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Positive for all ${\text{x}} \in {\text{R}}$.Thus, there does not exist any $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = - 2$.Therefore, ${\text{f}}$ is not onto.Hence, the function ${\text{f}}$ is neither one-one nor onto.8. Let ${\text{A}}$ and ${\text{B}}$ be sets. Show that ${\text{f}}:{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ such that $(a,b) = (b,a)$ isbijective function.Ans: ${\text{f}}\,{\text{:}}\,{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ is defined as ${\text{f}}(a,b) = (b,a)$Let $\left( {{a_1},{b_1}} \right),\,\,\left( {{a_2},{b_2}} \right) \in {\text{A \times B}}$ such that ${\text{f}}\left( {{a_1},{b_1}} \right) = {\text{f}}\left( {{a_2},{b_2}} \right)$$ \Rightarrow \left( {{b_1},{a_1}} \right) = \left( {{b_2},{a_2}} \right)$$ \Rightarrow {b_1} = {b_2}$ and \[{a_1} = {a_2}\]$ \Rightarrow \left( {{a_1},\;{b_1}} \right) = \left( {{a_2},\;{b_2}} \right)$Thus, ${\text{f}}$ is one-one.Now, let $(b,a) \in {\text{B A}}$ be any element.Then, there exists $(a,b) \in {\text{A B}}$ such that $f(a, b)=(b, a) .[$By definition of \[{\text{f}}]\]Thus, ${\text{f}}$ is onto.Therefore, the function ${\text{f}}$ is bijective.9. Let f: ${\text{N}} \to {\text{N}}$ be defined by ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{{n + 1}}{2},}&{{\text{ if }}n{\text{ is odd }}} \\ {\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}} \end{array}} \right.$ for all $n \in {\text{N}}$State whether the function ${\text{f}}$ is bijective. Justify your answer.Ans: ${\text{f}}:{\text{N}} \to {\text{N}}$ is defined as ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{{n + 1}}{2},}&{{\text{ if }}n\,\,{\text{is}}\,{\text{odd }}} \\ {\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}} \end{array}} \right.$ for all $n \in {\text{N}}$It can be observed that$f(1) = \dfrac{{1 + 1}}{2} = 1$ and $f(2) = \dfrac{2}{2} = 1$ (By definition of ${\text{f}}(n)$)$f(1) = f(2)$, where $1 \ne 2$Thus, ${\text{f}}$ is not one-one.Consider a natural number \[\left( n \right)\] in co-domain \[{\text{N}}.\]Case 1: $n$ is oddThus, $n = 2{\text{r}} + 1$ for some $r \in {\text{N}}$. Then, there exists $4r + 1 \in {\text{N}}$ such that$f(4r + 1) = \dfrac{{4r + 1 + 1}}{2} = 2r + 1$Case 2: $n$ is evenThus, $n = 2r$ for some $r \in {\text{N}}$. Then, there exists $4r \in {\text{N}}$ such that${\text{f}}(4r) = \dfrac{{4r}}{2} = 2r$Therefore, ${\text{f}}$ is onto.Hence, the function \[{\text{f}}\]is not a bijective function.10. Let ${\text{A}} = {\text{R}} - \{ 3\} $ and ${\text{B}} = {\text{R}} - \{ 1\} .$ Consider the function f: ${\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x)$ $ = \left( {\dfrac{{x - 2}}{{x - 3}}} \right).$ Is f one-one and onto? Justify your answer.Ans: ${\text{A}} = {\text{R}} - \{ 3\} ,{\text{B}} = {\text{R}} - \{ 1\} $ and ${\text{f}}:{\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x) = \left( {\dfrac{{x - 2}}{{x - 3}}} \right)$Let $x,y \in $

