# NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.3

## Chapter 1 Ex.1.3 Question 1

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

### Solution

The functions $$f:\left\{ {1,3,4} \right\} \to \left\{ {1,2,5} \right\}$$ and $$g:\left\{ {1,2,5} \right\} \to \left\{ {1,3} \right\}$$ are $$f = \left\{ {\left( {1,2} \right),\left( {3,5} \right),\left( {4,1} \right)} \right\}$$ and $$g = \left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {5,1} \right)} \right\}$$

\begin{align} gof\left( 1 \right) = g\left[ {f\left( 1 \right)} \right] = g\left( 2 \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 1 \right) = 2\,and\,g\left( 2 \right) = 3} \right]\\ gof\left( 3 \right) = g\left[ {f\left( 3 \right)} \right] = g\left( 5 \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 3 \right) = 5\,and\,g\left( 5 \right) = 1} \right]\\ gof\left( 4 \right) = g\left[ {f\left( 4 \right)} \right] = g\left( 1 \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 4 \right) = 1\,and\,g\left( 1 \right) = 3} \right] \end{align}

$$\therefore gof = \left\{ {\left( {1,3} \right),\left( {3,1} \right),\left( {4,3} \right)} \right\}$$

## Chapter 1 Ex.1.3 Question 2

Let $$f,g,h$$be functions from $$R{\text{ to }}R$$.Show that

\begin{align} \left( {f + g} \right)oh = foh + goh\\ \left( {f.g} \right)oh = \left( {foh} \right).\left( {goh} \right) \end{align}

### Solution

$$\left( {f + g} \right)oh = foh + goh$$

\begin{align} LHS &= \left[ {\left( {f + g} \right)oh} \right]\left( x \right)\\ &= \left( {f + g} \right)\left[ {h\left( x \right)} \right]= f\left[ {h\left( x \right)} \right] + g\left[ {h\left( x \right)} \right]\\ &= \left( {foh} \right)\left( x \right) + goh\left( x \right)\\ & = \left\{ {\left( {foh} \right) + \left( {goh} \right)} \right\}\left( x \right) = RHS \end{align}

$\therefore \left\{ {\left( {f + g} \right)oh} \right\}\left( x \right) = \left\{ {\left( {foh} \right) + \left( {goh} \right)} \right\}\left( x \right)\,\,\,\,for\,all\,x \in R$

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

$$\left( {f.g} \right)oh = \left( {foh} \right).\left( {goh} \right)$$

\begin{align} LHS& = \left[ {\left( {f.g} \right)oh} \right]\left( x \right)\\ &= \left( {f.g} \right)\left[ {h\left( x \right)} \right] = f\left[ {h\left( x \right)} \right].g\left[ {h\left( x \right)} \right]\\ &= \left( {foh} \right)\left( x \right).\left( {goh} \right)\left( x \right)\\ & = \left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right) = RHS \end{align}

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

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

## Chapter 1 Ex.1.3 Question 3

Find $$gof$$ and $$fog$$, if

(i) $$f\left( x \right) = \left| x \right|$$and $$g\left( x \right) = \left| {5x - 2} \right|$$

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

### Solution

(i) $$f\left( x \right) = \left| x \right|$$ and $$g\left( x \right) = \left| {5x - 2} \right|$$

\begin{align} \therefore gof\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left| {5\left| x \right| - 2} \right|\\ fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f\left( {\left| {5x - 2} \right|} \right) = \left| {\left| {5x - 2} \right|} \right| = \left| {5x - 2} \right| \end{align}

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

\begin{align} \therefore gof\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {8{x^3}} \right) = {\left( {8{x^3}} \right)^{\frac{1}{3}}} = 2x\\ fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f{\left( {{x^{\frac{1}{3}}}} \right)^3} = 8{\left( {{x^{\frac{1}{3}}}} \right)^3} = 8x \end{align}

## Chapter 1 Ex.1.3 Question 4

If $$f\left( x \right) = \frac{{\left( {4x + 3} \right)}}{{\left( {6x - 4} \right)}},x \ne \frac{2}{3}$$, show that $$fof\left( x \right) = x,\,\,\,for\,all\,x \ne \frac{2}{3}$$. What is the reverse of $$f$$?

