Question 1: State with reasons whether following functions have inverse:

$\text{(i) } f: \{1, 2 ,3, 4 \} \rightarrow \{10 \} \text{ with } f = \{ (1, 10), (2, 10), (3, 10), (4, 10) \}$

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

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

$\text{ (i) Given that } f: \{1, 2, 3, 4\} \rightarrow {10} \text{ with }f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

$\text{ We have, } f (1) = f (2) = f (3) = f (4) = 10$

$\Rightarrow f \text{ is not one-one. }$

$\Rightarrow f \text{ is not a bijection. }$

Therefore, $f$ does not have an inverse.

$\text{ (ii) Given that } g: \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} \text{ with } g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

$\text{ It is clear that } g(5) = g(7) = 4$

$\Rightarrow f \text{ is not one-one. }$

$\Rightarrow f \text{ is not a bijection. }$

Therefore, $f$ does not have an inverse.

$\text{ (iii) Given that } h: \{2, 3, 4, 5\} \rightarrow \{7, 9, 11, 13\} \text{ with } h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

Here, different elements of the domain have different images in the co-domain.

$\Rightarrow h \text{ is one-one. }$

Also, each element in the co-domain has a pre-image in the domain.

$\Rightarrow h\text{ is onto. }$

$\Rightarrow h \text{ is a bijection. }$

So, $h \text{ inverse exists. }$

$\Rightarrow h \text{ has an inverse and it is given by } h^{-1} = \{(7, 2), (9, 3), (11, 4), (13, 5)\}$

$\\$

$\text{Question 2: Find } f^{ -1} \text{ if it exists: } f : A \rightarrow B \text{ where }$

$\text{(i) } A= \{0, -1, -3, 2 \}; B = \{ -9, -3, 0, 6 \} \text{ and } f(x) = 3 x.$

$\text{(ii) } A= \{1, 3, 5, 7, 9 \}; B = \{ 0, 1, 9, 25, 49, 81 \} \text{ and } f(x) = x^2.$

$\text{ (i) Given as } A = \{0, -1, -3, 2\}; B = \{-9, -3, 0, 6\} \text{ and } f(x) = 3x.$

$\text{ Therefore, }f = \{(0, 0), (-1, -3), (-3, -9), (2, 6)\}$

Here, the different elements of the domain have different images in the co-domain.

Thus, this is one-one.

$\text{ Range of } f = \text{ Range of } f = B$

$\text{ Therefore, } f \text{ is a bijection and, Clearly, } f^{-1} \text{ exists. }$

$\text{ Hence, } f^{-1}= \{(0, 0), (-3, -1), (-9, -3), (6, 2)\}$

$\text{ (ii) Given as } A = \{1, 3, 5, 7, 9\}; B = \{0, 1, 9, 25, 49, 81\} \text{ and } f(x) = x^2$

$\text{ Therefore, } f = \{(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)\}$

Here, the different elements of the domain have different images in the co-domain.

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

This is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

$\Rightarrow f \text{ is not a bijection. }$

$\text{ Therefore, } f^{-1} \text{ does not exist. }$

$\\$

$\text{Question 3: Consider } f: {1, 2, 3} \rightarrow {a, b, c} \text{ and } g: {a, b, c} \rightarrow {apple, ball, cat} \text{ defined as } \\ \\ f(l) = a, f(2) = b, f( 3) = c, g (a)= apple, g(b) = ball \text{ and } g(c) = cat. \text{ Show that } \\ \\ f, g \text{ and } gof \text{ are invertible. Find } f^{- 1}, g^{- 1} \text{ and } (gof)^{- 1} \text{ and show that } \\ \\ (gof)^{- 1} = f^{- 1} o g^{-1} .$

$\text{Given as } f = \{(1, a), (2, b), (c, 3)\} \text{ and } g = \{(a, apple), (b, ball), (c, cat)\}$

Clearly, $f$ and $g$ are bijections.

Therefore, $f$ and $g$ are invertible.

$\text{Then, } f^{-1} = \{(a ,1) , (b , 2) , (3,c)\} \text{ and } g^{-1} = \{(apple, a), (ball , b), (cat , c)\}$

$\text{Therefore, } f^{-1}og^{-1}= \{apple, 1), (ball, 2), (cat, 3)\}$

$f: \{1,2,3\} \rightarrow {a, b, c} \text{ and } g: \{a, b, c\} \rightarrow \{apple, ball, cat\}$

$\text{Therefore } gof: \{1, 2, 3\} \rightarrow \{apple, ball, cat\}$

$(gof)(1) = g(f(1)) = g(a) = apple$

$(gof)(2) = g(f(2)) = g(b) = ball,$

$(gof)(3) = g(f(3)) = g(c) = cat$

$\therefore gof = \{(1, apple), (2, ball), (3, cat)\}$

Clearly, $gof$ is a bijection.

Therefore, $gof$ is invertible.

$(gof)^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\}$

$\text{Therefore } (gof)^{-1} = f^{-1}og^{-1}$

$\\$

$\text{Question 4: Let } A= \{ 1, 2, 3, 4 \}; B = \{ 3, 5, 7, 9 \}; C = \{ 7, 23, 47, 79 \} \text{ and } \\ \\ f: A \rightarrow B, g: B \rightarrow C \text{ be defined as } f(x) = 2x +1 \text{ and } g(x) = x^2 - 2. \\ \\ \text{ Express } (goj)^{-1} \text{ and } f^{-1}og^{-1} \text{ as the sets of ordered pairs and verify that } \\ \\ (gof)^{-1}=f^{-1}og^{-1}.$

$\text{Given } f(x) = 2x + 1$

$\Rightarrow f = \{(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)\} = \{(1, 3), (2, 5), (3, 7), (4, 9)\}$

