Q: Let $f:\{1,3,4\} \rightarrow\{1,2,5\}$ and $g:\{1,2,5\} \rightarrow\{1,3\}$ be given by $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(1,3),(2,3),(5,1)\}$. Write down $g o f$.

The functions $f:\{1,3,4\} \rightarrow\{1,2,5\}$ and $g:\{1,2,5\} \rightarrow\{1,3\}$ are $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(1,3),(2,3),(5,1)\}$ $g \circ f(1)=g[f(1)]=g(2)=3$ $[$ as $f(1)=2$ and $g(2)=3]$ $g \circ f(3)=g[f(3)]=g(5)=1$ $[$ as $f(3)=5$ and $g(5)=1]$ $g \circ f(4)=g[f(4)]=g(1)=3$ $[$ as $f(4)=1$ and $g(1)=3]$ $\therefore g o f=\{(1,3),(3,1),(4,3)\}$

Q: Find $g o f$ and $f o g$, if i. $\quad f(x)=|x|$ and $g(x)=|5 x-2|$ ii. $f(x)=8 x^{3}$ and $g(x)=x^{\frac{1}{3}}$

i. $\quad f(x)=|x|$ and $g(x)=|5 x-2|$ $\therefore g \circ f(x)=g(f(x))=g(|x|)=|5| x|-2|$ $f o g(x)=f(g(x))=f(|5 x-2|)=||5 x-2||=|5 x-2|$ ii. $\quad f(x)=8 x^{3}$ and $g(x)=x^{\frac{1}{3}}$ $$ \begin{aligned} & \therefore g o f(x)=g(f(x))=g\left(8 x^{3}\right)=\left(8 x^{3}\right)^{\frac{1}{3}}=2 x \\ & f o g(x)=f(g(x))=f\left(x^{\frac{1}{3}}\right)^{3}=8\left(x^{\frac{1}{3}}\right)^{3}=8 x \end{aligned} $$

Q: If $f(x)=\frac{(4 x+3)}{(6 x-4)}, x \neq \frac{2}{3}$, show that $f o f(x)=x$, for all $x \neq \frac{2}{3}$. What is the reverse of $f$ ?

$(f o f)(x)=f(f(x))=f\left(\frac{4 x+3}{6 x-4}\right)$ $$ =\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}=\frac{16 x+12+18 x-12}{24 x+18-24 x+16}=\frac{34 x}{34}=x $$ $\therefore$ fof $(x)=x \quad$ for all $x \neq \frac{2}{3}$ $\Rightarrow f o f=1$ Hence, the given function $f$ is invertible and the inverse of $f$ is $f$ itself.

Q: State with reason whether the following functions have inverse. i. $\quad f:\{1,2,3,4\} \rightarrow\{10\}_{\text {with }} f=\{(1,10),(2,10),(3,10),(4,10)\}$ ii. $g:\{5,6,7,8\} \rightarrow\{1,2,3,4\}$ with $g=\{(5,4),(6,3),(7,4),(8,2)\}$ iii. $\quad h:\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$

i. $\quad f:\{1,2,3,4\} \rightarrow\{10\}_{\text {with }} f=\{(1,10),(2,10),(3,10),(4,10)\}$ $f$ is a many one function as $f(1)=f(2)=f(3)=f(4)=10$ $\therefore f$ is not one-one. Function $f$ does not have an inverse. ii. $g:\{5,6,7,8\} \rightarrow\{1,2,3,4\}$ with $g=\{(5,4),(6,3),(7,4),(8,2)\}$ $g$ is a many one function as $g(5)=g(7)=4$ $\therefore g$ is not one-one. Function $g$ does not have an inverse. iii. $\quad h:\{2,3,4,5\} \rightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$ All distinct elements of the set $\{2,3,4,5\}$ have distinct images under $h$. $\therefore h$ is one-one. $h$ is onto since for every element $y$ of the set $\{7,9,11,13\}$, there exists an element $x$ in the set $\{2,3,4,5\}$, such that $h(x)=y$. $h$ is a one-one and onto function. Function $h$ has an inverse.