### Solution

\begin{align} \left( {fof} \right)\left( x \right) &= f\left( {f\left( x \right)} \right) = f\left( {\frac{{4x + 3}}{{6x - 4}}} \right)\\ & = \frac{{4\left( {\frac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\frac{{4x + 3}}{{6x - 4}}} \right) - 4}}\\& = \frac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}}\\& = \frac{{34x}}{{34}}\\&= x \end{align}

\begin{align}& \therefore fof\left( x \right) = x\,\,\,\,\,for\,\,all\,\,x \ne \frac{2}{3}\\ & \Rightarrow\; fof = 1 \end{align}

Hence, the given function $$f$$is invertible and the inverse of $$f$$is $$f$$itself.

## Chapter 1 Ex.1.3 Question 5

State with reason whether the following functions have inverse.

(i) $$f:\left\{ {1,2,3,4} \right\} \to \left\{ {10} \right\}$$with $$f = \left\{ {\left( {1,10} \right),\left( {2,10} \right),\left( {3,10} \right),\left( {4,10} \right)} \right\}$$

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

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

### Solution

(i) $$f:\left\{ {1,2,3,4} \right\} \to \left\{ {10} \right\}$$ with $$f = \left\{ {\left( {1,10} \right),\left( {2,10} \right),\left( {3,10} \right),\left( {4,10} \right)} \right\}$$

$$f$$ is a many one function as $$f\left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = f\left( 4 \right) = 10$$

$$\therefore f$$ is not one-one.

Function $$f$$does not have an inverse.

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

$$g$$ is a many one function as $$g\left( 5 \right) = g\left( 7 \right) = 4$$

$$\therefore g$$ is not one-one.

Function $$g$$does not have an inverse.

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

All distinct elements of the set $$\left\{ {2,3,4,5} \right\}$$have distinct images under $$h$$.

$$\therefore h$$ is one-one.

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

$$h$$is a one-one and onto function.

Function $$h$$ has an inverse.

## Chapter 1 Ex.1.3 Question 6

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

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

### Solution

$$f:\left[ { - 1,1} \right] \to R$$, given by $$f\left( x \right) = \frac{x}{{\left( {x + 2} \right)}}$$

For one-one

\begin{align} &f\left( x \right) = f\left( y \right)\\ & \Rightarrow \; \frac{x}{{x + 2}} = \frac{y}{{y + 2}}\\ & \Rightarrow \; xy + 2x = xy + 2y\\ &\Rightarrow \;2x = 2y\\& \Rightarrow \; x = y \end{align}

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

It is clear that $$f:\left[ { - 1,1} \right] \to R$$is onto.

$$\therefore$$ $$f:\left[ { - 1,1} \right] \to R$$is one-one and onto and therefore, the inverse of the function

$$f:\left[ { - 1,1} \right] \to R$$exists.

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

Let $$y$$ be an arbitrary element of range $$f$$.

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

\begin{align} &y = f\left( x \right)\,\,\,for\,\,same\,\,x \in \left[ { - 1,1} \right]\\& \Rightarrow\; y = \frac{x}{{x + 2}}\\ & \Rightarrow \;xy + 2y = x\\& \Rightarrow \;x\left( {1 - y} \right) = 2y\\ &\Rightarrow \;x = \frac{{2y}}{{1 - y}},y \ne 1 \end{align}

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

$$g\left( y \right) = \frac{{2y}}{{1 - y}},y \ne 1$$

Now,

\begin{align}&\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\frac{x}{{x + 2}}} \right) = \frac{{2\left( {\frac{x}{{x + 2}}} \right)}}{{1 - \frac{x}{{x + 2}}}} = \frac{{2x}}{{x + 2 - x}} = \frac{{2x}}{2} = x\\&\left( {fog} \right)\left( x \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{2y}}{{1 - y}}} \right) = \frac{{\frac{{2y}}{{1 - y}}}}{{\frac{{2y}}{{1 - y}} + 2}} = \frac{{2y}}{{2y + 2 - 2y}} = \frac{{2y}}{2} = y\\ &\therefore gof= {I_{\left[ { - 1,1} \right]}}\,\,\,\,\,\,\,and\,\,\,\,\,\,\,fog = {I_{Range\,f}}\\&\therefore {f^{ - 1}} &= g\\ &\Rightarrow\; {f^{ - 1}}\left( y \right) = \frac{{2y}}{{1 - y}},y \ne 1\end{align}

