NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 (Ex 1.2)

Q.1

$\begin{array}{l}\mathrm{Show }\mathrm{that }\mathrm{the}\mathrm{function}\mathrm{f}:\mathrm{R}{}_{*}\to {\mathrm{R}}_{*}\mathrm{definedby}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\\ \mathrm{one}–\mathrm{one}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{onto},\mathrm{\hspace{0.17em}}\mathrm{where}{\mathrm{R}}_{*}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{these}\mathrm{t}\mathrm{of}\mathrm{all}\mathrm{non}–\mathrm{zero}\\ \mathrm{realnumbers}.\mathrm{Is}\mathrm{there}\mathrm{sulttrue},\mathrm{if}\mathrm{the}\mathrm{domain}\mathrm{R}{}_{*}\mathrm{is}\hspace{0.17em}\hspace{0.17em}\\ \mathrm{replaced}\mathrm{by}\mathrm{Nwithco}–\mathrm{domain}\mathrm{being}\mathrm{same}\mathrm{as}{\mathrm{R}}_{*}?\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x},\mathrm{y}\in {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{such}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \mathrm{one}-\mathrm{one}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{f}:{\mathrm{R}}_{\mathrm{*}}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\end{array}$

Let f( x )=y where y be any element of R * ( Co−domain )

$\begin{array}{l}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{\mathrm{x}}=\mathrm{y}\hspace{0.17em}\hspace{0.17em}\mathrm{or}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\frac{1}{\mathrm{y}}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{f}\left(\frac{1}{\mathrm{y}}\right)=\frac{1}{\left(\frac{1}{\mathrm{y}}\right)}=\mathrm{y}\\ \mathrm{Since}\mathrm{Range}=\mathrm{Co}–\mathrm{domain}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Consider}\mathrm{function}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{For}\mathrm{any}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{we}\mathrm{see}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \mathrm{So}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{function}\\ \mathrm{Since}\mathrm{fractional}\mathrm{numbers}\mathrm{like}\frac{2}{3},\frac{2}{5}\mathrm{\hspace{0.17em}}\mathrm{etc}.\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{co}–\mathrm{domain}{\mathrm{R}}_{*}\mathrm{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{R}}_{\mathrm{*}}\mathrm{\hspace{0.17em}}\mathrm{have}\mathrm{no}\mathrm{pre}\mathrm{image}\mathrm{in}\mathrm{\hspace{0.17em}}\mathrm{domain}\mathrm{N}.\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{*}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\end{array}$

Q.2 (i) f: N→ N   given by  f(x)= x²

\begin{array}{l}(ii)f:Z\to Zgivenbyf(x)=x^2\\ (iii)f:R\to Rgivenbyf(x)=x^2\\ (iv)f:N\to Ngivenbyf(x)=x^3\\ (v)f:Z\to Zgivenbyf(x)=x^3\end{array}

Ans

\begin{array}{l}\text{(i)}\text{f:N}\to \text{N is given by,}\text{f}\left(x\right)=x^2\\ \text{To check for injectivity let f}\left(x\right)=\text{f}\left(y\right)\\ \Rightarrow x^2=y^2\\ \Rightarrow x=y\left(\begin{array}{l}There\text{are no negative}\\ \text{natural numbers}\text{.}\end{array}\right)\\ \therefore f\left(x\right)\text{is injective}\text{.}\\ \text{To check for onto}:\\ \text{Let}\text{f}\left(x\right)\text{=y}\\ \Rightarrow x^2\text{=y}\\ \Rightarrow \text{x=}\sqrt{y}\\ \therefore \text{f}\left(\sqrt{y}\right)\text{=y}\\ \text{Since range is not equal to co-domain}\text{.}\\ \text{f(x) is not onto or surjective}\\ \text{(ii) f: Z}\to \text{Z is given by,}\\ \text{f(x)}={\text{x}}^{\text{2}}\\ \text{For injectivity let}\text{f(x)}=\text{f(y)}\\ \Rightarrow x^2=y^2\\ \Rightarrow x=\pm y\\ \\ \therefore \text{f is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{(x)}=\text{y}\\ \Rightarrow {\text{x}}^{\text{2}}=\text{y}\\ \Rightarrow \text{x}=\sqrt{\text{y}}\\ \therefore f\left(\sqrt{y}\right)=y\\ \text{Hence, function f is neither injective nor surjective}\text{.}\\ \\ \left(iii\right)f:R\to R\text{is given by, f}\left(x\right)=x^2\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow x^2=y^2\\ \Rightarrow x=\pm y\left(\begin{array}{l}x\text{can not take more than one}\\ \text{value for injectivity}\text{.}\end{array}\right)\\ \therefore f\text{is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^2=y\\ \Rightarrow x=\sqrt{y}\\ \therefore f\left(\sqrt{y}\right)=y\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \left(iv\right)f:N\to N\text{is given by, f}\left(x\right)=x^3\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \\ \Rightarrow x^3=y^3\\ \Rightarrow x=y\\ \therefore f\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^3=y\\ \Rightarrow x=\sqrt[3]{y}\\ \therefore f\left(\sqrt[3]{y}\right)=y\left(y\in Nbut\sqrt[3]{y}\notin N\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\\ \\ \left(v\right)f:Z\to Z\text{is given by, f}\left(x\right)=x^3\\ For\text{injectivity:}\\ \text{f}\left(x\right)=f\left(y\right)\\ \Rightarrow x^3=y^3\\ \Rightarrow x=y\\ \therefore f\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\Rightarrow x^3=y\\ \Rightarrow x=\sqrt[3]{y}\\ \therefore f\left(\sqrt[3]{y}\right)=y\left(y\in Zbut\sqrt[3]{y}\notin Z\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\end{array}

