\displaystyle \textbf{Question 1: }\text{State with reasons whether the following functions have inverse:}
\displaystyle \text{(i) }f:\{1,2,3,4\}\rightarrow\{10\},\ f=\{(1,10),(2,10),(3,10),(4,10)\}
\displaystyle \text{(ii) }g:\{5,6,7,8\}\rightarrow\{1,2,3,4\},\ g=\{(5,4),(6,3),(7,4),(8,2)\}
\displaystyle \text{(iii) }h:\{2,3,4,5\}\rightarrow\{7,9,11,13\},\ h=\{(2,7),(3,9),(4,11),(5,13)\}.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }f:\{1,2,3,4\}\rightarrow\{10\},\ f=\{(1,10),(2,10),(3,10),(4,10)\}.
\displaystyle \text{Here }f(1)=f(2)=f(3)=f(4)=10.
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \therefore f^{-1}\text{ does not exist.}

\displaystyle \text{(ii) Given }g:\{5,6,7,8\}\rightarrow\{1,2,3,4\},\ g=\{(5,4),(6,3),(7,4),(8,2)\}.
\displaystyle \text{Here }g(5)=g(7)=4.
\displaystyle \therefore g\text{ is not one-one.}
\displaystyle \therefore g\text{ is not a bijection.}
\displaystyle \therefore g^{-1}\text{ does not exist.}

\displaystyle \text{(iii) Given }h:\{2,3,4,5\}\rightarrow\{7,9,11,13\},\ h=\{(2,7),(3,9),(4,11),(5,13)\}.
\displaystyle \text{Different elements of the domain have different images in the co-domain.}
\displaystyle \therefore h\text{ is one-one.}
\displaystyle \text{Also, every element of the co-domain has a pre-image in the domain.}
\displaystyle \therefore h\text{ is onto.}
\displaystyle \therefore h\text{ is a bijection.}
\displaystyle \therefore h^{-1}\text{ exists and }
\displaystyle h^{-1}=\{(7,2),(9,3),(11,4),(13,5)\}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find }f^{-1}\text{, if it exists, where }f:A\rightarrow B.
\displaystyle \text{(i) }A=\{0,-1,-3,2\},\ B=\{-9,-3,0,6\}\text{ and }f(x)=3x.
\displaystyle \text{(ii) }A=\{1,3,5,7,9\},\ B=\{0,1,9,25,49,81\}\text{ and }f(x)=x^2.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }A=\{0,-1,-3,2\},\ B=\{-9,-3,0,6\}\text{ and }f(x)=3x.
\displaystyle \therefore f=\{(0,0),(-1,-3),(-3,-9),(2,6)\}.
\displaystyle \text{Different elements of the domain have different images in the co-domain.}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Also, Range}(f)=B.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence }f^{-1}\text{ exists.}
\displaystyle \therefore f^{-1}=\{(0,0),(-3,-1),(-9,-3),(6,2)\}.
\displaystyle \\

\displaystyle \text{(ii) Given }A=\{1,3,5,7,9\},\ B=\{0,1,9,25,49,81\}\text{ and }f(x)=x^2.
\displaystyle \therefore f=\{(1,1),(3,9),(5,25),(7,49),(9,81)\}.
\displaystyle \text{Different elements of the domain have different images in the co-domain.}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{But }0\in B\text{ has no pre-image in }A.
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \therefore f^{-1}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Consider }f:\{1,2,3\}\rightarrow\{a,b,c\}\text{ and }g:\{a,b,c\}\rightarrow
\displaystyle \{\text{apple},\text{ball},\text{cat}\}\text{ defined by }f(1)=a,\ f(2)=b,\ f(3)=c,
\displaystyle g(a)=\text{apple},\ g(b)=\text{ball}\text{ and }g(c)=\text{cat}.
\displaystyle \text{Show that }f,\ g\text{ and }g\circ f\text{ are invertible. Find }f^{-1},\ g^{-1}\text{ and }(g\circ f)^{-1},
\displaystyle \text{and show that }(g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \text{Answer:}
\displaystyle f=\{(1,a),(2,b),(3,c)\}
\displaystyle \text{and }g=\{(a,\text{apple}),(b,\text{ball}),(c,\text{cat})\}.
\displaystyle \text{Different elements of the domain of }f\text{ have different images.}
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Also, every element of }\{a,b,c\}\text{ has a pre-image under }f.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle f^{-1}=\{(a,1),(b,2),(c,3)\}.
\displaystyle \text{Similarly, }g\text{ is one-one and onto.}
\displaystyle \therefore g\text{ is a bijection and hence is invertible.}
\displaystyle g^{-1}=\{(\text{apple},a),(\text{ball},b),(\text{cat},c)\}.
\displaystyle \\

\displaystyle \text{Now, }g\circ f:\{1,2,3\}\rightarrow\{\text{apple},\text{ball},\text{cat}\}.
\displaystyle (g\circ f)(1)=g(f(1))=g(a)=\text{apple}.
\displaystyle (g\circ f)(2)=g(f(2))=g(b)=\text{ball}.
\displaystyle (g\circ f)(3)=g(f(3))=g(c)=\text{cat}.
\displaystyle \therefore g\circ f=\{(1,\text{apple}),(2,\text{ball}),(3,\text{cat})\}.
\displaystyle \text{Different elements of its domain have different images.}
\displaystyle \therefore g\circ f\text{ is one-one.}
\displaystyle \text{Also, every element of its co-domain has a pre-image.}
\displaystyle \therefore g\circ f\text{ is onto.}
\displaystyle \therefore g\circ f\text{ is invertible and}
\displaystyle (g\circ f)^{-1}=\{(\text{apple},1),(\text{ball},2),(\text{cat},3)\}.
\displaystyle \\