## Chapter 1 Ex.1.3 Question 7

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

### Solution

$$f:R \to R$$given by$$f\left( x \right) = 4x + 3$$

For one-one

\begin{align} &f\left( x \right) = f\left( y \right)\\& \Rightarrow \;4x + 3 = 4y + 3\\ & \Rightarrow \;4x = 4y\\ & \Rightarrow \;x = y \end{align}

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

For onto

\begin{align}& y \in R,\,\,\,let\,\,y = 4x + 3\\& \Rightarrow \;x = \frac{{y - 3}}{4} \in R \end{align}

Therefore, for any $$y \in R$$, there exists $$x = \frac{{y - 3}}{4} \in R$$ such that

$$f\left( x \right) = f\left( {\frac{{y - 3}}{4}} \right) = 4\left( {\frac{{y - 3}}{4}} \right) + 3 = y$$

$$\therefore f$$is onto.

Thus, f is one-one and onto and therefore, $${f^{ - 1}}$$exists.

Let us define $$g:R \to R$$by $$g\left( x \right) = \frac{{y - 3}}{4}$$

Now,

\begin{align}& \left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {4x + 3} \right) = \frac{{\left( {4x + 3} \right) - 3}}{4} = x\\& \left( {fog} \right)\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{y - 3}}{4}} \right) = 4\left( {\frac{{y - 3}}{4}} \right) + 3 = y - 3 + 3 = y\\ &\therefore gof = fog = {{\rm{I}}_R} \end{align}

Hence, $$f$$is invertible and the inverse of $$f$$is given by

$${f^{ - 1}}\left( y \right) = g\left( y \right) = \frac{{y - 3}}{4}$$.

## Chapter 1 Ex.1.3 Question 8

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

### Solution

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

For one-one:

Let $$f\left( x \right) = f\left( y \right)$$

\begin{align}& \Rightarrow \;{x^2} + 4 = {y^2} + 4\\& \Rightarrow\; {x^2} = {y^2}\\ &\Rightarrow\; x = y \qquad \left[ {as\,\,x \in R} \right] \end{align}

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

For onto:

For $$y \in [4,\infty ),\,\,\,let\,\,y = {x^2} + 4$$

\begin{align}& \Rightarrow \; {x^2} = y - 4 \ge 0 \qquad \left[ {as\,\,y \ge 4} \right]\\ &\Rightarrow \; x = \sqrt {y - 4} \ge 0 \end{align}

Therefore, for any $$y \in R$$, there exists $$x = \sqrt {y - 4} \in R$$ such that

$$f\left( x \right) = f\left( {\sqrt {y - 4} } \right) = {\left( {\sqrt {y - 4} } \right)^2} + 4 = y - 4 + 4 = y$$

$$\therefore f$$ is an onto function.

Thus, $$f$$ is one-one and onto and therefore, $${f^{ - 1}}$$exists.

Let us define $$g:[4,\infty ) \to {R_ + }$$by

$$g\left( y \right) = \sqrt {y - 4}$$

Now, $$gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {{x^2} + 4} \right) = \sqrt {\left( {{x^2} + 4} \right) - 4} = \sqrt {{x^2}} = x$$

And $$fog\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\sqrt {y - 4} } \right) = {\left( {\sqrt {y - 4} } \right)^2} + 4 = \left( {y - 4} \right) + 4 = y$$

$$\therefore gof = fog = {{\rm{I}}_R}$$

Hence, $$f$$ is invertible and the inverse of $$f$$ is given by

$${f^{ - 1}}\left( y \right) = g\left( y \right) = \sqrt {y - 4}$$.