Q: Show that $f:[-1,1] \rightarrow R$, given by $f(x)=\frac{x}{(x+2)}$ is one-one. Find the inverse of the function $f:[-1,1] \rightarrow$ Range $f$. (Hint: For $y \in$ Range $f, y=f(x)=\frac{x}{x+2}$, for some $x$ in $[-1,1], \mathrm{i}, \mathrm{e} ., \quad x=\frac{2 y}{(1-y)}$

$f:[-1,1] \rightarrow R$, given by $f(x)=\frac{x}{(x+2)}$ For one-one $f(x)=f(y)$ $\Rightarrow \frac{x}{x+2}=\frac{y}{y+2}$ $\Rightarrow x y+2 x=x y+2 y$ $\Rightarrow 2 x=2 y$ $\Rightarrow x=y$ $\therefore f$ is a one-one function. It is clear that $f:[-1,1] \rightarrow R$ is onto. $f:[-1,1] \rightarrow R$ is one-one and onto and therefore, the inverse of the function $f:[-1,1] \rightarrow R$ exists. Let $g:$ Range $f \rightarrow[-1,1]$ be the inverse of $f$. Let $y$ be an arbitrary element of range $f$. Since $f:[-1,1] \rightarrow$ Range $f$ is onto, we have: $y=f(x)$ for same $x \in[-1,1]$ $\Rightarrow y=\frac{x}{x+2}$ $\Rightarrow x y+2 y=x$ $\Rightarrow x(1-y)=2 y$ $\Rightarrow x=\frac{2 y}{1-y}, y \neq 1$ Now, let us define $g:$ Range $f \rightarrow[-1,1]$ as $g(y)=\frac{2 y}{1-y}, y \neq 1$ Now, $(g \circ f)(x)=g(f(x))=g\left(\frac{x}{x+2}\right)=\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}=\frac{2 x}{x+2-x}=\frac{2 x}{2}=x$ $(f o g)(x)=f(g(y))=f\left(\frac{2 y}{1-y}\right)=\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}=\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}=y$ $\therefore g \circ f=I_{[-1,1]} \quad$ and $\quad f o g=I_{\text {Range } f}$ $\therefore f^{-1}=g$ $\Rightarrow f^{-1}(y)=\frac{2 y}{1-y}, y \neq 1$

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

$f: R \rightarrow R$ given by $f(x)=4 x+3$ For one-one $f(x)=f(y)$ $\Rightarrow 4 x+3=4 y+3$ $\Rightarrow 4 x=4 y$ $\Rightarrow x=y$ $\therefore f$ is a one-one function. For onto $y \in R$, let $y=4 x+3$ $\Rightarrow x=\frac{y-3}{4} \in R$ Therefore, for any $y \in R$, there exists $x=\frac{y-3}{4} \in R$ such that $f(x)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y$ $\therefore f$ is onto. Thus, f is one-one and onto and therefore, $f^{-1}$ exists. Let us define $g: R \rightarrow R$ by $g(x)=\frac{y-3}{4}$ Now, $(g \circ f)(x)=g(f(x))=g(4 x+3)=\frac{(4 x+3)-3}{4}=x$ $(f o g)(y)=f(g(y))=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y$ $\therefore g \circ f=f \circ g=\mathrm{I}_{R}$ Hence, $f$ is invertible and the inverse of $f$ is given by $f^{-1}(y)=g(y)=\frac{y-3}{4}$.

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

$f: R_{+} \rightarrow[4, \infty)$ given by $f(x)=x^{2}+4$ For one-one: Let $f(x)=f(y)$ $\Rightarrow x^{2}+4=y^{2}+4$ $\Rightarrow x^{2}=y^{2}$ $\Rightarrow x=y \quad[$ as $x \in R]$ $\therefore f$ is a one -one function. For onto: For $y \in[4, \infty)$, let $y=x^{2}+4$ $\Rightarrow x^{2}=y-4 \geq 0 \quad[$ as $y \geq 4]$ $\Rightarrow x=\sqrt{y-4} \geq 0$ Therefore, for any $y \in R$, there exists $x=\sqrt{y-4} \in R$ such that $f(x)=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=y-4+4=y$ $\therefore f$ is an onto function. Thus, $f$ is one-one and onto and therefore, $f^{-1}$ exists. Let us define $g:[4, \infty) \rightarrow R_{+}$by $g(y)=\sqrt{y-4}$ Now, $\operatorname{gof}(x)=g(f(x))=g\left(x^{2}+4\right)=\sqrt{\left(x^{2}+4\right)-4}=\sqrt{x^{2}}=x$ And $f \circ g(y)=f(g(y))=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=(y-4)+4=y$ $\therefore g \circ f=f o g=\mathrm{I}_{R}$ Hence, $f$ is invertible and the inverse of $f$ is given by $f^{-1}(y)=g(y)=\sqrt{y-4}$.