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Access PDF for Maths NCERT Chapter 1 Relation and Functions Exercise 1.2 Class 12Exercise 1.21. Show that the function ${\text{f}}:{{\mathbf{R}}_*} \to {{\mathbf{R}}_*}$ defined by ${\text{f(}}x) = \dfrac{1}{x}$ is one-one and onto, where ${{\mathbf{R}}_*}$ is the set of all non-zero real numbers. Is the result true, if the domain ${{\mathbf{R}}_*}$ is replaced by ${\text{N}}$ with co-domain being same as ${{\mathbf{R}}_*}$?Ans: Given that, f: ${{\text{R}}^*} \to {R_*}$ is defined by ${\text{f}}(x) = \dfrac{1}{x}$.Consider $x,\,\,y \in R*$ such that ${\text{f}}(x) = {\text{f}}(y)$$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{y}$$ \Rightarrow x = y$Thus, ${\text{f}}$ is one-one.It is clear that for $y \in R*$, there exists $x = \dfrac{1}{y} \in R*[$ as $y \ne 0]$ such that${\text{f}}(x) = \dfrac{1}{{\left( {\dfrac{1}{y}} \right)}} = {\text{y}}$Thus, ${\text{f}}$ is onto.Therefore, the given function ${\text{f}}$ is one-one and onto.Now, consider function ${\text{g}}:{\text{N}} \to {{\text{R}}_*}$ defined by ${\text{g}}(x) = \dfrac{1}{x}$We have, $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)\quad \Rightarrow \, = \dfrac{1}{{{x_1}}} = \dfrac{1}{{{x_2}}}$$ \Rightarrow {x_1} = {x_2}$Thus, \[{\text{g}}\] is one-one.It is clear that ${\text{g}}$ is not onto as for $1.2 \in = {{\text{R}}_*}$, there does not exist any $x$ in ${\text{N}}$ such that ${\text{g}}(x)$ $ = \dfrac{1}{{1.2}}$Therefore, the function ${\text{g}}$ is one-one but not onto.2. Check the injectivity and surjectivity of the following functions:(i) ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^2}$Ans: Here, ${\text{f}}\,{\text{:}}\,{\text{N}} \to {\text{N}}$ is given by ${\text{f}}(x) = {x^2}$For $x,y \in N,\,\,$${\text{f}}(x) = {\text{f}}(y) \Rightarrow {x^2} = {y^2} \Rightarrow x = y$Thus, ${\text{f}}$ is injective.Now, $2 \in {\text{N}}$. But, there does not exist any $x$ in ${\text{N}}$ such that ${\text{f}}(x) = {x^2} = 2$Thus, ${\text{f}}$ is not surjective.Therefore, the function ${\text{f}}$ is injective but not surjective.(ii) ${\text{f}}:{\text{Z}} \to {\text{Z}}$ given by ${\text{f}}(x) = {x^2}$Ans: Here, ${\text{f}}:{\text{Z}} \to {\text{Z}}$ is given by ${\text{f}}(x) = {x^2}$It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.Thus, ${\text{f}}$ is not injective.Now, $ - 2 \in {\text{Z}}$. But, there does not exist any element ${\text{x}} \in {\text{Z}}$ such that$f(x) = - 2$ or ${x^2} = - 2$Thus, ${\text{f}}$ is not surjective.Therefore, the function ${\text{f}}$ is neither injective nor surjective.(iii) f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$Ans: Here, f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.Thus, ${\text{f}}$ is not injective.Now, $ - 2 \in {\text{R}}$. But, there does not exist any element $x \in {\text{R}}$ such that