Q.3 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Ans

\begin{array}{l}Since,\\ f:R\to R\text{is given by,}\\ \text{f}\left(x\right)=\left[x\right]\\ We\text{​}\text{see that f}\left(0.1\right)=0,\text{f}\left(0.3\right)=0.\\ \therefore \text{f}\left(0.1\right)=\text{f}\left(0.3\right),but\text{0}\text{.1}\ne \text{0}\text{.3}\\ \text{So, f is not injective i}\text{.e}\text{., one-one}\text{.}\\ \text{Now, let 0}\text{.6}\in \text{R}\text{.}\\ \text{It is given that f}\left(\text{x}\right)\text{=}\left[x\right]\text{is always an integer}\text{. Thus, there does}\\ \text{not exist any element x}\in \text{R such that f}\left(\text{x}\right)\text{=0}\text{.6}\text{.}\\ \text{Then, f is not onto}\text{.}\\ \text{Therefore, the greatest integer function is neither one-one}\\ \text{nor onto}\text{.}\end{array}

Q.4 Show that the Modul us Function f :R→R  given by f(x)=|x|,is neither one–onen or onto, where|x|is  x,if x  is positiveor 0 and |x| is –x, if x  is negative.

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}\ge \mathrm{0}\\ -\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}(-1)=|-1|=1,\mathrm{\hspace{0.17em}}\mathrm{f}\left(1\right)=\left|1\right|=1,\mathrm{}\\ \therefore \mathrm{f}(-1)=\mathrm{f}\left(1\right),\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{but}-\mathrm{1}\ne \mathrm{1}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Let}\mathrm{}-\mathrm{1}\in \mathrm{R}\mathrm{.}\\ \mathrm{Since},\mathrm{f}\left(\mathrm{x}\right)\mathrm{=}\left|\mathrm{x}\right|\mathrm{is}\mathrm{always}\mathrm{positive}.\mathrm{So},\mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\\ \mathrm{any}\mathrm{element}\mathrm{in}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=-1.\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{​}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{modulus}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{f}:\mathrm{R}\to \mathrm{R},\mathrm{given}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}>0\\ \hspace{0.17em}\hspace{0.17em}0,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}=0\\ –1,\mathrm{\hspace{0.17em}}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{noronto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1,\mathrm{}\mathrm{if}\mathrm{x}>0\\ \hspace{0.17em}\hspace{0.17em}0,\mathrm{}\mathrm{if}\mathrm{x}=0\\ -1,\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(1\right)=1,\mathrm{\hspace{0.17em}}\mathrm{f}\left(3\right)=1,\mathrm{but}1\ne \mathrm{3}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \\ \mathrm{Since},\mathrm{range}\mathrm{of}\mathrm{f}\mathrm{is}(1,0,-1)\mathrm{for}\mathrm{element}-2\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\\ \mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{value}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=-2.\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{signum}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.6 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans

\begin{array}{l}\text{It is given that}\\ \text{A}=\left\{\text{1},\text{2},\text{3}\right\},\text{B}=\left\{\text{4},\text{5},\text{6},\text{7}\right\}\\ \text{f}:\text{A}\to \text{B is defined as}\text{f = {(1, 4), (2, 5), (3, 6)}}\text{.}\\ \therefore \text{f (1) = 4, f (2) = 5, f (3) = 6}\\ \text{Since every element has a unique value}\text{.}\\ \text{The given function f is one-one}\text{.}\end{array}

Q.7

\begin{array}{l}Ineachofthefollowingcases,statewhetherthefunctionis\\ one–one,ontoorbijective.Justifyyouranswer.\\ (i)f:R\to Rdefinedbyf(x)=3–4x\\ (ii)f:R\to Rdefinedbyf(x)=1+x^2\end{array}