## Chapter 1 Ex.1.3 Question 9

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

### Solution

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

Let $$y$$ be an arbitrary element of $$[ - 5,\infty )$$.

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

\begin{align} & \Rightarrow y = {\left( {3x + 1} \right)^2} - 1 - 5\\ & \Rightarrow y = {\left( {3x + 1} \right)^2} - 6\\ & \Rightarrow {\left( {3x + 1} \right)^2} = y + 6\\ & \Rightarrow 3x + 1 = \sqrt {y + 6} \qquad \left[ {as\,\,y \ge - 5 \Rightarrow y + 6 > 0} \right]\\& \Rightarrow x = \frac{{\sqrt {y + 6} - 1}}{3} \end{align}

$$\therefore f$$ is onto, thereby range$$f = [ - 5,\infty )$$.

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

We have,

\begin{align} \left( {gof} \right)\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {9{x^2} + 6x - 5} \right)\\&= g\left( {{{\left( {3x + 1} \right)}^2} - 6} \right)\\ &= \frac{{\sqrt {{{\left( {3x + 1} \right)}^2} - 6 + 6} - 1}}{3}\\&= \frac{{3x + 1 - 1}}{3} = x\\ \end{align}

And,

\begin{align} \left( {fog} \right)\left( y \right)& = f\left( {g\left( y \right)} \right) = f\left( {\frac{{\sqrt {y + 6} - 1}}{3}} \right)\\&= {\left[ {3\left( {\frac{{\sqrt {y + 6} - 1}}{3}} \right) + 1} \right]^2} - 6\\&= {\left( {\sqrt {y + 6} } \right)^2} - 6 = y + 6 - 6 = y \end{align}

$$\therefore gof = {{\rm{I}}_R}\,\,\,and\,\,\,fog = {{\rm{I}}_{[ - 5,\infty )}}$$

Hence, $$f$$ is invertible and the inverse of $$f$$ is given by

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

## Chapter 1 Ex.1.3 Question 10

Let $$f:X \to Y$$ be an invertible function. Show that $$f$$ has unique inverse.

(Hint: suppose $${g_1}$$ and $${g_2}$$ are two inverses of $$f$$. Then for all $$y \in Y$$, $$fo{g_1}\left( y \right) = {{\rm{I}}_Y}\left( y \right) = fo{g_2}\left( y \right)$$. Use one-one ness of $$f$$.

### Solution

Let $$f:X \to Y$$ be an invertible function.

Also suppose $$f$$ has two inverses ($${g_1}$$and $${g_2}$$)

Then, for all $$y \in Y$$,

\begin{align}&fo{g_1}\left( y \right) = {{\rm{I}}_Y}\left( y \right) = fo{g_2}\left( y \right)\\& \Rightarrow f\left( {{g_1}\left( y \right)} \right) = f\left( {{g_2}\left( y \right)} \right)\\ & \Rightarrow {g_1}\left( y \right) = {g_2}\left( y \right) \qquad \left[ {f\,{\rm{is}}\,{\rm{invertible}} \Rightarrow f\,{\rm{is}}\,{\rm{one - one}}} \right]\\& \Rightarrow {g_1} = {g_2} \qquad \qquad \;\; \left[ {g\,{\rm{is}}\,{\rm{one - one}}} \right] \end{align}

Hence, $$f$$ has unique inverse.

## Chapter 1 Ex.1.3 Question 11

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

### Solution

Function $$f:\left\{ {1,2,3} \right\} \to \left\{ {a,b,c} \right\}$$given by $$f\left( 1 \right) = a,f\left( 2 \right) = b,f\left( 3 \right) = c$$

If we define $$g:\left\{ {a,b,c} \right\} \to \left\{ {1,2,3} \right\}$$as$$g\left( a \right) = 1,g\left( b \right) = 2,g\left( c \right) = 3$$

\begin{align} \left( {fog} \right)\left( a \right) = f\left( {g\left( a \right)} \right) = f\left( 1 \right) = a\\ \left( {fog} \right)\left( b \right) = f\left( {g\left( b \right)} \right) = f\left( 2 \right) = b\\ \left( {fog} \right)\left( c \right) = f\left( {g\left( c \right)} \right) = f\left( 3 \right) = c \end{align}

