\displaystyle \textbf{Question 1: } \text{Show that the function }f:R\to R\text{ given by } \\ f(x)=ax+b,\text{ where }a,b\in R,\,a\neq 0\text{ is a bijection.} \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Injectivity : Let }x,y\text{ be any two real numbers. Then,}
\displaystyle f(x)=f(y)\Rightarrow ax+b=ay+b\Rightarrow ax=ay\Rightarrow x=y
\displaystyle \text{Thus, }f(x)=f(y)\Rightarrow x=y\text{ for all }x,y\in R\text{(domain).}
\displaystyle \text{So, }f\text{ is an injection.}
\displaystyle \text{Surjectivity : Let }y\text{ be an arbitrary element of }R\text{(co-domain). Then,}
\displaystyle f(x)=y\Rightarrow ax+b=y\Rightarrow x=\frac{y-b}{a}
\displaystyle \text{Clearly, }x=\frac{y-b}{a}\in R\text{ (domain) for all }y\in R\text{ (co-domain). Thus, for all }y\in R\text{ (co-domain) there} \text{exists }x=\frac{y-b}{a}\in R\text{ (domain) such that}
\displaystyle f(x)=f\left(\frac{y-b}{a}\right)=a\left(\frac{y-b}{a}\right)+b=y.
\displaystyle \text{This shows that every element in co-domain has its pre-image in domain. So, }f\text{ is a surjection.}
\displaystyle \text{Hence, }f\text{ is a bijection.}

\displaystyle \textbf{Question 2: } \text{Let }f:X\to Y\text{ be a function. Define a relation }R\text{ on }X\text{ given by } \\ R=\{(a,b):f(a)=f(b)\}. \text{ Show that }R\text{ is an equivalence relation on }X. \hspace{0.5cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe the following properties of relation }R.
\displaystyle \text{Reflexivity: For any }a\in X,\text{ we have}
\displaystyle f(a)=f(a)\Rightarrow (a,a)\in R\Rightarrow R\text{ is reflexive.}
\displaystyle \text{Symmetry: Let }a,b\in X\text{ be such that }(a,b)\in R.\text{ Then,}
\displaystyle (a,b)\in R\Rightarrow f(a)=f(b)\Rightarrow f(b)=f(a)\Rightarrow (b,a)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \text{Transitivity: Let }a,b,c\in X\text{ be such that }(a,b)\in R\text{ and }(b,c)\in R.\text{ Then,}
\displaystyle (a,b)\in R\text{ and }(b,c)\in R
\displaystyle \Rightarrow f(a)=f(b)\text{ and }f(b)=f(c)
\displaystyle \Rightarrow f(a)=f(c)
\displaystyle \Rightarrow (a,c)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}

\displaystyle \textbf{Question 3: } \text{Show that }f:N\to N\text{ defined by}
\displaystyle f(n)=\begin{cases}\frac{n+1}{2},&\text{if }n\text{ is odd}\\\frac{n}{2},&\text{if }n\text{ is even}\end{cases} 
\displaystyle \text{is many-one onto function.} \hspace{2.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle f(1)=\frac{1+1}{2}=1\text{ and }f(2)=\frac{2}{2}=1.
\displaystyle \text{Thus, }1,2\in N\text{ such that }1\neq 2\text{ but }f(1)=f(2).\text{ So, }f\text{ is a many-one function.}
\displaystyle \text{Surjectivity: Let }n\text{ be an arbitrary element of }N.
\displaystyle \text{If }n\text{ is an odd natural number, then }2n-1\text{ is also an odd natural number such that}
\displaystyle f(2n-1)=\frac{2n-1+1}{2}=n
\displaystyle \text{If }n\text{ is an even natural number, then }2n\text{ is also an even natural number such that}
\displaystyle f(2n)=\frac{2n}{2}=n.
\displaystyle \text{Thus, for every }n\in N\text{ (whether even or odd) there exists its pre-image in }N.\text{ So, }f\text{ is a surjection.}
\displaystyle \text{Hence, }f\text{ is a many-one onto function.}