\displaystyle \text{Also,}
\displaystyle (f^{-1}\circ g^{-1})(\text{apple})=f^{-1}(g^{-1}(\text{apple}))=f^{-1}(a)=1,
\displaystyle (f^{-1}\circ g^{-1})(\text{ball})=f^{-1}(g^{-1}(\text{ball}))=f^{-1}(b)=2,
\displaystyle (f^{-1}\circ g^{-1})(\text{cat})=f^{-1}(g^{-1}(\text{cat}))=f^{-1}(c)=3.
\displaystyle \therefore f^{-1}\circ g^{-1}=\{(\text{apple},1),(\text{ball},2),(\text{cat},3)\}.
\displaystyle \therefore (g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{1,2,3,4\},\ B=\{3,5,7,9\}\text{ and }C=\{7,23,47,79\}.
\displaystyle \text{Let }f:A\rightarrow B\text{ and }g:B\rightarrow C\text{ be defined by }f(x)=2x+1\text{ and }g(x)=x^2-2.
\displaystyle \text{Express }(g\circ f)^{-1}\text{ and }f^{-1}\circ g^{-1}\text{ as sets of ordered pairs and verify that}
\displaystyle (g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=2x+1.
\displaystyle f=\{(1,3),(2,5),(3,7),(4,9)\}.
\displaystyle \text{Also, }g(x)=x^2-2.
\displaystyle g=\{(3,7),(5,23),(7,47),(9,79)\}.
\displaystyle \text{Clearly, }f\text{ and }g\text{ are one-one and onto.}
\displaystyle \therefore f\text{ and }g\text{ are bijections and hence their inverses exist.}
\displaystyle f^{-1}=\{(3,1),(5,2),(7,3),(9,4)\}.
\displaystyle g^{-1}=\{(7,3),(23,5),(47,7),(79,9)\}.
\displaystyle \\

\displaystyle \text{Now, }f^{-1}\circ g^{-1}:C\rightarrow A.
\displaystyle (f^{-1}\circ g^{-1})(7)=f^{-1}(g^{-1}(7))=f^{-1}(3)=1.
\displaystyle (f^{-1}\circ g^{-1})(23)=f^{-1}(g^{-1}(23))=f^{-1}(5)=2.
\displaystyle (f^{-1}\circ g^{-1})(47)=f^{-1}(g^{-1}(47))=f^{-1}(7)=3.
\displaystyle (f^{-1}\circ g^{-1})(79)=f^{-1}(g^{-1}(79))=f^{-1}(9)=4.
\displaystyle \therefore f^{-1}\circ g^{-1}=\{(7,1),(23,2),(47,3),(79,4)\}.
\displaystyle \\

\displaystyle \text{Also, }g\circ f:A\rightarrow C.
\displaystyle (g\circ f)(x)=g(f(x))=g(2x+1)=(2x+1)^2-2.
\displaystyle =4x^2+4x-1.
\displaystyle (g\circ f)(1)=7,\quad (g\circ f)(2)=23,
\displaystyle (g\circ f)(3)=47,\quad (g\circ f)(4)=79.
\displaystyle \therefore g\circ f=\{(1,7),(2,23),(3,47),(4,79)\}.
\displaystyle \therefore (g\circ f)^{-1}=\{(7,1),(23,2),(47,3),(79,4)\}.
\displaystyle \text{Hence, }(g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the function }f:\mathbb Q\rightarrow\mathbb Q\text{ defined by }f(x)=3x+5
\displaystyle \text{ is invertible. Also, find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Q\rightarrow\mathbb Q,\text{ where }f(x)=3x+5.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb Q\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow 3x+5=3y+5.
\displaystyle \Rightarrow 3x=3y.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb Q\text{ be any element of the co-domain.}
\displaystyle \text{Let }f(x)=y.
\displaystyle \Rightarrow 3x+5=y.
\displaystyle \Rightarrow x=\frac{y-5}{3}.
\displaystyle \text{Since }y\in\mathbb Q,\ \frac{y-5}{3}\in\mathbb Q.
\displaystyle \therefore x\in\mathbb Q\text{ and every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=3y+5.
\displaystyle \Rightarrow 3y=x-5.
\displaystyle \Rightarrow y=\frac{x-5}{3}.
\displaystyle \therefore f^{-1}(x)=\frac{x-5}{3}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Consider }f:\mathbb R\rightarrow\mathbb R\text{ given by }f(x)=4x+3.
\displaystyle \text{Show that }f\text{ is invertible. Find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R,\text{ where }f(x)=4x+3.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow 4x+3=4y+3.
\displaystyle \Rightarrow 4x=4y.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb R\text{ be any element of the co-domain.}
\displaystyle \text{Let }f(x)=y.
\displaystyle \Rightarrow 4x+3=y.
\displaystyle \Rightarrow x=\frac{y-3}{4}.
\displaystyle \text{Since }\frac{y-3}{4}\in\mathbb R,\ x\in\mathbb R.
\displaystyle \therefore \text{every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=4y+3.
\displaystyle \Rightarrow 4y=x-3.
\displaystyle \Rightarrow y=\frac{x-3}{4}.
\displaystyle \therefore f^{-1}(x)=\frac{x-3}{4}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Consider }f:\mathbb R^+\rightarrow[4,\infty)\text{ given by }f(x)=x^2+4.
\displaystyle \text{Show that }f\text{ is invertible with inverse }f^{-1}(x)=\sqrt{x-4},
\displaystyle \text{where }\mathbb R^+\text{ is the set of all non-negative real numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R^+\rightarrow[4,\infty),\text{ where }f(x)=x^2+4.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R^+\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow x^2+4=y^2+4.
\displaystyle \Rightarrow x^2=y^2.
\displaystyle \Rightarrow x=\pm y.
\displaystyle \text{Since }x,y\geq 0,\text{ we get }x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in[4,\infty)\text{ be any element of the co-domain.}
\displaystyle \text{Let }f(x)=y.
\displaystyle \Rightarrow x^2+4=y.
\displaystyle \Rightarrow x^2=y-4.
\displaystyle \Rightarrow x=\sqrt{y-4}.
\displaystyle \text{Since }y\geq 4,\ \sqrt{y-4}\geq 0.
\displaystyle \therefore x=\sqrt{y-4}\in\mathbb R^+.
\displaystyle \therefore \text{every element of the co-domain has a pre-image in }\mathbb R^+.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=y^2+4.
\displaystyle \Rightarrow y^2=x-4.
\displaystyle \text{Since }y\in\mathbb R^+,\ y=\sqrt{x-4}.
\displaystyle \therefore f^{-1}(x)=\sqrt{x-4},\quad x\in[4,\infty).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }f(x)=\frac{4x+3}{6x-4},\ x\neq\frac{2}{3},\text{ show that }(f\circ f)(x)=x
\displaystyle \text{ for all }x\neq\frac{2}{3}.\text{ What is the inverse of }f?
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\mathbb R\setminus\left\{\frac{2}{3}\right\}.
\displaystyle \text{Then }f:A\rightarrow A\text{ is defined by }f(x)=\frac{4x+3}{6x-4}.
\displaystyle \text{First, we show that }f(x)\neq\frac{2}{3}\text{ for every }x\in A.
\displaystyle \text{Suppose }f(x)=\frac{2}{3}.
\displaystyle \Rightarrow \frac{4x+3}{6x-4}=\frac{2}{3}.
\displaystyle \Rightarrow 3(4x+3)=2(6x-4).
\displaystyle \Rightarrow 12x+9=12x-8,
\displaystyle \text{which is impossible.}
\displaystyle \therefore f(x)\neq\frac{2}{3},\text{ and hence }f(f(x))\text{ is defined for every }x\in A.
\displaystyle \\