Question 1

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

Accepted Answer

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

$g \circ f(1)=g[f(1)]=g(2)=3$

$[$ as $f(1)=2$ and $g(2)=3]$

$g \circ f(3)=g[f(3)]=g(5)=1$

$[$ as $f(3)=5$ and $g(5)=1]$

$g \circ f(4)=g[f(4)]=g(1)=3$

$[$ as $f(4)=1$ and $g(1)=3]$

$	herefore g o f=\{(1,3),(3,1),(4,3)\}$

Question 2

Let $f, g, h$ be functions from $R$ to $R$. Show that

$(f+g) o h=f o h+g o h$

$(f . g) o h=(f o h) .(g o h)$

Accepted Answer

$(f+g) o h=f o h+g o h$

LHS $=[(f+g) o h](x)$

$$

\begin{aligned}

& =(f+g)[h(x)]=f[h(x)]+g[h(x)] \

& =(f o h)(x)+g o h(x) \

& =\{(f o h)+(g o h)\}(x)=	ext { RHS }

\end{aligned}

$$

$	herefore\{(f+g)$ oh $\}(x)=\{($ foh $)+($ goh $)\}(x)$ for all $x \in R$

Hence, $(f+g) o h=f o h+g o h$

$(f . g) o h=(f o h) .(g o h)$

$L H S=[(f . g) o h](x)$

$$

\begin{aligned}

& =(f \cdot g)[h(x)]=f[h(x)] \cdot g[h(x)] \

& =(f o h)(x) \cdot(g o h)(x) \

& =\{(f o h) \cdot(g o h)\}(x)=R H S

\end{aligned}

$$

$	herefore[(f . g) o h](x)=\{(f$ oh $) \cdot(g o h)\}(x)$ for all $x \in R$

Hence, $(f . g) o h=(f o h) .(g o h)$

Question 3

Find $g o f$ and $f o g$, if

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

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

Accepted Answer

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

$	herefore g \circ f(x)=g(f(x))=g(|x|)=|5| x|-2|$

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

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

$$

\begin{aligned}

& 	herefore g o f(x)=g(f(x))=g\left(8 x^{3}ight)=\left(8 x^{3}ight)^{\frac{1}{3}}=2 x \

& f o g(x)=f(g(x))=f\left(x^{\frac{1}{3}}ight)^{3}=8\left(x^{\frac{1}{3}}ight)^{3}=8 x

\end{aligned}

$$

Question 4

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

Accepted Answer

$(f o f)(x)=f(f(x))=f\left(\frac{4 x+3}{6 x-4}ight)$

$$

=\frac{4\left(\frac{4 x+3}{6 x-4}ight)+3}{6\left(\frac{4 x+3}{6 x-4}ight)-4}=\frac{16 x+12+18 x-12}{24 x+18-24 x+16}=\frac{34 x}{34}=x

$$

$	herefore$ fof $(x)=x \quad$ for all $x 
eq \frac{2}{3}$

$\Rightarrow f o f=1$

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

Question 5

State with reason whether the following functions have inverse.

i. $\quad f:\{1,2,3,4\} ightarrow\{10\}_{	ext {with }} f=\{(1,10),(2,10),(3,10),(4,10)\}$

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

iii. $\quad h:\{2,3,4,5\} ightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$

Accepted Answer

i. $\quad f:\{1,2,3,4\} ightarrow\{10\}_{	ext {with }} f=\{(1,10),(2,10),(3,10),(4,10)\}$

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

$	herefore f$ is not one-one.

Function $f$ does not have an inverse.

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

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

$	herefore g$ is not one-one.

Function $g$ does not have an inverse.

iii. $\quad h:\{2,3,4,5\} ightarrow\{7,9,11,13\}$ with $h=\{(2,7),(3,9),(4,11),(5,13)\}$

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

$	herefore h$ is one-one.

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

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

Function $h$ has an inverse.

Question 6

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

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

Accepted Answer

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

For one-one

$f(x)=f(y)$

$\Rightarrow \frac{x}{x+2}=\frac{y}{y+2}$

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

$\Rightarrow 2 x=2 y$

$\Rightarrow x=y$

$	herefore f$ is a one-one function.

It is clear that $f:[-1,1] ightarrow R$ is onto.

$f:[-1,1] ightarrow R$ is one-one and onto and therefore, the inverse of the function $f:[-1,1] ightarrow R$ exists.