$\text{Also given } g(x) = x^2 - 2$

$\Rightarrow g = \{(3, 32 - 2), (5, 52 - 2), (7, 72 - 2), (9, 92 - 2)\} = \{(3, 7), (5, 23), (7, 47), (9, 79)\}$

So, $f$ and $g$ are bijections and,

$\text{Hence, } f^{-1}: B \rightarrow A \text{ and } g^{-1}: C \rightarrow B \text{ exist. }$

$\text{Therefore, } f^{-1}= \{(3, 1), (5, 2), (7, 3), (9, 4)\}$

$\text{And } g^{-1}= \{(7, 3), (23, 5), (47, 7), (79, 9)\}$

$\text{Now, } (f^{-1}og^{-1}): C \rightarrow A$

$f^{-1}og^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\}$

$\text{Also, } f: A \rightarrow B \text{ and } g: B \rightarrow C$

$\Rightarrow gof: A \rightarrow C, (gof)^{-1} : C \rightarrow A$

$\text{Therefore, } f^{-1}og^{-1} \text{ and } (gof)^{-1} \text{ have same domains. }$

$(gof)(x) = g(f(x)) =g(2x + 1) =(2x +1 )^2 - 2$

$\Rightarrow (gof)(x) = 4x^2 + 4x + 1 - 2$

$\Rightarrow (gof)(x) = 4x^2 + 4x - 1$

$\text{Now, } (gof)(1) = g(f(1)) = 4 \times 1+ 4 \times 1 - 1 = 7$

$(gof)(2) = g(f(2)) = 4 \times 4 + 4\times 2 - 1 = 23,$

$(gof)(3) = g(f(3)) = 4 \times 9 + 4 \times 3 - 1 = 47$

$(gof)(4) = g(f(4)) = 4 \times 16 + 4 \times 4 - 1 = 79$

$\text{Therefore, } gof = \{(1, 7), (2, 23), (3, 47), (4, 79)\}$

$\Rightarrow (gof)^{-1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\}$

$\text{Therefore } (gof)^{-1} = f^{-1}og^{-1}$

$\\$

Question 5: Show that the function $f: Q \rightarrow Q$ defined by $f(x) = 3x + 5$ is invertible. Also, find $f^{-1}.$

$\text{Given } f: Q \rightarrow Q, \text{ defined by } f(x) = 3x + 5$

Let us show that the given function is invertible.

Injection of $f:$

Let $x$ and $y$ be any two elements of the domain $(Q),$

$\text{Such that } f(x) = f(y) \Rightarrow 3x + 5 = 3y + 5 \Rightarrow 3x = 3y \Rightarrow x = y$

Therefore, $f$ is one-one.

Surjection of $f:$

Let $y$ be in the co-domain $(Q),$

$\displaystyle \text{Such that } f(x) = y \Rightarrow 3x + 5 = y \Rightarrow 3x = y - 5 \Rightarrow x = \frac{y - 5}{3} \text{ in (domain) }$

Therefore,  $f$ is onto.

Therefore, $f$ is a bijection and, hence, it is invertible.

Let us find $f^{-1}$

$\displaystyle \text{Let } f^{-1}(x) = y \Rightarrow x = f(y) \Rightarrow x = 3y + 5 \Rightarrow x - 5 = 3y \Rightarrow y = \frac{x - 5}{3}$

$\displaystyle\text{Therefore, } f-1(x) = \frac{x - 5}{3}$

$\\$

Question 6: Consider $f: R \rightarrow R$ given by $f (x) = 4x + 3$. Show that $f$ is invertible. Find the inverse of $f.$

$\text{Given function } f: R \rightarrow R \text{ given by } f(x) = 4x + 3$

Let us show that the given function is invertible.

Consider injection of $f:$

Let $x$ and $y$ be any two elements of domain $(R),$

$\text{Such that } f(x) = f(y) \Rightarrow 4x + 3 = 4y + 3 \Rightarrow 4x = 4y \Rightarrow x = y$

Therefore, $f$ is one-one.

Surjection of $f:$

Let $y$ be in the co-domain $(R),$

$\displaystyle \text{Such that } f(x) = y. \Rightarrow 4x + 3 = y \Rightarrow 4x = y - 3 \Rightarrow x = \frac{y - 3}{4} \text{ in R (domain) }$

$\Rightarrow f \text{ is onto. }$

Therefore, $f$ is a bijection and, hence, is invertible.

$\text{Let us find } f^{-1}$

$\text{ Let } f^{-1}(x) = y$

$\displaystyle \Rightarrow x = f(y) \Rightarrow x = 4y + 3 \Rightarrow x - 3 = 4y \Rightarrow y = \frac{x - 3}{4}$

$\displaystyle \text{Therefore, } f^{-1}(x) = \frac{x - 3}{4}$

$\\$

Question 7:  Consider $f: R+ \rightarrow [4, \infty)$ given by $f (x) = x^2 + 4$. Show that $f$ is invertible with inverse $f^{- 1}$ of f given by $f^{-1} (x) = \sqrt{x - 4}$, where $R^+$ is the set of all non-negative real numbers.

$\text{Given function } f: R \rightarrow R+ \rightarrow [4, \infty) \text{ given by } f(x) = x^2 + 4.$

Let us show that $f$ is invertible,

Consider injection of $f:$

Let $x$ and $y$ be any two elements of the domain $(Q),$

$\text{Such that } f(x) = f(y) \Rightarrow x^2 + 4 = y^2 + 4 \Rightarrow x^2 = y^2 \Rightarrow x = y ( \text{ as co-domain as } R+)$

Therefore, $f$ is one-one

Now surjection of $f:$

Let $y$ be in the co-domain $(Q),$

$\text{Such that } f(x) = y \Rightarrow x^2 + 4 = y \Rightarrow x^2 = y - 4 \Rightarrow x = \sqrt{y - 4} \text{ in } R \\ \\ \text{Therefore } f \text{ is onto. }$

Therefore, $f$ is a bijection and, hence, it is invertible.