\displaystyle \text{Now,}
\displaystyle (f\circ f)(x)=f(f(x))
\displaystyle =f\left(\frac{4x+3}{6x-4}\right)
\displaystyle =\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}
\displaystyle =\frac{\frac{16x+12+18x-12}{6x-4}}{\frac{24x+18-24x+16}{6x-4}}
\displaystyle =\frac{34x}{34}
\displaystyle =x.
\displaystyle \therefore (f\circ f)(x)=x\text{ for all }x\neq\frac{2}{3}.
\displaystyle \therefore f\circ f=I_A.
\displaystyle \text{Hence, }f\text{ is invertible and }f^{-1}=f.
\displaystyle \therefore f^{-1}(x)=\frac{4x+3}{6x-4},\quad x\neq\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Consider }f:\mathbb R^+\rightarrow[-5,\infty)\text{ given by }f(x)=9x^2+6x-5.
\displaystyle \text{Show that }f\text{ is invertible with }f^{-1}(x)=\frac{\sqrt{x+6}-1}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R^+\rightarrow[-5,\infty),\text{ where }f(x)=9x^2+6x-5.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x_1,x_2\in\mathbb R^+\text{ such that }f(x_1)=f(x_2).
\displaystyle \Rightarrow 9x_1^2+6x_1-5=9x_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)\left[9(x_1+x_2)+6\right]=0.
\displaystyle \Rightarrow x_1-x_2=0\text{ or }9(x_1+x_2)+6=0.
\displaystyle \text{Since }x_1,x_2\geq 0,\ 9(x_1+x_2)+6>0.
\displaystyle \therefore x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in[-5,\infty)\text{ be any element of the co-domain.}
\displaystyle \text{Choose }x=\frac{\sqrt{y+6}-1}{3}.
\displaystyle \text{Since }y\geq-5,\ \sqrt{y+6}\geq 1.
\displaystyle \therefore x=\frac{\sqrt{y+6}-1}{3}\geq 0.
\displaystyle \therefore x\in\mathbb R^+.
\displaystyle \text{Also, }3x+1=\sqrt{y+6}.
\displaystyle \Rightarrow (3x+1)^2=y+6.
\displaystyle \Rightarrow 9x^2+6x-5=y.
\displaystyle \Rightarrow f(x)=y.
\displaystyle \therefore \text{every element of the co-domain has a pre-image in }\mathbb R^+.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=9y^2+6y-5.
\displaystyle \Rightarrow x+6=9y^2+6y+1.
\displaystyle \Rightarrow x+6=(3y+1)^2.
\displaystyle \text{Since }y\in\mathbb R^+,\ 3y+1>0.
\displaystyle \Rightarrow 3y+1=\sqrt{x+6}.
\displaystyle \Rightarrow y=\frac{\sqrt{x+6}-1}{3}.
\displaystyle \therefore f^{-1}(x)=\frac{\sqrt{x+6}-1}{3},\quad x\in[-5,\infty).
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }f:\mathbb R\rightarrow\mathbb R\text{ is defined by }f(x)=x^3-3,\text{ prove that }f^{-1}
\displaystyle \text{ exists and find a formula for }f^{-1}.\text{ Hence, find }f^{-1}(24)\text{ and }f^{-1}(5).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R,\text{ where }f(x)=x^3-3.
\displaystyle \text{To prove that }f^{-1}\text{ exists, we show that }f\text{ is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow x^3-3=y^3-3.
\displaystyle \Rightarrow x^3=y^3.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb R\text{ be any element of the co-domain.}
\displaystyle \text{Let }f(x)=y.
\displaystyle \Rightarrow x^3-3=y.
\displaystyle \Rightarrow x^3=y+3.
\displaystyle \Rightarrow x=\sqrt[3]{y+3}.
\displaystyle \text{Since }\sqrt[3]{y+3}\in\mathbb R,\text{ every }y\in\mathbb R\text{ has a pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence }f^{-1}\text{ exists.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=y^3-3.
\displaystyle \Rightarrow y^3=x+3.
\displaystyle \Rightarrow y=\sqrt[3]{x+3}.
\displaystyle \therefore f^{-1}(x)=\sqrt[3]{x+3}.
\displaystyle f^{-1}(24)=\sqrt[3]{24+3}=\sqrt[3]{27}=3.
\displaystyle f^{-1}(5)=\sqrt[3]{5+3}=\sqrt[3]{8}=2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A function }f:\mathbb R\rightarrow\mathbb R\text{ is defined by }f(x)=x^3+4.
\displaystyle \text{Is it a bijection or not? In case it is a bijection, find }f^{-1}(3).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R,\text{ where }f(x)=x^3+4.
\displaystyle \text{To determine whether }f\text{ is a bijection, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow x^3+4=y^3+4.
\displaystyle \Rightarrow x^3=y^3.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb R\text{ be any element of the co-domain.}
\displaystyle \text{Let }f(x)=y.
\displaystyle \Rightarrow x^3+4=y.
\displaystyle \Rightarrow x^3=y-4.
\displaystyle \Rightarrow x=\sqrt[3]{y-4}.
\displaystyle \text{Since }\sqrt[3]{y-4}\in\mathbb R,\text{ every }y\in\mathbb R\text{ has a pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=y^3+4.
\displaystyle \Rightarrow y^3=x-4.
\displaystyle \Rightarrow y=\sqrt[3]{x-4}.
\displaystyle \therefore f^{-1}(x)=\sqrt[3]{x-4}.
\displaystyle f^{-1}(3)=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }f:\mathbb Q\rightarrow\mathbb Q\text{ and }g:\mathbb Q\rightarrow\mathbb Q\text{ are defined by}
\displaystyle f(x)=2x\text{ and }g(x)=x+2,\text{ show that }f\text{ and }g\text{ are bijective maps.}
\displaystyle \text{Verify that }(g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{First, we show that }f\text{ is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb Q\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow 2x=2y.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb Q\text{ be any element of the co-domain.}
\displaystyle \text{Choose }x=\frac{y}{2}.
\displaystyle \text{Since }y\in\mathbb Q,\ \frac{y}{2}\in\mathbb Q.
\displaystyle \text{Also, }f(x)=2\left(\frac{y}{2}\right)=y.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=2y.
\displaystyle \Rightarrow y=\frac{x}{2}.
\displaystyle \therefore f^{-1}(x)=\frac{x}{2}.
\displaystyle \\