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Read the NoSweatShakespeare Modern Twelfth Night ebook for free! Chose the Act & Scene from the list below to read Twelfth Night translated into modern English.What’s so special about NoSweatShakespeare’s modern English translation of Twelfth Night?Translated as an easy to read, exciting teenage novelFollows the acts and scenes of the original Twelfth Night textAllows you to master the plot, characters, ideas and language of Twelfth NightRead scenes from Twelfth Night in modern English, or in Shakespeare’s original text: Twelfth Night in modern English| Twelfth Night original text|Modern Twelfth Night Act 1, Scene 1|Twelfth Night original text, Act 1, Scene 1Modern Twelfth Night Act 1, Scene 2|Twelfth Night original text, Act 1, Scene 2Modern Twelfth Night Act 1, Scene 3|Twelfth Night original text, Act 1, Scene 3Modern Twelfth Night Act 1, Scene 4|Twelfth Night original text, Act 1, Scene 4Modern Twelfth Night Act 1, Scene 5|Twelfth Night original text, Act 1, Scene 5|Modern Twelfth Night Act 2, Scene 1|Twelfth Night original text, Act 2, Scene 1Modern Twelfth Night Act 2, Scene 2|Twelfth Night original text, Act 2, Scene 2Modern Twelfth Night Act 2, Scene 3|Twelfth Night original text, Act 2, Scene 3Modern Twelfth Night Act 2, Scene 4|Twelfth Night original text, Act 2, Scene 4Modern Twelfth Night Act 2, Scene 5|Twelfth Night original text, Act 2, Scene 5|Modern Twelfth Night Act 3, Scene 1|Twelfth Night original text, Act 3, Scene 1Modern Twelfth Night Act 3, Scene 2|Twelfth Night original text, Act 3, Scene 2Modern Twelfth Night Act 3, Scene 3|Twelfth Night original text, Act 3, Scene 3Modern Twelfth Night Act 3, Scene 4|Twelfth Night original text, Act 3, Scene 4|Modern Twelfth Night Act 4, Scene 1|Twelfth Night original text, Act 4, Scene 1Modern Twelfth Night Act 4, Scene 2|Twelfth Night original text, Act 4, Scene 2Modern Twelfth Night Act 4, Scene 3|Twelfth Night original text, Act 4, Scene 3|Modern Twelfth Night Act 5, Scene 1|Twelfth Night original text, Act 5, Scene 1. Keywords: text twist 2, text twist 2 online, Text Twist 2 Unblocked, Text Twist 2 Game, Text Twist 2 official

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A such that $f(x) = f(y)$$ \Rightarrow \dfrac{{x - 2}}{{x - 3}} = \dfrac{{y - 2}}{{y - 3}}$By cross multiplication,$ \Rightarrow (x - 2)(y - 3) = (y - 2)(x - 3)$Expand brackets,$ \Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6$$ \Rightarrow - 3x - 2y = - 2x - 3y \Rightarrow x = y$Thus, ${\text{f}}$ is one-one.Let $y \in {\text{B}} = {\text{R}} - \{ 1\} .$ Then, $y \ne 1$.The function ${\text{f}}$ is onto if there exists $x \in {\text{A}}$ such that ${\text{f}}(x) = y$.Now, $f(x) = y$$ \Rightarrow \dfrac{{x - 2}}{{y - 3}} = y$By cross multiplication,$ \Rightarrow x - 2 = xy - 3y \Rightarrow x(1 - y) = - 3y + 2$$ \Rightarrow x = \dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$$[y \ne 1]$Thus, for any $y \in {\text{B}}$, there exists $\dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$ such that${\text{f}}\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) = \dfrac{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 2}}{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 3}}$Take LCM,$ = \dfrac{{2 - 3y - 2 + 2y}}{{2 - 3y - 3 + 3y}}$$ = \dfrac{{ - y}}{{ - 1}} = y$Thus, ${\text{f}}$ is onto.Therefore, the function ${\text{f}}$ is one-one and onto.11. Let ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ be defined as ${\text{f}}(x) = {x^4}$. Choose the correct answer.