$\text{Let us find } f^{-1}$

$\text{Let } f^{-1}(x) = y \Rightarrow x = f(y) \Rightarrow x = y^2 + 4 \Rightarrow x - 4 = y^2 \Rightarrow y = \sqrt{x - 4}$

$\text{Therefore, } f^{-1}(x) = \sqrt{x - 4}$

$\\$

$\displaystyle \text{Question 8: If } f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3}, \text{ show that } fof (x) = x \text{ for all } x \neq \frac{2}{3}. \\ \\ \text{ What is the inverse of } f?$

$\displaystyle \text{Given } f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3}$

$\displaystyle \text{Let us show } fof(x) = x$

$\displaystyle (fof)(x) = f(f(x)) \\ \\ = f\Big(\frac{4x+3}{6x-4}\Big) = \frac{4 \Big(\frac{4x+3}{6x-4} \Big)+3}{6\Big(\frac{4x+3}{6x-4} \Big)-4} = \frac{16x+12+18x-12}{24x+18-24x+16} = \frac{34x}{34} = x$

$\displaystyle \text{So, } fof(x) = x \text{ for all } x \neq \frac{2}{3}$

$\displaystyle \Rightarrow fof = 1$

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

$\\$

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

$\displaystyle \text{Given } f: R^{+} \rightarrow [-5, \infty) \text{ given by } f(x) = 9x^2 + 6x - 5.$

$\displaystyle (i) \text{ Let } f(x_1) = f(x_2)$

$\displaystyle \Rightarrow 9{x_1}^2 + 6x_1 - 5 = 9{x_2}^2 + 6x_2 - 5$

$\displaystyle \Rightarrow 9(x_1-x_2)(x_1+x_2)+ 6 (x_1-x_2) =0$

$\displaystyle \Rightarrow (x_1-x_2)[9(x_1+x_2)+6]=0$

$\displaystyle \Rightarrow x_1 - x_2 = 0 \text{ or }9(x_1+x_2) + 6 = 0$

$\displaystyle \Rightarrow x_1 = x_2 \text{ or } 9(x_1+x_2) = - 6 \Rightarrow (x_1+x_2) = -\frac{6}{9} \ \ \ \text{[This is not possible]}$

$\displaystyle \therefore x_1 = x_2$

$\displaystyle \text{Hence, we can say } f(x_1) = f(x_2) \Rightarrow x_1 = x_ 2$

$\displaystyle \therefore f \text{ is one-one }$

$\displaystyle \text{(ii) Let } y \in [-5, \infty] \text{So that } y = f(x) \text{ for some } x \in R_{+}$

$\displaystyle \Rightarrow 9x^2+6x-5=y$

$\displaystyle \Rightarrow 9x^2+6x-5-y=0$

$\displaystyle \Rightarrow 9x^2+6x-(5+y) = 0$

$\displaystyle \Rightarrow x = \frac{-6 \pm \sqrt{36+4(9)(5+y)}}{2 \times 9}$

$\displaystyle \Rightarrow x = \frac{-6 \pm 6\sqrt{1+5+y}}{18} = \frac{-1 \pm 1\sqrt{6+y}}{3}$

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

$\displaystyle \text{Therefore } x = \frac{-1 + \sqrt{6+y}}{3} \in R_{+}$

$f$ is onto.

Since function is one-one and onto, so it is invertible.

$\\$

$\displaystyle \text{Question 10: If } f: R \rightarrow R \text{ be defined by } f(x) x^3 - 3, \text{ then prove that } f^{-1} \\ \\ \text{ exists and find a formula for } f^{-1}. \text{ Hence, find }f^{-1}(24) \text{ and } f^{-1}(5).$

$\text{Given function } f: R \rightarrow R \text{ be defined by } f(x) = x^3 - 3$

Let us prove that $f^{-1}$ exists

Injectivity of $f:$

Let $x$ and $y$ be two elements in domain $(R),$

$\text{Such that, } x^3 - 3 = y^3 - 3 \Rightarrow x^3 = y^3 \Rightarrow x = y$

Therefore, $f$ is one-one.

Surjectivity of $f:$

Let $y$ be in the co-domain $(R)$

$\text{Such that } f(x) = y \Rightarrow x^3 - 3 = y \Rightarrow x^3 = y + 3 \Rightarrow x = \sqrt[3]{y + 3} \text{ in } R$

Therefore $f$ is onto.

Therefore, $f$ is a bijection and, hence, it is invertible.

Find $f^{-1}$

$\text{Let } f^{-1}(x) = y$

$\Rightarrow x = f(y) \Rightarrow x = y^3 - 3 \Rightarrow x + 3 = y^3 \Rightarrow y = \sqrt[3]{x + 3} = f^{-1}(x)$

$\text{Therefore, } f-1(x) = \sqrt[3]{x + 3}$

$f^{-1}(24) = \sqrt[3]{(24 + 3)} = \sqrt[3]{27} = \sqrt[3]{3^3} = 3$

$f^{-1}(5) = \sqrt[3]{(5 + 3)} = \sqrt[3]{8} = \sqrt[3]{2^3} = 2$

$\\$

Question 11: A function $f: R \rightarrow R$ is defined as $f(x) = x^3 + 4$. Is it a bijection or not? In case it is a bijection, find $f^{-1}(3).$

$\text{Given function } f: R \rightarrow R \text{ is defined as } f(x) = x^3 + 4$

Injectivity of $f:$

Let $x$ and $y$ be two elements of domain $(R).$

$\text{Such that } f(x) = f(y) \Rightarrow x^3 + 4 = y^3 + 4 \Rightarrow x^3 = y^3 \Rightarrow x = y$

Therefore, $f$ is one-one.