\displaystyle \text{Next, we show that }g\text{ is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb Q\text{ such that }g(x)=g(y).
\displaystyle \Rightarrow x+2=y+2.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore g\text{ is one-one.}
\displaystyle \text{Now, let }y\in\mathbb Q\text{ be any element of the co-domain.}
\displaystyle \text{Choose }x=y-2.
\displaystyle \text{Since }y\in\mathbb Q,\ y-2\in\mathbb Q.
\displaystyle \text{Also, }g(x)=(y-2)+2=y.
\displaystyle \therefore g\text{ is onto.}
\displaystyle \therefore g\text{ is a bijection and hence is invertible.}
\displaystyle \text{To find }g^{-1},\text{ let }g^{-1}(x)=y.
\displaystyle \Rightarrow x=g(y)=y+2.
\displaystyle \Rightarrow y=x-2.
\displaystyle \therefore g^{-1}(x)=x-2.
\displaystyle \\

\displaystyle \text{Now,}
\displaystyle (f^{-1}\circ g^{-1})(x)=f^{-1}(g^{-1}(x))
\displaystyle =f^{-1}(x-2)
\displaystyle =\frac{x-2}{2}.\qquad\cdots(1)
\displaystyle \text{Also,}
\displaystyle (g\circ f)(x)=g(f(x))=g(2x)=2x+2.
\displaystyle \text{Let }(g\circ f)^{-1}(x)=y.
\displaystyle \Rightarrow x=(g\circ f)(y)=2y+2.
\displaystyle \Rightarrow 2y=x-2.
\displaystyle \Rightarrow y=\frac{x-2}{2}.
\displaystyle \therefore (g\circ f)^{-1}(x)=\frac{x-2}{2}.\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle (g\circ f)^{-1}(x)=(f^{-1}\circ g^{-1})(x).
\displaystyle \therefore (g\circ f)^{-1}=f^{-1}\circ g^{-1}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Let }A=\mathbb R-\{3\}\text{ and }B=\mathbb R-\{1\}.
\displaystyle \text{Consider }f:A\rightarrow B\text{ defined by }f(x)=\frac{x-2}{x-3}.
\displaystyle \text{Show that }f\text{ is one-one and onto and hence find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:A\rightarrow B,\text{ where }A=\mathbb R-\{3\},\ B=\mathbb R-\{1\}
\displaystyle \text{and }f(x)=\frac{x-2}{x-3}.
\displaystyle \text{Let }x_1,x_2\in A\text{ such that }f(x_1)=f(x_2).
\displaystyle \Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}.
\displaystyle \Rightarrow (x_1-2)(x_2-3)=(x_2-2)(x_1-3).
\displaystyle \Rightarrow x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6.
\displaystyle \Rightarrow -x_1+x_2=0.
\displaystyle \Rightarrow x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in B\text{ be any element of the co-domain.}
\displaystyle \text{Since }y\neq 1,\text{ choose }x=\frac{2-3y}{1-y}.
\displaystyle \text{We first show that }x\in A.
\displaystyle \text{If }x=3,\text{ then }\frac{2-3y}{1-y}=3.
\displaystyle \Rightarrow 2-3y=3-3y,
\displaystyle \text{which gives }2=3,\text{ an impossibility.}
\displaystyle \therefore x\neq 3,\text{ so }x\in A.
\displaystyle \text{Also,}
\displaystyle f(x)=\frac{x-2}{x-3}
\displaystyle =\frac{\frac{2-3y}{1-y}-2}{\frac{2-3y}{1-y}-3}
\displaystyle =\frac{\frac{-y}{1-y}}{\frac{-1}{1-y}}
\displaystyle =y.
\displaystyle \therefore \text{every }y\in B\text{ has a pre-image in }A.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence }f^{-1}\text{ exists.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=\frac{y-2}{y-3}.
\displaystyle \Rightarrow x(y-3)=y-2.
\displaystyle \Rightarrow xy-3x=y-2.
\displaystyle \Rightarrow y(x-1)=3x-2.
\displaystyle \Rightarrow y=\frac{3x-2}{x-1}.
\displaystyle \therefore f^{-1}(x)=\frac{3x-2}{x-1},\quad x\in B.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Consider the function }f:\mathbb R^+\rightarrow[-9,\infty)\text{ given by}
\displaystyle f(x)=5x^2+6x-9.\text{ Prove that }f\text{ is invertible and find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R^+\rightarrow[-9,\infty),\text{ where }f(x)=5x^2+6x-9.
\displaystyle \text{To prove that }f\text{ is invertible, we show that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R^+\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow 5x^2+6x-9=5y^2+6y-9.
\displaystyle \Rightarrow 5(x^2-y^2)+6(x-y)=0.
\displaystyle \Rightarrow (x-y)\left[5(x+y)+6\right]=0.
\displaystyle \text{Since }x,y\geq 0,\ 5(x+y)+6>0.
\displaystyle \therefore x-y=0.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in[-9,\infty)\text{ be any element of the co-domain.}
\displaystyle \text{Choose }x=\frac{-3+\sqrt{54+5y}}{5}.
\displaystyle \text{Since }y\geq-9,\ 54+5y\geq 9.
\displaystyle \Rightarrow \sqrt{54+5y}\geq 3.
\displaystyle \therefore x=\frac{-3+\sqrt{54+5y}}{5}\geq 0.
\displaystyle \therefore x\in\mathbb R^+.
\displaystyle \text{Also, }5x+3=\sqrt{54+5y}.
\displaystyle \Rightarrow (5x+3)^2=54+5y.
\displaystyle \Rightarrow 25x^2+30x+9=54+5y.
\displaystyle \Rightarrow 5x^2+6x-9=y.
\displaystyle \Rightarrow f(x)=y.
\displaystyle \therefore \text{every }y\in[-9,\infty)\text{ has a pre-image in }\mathbb R^+.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(y)=x.
\displaystyle \Rightarrow y=f(x)=5x^2+6x-9.
\displaystyle \Rightarrow 5x^2+6x-(9+y)=0.
\displaystyle \Rightarrow x=\frac{-6\pm\sqrt{36+20(9+y)}}{10}.
\displaystyle \Rightarrow x=\frac{-3\pm\sqrt{54+5y}}{5}.