\displaystyle \textbf{Question 4: } \text{If the function }f:R\to R\text{ be given by }f(x)=x^{2}+2\text{ and } \\ g:R-\{1\}\to R\text{ be given by } g(x)=\frac{x}{x-1}.\text{ Find }fog\text{ and }gof. \hspace{2.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Clearly, range }f=\text{domain }g\text{ and, range }g=\text{domain }f.\text{ So, }fog\text{ and }gof\text{ both exist.}
\displaystyle \text{Now, }(fog)(x)=f(g(x))=f\left(\frac{x}{x-1}\right)=\left(\frac{x}{x-1}\right)^{2}+2=\frac{x^{2}}{(x-1)^{2}}+2
\displaystyle \text{and, }(gof)(x)=g(f(x))=g(x^{2}+2)=\frac{x^{2}+2}{(x^{2}+2)-1}=\frac{x^{2}+2}{x^{2}+1}
\displaystyle \text{Hence, }gof:R\to R\text{ and }fog:R-\{1\}\to R\text{ are given by}
\displaystyle (gof)(x)=\frac{x^{2}+2}{x^{2}+1}\text{ and }(fog)(x)=\frac{x^{2}}{(x-1)^{2}}+2

\displaystyle \textbf{Question 5: } \text{If }f(x)=e^{x}\text{ and }g(x)=\log_{e}x\,(x>0),\text{ find }fog\text{ and }gof. \\ \text{Is }fog=gof? \hspace{8.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle \text{Domain }(f)=R,\text{ Range }(f)=(0,\infty),\text{ Domain }(g)=(0,\infty)\text{ and, Range }(g)=R.
\displaystyle \text{Computation of }fog:\text{ We observe that}
\displaystyle \text{Range }(g)=\text{Domain }(f)
\displaystyle \therefore fog\text{ exists and }fog:\text{ Domain }(g)\to R\text{ i.e. }fog:(0,\infty)\to R\text{ such that}
\displaystyle fog(x)=f(g(x))=f(\log_{e}x)=e^{\log_{e}x}=x
\displaystyle \text{Thus, }fog:(0,\infty)\to R\text{ is defined as }fog(x)=x.
\displaystyle \text{Computation of }gof:\text{ We have,}
\displaystyle \text{Range }(f)=(0,\infty)=\text{Domain }(g)
\displaystyle \therefore gof\text{ exists and }gof:\text{ Domain }(f)\to R\text{ i.e. }gof:R\to R\text{ such that}
\displaystyle gof(x)=g(f(x))=g(e^{x})=\log_{e}e^{x}=x\log_{e}e=x
\displaystyle \text{Thus, }gof:R\to R\text{ is defined as }gof(x)=x.
\displaystyle \text{We observe that Domain }(gof)\neq \text{Domain }(fog).
\displaystyle \therefore gof\neq fog.

\displaystyle \textbf{Question 6: } \text{If }f(x)=\sqrt{x}\,(x>0)\text{ and }g(x)=x^{2}-1\text{ are two real functions, find } \\ fog\text{ and }gof.\text{ Is }fog=gof? \hspace{7.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{We observe that}
\displaystyle \text{Domain }(f)=[0,\infty),\text{ Range }(f)=[0,\infty),\text{ Domain }(g)=R
\displaystyle \text{and, Range }(g)=[-1,\infty)\quad [\because x^{2}\ge 0\text{ for all }x\in R\ \therefore x^{2}-1\ge -1\text{ for all }x\in R]
\displaystyle \text{Computation of }gof:\text{ We observe that Range }(f)=[0,\infty)\subseteq \text{Domain }(g).
\displaystyle \therefore gof\text{ exists and }gof:[0,\infty)\to R\text{ such that}
\displaystyle gof(x)=g(f(x))=g(\sqrt{x})=(\sqrt{x})^{2}-1=x-1
\displaystyle \text{Thus, }gof:[0,\infty)\to R\text{ is defined as }gof(x)=x-1.
\displaystyle \text{Computation of }fog:\text{ We observe that Range }(g)=[-1,\infty)\nsubseteq \text{Domain }(f)
\displaystyle \therefore \text{Domain }(fog)=\{x:x\in \text{Domain }(g)\text{ and }g(x)\in \text{Domain }(f)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }g(x)\in [0,\infty)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }x^{2}-1\in [0,\infty)\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\in R\text{ and }x^{2}-1\ge 0\}
\displaystyle \Rightarrow \text{Domain }(fog)=\{x:x\le -1\text{ or }x\ge 1\}
\displaystyle \Rightarrow \text{Domain }(fog)=(-\infty,-1]\cup [1,\infty)
\displaystyle \text{Also, }fog(x)=f(g(x))=f(x^{2}-1)=\sqrt{x^{2}-1}
\displaystyle \text{Thus, }fog:(-\infty,-1]\cup [1,\infty)\to R\text{ is defined as }fog(x)=\sqrt{x^{2}-1}.
\displaystyle \text{We find that }fog\text{ and }gof\text{ have distinct domains. Also, their formulas are not same.}
\displaystyle \text{Hence, }fog\neq gof.