Let $g:$ Range $f ightarrow[-1,1]$ be the inverse of $f$.

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

Since $f:[-1,1] ightarrow$ Range $f$ is onto, we have:

$y=f(x)$ for same $x \in[-1,1]$

$\Rightarrow y=\frac{x}{x+2}$

$\Rightarrow x y+2 y=x$

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

$\Rightarrow x=\frac{2 y}{1-y}, y 
eq 1$

Now, let us define $g:$ Range $f ightarrow[-1,1]$ as

$g(y)=\frac{2 y}{1-y}, y 
eq 1$

Now,

$(g \circ f)(x)=g(f(x))=g\left(\frac{x}{x+2}ight)=\frac{2\left(\frac{x}{x+2}ight)}{1-\frac{x}{x+2}}=\frac{2 x}{x+2-x}=\frac{2 x}{2}=x$

$(f o g)(x)=f(g(y))=f\left(\frac{2 y}{1-y}ight)=\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}=\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}=y$

$	herefore g \circ f=I_{[-1,1]} \quad$ and $\quad f o g=I_{	ext {Range } f}$

$	herefore f^{-1}=g$

$\Rightarrow f^{-1}(y)=\frac{2 y}{1-y}, y 
eq 1$

Question 7

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

Accepted Answer

$f: R ightarrow R$ given by $f(x)=4 x+3$

For one-one

$f(x)=f(y)$

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

$\Rightarrow 4 x=4 y$

$\Rightarrow x=y$

$	herefore f$ is a one-one function.

For onto

$y \in R$, let $y=4 x+3$

$\Rightarrow x=\frac{y-3}{4} \in R$

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

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

$	herefore f$ is onto.

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

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

Now,

$(g \circ f)(x)=g(f(x))=g(4 x+3)=\frac{(4 x+3)-3}{4}=x$

$(f o g)(y)=f(g(y))=f\left(\frac{y-3}{4}ight)=4\left(\frac{y-3}{4}ight)+3=y-3+3=y$

$	herefore g \circ f=f \circ g=\mathrm{I}_{R}$

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

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

Question 8

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

Accepted Answer

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

For one-one:

Let $f(x)=f(y)$

$\Rightarrow x^{2}+4=y^{2}+4$

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

$\Rightarrow x=y \quad[$ as $x \in R]$

$	herefore f$ is a one -one function.

For onto:

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

$\Rightarrow x^{2}=y-4 \geq 0 \quad[$ as $y \geq 4]$

$\Rightarrow x=\sqrt{y-4} \geq 0$

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

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

$	herefore f$ is an onto function.

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

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

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

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

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

$	herefore g \circ f=f o g=\mathrm{I}_{R}$

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

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

Question 9

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

Accepted Answer

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

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

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

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

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

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

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

$\Rightarrow x=\frac{\sqrt{y+6}-1}{3}$

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

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

We have,

$$

\begin{aligned}

(g \circ f)(x)=g(f(x)) & =g\left(9 x^{2}+6 x-5ight) \

& =g\left((3 x+1)^{2}-6ight) \

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

& =\frac{3 x+1-1}{3}=x

\end{aligned}

$$

And,

$$

\begin{aligned}

& \begin{aligned}

(f \circ g)(y)=f(g(y)) & =f\left(\frac{\sqrt{y+6}-1}{3}ight) \

= & {\left[3\left(\frac{\sqrt{y+6}-1}{3}ight)+1ight]^{2}-6 } \

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

\end{aligned} \

& 	herefore g \circ f=\mathrm{I}_{R} 	ext { and } f o g=\mathrm{I}_{[-5, \infty)}

\end{aligned}

$$

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

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

Question 10

Let $f: X \rightarrow Y$ 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 $y \in Y$, $f o g_{1}(y)=\mathrm{I}_{Y}(y)=f o g_{2}(y)$ . Use one-one ness of $f$.

Accepted Answer

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

Also suppose $f$ has two inverses ( $g_{1}$ and $g_{2}$ )

Then, for all $y \in Y$,

$f o g_{1}(y)=\mathrm{I}_{Y}(y)=f o g_{2}(y)$

$\Rightarrow f\left(g_{1}(y)\right)=f\left(g_{2}(y)\right)$

$\Rightarrow g_{1}(y)=g_{2}(y) \quad[f$ is invertible $\Rightarrow f$ is one-one $]$

$\Rightarrow g_{1}=g_{2} \quad[g$ is one-one $]$

Hence, $f$ has unique inverse.