\displaystyle \text{Since }x\in\mathbb R^+,\text{ the negative sign is rejected.}
\displaystyle \therefore f^{-1}(y)=\frac{-3+\sqrt{54+5y}}{5},\quad y\in[-9,\infty).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Let }f:\mathbb N\rightarrow\mathbb N\text{ be defined by }f(x)=9x^2+6x-5.
\displaystyle \text{Show that }f:\mathbb N\rightarrow S,\text{ where }S\text{ is the range of }f,\text{ is invertible.}
\displaystyle \text{Find }f^{-1}\text{ and hence find }f^{-1}(43)\text{ and }f^{-1}(163).
\displaystyle \text{Answer:}
\displaystyle \text{Assume }\mathbb N=\{1,2,3,\ldots\}.
\displaystyle \text{Given }f:\mathbb N\rightarrow S,\text{ where }f(x)=9x^2+6x-5
\displaystyle \text{and }S=\{f(x):x\in\mathbb N\}.
\displaystyle \text{Let }x_1,x_2\in\mathbb N\text{ such that }f(x_1)=f(x_2).
\displaystyle \Rightarrow 9x_1^2+6x_1-5=9x_2^2+6x_2-5.
\displaystyle \Rightarrow 9(x_1^2-x_2^2)+6(x_1-x_2)=0.
\displaystyle \Rightarrow (x_1-x_2)\left[9(x_1+x_2)+6\right]=0.
\displaystyle \text{Since }x_1,x_2\in\mathbb N,\ 9(x_1+x_2)+6>0.
\displaystyle \therefore x_1-x_2=0.
\displaystyle \Rightarrow x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Also, }S\text{ is the range of }f.
\displaystyle \therefore \text{every element of }S\text{ has a pre-image in }\mathbb N.
\displaystyle \therefore f:\mathbb N\rightarrow S\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(y)=x.
\displaystyle \Rightarrow y=f(x)=9x^2+6x-5.
\displaystyle \Rightarrow 9x^2+6x-(y+5)=0.
\displaystyle \Rightarrow x=\frac{-6\pm\sqrt{36+36(y+5)}}{18}.
\displaystyle \Rightarrow x=\frac{-6\pm6\sqrt{y+6}}{18}.
\displaystyle \Rightarrow x=\frac{-1\pm\sqrt{y+6}}{3}.
\displaystyle \text{Since }x\in\mathbb N,\text{ the negative sign is rejected.}
\displaystyle \therefore f^{-1}(y)=\frac{-1+\sqrt{y+6}}{3},\quad y\in S.
\displaystyle f^{-1}(43)=\frac{-1+\sqrt{43+6}}{3}
\displaystyle =\frac{-1+7}{3}=2.
\displaystyle f^{-1}(163)=\frac{-1+\sqrt{163+6}}{3}
\displaystyle =\frac{-1+13}{3}=4.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Let }f:\mathbb R-\left\{-\frac{4}{3}\right\}\rightarrow\mathbb R-\left\{\frac{4}{3}\right\}
\displaystyle \text{ be defined by }f(x)=\frac{4x}{3x+4}.
\displaystyle \text{Show that }f\text{ is one-one and onto. Hence, find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R-\left\{-\frac{4}{3}\right\}\rightarrow\mathbb R-\left\{\frac{4}{3}\right\},
\displaystyle \text{where }f(x)=\frac{4x}{3x+4}.
\displaystyle \text{First, we show that }f\text{ is one-one.}
\displaystyle \text{Let }x_1,x_2\in\mathbb R-\left\{-\frac{4}{3}\right\}\text{ such that }f(x_1)=f(x_2).
\displaystyle \Rightarrow \frac{4x_1}{3x_1+4}=\frac{4x_2}{3x_2+4}.
\displaystyle \Rightarrow 4x_1(3x_2+4)=4x_2(3x_1+4).
\displaystyle \Rightarrow 12x_1x_2+16x_1=12x_1x_2+16x_2.
\displaystyle \Rightarrow 16x_1=16x_2.
\displaystyle \Rightarrow x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in\mathbb R-\left\{\frac{4}{3}\right\}\text{ be any element of the co-domain.}
\displaystyle \text{We find }x\text{ such that }f(x)=y.
\displaystyle \frac{4x}{3x+4}=y.
\displaystyle \Rightarrow 4x=3xy+4y.
\displaystyle \Rightarrow x(4-3y)=4y.
\displaystyle \Rightarrow x=\frac{4y}{4-3y}.
\displaystyle \text{Since }y\neq\frac{4}{3},\ 4-3y\neq 0,\text{ so }x\in\mathbb R.
\displaystyle \text{Also, suppose }x=-\frac{4}{3}.
\displaystyle \Rightarrow \frac{4y}{4-3y}=-\frac{4}{3}.
\displaystyle \Rightarrow 12y=-16+12y,
\displaystyle \text{which gives }0=-16,\text{ an impossibility.}
\displaystyle \therefore x\neq-\frac{4}{3}.
\displaystyle \therefore x=\frac{4y}{4-3y}\in\mathbb R-\left\{-\frac{4}{3}\right\}.
\displaystyle \text{Further,}
\displaystyle f\left(\frac{4y}{4-3y}\right)
\displaystyle =\frac{4\left(\frac{4y}{4-3y}\right)}{3\left(\frac{4y}{4-3y}\right)+4}
\displaystyle =\frac{\frac{16y}{4-3y}}{\frac{12y+16-12y}{4-3y}}
\displaystyle =\frac{16y}{16}
\displaystyle =y.
\displaystyle \therefore \text{every element of the co-domain has a pre-image in the domain.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=\frac{4y}{3y+4}.
\displaystyle \Rightarrow x(3y+4)=4y.
\displaystyle \Rightarrow 3xy+4x=4y.
\displaystyle \Rightarrow y(4-3x)=4x.
\displaystyle \Rightarrow y=\frac{4x}{4-3x}.
\displaystyle \therefore f^{-1}(x)=\frac{4x}{4-3x},\quad x\neq\frac{4}{3}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }f:\mathbb R\rightarrow(-1,1)\text{ is defined by}
\displaystyle f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}},\text{ show that }f\text{ is invertible and find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow(-1,1),\text{ where }f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R\text{ 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^{2x}-1}{10^{2x}+1}=\frac{10^{2y}-1}{10^{2y}+1}.
\displaystyle \Rightarrow (10^{2x}-1)(10^{2y}+1)=(10^{2y}-1)(10^{2x}+1).
\displaystyle \Rightarrow 10^{2x+2y}+10^{2x}-10^{2y}-1
\displaystyle =10^{2x+2y}+10^{2y}-10^{2x}-1.
\displaystyle \Rightarrow 2\cdot10^{2x}=2\cdot10^{2y}.