Surjectivity of $f:$

Let $y$ be in the co-domain $(R).$

$\text{Such that } f(x) = y. \Rightarrow x^3 + 4 = y \Rightarrow x^3 = y - 4 \Rightarrow x = \sqrt[3]{y - 4} \text{ in R (domain) }$

Therefore, $f$ is onto.

Therefore, $f$ is a bijection and, hence, is invertible.

$\text{ Let } f^{-1}(x) = y$

$\Rightarrow x = f(y) \Rightarrow x = y^3 + 4 \Rightarrow x - 4 = y^3 \Rightarrow y = \sqrt[3]{x - 4}$

$\text{Therefore, } f^{-1}(x) = \sqrt[3]{x - 4}$

$f^{-1} (3) = \sqrt[3]{3-4} = \sqrt[3]{-1} = -1$

$\\$

$\displaystyle \text{Question 12: If } f: Q \rightarrow Q, g : Q \rightarrow Q \text{ are two functions defined by } \\ \\ f(x) = 2x \text{ and } g(x) = x+2, \text{ show that } f \text{ and } g \text{ are bijective maps. Verify that } \\ \\ ( gof)^{-1} = f^{-1}og^{-1}.$

$\textbf{ Injectivity test of } f:$

$\text{ Let } x \text{ and } y \text{ be two elements of domain } (Q), \text{ such that }$

$\Rightarrow f(x)=f(y)$

$\Rightarrow 2x=2y$

$\Rightarrow x=y$

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

$\textbf{ Surjectivity test of } f:$

$\text{ Let } y \text{ be the co-domain } (Q), \text{ such that } f(x)=y.$

$\Rightarrow 2x=y$

$\displaystyle \Rightarrow x= \frac{y}{2} \in Q$

$\therefore f \text{ is onto. }$

$\text{ Since, } f \text{ is one-one and onto, then it is bijection. }$     $\therefore f^{-1} \text{ is invertible. }$

$\text{ Find } f^{-1}:$

$\text{ Let } f^{-1} (x)=y$

$\Rightarrow x=f(y)$

$\Rightarrow x=2y$

$\displaystyle \therefore y=\frac{x}{2} \cdots \cdots ( 1 )$

$\displaystyle \text{ From ( 1 ), So, } f^{-1} = \frac{x}{2}$

$\textbf{ Injectivity test of } g:$

$\text{ Let } x \text{ and } y \text{ be two elements of domain } (Q), \text{ such that }$

$\Rightarrow g(x)=g(y)$

$\Rightarrow x+2=y+2$

$\Rightarrow x=y$

$\therefore g \text{ is one-one. }$

$\textbf{ Surjectivity test of } g:$

$\text{ Let } y \text{ be in the co-domain } (Q), \text{ such that } g(x)=y.$

$\Rightarrow x+2=y$

$\Rightarrow x=2-y \in Q \text{ (domain) }$

$\therefore g \text{ is onto. }$

$\text{ Since, } g \text{ is one-one and onto, then it is bijection. }$     $\therefore g^{-1} \text{ is invertible. }$

$\text{ Find } g^{-1} :$

$\text{ Let } g^{-1} (x)=y$

$\Rightarrow x=g(y)$

$\Rightarrow x=y+2$

$\Rightarrow y=x-2 \hspace{1.0cm} \cdots \cdots ( 2 )$

From ( 2 ),

$\Rightarrow g^{-1} (x)=x-2$

$\text{ Verification of } (gof)^{-1} = f^{-1}og^{-1} :$

$f(x)=2x \text{ and } g(x)=x+2$

$\displaystyle f^{-1}(x) = \frac{x}{2} \text{ and } g^{-1}(x)= x-2$

Now,

$\displaystyle \Rightarrow (f^{-1} o g^{-1})(x) = f^{-1} (( g^{-1}(x)) = f^{-1}(x-2) = \frac{x-2}{2} \hspace{1.0cm} \cdots \cdots (3)$

$\Rightarrow (gof)(x) = f(f(x)) = g(2x) = 2x + 2 \hspace{1.0cm} \cdots \cdots (4)$

$\text{ Let } (gof)^{-1} (x) = y$

$\Rightarrow x = (gof)(y)$

$\Rightarrow 2y = x-2$

$\displaystyle \Rightarrow y = \frac{x-2}{2}$

From (4)

$\displaystyle \Rightarrow (gof)^{-1} (x) = \frac{x-2}{2} \hspace{1.0cm} \cdots \cdots (5)$

From (3) and (5)

$\Rightarrow (gof)^{-1} = f^{-1} o g^{-1}$

$\\$

$\displaystyle \text{Question 13: Let } A = R - \{ 3 \} \text{ and } B = R - \{ 1\}. \text{ Consider the function } \\ \\ f: A \rightarrow B \text{ defined by } f(x) = \frac{x-2}{x-3}. \\ \\ \text{ Show that f is one-one and onto and hence find } f^{-1}.$

Given;

$f: A \rightarrow B, A = R - \{3\} \text{ and } B = R - \{1\}$

$f(x_1) = f(x_2)$

$\displaystyle \Rightarrow \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}$

$\Rightarrow (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$

$\Rightarrow x_1x_2 - 2x_2 - 3x_1 + 6 = x_2x_1 - 2x_1 - 3x_2 + 6$

$\Rightarrow -2x_2 - 3x_1 = - 2x_1 - 3x_2$

$\Rightarrow x_1 = x_2$

$\text{Hence one-one Let } y \in B$

$\text{Now, } f(x) = y$

$\displaystyle \Rightarrow \frac{x-2}{x-3} = y$

$\Rightarrow x - 2 = xy - 3y$

$\Rightarrow x - xy = 2 - 3y$

$\Rightarrow x(1 - y) = 2 - 3y$

$\displaystyle \Rightarrow x = \frac{2-3y}{1-y}$

Hence $f$ is onto function.

Thus $f$ is one=one onto function.