\displaystyle \textbf{Question 7: } \text{Let }f:N\cup\{0\}\to N\cup\{0\}\text{ be defined by}
\displaystyle f(n)=\begin{cases}n+1,&\text{if }n\text{ is even}\\n-1,&\text{if }n\text{ is odd}\end{cases}
\displaystyle \text{Show that }f\text{ is invertible and }f=f^{-1}. \hspace{7.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle   f\text{ is a bijection. So, it is invertible.}
\displaystyle \text{In order to find }f^{-1},\text{ let }n,m\in N\cup\{0\}\text{ such that}
\displaystyle f(n)=m
\displaystyle \Rightarrow n+1=m,\text{ if }n\text{ is even}
\displaystyle \Rightarrow n-1=m,\text{ if }n\text{ is odd}
\displaystyle \Rightarrow n=\begin{cases}m-1,&\text{if }m\text{ is odd}\\m+1,&\text{if }m\text{ is even}\end{cases}
\displaystyle \text{[If }n\text{ is even, then }n+1=m\text{ is odd}\quad \text{If }n\text{ is odd, then }n-1=m\text{ is even]}
\displaystyle \Rightarrow f^{-1}(m)=\begin{cases}m-1,&\text{if }m\text{ is odd}\\m+1,&\text{if }m\text{ is even}\end{cases}
\displaystyle \text{Hence, }f^{-1}(n)=\begin{cases}n+1,&\text{if }n\text{ is even}\\n-1,&\text{if }n\text{ is odd}\end{cases}
\displaystyle \text{Clearly, }f=f^{-1}.

\displaystyle \textbf{Question 8: } \text{Let }f:R\to R\text{ be defined as }f(x)=10x+7.\text{ Find the function } \\ g:R\to R\text{ such that} gof=fog=I_{R}. \hspace{7.0cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle fog=I_{R}
\displaystyle \Rightarrow fog(x)=I_{R}(x)\text{ for all }x\in R
\displaystyle \Rightarrow f(g(x))=x\text{ for all }x\in R
\displaystyle \Rightarrow 10g(x)+7=x\text{ for all }x\in R
\displaystyle \Rightarrow g(x)=\frac{x-7}{10}\text{ for all }x\in R

\displaystyle \textbf{Question 9: } \text{Let }f:N\to R\text{ be a function defined as }f(x)=4x^{2}+12x+15.\text{ Show that }f:N\to \text{Range}(f)\text{ is invertible. Find the inverse of }f. \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{In order to prove that }f\text{ is invertible, it is sufficient to show that }f:N\to \text{Range}(f)\text{ is a bijection.}
\displaystyle f\text{ is one-one: For any }x,y\in N,\text{ we find that}
\displaystyle f(x)=f(y)
\displaystyle \Rightarrow 4x^{2}+12x+15=4y^{2}+12y+15
\displaystyle \Rightarrow 4(x^{2}-y^{2})+12(x-y)=0
\displaystyle \Rightarrow (x-y)(4x+4y+3)=0
\displaystyle \Rightarrow x-y=0\quad [\because 4x+4y+3\neq 0\text{ for any }x,y\in N]
\displaystyle \Rightarrow x=y
\displaystyle \text{So, }f:N\to \text{Range}(f)\text{ is one-one.}
\displaystyle \text{Obviously, }f:N\to \text{Range}(f)\text{ is onto. Hence, }f:N\to \text{Range}(f)\text{ is invertible.}
\displaystyle \text{Let }f^{-1}\text{ denote the inverse of }f.\text{ Then,}
\displaystyle f(f^{-1}(x))=x\text{ for all }x\in \text{Range}(f)
\displaystyle \Rightarrow 4(f^{-1}(x))^{2}+12f^{-1}(x)+15=x\text{ for all }x\in \text{Range}(f)
\displaystyle \Rightarrow 4(f^{-1}(x))^{2}+12f^{-1}(x)+15-x=0
\displaystyle \Rightarrow f^{-1}(x)=\frac{-12\pm \sqrt{144-16(15-x)}}{8}
\displaystyle \Rightarrow f^{-1}(x)=\frac{-12\pm \sqrt{16x-96}}{8}=\frac{-3\pm \sqrt{x-6}}{2}
\displaystyle \Rightarrow f^{-1}(x)=\frac{-3+\sqrt{x-6}}{2}\quad [\because f^{-1}(x)\in N\therefore f^{-1}(x)>0]

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