\displaystyle \Rightarrow 10^{2x}=10^{2y}.
\displaystyle \Rightarrow 2x=2y.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in(-1,1)\text{ be any element of the co-domain.}
\displaystyle \text{We find }x\in\mathbb R\text{ such that }f(x)=y.
\displaystyle \frac{10^x-10^{-x}}{10^x+10^{-x}}=y.
\displaystyle \Rightarrow \frac{10^{2x}-1}{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 \text{Since }-1<y<1,\ 1+y>0\text{ and }1-y>0.
\displaystyle \therefore \frac{1+y}{1-y}>0.
\displaystyle \Rightarrow 2x=\log_{10}\left(\frac{1+y}{1-y}\right).
\displaystyle \Rightarrow x=\frac{1}{2}\log_{10}\left(\frac{1+y}{1-y}\right)\in\mathbb R.
\displaystyle \therefore \text{every element of the co-domain has a pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=\frac{10^y-10^{-y}}{10^y+10^{-y}}.
\displaystyle \Rightarrow x=\frac{10^{2y}-1}{10^{2y}+1}.
\displaystyle \Rightarrow x(10^{2y}+1)=10^{2y}-1.
\displaystyle \Rightarrow 10^{2y}(1-x)=1+x.
\displaystyle \Rightarrow 10^{2y}=\frac{1+x}{1-x}.
\displaystyle \Rightarrow 2y=\log_{10}\left(\frac{1+x}{1-x}\right).
\displaystyle \Rightarrow y=\frac{1}{2}\log_{10}\left(\frac{1+x}{1-x}\right).
\displaystyle \therefore f^{-1}(x)=\frac{1}{2}\log_{10}\left(\frac{1+x}{1-x}\right),\quad -1<x<1.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }f:\mathbb R\rightarrow(0,2)\text{ is defined by}
\displaystyle f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1,\text{ show that }f\text{ is invertible and find }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow(0,2),\text{ where }f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1.
\displaystyle \text{To show that }f\text{ is invertible, we prove that it is one-one and onto.}
\displaystyle \text{Let }x,y\in\mathbb R\text{ 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^{2x}-1}{e^{2x}+1}=\frac{e^{2y}-1}{e^{2y}+1}.
\displaystyle \Rightarrow (e^{2x}-1)(e^{2y}+1)=(e^{2y}-1)(e^{2x}+1).
\displaystyle \Rightarrow e^{2x+2y}+e^{2x}-e^{2y}-1
\displaystyle =e^{2x+2y}+e^{2y}-e^{2x}-1.
\displaystyle \Rightarrow 2e^{2x}=2e^{2y}.
\displaystyle \Rightarrow e^{2x}=e^{2y}.
\displaystyle \Rightarrow 2x=2y.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in(0,2)\text{ be any element of the co-domain.}
\displaystyle \text{We find }x\in\mathbb R\text{ such that }f(x)=y.
\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}+1=y.
\displaystyle \Rightarrow \frac{e^{2x}-1}{e^{2x}+1}=y-1.
\displaystyle \Rightarrow e^{2x}-1=(y-1)(e^{2x}+1).
\displaystyle \Rightarrow e^{2x}-1=ye^{2x}+y-e^{2x}-1.
\displaystyle \Rightarrow e^{2x}(2-y)=y.
\displaystyle \Rightarrow e^{2x}=\frac{y}{2-y}.
\displaystyle \text{Since }0<y<2,\ y>0\text{ and }2-y>0.
\displaystyle \therefore \frac{y}{2-y}>0.
\displaystyle \Rightarrow 2x=\log_e\left(\frac{y}{2-y}\right).
\displaystyle \Rightarrow x=\frac{1}{2}\log_e\left(\frac{y}{2-y}\right)\in\mathbb R.
\displaystyle \therefore \text{every element of }(0,2)\text{ has a pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=\frac{e^y-e^{-y}}{e^y+e^{-y}}+1.
\displaystyle \Rightarrow \frac{e^{2y}-1}{e^{2y}+1}=x-1.
\displaystyle \Rightarrow e^{2y}-1=(x-1)(e^{2y}+1).
\displaystyle \Rightarrow e^{2y}-1=xe^{2y}+x-e^{2y}-1.
\displaystyle \Rightarrow e^{2y}(2-x)=x.
\displaystyle \Rightarrow e^{2y}=\frac{x}{2-x}.
\displaystyle \Rightarrow 2y=\log_e\left(\frac{x}{2-x}\right).
\displaystyle \Rightarrow y=\frac{1}{2}\log_e\left(\frac{x}{2-x}\right).
\displaystyle \therefore f^{-1}(x)=\frac{1}{2}\log_e\left(\frac{x}{2-x}\right),\quad 0<x<2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Let }f:[-1,\infty)\rightarrow[-1,\infty)\text{ be given by}
\displaystyle f(x)=(x+1)^2-1.
\displaystyle \text{Show that }f\text{ is invertible. Also, find }S=\{x:f(x)=f^{-1}(x)\}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:[-1,\infty)\rightarrow[-1,\infty),\text{ where }f(x)=(x+1)^2-1.
\displaystyle \text{First, we show that }f\text{ is one-one.}
\displaystyle \text{Let }x,y\in[-1,\infty)\text{ such that }f(x)=f(y).
\displaystyle \Rightarrow (x+1)^2-1=(y+1)^2-1.
\displaystyle \Rightarrow (x+1)^2=(y+1)^2.
\displaystyle \text{Since }x+1\geq0\text{ and }y+1\geq0,
\displaystyle \Rightarrow x+1=y+1.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in[-1,\infty)\text{ be any element of the co-domain.}
\displaystyle \text{We find }x\in[-1,\infty)\text{ such that }f(x)=y.
\displaystyle (x+1)^2-1=y.
\displaystyle \Rightarrow (x+1)^2=y+1.
\displaystyle \text{Since }x+1\geq0,
\displaystyle \Rightarrow x+1=\sqrt{y+1}.
\displaystyle \Rightarrow x=\sqrt{y+1}-1.
\displaystyle \text{Since }y\geq-1,\ \sqrt{y+1}\geq0.
\displaystyle \therefore x=\sqrt{y+1}-1\geq-1.
\displaystyle \therefore x\in[-1,\infty).
\displaystyle \text{Also,}
\displaystyle f\left(\sqrt{y+1}-1\right)
\displaystyle =\left(\sqrt{y+1}\right)^2-1
\displaystyle =y.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection and hence is invertible.}
\displaystyle \\