And,

\begin{align} \left( {gof} \right)\left( 1 \right) = g\left( {f\left( 1 \right)} \right) = g\left( a \right) = 1\\ \left( {gof} \right)\left( 2 \right) = g\left( {f\left( 2 \right)} \right) = g\left( b \right) = 2\\ \left( {gof} \right)\left( 3 \right) = g\left( {f\left( 3 \right)} \right) = g\left( c \right) = 3 \end{align}

$$\therefore gof = {{\rm{I}}_X}\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,fog = {{\rm{I}}_Y}$$ where $$X = \left\{ {\left( {1,2,3} \right)} \right\}\,\,\,{\rm{and}}\,\,\,Y = \left\{ {a,b,c} \right\}$$

Thus, the inverse of $$f$$exists and$${f^{ - 1}} = g$$.

$$\therefore {f^{ - 1}}:\left\{ {a,b,c} \right\} \to \left\{ {1,2,3} \right\}$$ is given by, $${f^{ - 1}}\left( a \right) = 1,{f^{ - 1}}\left( b \right) = 2,{f^{ - 1}}\left( c \right) = 3$$

We need to find the inverse of $${f^{ - 1}}$$ i.e., inverse of $$g$$.

If we define $$h:\left\{ {1,2,3} \right\} \to \left\{ {a,b,c} \right\}$$as $$h\left( 1 \right) = a,h\left( 2 \right) = b,h\left( 3 \right) = c$$

\begin{align} \left( {goh} \right)\left( 1 \right) = g\left( {h\left( 1 \right)} \right) = g\left( a \right) = 1\\ \left( {goh} \right)\left( 2 \right) = g\left( {h\left( 2 \right)} \right) = g\left( b \right) = 2\\ \left( {goh} \right)\left( 3 \right) = g\left( {h\left( 3 \right)} \right) = g\left( c \right) = 3 \end{align}

And,

\begin{align} \left( {hog} \right)\left( a \right) = h\left( {g\left( a \right)} \right) = h\left( 1 \right) = a\\ \left( {hog} \right)\left( b \right) = h\left( {g\left( b \right)} \right) = h\left( 2 \right) = b\\ \left( {hog} \right)\left( c \right) = h\left( {g\left( c \right)} \right) = h\left( 3 \right) = c \end{align}

$$\therefore goh = {{\rm{I}}_X}\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,hog = {{\rm{I}}_Y}$$ where $$X = \left\{ {\left( {1,2,3} \right)} \right\}\,\,\,{\rm{and}}\,\,\,Y = \left\{ {a,b,c} \right\}$$

Thus, the inverse of $$g$$exists and$${g^{ - 1}} = h \Rightarrow {\left( {{f^{ - 1}}} \right)^{ - 1}} = h$$.

It can be noted that $$h = f$$.

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

## Chapter 1 Ex.1.3 Question 12

Let $$f:X \to Y$$ be an invertible function. Show that the inverse of $${f^{ - 1}}$$ is $$f$$ i.e.,$${\left( {{f^{ - 1}}} \right)^{ - 1}} = f$$.

### Solution

Let $$f:X \to Y$$ be an invertible function.

Then there exists a function $$g:Y \to X$$ such that $$gof = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fog = {{\rm{I}}_Y}$$

Here, $${f^{ - 1}} = g$$

Now, $$gof = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fog = {{\rm{I}}_Y}$$

$$\Rightarrow {f^{ - 1}}of = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fo{f^{ - 1}} = {{\rm{I}}_Y}$$

Hence, $${f^{ - 1}}:Y \to X$$ is invertible and $${f^{ - 1}}$$is $$f$$ i.e.,$${\left( {{f^{ - 1}}} \right)^{ - 1}} = f$$.

