1}}} \right)^{ - 1}} = {\text{f}}$.

Ans: Function ${\text{f}}:\{ 1,2,3\}  \to \{ {\text{a}},{\text{b}},{\text{c}}\} $ given by ${\text{f}}(1) = {\text{a}},{\text{f}}(2) = {\text{b}}$ and ${\text{f}}(3) = {\text{c}}$.

So $f = \{ (a,1),(b,2),(c,3)\} $

Hence ${f^{ - 1}}(a) = 1,{f^{ - 1}}(b) = 2$ and ${f^{ - 1}}(c) = 3$

Now, The ${f^{ - 1}} = \{ (a,1),(b,2),(c,3)\} $

Therefore,
Inverse of ${{\text{f}}^{ - 1}} = {\left( {{{\text{f}}^{ - 1}}} \right)^{ - 1}} = \{ (1,{\text{a}}),(2,\;{\text{b}}),(3,{\text{c}})\}  = {\text{f}}$

Hence,${\left( {{f^{ - 1}}} \right)^{ - 1}} = f$.

12. Let ${\text{f}}:{\text{X}} \to {\text{Y}}$ be an invertible function. Show that the inverse of ${{\text{f}}^{ - 1}}$ is ${\text{f}}$,

i.e., ${\left( {{{\text{f}}^{ - 1}}} \right)^{ - 1}} = {\text{f}}$

Ans: Consider $f:{\text{X}} \to {\text{Y}}$ be an invertible function, 

So $f$ is one-one and onto 

Now $g:y \to X$ where $g$ is also one-one and onto $g^\circ f(x) = {{\text{I}}_z}$

$f^\circ g(y) = {1_y}$

So $g = {f^i}$

We know that,

${f^{ - 1}}o{\left( {{f^{ - 1}}} \right)^{ - 1}} = {I_{{\text{and }}}}f0\left[ {{f^{ - 1}}o{{\left( {{f^{ - 1}}} \right)}^{ - 1}}} \right] = fo{\text{I}}$

It can be written as $\left[ {fo{f^{ - 1}}} \right]o{\left( {{f^{ - 1}}} \right)^{ - 1}} = f$

By further simplification,

 $Io{\left( {{f^{ - 1}}} \right)^{ - 1}} = f$

We get ${\left( {{f^{ - 1}}} \right)^{ - 1}} = f$.

13. If ${\text{f}}:{\text{R}} \to {\text{R}}$ be given by $(x) = {\left( {3 - {x^3}} \right)^{\dfrac{1}{3}}}$, then fof $({\text{x}})$ is

a. $\dfrac{1}{{{x^3}}}$

b. ${{\text{x}}^3}$

c. ${\text{X}}$

d. $\left( {3 - {x^3}} \right)$

Ans: The function f: ${\text{R}} \to {\text{R}}$ be given by $f(x) = {\left( {3 - {x^3}} \right)^{\dfrac{1}{3}}}$, then

$fof(x) = f(f(x)$

Substitute the values,

$ = f\left( {{{\left( {3 - {x^\prime }} \right)}^{\dfrac{1}{7}}}} \right)$

$ = {\left[ {3 - {{\left( {{{\left( {3 - {x^2}} \right)}^{\dfrac{1}{4}}}} \right)}^2}} \right]^{\dfrac{1}{3}}}$

$ = {\left[ {3 - \left( {3 - {x^2}} \right)} \right]^{\dfrac{1}{3}}}$

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

Option (C) is correct.

14. Let f: ${\text{R}} - \left\{ { - \dfrac{4}{3}} \right\} \to {\text{R}}$ be a function as ${\text{f}}({\text{x}}) = \dfrac{{4x}}{{3x + 4}}$. The inverse of ${\text{f}}$ is map ${\text{g}}$ :

Range ${\text{f}} \to {\text{R}} - \left\{ { - \dfrac{4}{3}} \right\}$ given by

a. $g(y) = \dfrac{{3y}}{{3 - 4y}}$

b. $g(y) = \dfrac{{4y}}{{4 - 3y}}$

c. $g(y) = \dfrac{{4y}}{{4 - 3y}}$

d. $g(y) = \dfrac{{3y}}{{4 - 3y}}$

Ans: Let $f:R - \left\{ {\dfrac{{ - 4}}{3}} \right\} \to R$ be a function defined as $f(x) = \dfrac{{4x}}{{3x + 4}}$. And Range $f \to R - \left\{ {\dfrac{{ - 4}}{3}} \right\}$

$y = f(x) = \dfrac{{4x}}{{3x + 4}}$

$y(3x + 4) = 4x$

$3xy + 4y = 4x$

$x(3y - 4) =  - 4y$

$x = 4y/(4 - 3y)$

Therefore, ${f^{ - 1}}(y) = g(y) = \dfrac{{4y}}{{(4 - 3y)}}$. 