\displaystyle \text{To find }f^{-1},\text{ let }f^{-1}(x)=y.
\displaystyle \Rightarrow x=f(y)=(y+1)^2-1.
\displaystyle \Rightarrow (y+1)^2=x+1.
\displaystyle \text{Since }y\geq-1,\ y+1\geq0.
\displaystyle \Rightarrow y+1=\sqrt{x+1}.
\displaystyle \Rightarrow y=\sqrt{x+1}-1.
\displaystyle \therefore f^{-1}(x)=\sqrt{x+1}-1,\quad x\geq-1.
\displaystyle \\

\displaystyle \text{Now, }f(x)=f^{-1}(x).
\displaystyle \Rightarrow (x+1)^2-1=\sqrt{x+1}-1.
\displaystyle \Rightarrow (x+1)^2=\sqrt{x+1}.
\displaystyle \text{Let }t=\sqrt{x+1},\text{ where }t\geq0.
\displaystyle \text{Then }x+1=t^2.
\displaystyle \Rightarrow t^4=t.
\displaystyle \Rightarrow t(t^3-1)=0.
\displaystyle \Rightarrow t=0\text{ or }t=1.
\displaystyle \text{If }t=0,\text{ then }x+1=0\Rightarrow x=-1.
\displaystyle \text{If }t=1,\text{ then }x+1=1\Rightarrow x=0.
\displaystyle \therefore S=\{-1,0\}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Let }A=\{x\in\mathbb R:-1\leq x\leq1\}\text{ and let}
\displaystyle f:A\rightarrow A,\quad g:A\rightarrow A\text{ be defined by}
\displaystyle f(x)=x^2\text{ and }g(x)=\sin\left(\frac{\pi x}{2}\right).
\displaystyle \text{Show that }g^{-1}\text{ exists but }f^{-1}\text{ does not exist. Also, find }g^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }f:A\rightarrow A,\text{ where }f(x)=x^2.
\displaystyle f\left(-\frac12\right)=\frac14
\displaystyle \text{and }f\left(\frac12\right)=\frac14.
\displaystyle \text{But }-\frac12\neq\frac12.
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \therefore f\text{ is not a bijection and hence }f^{-1}\text{ does not exist.}
\displaystyle \\

\displaystyle \text{Now, consider }g:A\rightarrow A,\text{ where }g(x)=\sin\left(\frac{\pi x}{2}\right).
\displaystyle \text{First, we show that }g\text{ is one-one.}
\displaystyle \text{Let }x,y\in A\text{ such that }g(x)=g(y).
\displaystyle \Rightarrow \sin\left(\frac{\pi x}{2}\right)=\sin\left(\frac{\pi y}{2}\right).
\displaystyle \text{Since }x,y\in[-1,1],
\displaystyle \frac{\pi x}{2},\frac{\pi y}{2}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
\displaystyle \text{The sine function is one-one on }\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
\displaystyle \therefore \frac{\pi x}{2}=\frac{\pi y}{2}.
\displaystyle \Rightarrow x=y.
\displaystyle \therefore g\text{ is one-one.}
\displaystyle \\