(A) ${\text{f}}$ is one-one onto(B) ${\text{f}}$ is many-one onto(C) ${\text{f}}$ is one-one but not onto(D) ${\text{f}}$ is neither one-one nor ontoAns: Here, ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ is defined as ${\text{f}}(x) = {x^4}$.Let $x,y \in R$ such that ${\text{f}}(x) = {\text{f}}(y)$.$ \Rightarrow {x^4} = {y^4}$$ \Rightarrow x = \pm y$${\text{f}}(x) = {\text{f}}(y)$ does not imply that $x = y$For example, ${\text{f}}(1) = {\text{f}}( - 1) = 1$Thus, ${\text{f}}$ is not one-one.Consider an element \[2\] in co-domain ${\mathbf{R}}$. It is clear that there does not exist any $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = 2$Thus, ${\text{f}}$ is not onto.Hence, the function ${\text{f}}$ is neither one-one nor onto.The correct answer is \[{\text{D}}.\]12. Let ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ be defined as ${\text{f}}(x) = 3x$. Choose the correct answer.(A) f is one – one and onto (B) f is many – one and onto (C) f is one – one but not onto (D) f is neither one – one nor ontoAns: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3x$.Let $x,\,\,y \in {\text{R}}$ such that ${\text{f}}(x) = {\text{f}}(y)$.$ \Rightarrow 3x = 3y$$

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That the Signum Function f: ${\text{R}} \to {\text{R}}$, given by ${\text{f}}(x) = \left\{ {\begin{array}{*{20}{l}} 1&{{\text{ if }}x > 0} \\ {0,}&{{\text{ if }}x = 0} \\ { - 1,}&{{\text{ if }}x is neither one-one nor onto.Ans: Here, f: ${\text{R}} \to {\text{R}}$, given by ${\text{f}}(x) = \left\{ {\begin{array}{*{20}{l}} 1&{{\text{ if }}x > 0} \\ {0,}&{{\text{ if }}x = 0} \\ { - 1,}&{{\text{ if }}x It is seen that ${\text{f}}(1) = {\text{f}}(2) = 1$, but $1 \ne 2$.Thus, ${\text{f}}$ is not one-one.Now, as ${\text{f}}(x)$ takes only \[3\] values $(1,\,\,0$, or $ - 1)$ for the element $ - 2$ in co-domain${\text{R}}$, there does not exist any $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = - 2$.Thus, ${\text{f}}$ is not ontoTherefore, the Signum function is neither one-one nor onto.6. Let ${\text{A}} = \{ 1,2,3\} ,\,\,{\text{B}} = \{ 4,5,6,7\} $ and let ${\text{f}} = \{ (1,4),\,\,(2,5),\,\,(3,6)\} $ be a function from ${\text{A}}$ to ${\text{B}}$. Show that ${\text{f}}$ is one-one.Ans: Given that, ${\text{A}} = \{ 1,2,3\} $${\text{B}} = \{ 4,5,6,7\} $${\text{f}}\,{\text{:}}\,{\text{A}} \to {\text{B}}$ is defined as ${\text{f}} = \{ (1,4),\,\,(2,5),\,\,(3,6)\} $Thus, ${\text{f}}(1) = 4,\,\,{\text{f}}(2) = 5,\,\,{\text{f}}(3) = 6$It is seen that the images of distinct elements of ${\text{A}}$ under ${\text{f}}$ are distinct.Therefore, the function ${\text{f}}$ is one-one.7. In each of the following cases, state whether the function is one-one,onto or bijective. Justify your answer.(i) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 3 - 4x$Ans: Here, f: ${\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3 - 4x$.Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$$ \Rightarrow 3 - 4{x_1} = 3 - 4{x_2}$$ \Rightarrow - 4{x_1} = - 4{x_2}$$ \Rightarrow {x_1} = {x_2}$Thus, ${\text{f}}$ is one-one.For any real number \[\left( y \right)\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{{3 - y}}{4}} \right) = 3 - 4\left( {\dfrac{{3 - y}}{4}} \right) = y$Thus, ${\text{f}}$ is onto.Therefore, the function ${\text{f}}$ is bijective.