$\displaystyle \text{If } f^{-1} \text{ is inverse function of } f \text{ then } f^{-1} (y) = \frac{2-3y}{1-y}$

$\\$

$\displaystyle \text{Question 14: Consider the function } f : R^+ \rightarrow [ -9, \infty) \text{ given by } \\ \\ f(x) = 5x^2 + 6x - 9. \text{ Prove that } f \text{ is invertible with } f^{-1}(y) = \frac{\sqrt{54+5y}}{5}.$

$\text{Here } f : R^+ \rightarrow [-9, \infty) \text{ as } f(x) = 5x^2 + 6x - 9$

First we shall show that $f$ is one-one.

$\text{Let } f(x) = f(y) \text{ for } x, y \in R$

$\Rightarrow 5x^2 + 6x - 9 =5y^2 + 6y - 9$

$\Rightarrow 5(x^2-y^2) + 6(x-y) = 0$

$\Rightarrow (x-y) [ 5(x+y) + 6] = 0$

$\Rightarrow x=y \ \ \ \ [\because \text{ for } x, y \in R^+ , 5(x+y) + 6 \neq 0 ]$

$\therefore f \text{ is one=one }$

$\text{Let } y \in [-9, \infty) \text{ be such that } f(x) = y$

$\Rightarrow 5x^2 + 6x - 9 = y$

$\Rightarrow 5x^2 + 6x - (9+y) = 0$

$\displaystyle \Rightarrow x = \frac{-6 \pm \sqrt{6^2 + 4 \times 5 ( 9+y)}}{2 \times 5} = \frac{-6 \pm \sqrt{216 + 20 y}}{10} = \frac{-3 \pm \sqrt{54+5y}}{5}$

$\text{Taking only } +ve \text{sign ( as for } -ve \text{ sign } , x \notin R^+)$

$\displaystyle \text{We get } x = \frac{-3+\sqrt{54+5y}}{5} \in R^+ \text{ for which } f(x) = y$

$\therefore f \text{ is onto }$

Therefore $f$ is both one-one and onto

$\text{Therefore } f \text{ is invertible and } f^{-1} \text{ is given by }$

$\displaystyle f^{-1}(y) = x = \frac{-3\pm \sqrt{54+5y}}{5}$

$\\$

$\displaystyle \text{Question 15: Let } f: N \rightarrow N \text{ be a function defined as } f(x) = 9x^2 + 6x - 5. \\ \\ \text{ Show that } f: N \rightarrow S, \text{ where } S \text{ is the range of } f, \text{ is invertible. Find the inverse of } \\ \\ f \text{ and hence find } f^{-1}(43) \text{ and } f^{-1}(163).$

$f(x) = 9x^2 + 6x - 5$

$\text{Let } x_1 , x_2 \in N \text{ such that } f(x_1) = f(x_2)$

$\Rightarrow 9{x_1}^2 + 6x_1 - 5 = 9{x_2}^2 + 6x_2 - 5$

$\Rightarrow 9({x_1}^2 - {x_2}^2) + 6 ( x_1 - x_2) = 0$

$\Rightarrow (x_1-x_2) [ 9 (x_1+x_2) + 6 ] = 0$

$\Rightarrow (x_1-x_2) = 0 \ \ \ \because 9 (x_1+x_2) + 6 \neq 0 \ \ \ [ \because x_1, x_2 \in N ]$

$\therefore x_1 = x_ 2$

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

$\text{Now, Let } y = 9x^2 + 6x - 5$

$\Rightarrow 9x^2 + 6x - (5+y) = 0$

$\displaystyle \Rightarrow x = \frac{-6 \pm \sqrt{36 + 4 \times 9 ( 5+y)}}{18} = \frac{-6 \pm 6 \sqrt{1 + 5 + y}}{18} = \frac{-1 + \sqrt{y+6}}{3}$

$\therefore \text{for all } y \in S, \text{ there exists } x \in N \text{ such that }$

$\displaystyle f(x) = f\Bigg( \frac{-1 + \sqrt{y+6}}{3} \Bigg) = 9 \Bigg(\frac{-1 + \sqrt{y+6}}{3}\Bigg)^2 + 6\Bigg(\frac{-1 + \sqrt{y+6}}{3}\Bigg) - 5$

$\displaystyle = 9\Bigg(\frac{1 - 2 \sqrt{6+y} + 6 + y}{9} \Bigg)- 2 + 2 \sqrt{6+y} - 5$

$= 1 - 2 \sqrt{6+y} + 6 + y + 2 \sqrt{6+y} - 7 = y$

$\therefore f \text{ is onto. Hence } f \text{ is invertible and }$

$\displaystyle f^{-1} (x) = \frac{-1 + \sqrt{6+x}}{3}$

$\displaystyle \therefore f^{-1} (43) = \frac{-1 + \sqrt{6+43}}{3} = 2$

$\displaystyle \text{And } f^{-1} (163) = \frac{-1 + \sqrt{6+163}}{3} = 4$

$\\$

$\displaystyle \text{Question 16: Let } f: R - \Big\{ -\frac{4}{3} \Big\} \rightarrow R \text{ be a function defined as } \\ \\ f(x) = \frac{4x}{3x+4}. \text{ Show that } f : R - \Big\{ -\frac{4}{3} \Big\} \rightarrow R \text{ and } (f) \text{ is one-one and onto.} \\ \\ \text{ Hence, find } f^{-1}.$

For one-one

$\displaystyle \text{Let } x_1 , x_2 \in R - \Big\{ -\frac{4}{3} \Big\} \text{ such that } f(x_1) = f(x_2)$

$\displaystyle \Rightarrow \frac{4x_1}{3x_1+4} = \frac{4x_2}{3x_2+4}$

$\Rightarrow 12x_1x_2 + 16x_2 = 12x_1x_2 + 16 x_2$

$\Rightarrow x_1 = x_2$

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

$\displaystyle \text{Clearly, } f : R - \Big\{ -\frac{4}{3} \Big\} \rightarrow R$