Ans

\begin{array}{l}\text{(i) f: R}\to \text{R is defined as f(x) = 3}-\text{4x}\text{.}\\ \text{For one-one}\\ \text{Let}\text{f(x)}=\text{f(y)}\\ \Rightarrow \text{3}-\text{4x}=\text{3}-\text{4y}\\ \\ \Rightarrow \text{x}=\text{y}\\ \therefore \text{f is one-one}\text{.}\\ \text{for onto}\\ \text{Let f(x)=y}\\ \Rightarrow \text{3}-\text{4x=y}\\ \Rightarrow \text{x=}\frac{\text{3}-y}{4}\\ \Rightarrow \text{f}\left(\frac{\text{3}-y}{4}\right)\text{=3}-\text{4}\left(\frac{\text{3}-y}{4}\right)\text{=y}\\ \text{Since Range = Co-domain}\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is bijective}\text{.}\\ \\ \text{(ii) f: R}\to {\text{R is defined as f(x)=1+x}}^{\text{2}}\\ \text{For one-one}\\ \text{Let f(x)=f(y )}\\ \Rightarrow {\text{1+x}}^{\text{2}}={\text{1+y}}^{\text{2}}\\ \Rightarrow {\text{x}}^{\text{2}}={\text{y}}^{\text{2}}\\ \Rightarrow \text{x}=\pm \text{y}\\ \therefore \text{f is not one-one}\text{.}\\ \text{For onto:}\\ \text{Let}\text{f}\left(x\right)=y\\ \Rightarrow {\text{1+x}}^{\text{2}}=\text{y}\\ \Rightarrow \text{x}=\pm \sqrt{y-1}\\ \text{Since Range is not equal to Co-domain}\text{.}\\ \text{Then given function f(x) is not onto}\text{.}\end{array}

Q.8

$\begin{array}{l}\mathrm{Let}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{be}\mathrm{sets}.\mathrm{Show}\mathrm{that}\mathrm{f}:\mathrm{A}\times \mathrm{B}\to \mathrm{B}\times \mathrm{A}\hspace{0.17em}\mathrm{such}\mathrm{that}\\ \mathrm{f}(\mathrm{a},\mathrm{b})=\mathrm{f}(\mathrm{b},\mathrm{a})\hspace{0.17em}\mathrm{is}\mathrm{bijective}\mathrm{function}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{A}\times \mathrm{B}\to \mathrm{B}\times \mathrm{A}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}(\mathrm{a},\mathrm{b})\; =\; (\mathrm{b},\mathrm{a}).\hspace{0.17em}\\ \mathrm{Let}\mathrm{}({\mathrm{a}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_1),({\mathrm{a}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_2)\in \mathrm{A}\times \mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}({\mathrm{a}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_1)=\mathrm{f}({\mathrm{a}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_2)\\ \Rightarrow ({\mathrm{b}}_1,\hspace{0.17em}\hspace{0.17em}{\mathrm{a}}_1)=({\mathrm{b}}_2,\hspace{0.17em}\hspace{0.17em}{\mathrm{a}}_2)\\ \Rightarrow {\mathrm{b}}_1={\mathrm{b}}_2\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}{\mathrm{a}}_1={\mathrm{a}}_2\\ \Rightarrow ({\mathrm{a}}_1,{\mathrm{b}}_1)=({\mathrm{a}}_2,{\mathrm{b}}_2)\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}(\mathrm{b},\mathrm{a})\in \mathrm{B}\times \mathrm{A}\mathrm{be}\mathrm{any}\mathrm{element}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}(\mathrm{a},\mathrm{b})\in \mathrm{A}\times \mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{a},\mathrm{b})\; =\; (\mathrm{b},\mathrm{a}).\\ [\mathrm{By}\mathrm{definition}\mathrm{of}\mathrm{f}]\\ \therefore \mathrm{\hspace{0.17em}}\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{bijective}\mathrm{.}\end{array}$

Q.9

\begin{array}{l}Letf:N\to Nbedefinedby\\ f\left(n\right)=\{\begin{array}{l}\frac{n+1}{2},ifnisodd\\ \frac{n}{2},ifniseven\end{array}foralln\in N.\\ Statewhetherthefunctionfisbijective.\\ Justifyyouranswer.\end{array}

Ans

\begin{array}{l}Letf:N\to Nisdefinedas\\ \\ f\left(n\right)=\{\begin{array}{l}\frac{n+1}{2},ifnisodd\\ \frac{n}{2},ifniseven\end{array}foralln\in N.\\ \text{It can be observed that:}\\ f\left(1\right)=\frac{1+1}{2}=1,f\left(2\right)=\frac{2}{2}=1\left[By\text{definition of f}\right]\\ \therefore f\left(1\right)=f\left(2\right),where1\ne 2.\\ \therefore \text{f is not one-one}\text{.}\\ \text{Consider a natural number (n) in co-domain N}\text{.}\\ \text{Case I: n is odd}\\ \therefore \text{n = 2m + 1 for some m}\in \text{N}\text{. Then,there exists 4m + 1}\in \text{N}\\ \text{such that}\\ \text{f}\left(4m+1\right)=\frac{4m+1+1}{2}=2m+1\\ CaseII:\text{n is even}\\ \therefore \text{n=2}m\text{for some}m\in \text{N}\text{. Then,there exists 4}m\in \text{N such that}\\ \text{f}\left(4m\right)=\frac{4m}{2}=2m\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is not a bijective function}\text{.}\end{array}