Question 11

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

Accepted Answer

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

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

$(f o g)(a)=f(g(a))=f(1)=a$

$(f o g)(b)=f(g(b))=f(2)=b$

$($ fog $)(c)=f(g(c))=f(3)=c$

And,

$(g \circ f)(1)=g(f(1))=g(a)=1$

$(g \circ f)(2)=g(f(2))=g(b)=2$

$(g \circ f)(3)=g(f(3))=g(c)=3$

$	herefore g \circ f=\mathrm{I}_{X} \quad$ and $\quad f o g=\mathrm{I}_{Y} \quad$ where $X=\{(1,2,3)\}$ and $Y=\{a, b, c\}$

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

$	herefore f^{-1}:\{a, b, c\} ightarrow\{1,2,3\}$ is given by, $f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3$

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

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

$(g o h)(1)=g(h(1))=g(a)=1$

$(g \circ h)(2)=g(h(2))=g(b)=2$

$(g \circ h)(3)=g(h(3))=g(c)=3$

And,

$(h o g)(a)=h(g(a))=h(1)=a$

$(h o g)(b)=h(g(b))=h(2)=b$

$(h o g)(c)=h(g(c))=h(3)=c$

$	herefore g o h=\mathrm{I}_{X} \quad$ and $h o g=\mathrm{I}_{Y} \quad$ where $X=\{(1,2,3)\}$ and $Y=\{a, b, c\}$

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

It can be noted that $h=f$.

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

Question 12

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

Accepted Answer

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

Then there exists a function $g: Y \rightarrow X$ such that $g \circ f=\mathrm{I}_{X}$ and $f o g=\mathrm{I}_{Y}$

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

Now, $g o f=\mathrm{I}_{X}$ and $f o g=\mathrm{I}_{Y}$

$\Rightarrow f^{-1}$ of $=\mathrm{I}_{X}$ and $f o f^{-1}=\mathrm{I}_{Y}$

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

Question 13

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

A. $\frac{1}{x^{3}}$

B. $x^{3}$

C. $x$

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

Accepted Answer

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

$f(x)=\left(3-x^{3}ight)^{\frac{1}{3}}$

$	herefore f o f(x)=f(f(x))=f\left(\left(3-x^{3}ight)^{\frac{1}{3}}ight)=\left[3-\left(\left(3-x^{3}ight)^{\frac{1}{3}}ight)^{3}ight]^{\frac{1}{3}}$

$$

=\left[3-\left(3-x^{3}ight)ight]^{\frac{1}{3}}=\left(x^{3}ight)^{\frac{1}{3}}=x

$$

$	herefore f o f(x)=x$

The correct answer is C .

Question 14

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

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

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

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

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

Accepted Answer

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

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

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

$\Rightarrow y=\frac{4 x}{3 x+4}$

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

$\Rightarrow x(4-3 y)=4 y$

$\Rightarrow x=\frac{4 y}{4-3 y}$

Define $f: R-\left\{-\frac{4}{3}ight\} ightarrow R \quad$ as $\quad g(y)=\frac{4 y}{4-3 y}$

Now,

$$

\begin{aligned}

(g o f)(x) & =g(f(x))=g\left(\frac{4 x}{3 x+4}ight) \

& =\frac{4\left(\frac{4 x}{3 x+4}ight)}{4-3\left(\frac{4 x}{3 x+4}ight)}=\frac{16 x}{12 x+16-12 x} \

& =\frac{16 x}{16}=x

\end{aligned}

$$

And

$$

\begin{aligned}

& (f o g)(x)=(g(x))=f\left(\frac{4 y}{4-3 y}ight) \

& =\frac{4\left(\frac{4 y}{4-3 y}ight)}{3\left(\frac{4 y}{4-3 y}ight)+4}=\frac{16 y}{12 y+16-12 y} \

& =\frac{16 y}{16}=y \

& 	herefore g o f=\mathrm{I}_{R-\left\{-\frac{4}{3}ight\}} 	ext { and } f o g=\mathrm{I}_{	ext {Range } f}

\end{aligned}

$$

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

Hence, the inverse of $f$ is the map $g:$ Range $f ightarrow R-\left\{-\frac{4}{3}ight\}$, which is given by $g(y)=\frac{4 y}{4-3 y}$. The correct answer is B.

\section*{EXERCISE 1.4}