(ii) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 1 + {x^2}$Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 1 + {x^2}$Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}({x_1}) = {\text{f}}\left( {{x_2}} \right)$$ \Rightarrow 1 + {({x_1})^2} = 1 + {({x_2})^2}$$ \Rightarrow {({x_1})^2} = {({x_2})^2}$$ \Rightarrow {x_1} = {x_2}$Thus, ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$ does not imply that ${x_1} = {x_2}$.For example, ${\text{f}}(1) = {\text{f}}( - 1) = 2$Therefore, ${\text{f}}$ is not one-one.Consider an element $ - 2$ in co-domain ${\text{R}}$.It is seen that ${\text{f}}(x) = 1 + {x^2}$ is

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M as$$ M = {\text{Def}}_{0} + 2 \cdot d \cdot {\text{Ha}} \to {\text{Ha}} = M/\left( {2 \cdot d} \right) $$ (19) Considering stage 1 beam enlargement and substituting Ha, we have$$ {\text{Def}}_{1h} = {\text{Def}}_{1v} = {\text{Def}}_{1} = {\text{Def}}_{0} + 2 \cdot ({\text{OPL}}^{\prime } - d) \cdot M/\left( {2 \cdot d} \right) $$ (20) $$ {\text{Def}}_{1} = {\text{Def}}_{0} + \left( {1/\cos \left( \Theta \right) - 1} \right) \cdot M $$ (21) Stage 2 beam enlargement affects only the vertical axis, thus$$ {\text{Def}}_{2v} = {\text{Def}}_{1} /\cos \left( \Theta \right) $$ (22) where Def2h remains unchanged. We can now calculate a new Def for the two stages combined by calculating the area of the ellipse, then normalizing it to an equal circular area, resulting in$$ {\text{Def}} = \frac{1}{{\sqrt {\cos \left( \Theta \right)} }} \cdot {\text{Def}}_{1} = \frac{1}{{\sqrt {\cos \left( \Theta \right)} }} \cdot \left( {{\text{Def}}_{0} + \left( {\frac{1}{\cos \left( \Theta \right)} - 1} \right) \cdot M } \right) $$ (23) The beam speed is extracted from Fig. 2 as$$ v = {\text{d}}h/{\text{d}}\Theta = \frac{{d \left( {d \cdot \tan \left( \Theta \right) } \right)}}{d\Theta } = \frac{d}{{{\text{cos}}^{2} \left( \Theta \right)}} $$ (24) Using the Ev in Eqs. 2, 22 and 23 and applying Lambert’s cosine law, this yields (Fig. 3)$$ {\text{Ev}} = \frac{{P_{{\text{l}}} }}{{v \cdot {\text{Def}} \cdot L_{h} }} = \frac{{P_{{\text{l}}} \cdot \cos (\Theta )}}{{\frac{2 \cdot \pi \cdot R \cdot f \cdot d}{{\cos^{2} (\Theta )}} \cdot {\text{Def}} \cdot L_{h} }} = P_{{\text{l}}} \cdot \frac{{\cos^{3} (\Theta )}}{{L_{h} \cdot v_{0} \cdot \left( {\frac{{{\text{Def}}_{0} + \left( {\frac{1}{\cos (\Theta )} - 1} \right) \cdot M}}{{\sqrt {\cos (\Theta )} }}} \right)}} $$ (25) $$ {\text{Ev}} = \frac{{P_{{\text{l}}} }}{{L_{h} \cdot V_{0} }} \cdot \frac{{\cos^{3} (\Theta ) \cdot \sqrt {\cos (\Theta )} }}{{\left( {{\text{Def}}_{0} + \left( {\frac{1}{\cos (\Theta )} - 1} \right) \cdot M} \right)}} $$ (26) Fig. 3Def and Ev. Keywords: text twist 2, text twist 2 online, Text Twist 2 Unblocked, Text Twist 2 Game, Text Twist 2 official Download Text Twist 2 [NL] Download do Text Twist 2 [PT] Скачать Text Twist 2 [RU] Descargar Text Twist 2 [ES] Scarica Text Twist 2 [IT] Ladda ner Text Twist 2 [SV] T l charger Text Twist 2 [FR] Text Twist 2 다운로드 [KO] Unduh Text Twist 2 [ID]