$\text{Let } f(x) = y$

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

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

$\displaystyle \text{As } y \in R - \Big\{ \frac{4}{3} \Big\}, \frac{4y}{4-3y} \in R$

$\displaystyle \text{Also } \frac{4y}{4-3y} \neq - \frac{4}{3} \text{ because } \frac{4y}{4-3y} = - \frac{4}{3}$

$\Rightarrow 12 y = -16 + 12y$

$\Rightarrow 0 = - 16 \text{ which is not possible. }$

$\displaystyle \text{Thus } x = \frac{4y}{4-3y} \in R - \Big\{ -\frac{4}{3} \Big\} \text{ such that }$

$\displaystyle f(x) = f( \frac{4x}{3x+4} ) = \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y})+4} = \frac{16y}{12y+16-12y} = \frac{16y}{16} = y$

$\displaystyle \text{So every element in } R - \Big\{ \frac{4}{3} \Big\} \text{ has a pre-image in } R - \Big\{ -\frac{4}{3} \Big\}.$

$\text{Hence, } f \text{ is onto. }$

$\displaystyle \text{Now, } x = \frac{4y}{4-3y}$

$\text{Replacing } x \text{ by } f^{-1} \text{ and } y \text{ by } x, \text{ we have }$

$\displaystyle f^{-1} =\frac{4x}{4-3x}$

$\\$

$\displaystyle \text{Question 17: If } f: R \rightarrow (-1, 1) \text{ defined by } f(x) = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \text{ is invertible, find } f^{-1}.$

Injectivity of $f:$

Let $x$ and $y$ be two elements of domain (R), such that

$f(x) = f(y)$

$\displaystyle \Rightarrow \frac{10^x - 10^{-x}}{10^x + 10^{-x}}= \frac{10^y - 10^{-y}}{10^y + 10^{-y}}$

$\displaystyle \Rightarrow \frac{10^x - 10^{-x}}{10^x + 10^{-x}}= \frac{10^y - 10^{-y}}{10^y + 10^{-y}}$

$\displaystyle \Rightarrow \frac{10^{-x}(10^{2x} - 1)}{10^{-x}(10^{2x} + 1)} = \frac{10^{-y}(10^{2y} - 1)}{10^{-y}(10^{2y} + 1)}$

$\displaystyle \Rightarrow \frac{(10^{2x} - 1)}{(10^{2x} + 1)} = \frac{(10^{2y} - 1)}{(10^{2y} + 1)}$

$\Rightarrow 10^{2x+2y} + 10^{2x} - 10^{2y} - 1 = 10^{2x+2y} - 10^{2x} + 10^{2y} - 1$

$\Rightarrow 2 \times 10^{2x} = 2 \times 10^{2y}$

$\Rightarrow 2x = 2y$

$\Rightarrow x = y$

Hence, f is one-one.

Surjectivity of f:

Let y be in the co-domain such that f(x) = y.

$\displaystyle \frac{10^x - 10^{-x}}{10^x + 10^{-x}}= y$

$\displaystyle \Rightarrow \frac{10^{-x}(10^{2x} - 1)}{10^{-x}(10^{2x} + 1)} = y$

$\displaystyle \Rightarrow \frac{10^{-x}(10^{2x} - 1)}{10^{-x}(10^{2x} + 1)} = y$

$\displaystyle \Rightarrow 10^{2x} - 1 = y10^{2x} + y$

$\displaystyle \Rightarrow 10^{2x} (1-y) = 1 + y$

$\displaystyle \Rightarrow 10^{2x} = \frac{1+y}{1-y}$

$\displaystyle \Rightarrow 2x = \log_e \Big( \frac{1+y}{1-y} \Big)$

$\displaystyle \Rightarrow x = \frac{1}{2} \log_e \Big( \frac{1+y}{1-y} \Big) \in R ( \text{ domain } )$

Hence, $f$ is onto

Finding $f^{-1}$

$\text{Let } f^{-1}(x) = y$

$\displaystyle \Rightarrow f(y) = x$

$\displaystyle \Rightarrow \frac{10^y - 10^{-y}}{10^y + 10^{-y}}= x$

$\displaystyle \Rightarrow \frac{10^{-y}(10^{2y} - 1)}{10^{-y}(10^{2y} + 1)} = x$

$\displaystyle \Rightarrow 10^{2y} - 1 = x \times 10^{2y} + x$

$\displaystyle \Rightarrow 10^{2y} = \frac{1+x}{1-x}$

$\displaystyle \Rightarrow 2y = \log_e \Big( \frac{1+x}{1-x} \Big)$

$\displaystyle \Rightarrow y = \frac{1}{2} \log_e \Big( \frac{1+x}{1-x} \Big) \in R \text{ (domain) }$

$\displaystyle \Rightarrow y = \frac{1}{2} \log_e \Big( \frac{1+x}{1-x} \Big) = f^{-1}(x)$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{1}{2} \log_e \Big( \frac{1+x}{1-x} \Big)$

$\\$

$\displaystyle \text{Question 18: If } f: R \rightarrow (0, 2) \text{ defined by } f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}+1 \text{ is invertible, find } f^{-1}.$

Injectivity of f:

Let x and y be two elements of domain (R), such that

f(x) = f(y)

$\displaystyle \Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}}+1= \frac{e^y - e^{-y}}{e^y + e^{-y}}+1$

$\displaystyle \Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}}= \frac{e^y - e^{-y}}{e^y + e^{-y}}$

$\displaystyle \Rightarrow \frac{e^{-x}(e^{2x} - 1)}{e^{-x}(e^{2x} + 1)} = \frac{e^{-y}(e^{2y} - 1)}{e^{-y}(e^{2y} + 1)}$

$\displaystyle \Rightarrow \frac{(e^{2x} - 1)}{(e^{2x} + 1)} = \frac{(e^{2y} - 1)}{(e^{2y} + 1)}$

$\Rightarrow e^{2x+2y} + e^{2x} - e^{2y} - 1 = e^{2x+2y} - e^{2x} + e^{2y} - 1$

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

$\Rightarrow 2x = 2y$

$\Rightarrow x = y$

Hence, f is one-one.