\displaystyle \text{Now, let }y\in A=[-1,1]\text{ be any element of the co-domain.}
\displaystyle \text{Choose }x=\frac{2}{\pi}\sin^{-1}y.
\displaystyle \text{Since }y\in[-1,1],
\displaystyle \sin^{-1}y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
\displaystyle \therefore -1\leq\frac{2}{\pi}\sin^{-1}y\leq1.
\displaystyle \therefore x\in A.
\displaystyle \text{Also,}
\displaystyle g(x)=\sin\left(\frac{\pi}{2}\cdot\frac{2}{\pi}\sin^{-1}y\right)
\displaystyle =\sin(\sin^{-1}y)
\displaystyle =y.
\displaystyle \therefore \text{every }y\in A\text{ has a pre-image in }A.
\displaystyle \therefore g\text{ is onto.}
\displaystyle \therefore g\text{ is a bijection and hence }g^{-1}\text{ exists.}
\displaystyle \\

\displaystyle \text{To find }g^{-1},\text{ let }g^{-1}(x)=y.
\displaystyle \Rightarrow x=g(y)=\sin\left(\frac{\pi y}{2}\right).
\displaystyle \Rightarrow \sin^{-1}x=\frac{\pi y}{2}.
\displaystyle \Rightarrow y=\frac{2}{\pi}\sin^{-1}x.
\displaystyle \therefore g^{-1}(x)=\frac{2}{\pi}\sin^{-1}x,\quad -1\leq x\leq1.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Let }f:\mathbb R\rightarrow\mathbb R\text{ be defined by }f(x)=\cos(x+2).
\displaystyle \text{Is }f\text{ invertible? Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{The function }f:\mathbb R\rightarrow\mathbb R\text{ is defined by }f(x)=\cos(x+2).
\displaystyle \text{To test whether }f\text{ is one-one, consider}
\displaystyle x=\frac{\pi}{2}-2\text{ and }y=\frac{3\pi}{2}-2.
\displaystyle \text{Clearly, }x\neq y.
\displaystyle \text{But}
\displaystyle f(x)=\cos\left(\frac{\pi}{2}-2+2\right)
\displaystyle =\cos\frac{\pi}{2}=0,
\displaystyle \text{and}
\displaystyle f(y)=\cos\left(\frac{3\pi}{2}-2+2\right)
\displaystyle =\cos\frac{3\pi}{2}=0.
\displaystyle \therefore f(x)=f(y)\text{ for }x\neq y.
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \\

\displaystyle \text{Also, the range of the cosine function is }[-1,1].
\displaystyle \therefore \text{Range}(f)=[-1,1]\neq\mathbb R.
\displaystyle \therefore f\text{ is not onto }\mathbb R.
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \therefore f\text{ is not invertible.}
\displaystyle \\

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

\displaystyle \text{(a) }f_1:A\rightarrow B,\quad f_1=\{(1,a),(2,b),(3,c),(4,d)\}.
\displaystyle f_1^{-1}=\{(a,1),(b,2),(c,3),(d,4)\}.
\displaystyle \\

\displaystyle \text{(b) }f_2:A\rightarrow B,\quad f_2=\{(1,b),(2,a),(3,c),(4,d)\}.
\displaystyle f_2^{-1}=\{(b,1),(a,2),(c,3),(d,4)\}.
\displaystyle \\

\displaystyle \text{(c) }f_3:A\rightarrow B,\quad f_3=\{(1,a),(2,b),(3,d),(4,c)\}.
\displaystyle f_3^{-1}=\{(a,1),(b,2),(d,3),(c,4)\}.
\displaystyle \\

\displaystyle \text{(d) }f_4:A\rightarrow B,\quad f_4=\{(1,b),(2,a),(3,d),(4,c)\}.
\displaystyle f_4^{-1}=\{(b,1),(a,2),(d,3),(c,4)\}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Let }A\text{ and }B\text{ be two finite sets. Assume that there is an injective map}
\displaystyle \text{from }A\text{ to }B\text{ and an injective map from }B\text{ to }A.
\displaystyle \text{Prove that there is a bijection from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:A\rightarrow B\text{ and }g:B\rightarrow A\text{ be injective maps.}
\displaystyle \text{Since }f\text{ is injective and }A,B\text{ are finite sets,}
\displaystyle |A|\leq |B|.
\displaystyle \text{Since }g\text{ is injective,}
\displaystyle |B|\leq |A|.
\displaystyle \therefore |A|=|B|.
\displaystyle \text{Now, }f:A\rightarrow B\text{ is injective and }A\text{ and }B\text{ have the same number of elements.}
\displaystyle \text{Therefore, every element of }B\text{ must be the image of some element of }A.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is both one-one and onto.}
\displaystyle \therefore f:A\rightarrow B\text{ is a bijection.}
\displaystyle \text{Hence, there exists a bijection from }A\text{ to }B.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }f:A\rightarrow A\text{ and }g:A\rightarrow A\text{ are two bijections, prove that}
\displaystyle \text{(i) }f\circ g\text{ is an injection.}
\displaystyle \text{(ii) }f\circ g\text{ is a surjection.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }f\text{ and }g\text{ are bijections, both are one-one and onto.}
\displaystyle \text{Also, }f\circ g:A\rightarrow A.
\displaystyle \\

\displaystyle \text{(i) To prove that }f\circ g\text{ is one-one, let }x,y\in A\text{ such that}
\displaystyle (f\circ g)(x)=(f\circ g)(y).
\displaystyle \Rightarrow f(g(x))=f(g(y)).
\displaystyle \text{Since }f\text{ is one-one,}
\displaystyle \Rightarrow g(x)=g(y).
\displaystyle \text{Since }g\text{ is one-one,}
\displaystyle \Rightarrow x=y.
\displaystyle \therefore f\circ g\text{ is an injection.}
\displaystyle \\

\displaystyle \text{(ii) To prove that }f\circ g\text{ is onto, let }z\in A\text{ be arbitrary.}
\displaystyle \text{Since }f:A\rightarrow A\text{ is onto, there exists }y\in A\text{ such that}
\displaystyle f(y)=z.
\displaystyle \text{Since }g:A\rightarrow A\text{ is onto, there exists }x\in A\text{ such that}
\displaystyle g(x)=y.
\displaystyle \therefore z=f(y)
\displaystyle =f(g(x))
\displaystyle =(f\circ g)(x).
\displaystyle \therefore \text{every }z\in A\text{ has a pre-image }x\in A.
\displaystyle \therefore f\circ g\text{ is a surjection.}
\displaystyle \\

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