## Chapter 1 Ex.1.3 Question 13

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

(A) $$\frac{1}{{{x^3}}}$$

(B) $${x^3}$$

(C) $$x$$

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

### Solution

$$f:R \to R$$ is given by $$f\left( x \right) = {\left( {3 - {x^3}} \right)^{\frac{1}{3}}}$$

\begin{align} f\left( x \right)& = {\left( {3 - {x^3}} \right)^{\frac{1}{3}}}\\ \therefore fof\left( x \right) &= f\left( {f\left( x \right)} \right) = f\left( {{{\left( {3 - {x^3}} \right)}^{\frac{1}{3}}}} \right) = {\left[ {3 - {{\left( {{{\left( {3 - {x^3}} \right)}^{\frac{1}{3}}}} \right)}^3}} \right]^{\frac{1}{3}}}\\ &= {\left[ {3 - \left( {3 - {x^3}} \right)} \right]^{\frac{1}{3}}} = {\left( {{x^3}} \right)^{\frac{1}{3}}} = x\\ \therefore fof\left( x \right)& = x \end{align}

## Chapter 1 Ex.1.3 Question 14

If $$f:R - \left\{ { - \frac{4}{3}} \right\} \to R$$ be a function defined as$$f\left( x \right) = \frac{{4x}}{{3x + 4}}$$. The inverse of $$f$$ is the map $$g:Range\,f \to R - \left\{ { - \frac{4}{3}} \right\}$$given by :

(A) $$g\left( y \right) = \frac{{3y}}{{3 - 4y}}$$

(B) $$g\left( y \right) = \frac{{4y}}{{4 - 3y}}$$

(C) $$g\left( y \right) = \frac{{4y}}{{3 - 4y}}$$

(D) $$g\left( y \right) = \frac{{3y}}{{4 - 3y}}$$

### Solution

It is given that $$f:R - \left\{ { - \frac{4}{3}} \right\} \to R$$ is defined as $$f\left( x \right) = \frac{{4x}}{{3x + 4}}$$

Let $$y$$ be an arbitrary element of Range $$f$$.

Then, there exists $$x \in R - \left\{ { - \frac{4}{3}} \right\}$$such that $$y = f\left( x \right)$$.

\begin{align} & \Rightarrow \;y = \frac{{4x}}{{3x + 4}}\\& \Rightarrow\;3xy + 4y = 4x\\& \Rightarrow\; x\left( {4 - 3y} \right) = 4y\\& \Rightarrow \;x = \frac{{4y}}{{4 - 3y}} \end{align}

Define $$f:R - \left\{ { - \frac{4}{3}} \right\} \to R$$ as $$g\left( y \right) = \frac{{4y}}{{4 - 3y}}$$

Now,

\begin{align}\left( {gof} \right)\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {\frac{{4x}}{{3x + 4}}} \right)\\&= \frac{{4\left( {\frac{{4x}}{{3x + 4}}} \right)}}{{4 - 3\left( {\frac{{4x}}{{3x + 4}}} \right)}} = \frac{{16x}}{{12x + 16 - 12x}}\\ &= \frac{{16x}}{{16}} = x\end{align}

And

\begin{align} \left( {fog} \right)\left( x \right) &= \left( {g\left( x \right)} \right) = f\left( {\frac{{4y}}{{4 - 3y}}} \right)\\ &= \frac{{4\left( {\frac{{4y}}{{4 - 3y}}} \right)}}{{3\left( {\frac{{4y}}{{4 - 3y}}} \right) + 4}} = \frac{{16y}}{{12y + 16 - 12y}}\\& = \frac{{16y}}{{16}} = y \end{align}

$$\therefore gof = {{\rm{I}}_{R - \left\{ { - \frac{4}{3}} \right\}}}{\rm{ and }}fog = {{\rm{I}}_{Range\,f}}{\rm{ }}$$

Thus, $$g$$ is the inverse of $$f$$ i.e., $${f^{ - 1}} = g$$

Hence, the inverse of $$f$$ is the map $$g:Range\,f \to R - \left\{ { - \frac{4}{3}} \right\}$$, which is given by $$g\left( y \right) = \frac{{4y}}{{4 - 3y}}$$.

The correct answer is $$B$$.

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