Graph of $y=|x-h|+k$ and $y=-|x-h|+k$ : Exercise (6.1) - Solutions

Graph of the Function $y = |x − h| + k$

The graph of the absolute value function$y = |x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = |x|$.

Graph of the Function $y = -|x − h| + k$

The graph of the absolute value function$y = -|x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = -|x|$.

1.           Compare the graphs of the following functions to the graph of $y=|x|$.

             (a)    $y=|x-3|-2$

             (b)    $y=|x+1|+3$

             (c)    $y=|x-2|+3$

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(a) The graph of $y=|x-3|-2$ is the translation of positive 3 units horizontally and negative 2 units vertically of the graph $y=|x| .$

(b) The graph of $y=|x+1|+3$ is the translation of negative 1 unit horizontally and positive 3 units vertically of the graph $y=|x| .$

(c) The graph of $y=|x-2|+3$ is the translation of positive 2 units horizontally and positive 3 units vertically of the graph $y=|x| .$

2.           Compare the graphs of the following functions to the graph of $y=-|x|$.

             (a)    $y=-|x+3|+2$

             (b)    $y=-|x-4|+1$

             (c)    $y=-|x+4|-1$

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(a) The graph of $y=-|x+3|+2$ is the translation of negative 3 units horizontally and positive 2 units vertically of the graph $y=-|x| .$

(b) The graph of $y=-|x-4|+1$ is the translation of positive 4 units horizontally and positive 1 unit vertically of the graph $y=-|x| .$

(c) The graph of $y=y=-|x+4|-1$ is the translation of negative 4 units horizontally and negative 1 unit vertically of the graph $y=-|x| .$

Domain and Range of a Function - Exercise (4.3) :Solutions

How to Determine the Domain of a Function

When the domain of a function is not specified, then assume that it is the set of all possible real numbers for which the function makes sense.

ပေးထားသော function တစ်ခုအတွက် domain သတ်မှတ်ပေးထားခြင်း မရှိလျှင် function ၏ သတ်မှတ်ချက်ကို ပြေလည်စေသော အစုဝင်အားလုံးပါသည် ကိန်းစစ်အစုသည် အဆိုပါ function ၏ domain ဖြစ်သည်ဟု မှတ်ယူရမည်။

Equality of Functions

Two functions $f$ and $g$ are equal (and we write $f=g$) if and only if

1. $f$ and $g$ have the same domain, and

2. $f(c)$ = $g(c)$ for each element $c$ in the domain.

Function နှစ်ခု $f$ နှင့် $g$ ၏ domain များတူညီကြပြီး domain ထဲရှိ အစုဝင် $c$ တိုင်းအတွက် $f(c)= g(c)$ [$c$ ၏ image တူညီလျှင်] ဖြစ်လျှင် $f$ နှင့် $g$ သည် တူညီသော function များ ဖြစ်ကြသည်။

1.           Determine whether each relation is a function or not. If it is a function, state the domain and range.

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$\begin{array}{ll} \text{(a)} & \text{The relation}\ R\ \text{is a function.}\\\\ & \text{dom}\ (R ) = \{-1, 0, 1\}\\\\ & \text{ran}\ (R) = \{0, 1\}\\\\ \text{(b)} & \text{The relation}\ R\ \text{is not a function.}\\\\ \text{(c)} & \text{The relation}\ R\ \text{is not a function.} \end{array}$

2.           Consider the following relations. Determine whether each relation is a function or not. If it is a function, write down the domain and range.

              $\begin{array}{l} \text{(a)}\ \ \{(1,3),(2,5),(3,7),(4,9)\} \\\\ \text{(b)}\ \ \{(-2,5),(-1,3),(0,1),(-1,1)\}\\\\ \text{(c)}\ \ \{(1,3),(2,3),(3,2),(2,1)\} \\\\ \text{(d)}\ \ \{(2,4),(3,6),(4,6),(7,14)\} \\\\ \text{(e)}\ \ \{(0,0),(1,1),(3,3),(4,4)\} \\\\ \text{(f)}\ \ \{(2, a),(4, c),(5, a),(4, e)\} \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\{(1,3),(2,5),(3,7),(4,9)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{1,\ 2,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{3,\ 5,\ 7,\ 9\}\\\\\text{(b)}\;\;\{(-2,5),(-1,3),(0,1),(-1,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }-\text{1 of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(c)}\;\;\{(1,3),(2,3),(3,2),(2,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }2\text{ of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(d)}\;\;\{(2,4),(3,6),(4,6),(7,14)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{2,\ 3,\ 4,\ 7\}\\\ \ \ \ \ \text{Range}=\{4,\ 6,\ 14\}\\\\\text{(e)}\;\;\{(0,0),(1,1),(3,3),(4,4)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{0,\ 1,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{0,\ 1,\ 3,\ 4\}\\\\\text{(f)}\;\;\{(2,a),(4,c),(5,a),(4,e)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }4\text{ of domain is related to two}\\\ \ \ \ \ \text{elements }c\text{ and }e\text{.}\end{array}$

3.           Let $f$ be a function from $\mathbb{R} \rightarrow \mathbb{R}$. Which of the following statements are true?

              (a)     If $f(x)=5-x,$ the image of -3 under $f$ is 8.