Q.10

\begin{array}{l}LetA=R-\left\{3\right\}andB=R-\left\{1\right\}.Considerthefunction\\ f:A\to Bdefinedbyf\left(x\right)=\frac{x–2}{x–3}.Isfone-oneandonto?\\ Justifyyouranswer.\end{array}

Ans

\begin{array}{l}A=R-\left\{3\right\}andB=R-\left\{1\right\}\\ \text{f: A}\to \text{B is defined as f}\left(x\right)=\frac{x-2}{x-3}\\ Let\text{x,y}\in \text{A such that}\\ \text{f}\left(x\right)=f\left(y\right).\\ \Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}\\ \Rightarrow \left(x-2\right)\left(y-3\right)=\left(x-3\right)\left(y-2\right)\\ \Rightarrow \bcancel{xy}-3x-2y+\bcancel{6}=\bcancel{xy}-2x-3y+\bcancel{6}\\ \Rightarrow -3x-2y=-2x-3y\\ \Rightarrow x=y\\ \therefore f\text{if one-one}\text{.}\\ \text{y}\in \text{B}=\text{R-}\left\{1\right\}.Then,y\ne 1\\ \text{The function f is onto if there exists x}\in \text{A such that}\\ \text{f(x) = y}\\ \Rightarrow \frac{x-2}{x-3}=y\\ \Rightarrow x-2=y\left(x-3\right)\\ \Rightarrow x-2=yx-3y\\ \Rightarrow x-yx=2-3y\\ \Rightarrow x\left(1-y\right)=2-3y\\ \Rightarrow x=\frac{2-3y}{\left(1-y\right)}\in A\left[y\ne 1\right]\\ \text{Thus, for any y}\in \text{B, there exists}\frac{2-3y}{\left(1-y\right)}\in Asuchthat\\ \\ f\left(\frac{2-3y}{1-y}\right)=\frac{\frac{2-3y}{\left(1-y\right)}-2}{\frac{2-3y}{\left(1-y\right)}-3}\\ =\frac{\frac{2-3y-2\left(1-y\right)}{1-y}}{\frac{2-3y-3\left(1-y\right)}{1-y}}\\ =\frac{2-3y-2+2y}{2-3y-3+3y}\\ =\frac{-y}{-1}\\ f\left(\frac{2-3y}{1-y}\right)=y\\ \therefore f\text{is onto}\text{.}\\ \text{Hence, function f is one-one and onto}\text{.}\end{array}

Q.11

$\begin{array}{l}\mathrm{Letf}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\mathrm{fismany}–\mathrm{one}\mathrm{onto}.\\ (\mathrm{C})\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{D}\right)\mathrm{f}\mathrm{is}\mathrm{n}\mathrm{either}\mathrm{one}–\mathrm{one}\\ \mathrm{noronto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right).\\ \mathrm{Then},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^4={\mathrm{y}}^4\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=\pm \mathrm{y}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{\hspace{0.17em}}\mathrm{but}\mathrm{\hspace{0.17em}}\mathrm{x}\ne \mathrm{y}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Consider}\mathrm{an}\mathrm{element}3\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{there}\\ \mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{x})\; =\; 3\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{asf}\left(\mathrm{x}\right)=3\mathrm{x}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{isone}–\mathrm{one}\mathrm{onto}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{many}–\mathrm{one}\mathrm{on}\mathrm{to}\\ \left(\mathrm{C}\right)\mathrm{f}\hspace{0.17em}\hspace{0.17em}\mathrm{isone}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\left(\mathrm{D}\right)\mathrm{f}\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{Rbedefinedasf}\left(\mathrm{x}\right)=3\mathrm{x}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}(\mathrm{x})\; =\mathrm{f}(\mathrm{y})\mathrm{.}\\ \Rightarrow 3\mathrm{x}=3\mathrm{y}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{number}\left(\mathrm{y}\right)\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\mathrm{there}\mathrm{exists}\\ \frac{\mathrm{y}}{3}\hspace{0.17em}\hspace{0.17em}\mathrm{in}\mathrm{R}\mathrm{such}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\frac{\mathrm{y}}{3}\right)=3\left(\frac{\mathrm{y}}{3}\right)=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \\ \mathrm{Therefore},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$