Surjectivity of f:

Let y be in the co-domain (0, 2) such that f(x) = y.

$\displaystyle \frac{e^x - e^{-x}}{e^x + e^{-x}}+1 = y$

$\displaystyle \Rightarrow \frac{e^{-x}(e^{2x} - 1)}{e^{-x}(e^{2x} + 1)} + 1 = y$

$\displaystyle \Rightarrow \frac{e^{-x}(e^{2x} - 1)}{e^{-x}(e^{2x} + 1)} = y - 1$

$\displaystyle e^{2x} - 1 = (y-1)(e^{2x} + 1)$

$\displaystyle \Rightarrow e^{2x} - 1 = ye^{2x} + y - e^{2x} - 1$

$\displaystyle \Rightarrow e^{2x} = ye^{2x} + y - e^{2x}$

$\displaystyle \Rightarrow e^{2x} (2 - y) = y$

$\displaystyle \Rightarrow e^{2x} = \frac{y}{2-y}$

$\displaystyle \Rightarrow 2x = \log_e \Big( \frac{y}{2-y} \Big)$

$\displaystyle \Rightarrow x = \frac{1}{2} \log_e \Big( \frac{y}{2-y} \Big) \in R ( \text{ domain } )$

Hence, $f$ is onto

Therefore $f$ is bijection and hence, it is invertible.

Finding $f^{-1}$

$\text{Let } f^{-1}(x) = y$

$\displaystyle \Rightarrow f(y) = x$

$\displaystyle \Rightarrow \frac{e^y - e^{-y}}{e^y + e^{-y}}+1 = x$

$\displaystyle \Rightarrow \frac{e^{-y}(e^{2y} - 1)}{e^{-y}(e^{2y} + 1)} + 1 = x$

$\displaystyle \Rightarrow e^{2y} - 1 = (x-1) (e^{2y} + 1)$

$\displaystyle \Rightarrow e^{2y} - 1 = x e^{2y} + x - e^{2y} - 1$

$\displaystyle \Rightarrow e^{2y} = x e^{2y} + x - e^{2y}$

$\displaystyle \Rightarrow e^{2y} (2-x) = x$

$\displaystyle \Rightarrow e^{2y} = \frac{x}{2-x}$

$\displaystyle \Rightarrow 2y = \log_e \Big( \frac{x}{2-x} \Big)$

$\displaystyle \Rightarrow y = \frac{1}{2} \log_e \Big( \frac{x}{2-x} \Big) \in R \text{ (domain) }$

$\displaystyle \Rightarrow y = \frac{1}{2} \log_e \Big( \frac{x}{2-x} \Big) = f^{-1}(x)$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{1}{2} \log_e \Big( \frac{x}{2-x} \Big)$

$\\$

$\displaystyle \text{Question 19: Let } f: [-1, \infty) \rightarrow [-1, \infty) \text{ is given by } f(x) = (x+1)^2 - 1. \\ \\ \text{ Show that } f \text{ is invertible. Also, find the set } S = \{ x : f(x) = f^{-1}(x) \}.$

Injectivity test:

$Let x and y \in [-1,\infty ), such that$

$f(x)=f(y)$

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

$\Rightarrow (x+1)^2 =(y+1)^2$

$\Rightarrow x+1=y+1$

$\Rightarrow x=y$

Therefore $f$ is an injection.

Surjectivity test:

$Let y \in [-1,\infty).$

$Then, f(x)=y$

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

$\Rightarrow x+1= \sqrt{y+1}$

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

$\text{Clearly, } x = \sqrt{y+1} - 1 \text{ is real for all } y \geq - 1$

$\text{ Thus, every element } y \in [-1,\infty) \text{ has its pre-image } x \in [-1,\infty) \text{ given by } x= \sqrt{y+1} -1$

Therefore $f$ is a surjection.

Since, $f$ is an injection and surjection, then it is bijection.

Hence, $f$ is a invertible.

$Let f^{-1} (x)=y \cdots \cdots ( 1 )$

$\Rightarrow f(y)=x$

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

$\Rightarrow (y+1)^2 =x+1$

$\Rightarrow y+1= \pm \sqrt{x+1}$

$\Rightarrow y = \pm \sqrt{x+1}-1$

From (1)

$\Rightarrow f^{-1} (x) = \pm \sqrt{x+1}-1$

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

$\Rightarrow (x+1)^2 - 1 = \pm \sqrt{x+1}-1$

$\Rightarrow (x+1)^2 = \pm \sqrt{x+1}$

$\Rightarrow (x+1)^4 = x + 1$

$\Rightarrow (x+1) [ (x+1)^3 - 1 ] = 0$

$\Rightarrow x+1 = 0 \text{ or } (x+1)^3 - 1 = 0$

$\Rightarrow x = -1 \text{ or } (x+1)^3 = 1$

$\Rightarrow x = -1 \text{ or }x+1 = -1$

$\Rightarrow x= -1 \text{ or } x = 0$

$\Rightarrow S = \{ 0, -1 \}$

$\\$

$\displaystyle \text{Question 20: Let } A = \{ x \in R | -1 \leq x \leq 1 \} \text{ and let } f: A \rightarrow A, g : A \rightarrow A \\ \\ \text{ be two functions defined by } f(x) = x^2 \text{ and } g(x) = \sin \frac{\pi x }{2} . \text{ Show that } g^{-1} \text{ exists but } \\ \\ f^{-1} \text{ does not exist. Also, find } g^{-1}.$