Option (B) is the correct answer.

List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 1 

Exercise 1.3 includes the questions related to the composition of functions and invertible functions.

Introduction About Types of Functions 

We are already well aware of identity function, rational function, polynomial function, constant function, etc. Also, we know how to perform addition, subtraction, multiplication and division between two functions.  

Types of functions

• One-one function (injective)

• Many-one functions 

• Onto function (surjective)

• One-one and onto (bijective)

Example for One-one function: Number of students in the class and their roll numbers. Because each student will have a specified roll number.

Example for Many-one function: Number of students in the class and their ranks, because more than two students can be in the same rank.  

Example for onto function: Students in a class and their hobbies. Because each student may have different hobbies or more than one student may have the same hobby. 

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The above image explains the types of functions in detail.

Ncert Solutions for Class 12 Maths Chapter 1 Includes the Following Topics -

• Introduction

• Types of Relations

• Types of Functions

Introduction to the Topic:

A collection of pairs that consists of one object from each set can be called as a relation between two sets. For example, if the object X is from the first set while the object Y is from the second set, then these both objects can be called as related to each other if they are in the ordered pair.

To make your understanding of relation and function Class 12 Exercise 1.3 Some of the relations from X to Y are as shown in the example below:

• {(x, y) Î X × Y: x is the brother of y},

• {(x, y) Î X × Y: x is the sister of y},

Types of Relations:

In Class 12 Maths Exercise 1.3 Solution consists of different types of relations. As we studied earlier, Set X is a subset of X x X. Hence, the empty set and X x X are closely related to each other. For reference, let's take a relation R in the set X={ 1,2,3,4} given by R = { (X,Y): X-Y = 10. Now, this will be considered as an empty set as no pair (x,y) satisfies the condition x-y=10. Similarly, R = { (x,y) : | x-y| > 0} is the whole set X x X as everything here satisfies the equation. 

Here, both the empty relations and the universal relations can also be called trivial relations.

Now let us study an example to make your understanding more precise.

Question: X is the arrangement of all students of a young boys' school. Show that the relation R in X given by R = {(x, y) : x is sister of y} is an empty relation and R = {(x, y) : the distinction between heights of x and y is under 3 meters} is the universal relation. 

The school is a boys' school - hence no student can be female and be the sister of any student. Therefore, R is equal to the null set showing that R is the empty relation. It also shows that the difference between the heights of both the students is less than 3 meters, which shows that R= X x X, which is the universal relation.

Types of Functions:

In the above paragraph, we studied types of relations now to make your basics of the chapter clear. We will see the types of functions. Have a look: 

The notion of a function alongside some extraordinary capacities like personality work, consistent capacity, polynomial capacity, standard capacity, modulus, signum work, and so on alongside their diagrams have been given in Class XI. 

We have also studied subtraction, Addition, multiplication, and division related to it as the concept of function holds the maximum amount of importance in Maths and numerous other exercises as well. Now we will study about function from where we finished earlier. 

That is all for the theory part, now let us study a few of the examples to make your understanding more undoubtedly.

Show that the function f : N N, given by f (1) = f (2) = 1 and f (x) = x – 1, for every x > 2, is onto but not one-one.

Solution: f is not one-one, as f (1) = f (2) = 1. But f is onto, as given  y N, y 1, we can now select x as y + 1 such that f (y + 1) = y + 1 – 1 = y. For 1 N, we have f (1) = 1 as well.

Overview of the Chapter:

A student needs to get precise knowledge about functions and relations in math as well as in life as well. The concept is extremely helpful in life learning as well. Hence, the chapter shows realistic examples of relations so that a student can grab it more easily. What is more? Even the exercise 1.3 Class 12 Maths Solutions consist of all the NCERT Solutions that could be asked in examinations and as per the NCERT guidelines.

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Class 12 Maths Chapter 1.

It explains easy methods and helps students with all the difficult questions that can be asked in the examinations. The students can likewise contact the specialists on the off chance that they are confronting issues with the arrangements that are given in the PDF. The PDF solutions contain all the end questions and solutions that are significant and, as indicated by the CBSE rules. So if you are someone struggling with the solutions for Exercise

1.3 Maths Class 12

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NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Students can also download the following additional study material -

• Chapter 1: Important Questions

• Chapter 1: Revision Notes

• Chapter 1: Formulas

• Chapter 1: NCERT Exemplar Solutions

• Chapter 1: RD Sharma Solutions

• Chapter 1: RS Aggarwal Solutions