$f \text{ is not one-one because }$

$f(-1) = 1 \text{ and } f(1) = 1$

$\Rightarrow -1 \text{ and } 1 \text{ have the same image under } f$

$\Rightarrow f \text{ is not a bijection }$

$\text{ So, } f^{-1} \text{ does not exist. }$

$\text{ Injectivity of } g:$

$\text{ Let } x \text{ and } y \text{ be two elements in the domain } (A), \text{ such that }$

$\displaystyle \Rightarrow \sin \Big(\frac{\pi x}{2}\Big) = \sin \Big(\frac{\pi y}{2}\Big)$

$\displaystyle \Rightarrow \Big(\frac{\pi x}{2}\Big) = \Big(\frac{\pi y}{2}\Big)$

$\Rightarrow x = y$

$\text{ So, } g \text{ is one-one. }$

$\text{ Surjectivity of } g:$

$\displaystyle \text{ Range of } g = \Big[ \sin \Big(\frac{\pi (-1)}{2}\Big) , \sin \Big(\frac{\pi (1)}{2}\Big) \Big] \\ \\ =\Big [ \sin \Big(\frac{-\pi }{2}\Big) , \sin \Big(\frac{\pi }{2}\Big) \Big] = [ -1, 1 ] = A ( \text{ co-domain of } g)$

$\Rightarrow g \text{ is onto }$

$\Rightarrow g \text{ is a bijection }$

$\text{ Therefore, } g^{-1} \text{ exists. }$

$\text{ Let } g^{-1} (x) = y$

$\displaystyle \Rightarrow g(y) = x$

$\displaystyle \Rightarrow \sin \Bigg(\frac{xy}{2} \Bigg) = x$

$\displaystyle \Rightarrow y = \frac{2}{\pi} \sin^{-1} x$

$\displaystyle \Rightarrow g^{-1}(x) = \frac{2}{\pi} \sin^{-1} x$

$\\$

$\displaystyle \text{Question 21: Let } f \text{ be a function from } R \text{ to } R \text{ such that } f(x) = \cos ( x+2). \\ \\ \text{ Is } f \text{ invertible? Justify your answer. }$

Injectivity:

Let $x$  and $y$  be two elements in the domain (R), such that

$f(x) = f(y)$

$\Rightarrow \cos( x+2) = \cos (y+2)$

$\Rightarrow x+2 = y + 2 \ \ \ \text{ or } \ \ \ x + 2 = 2\pi - ( y+2)$

$\Rightarrow x = y \ \ \ \text{ or } \ \ \ x + 2 = 2 \pi - y - 2$

$\Rightarrow x = y \ \ \ \text{ or } \ \ \ x = 2 \pi - y - 4$

Hence, we cannot say that $x = y$

$\displaystyle \text{For example: } \cos \frac{\pi}{2} = \cos \frac{3\pi}{2} = 0$

$\displaystyle \text{Hence, } \frac{\pi}{2} \text{ and } \frac{3\pi}{2} \text{ have the same image } 0.$

Therefore $f$  is not one-one.

Therefore $f$  is not a bijection.

Thus, $f$  is not invertible.

$\\$

$\displaystyle \text{Question 22: If } A = \{ 1, 2, 3, 4 \} \text{ and } B = \{ a, b, c, d \}. \text{ Define any four bijections from } \\ \\ A \text{ to } B. \text{ Also, give their inverse functions. }$

$A = \{1, 2, 3, 4\} \ \ \ B = \{a, b, c, d\}$

(a) $f_1 = \{(1, a),(2, b), (3, c), (4, d)\}$

$f_1^{-1} = \{(a, 1), (b, 2), (c, 3), (d, 4)\}$

(b)$f_2 = \{(1, b), (2, a), (3, c), (4, d)\}$

$f_2^{-1} = \{(b, 1), (a, 2), (c, 3), (d, 4)\}$

(c) $f_3 =\{(1, a), (2, b), (4, c), (3, d)\}$

$f_3^{-1} = \{(a, 1), (b, 2), (c, 4), (d,3)\}$

(d) $f_4 = \{(1, b), (2, a), (4, c),(3, d)\}$

$f_4^{-1} = \{(b, 1), (a, 2), (c, 4), (d, 3)\}$

$\\$

Question 23: Let  A and B be two sets each with finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.

Since, there is an injective mapping from A to B, each element of A has unique image in B.

Similarly, there is also an injective mapping from B to A, each element of B has unique image in A or in other words there is one to one onto mapping from A to B.

Thus, there is bijective mapping from A to B.

$\\$

$\displaystyle \text{Question 24: If } f: A \rightarrow A, g: A \rightarrow A \text{ are two bijections, then prove that }$

(i) $fog$ is an injection

(ii) $fog$ is a surjection.

(i) $fog$ is an injection

Injectivity of $fog$

Let $x$ and $y$ be two elements of the domain (A), such that

$(fog)(x) = (fog)(y)$

$\Rightarrow f(g(x)) = f(g(y))$

$\Rightarrow g(x) = g(y)$  ( As, $f$ is one-one)

$\Rightarrow x = y$   ( As $g$ is one-one)

Hence, $fog$ is an injection.

(ii) $fog$ is a surjection.

$\text{Given: } A \rightarrow A, g : A \rightarrow A \text{are two bijections. }$

$\text{Then } fog : A \rightarrow A$

Surjectivity of $fog:$

Let $z$ be an element in the co-domain of $fog (A).$

Now, $z \in A$ (co-domain of $f$) and $f$is a surjection.

$\text{So, } z = f(y), \text{ where } y \in A ( \text{ domain of } f )$

Now $y \in A$ (co-domain of $g$) and $g$ is a surjection.

$So, y = g(x), \text{ where } x \in A (\text{ domain of } g)$

$\text{Therefore } z = f(y) = f(g(x)) = (fog)(x), \text{ where } x \in A (\text{ domain of } fog)$

Hence, $fog$ is a surjection.

$\\$