              (b)     If $f(x)=x^{2}+9,$ the image of -3 under $f$ is zero.

              (c)     If $f(x)=3 x+4,$ then $f(a)=a$ implies that $a=-2$.

              (d)     If $f(x)=x+3,$ there is only one value $a \in \mathbb{R}$ such that $f(a)=0$.

              (e)     If $f(x)=x^{2}-1,$ then there are exactly two values $a \in \mathbb{R}$ such that $f(a)=0$.

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$\begin{array}{l}\ \ \ \ \ \ \ \ f:\mathbb{R}\to \mathbb{R}\\\\\text{(a)}\ \ f(x)=5-x\\\ \ \ \ \ f(-3)=5-(-3)=8\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(b)}\ \ f(x)={{x}^{2}}+9\\\ \ \ \ \ f(-3)={{(-3)}^{2}}+9=18\\\ \ \ \ \ \therefore \ \text{ The statements is false}\text{.}\\\\\text{(c)}\ \ f(x)=3x+4\\\ \ \ \ \ f(a)=a\\\ \ \ \ \ 3a+4=a\\\ \ \ \ \ 2a=-4\\\ \ \ \ \ a=-2\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(d)}\ \ f(x)=x+3\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ a+3=0\\\ \ \ \ \ a=-3\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(e)}\ \ f(x)={{x}^{2}}-1\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ {{a}^{2}}-1=0\\\ \ \ \ \ {{a}^{2}}=1\\\ \ \ \ \ a=\pm 1\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\end{array}$

4.           Illustrate the function $f: x \mapsto x+2$ with an arrow diagram for the domain $\{3,5,7,9,10\} .$ Write down the range of $f$.

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$\begin{array}{l}\ \ \ f:x\mapsto x+2\\\\\ \ \ \therefore \ f(x)=x+2\\\\\ \ \ \operatorname{dom}(f)=\{3,5,7,9,10\}\\\\\ \ \ f(3)=3+2=5\\\\\ \ \ f(5)=5+2=7\\\\\ \ \ f(7)=7+2=9\\\\\ \ \ f(9)=9+2=11\\\\\ \ \ f(10)=10+2=12\\\\\ \ \ \operatorname{ran}(f)=\{5,7,9,11,12\}\end{array}$

5.           Let the domain of function $h: x \mapsto 0$ be $\{2,4,6,7\} .$ What is the range of $h ?$ Draw an arrow diagram for $h$.

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$\begin{array}{l}h:x\mapsto 0\\\\\operatorname{dom}(h)=\{2,4,6,7\}\\\\\therefore \ \ h:2\mapsto 0\\\\\ \ \ h:4\mapsto 0\\\\\ \ \ h:6\mapsto 0\\\\\ \ \ h:7\mapsto 0\\\\\operatorname{ran}(h)=\{0\}\end{array}$

6.           Let the domain of function $f: x \mapsto 3 x$ be the set of natural numbers less than $5 .$ State the domain and range.

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$\begin{array}{l} f:x\mapsto 3x\\\\ \operatorname{dom}(f)=\text{the set of natural numbers less}\ \text{than 5}\text{.}\\\\ \operatorname{dom}(f)=\{1,\ 2,\ 3,\ 4\}\\\\ f:1\mapsto 3(1)=3\\f:2\mapsto 3(2)=6\\\\ f:3\mapsto 3(3)=9\\f:4\mapsto 3(4)=12\\\\ \operatorname{ran}(f)=\{3,\ 6,\ 9,\ 12\}\ end{array}$

7.           Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be given by

              (a)     $g(x)=3-4 x$ .Find $g(1), \quad g(3), \quad g(-2), \quad g(x+3), \quad g\left(\displaystyle \frac{1}{2}\right). $

              (b)    $g(x)=2 x-5 .$ Find $g(3), g\left(\displaystyle\frac{1}{2}\right), g(0), g(-4), g(4) .$ If $g(a)=99,$ find $a$.

              (c)    $g(x)=\displaystyle\frac{x+5}{2} .$ Find the images of $3,0,-3 .$ Find $x$ if $g(x)=0$.

              (d)    $g(x)=3 x-1 .$ Find $x$ such that $g(x)=20$.

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$\begin{array}{ll} \text{(a)} & g(x) = 3-4x\\ & g(1) = 3-4(1) = -1\\ & g(3) = 3-4(3) = -9\\ & g(-2) = 3-4(-2) = 11\\ & g(x+3) = 3-4(x+3) = -4x-9\\ & g \left({\displaystyle\frac{1}{2}} \right) = 3- 4\left(\displaystyle\frac{1}{2}\right) = 1\\\\ \text{(b)} & g(x) = 2x-5\\ & g(3) = 2(3)-5 = 1\\ & g \left(\displaystyle\frac{1}{2}\right) = 2 \left(\displaystyle\frac{1}{2}\right)-5 = -4\\ & g(0) = 2(0)-5 = -5\\ & g(-4) = 2(-4)-5 = -13\\ & g(4) = 2(4)-5 = 3\\ & g(a) = 99\\ & 2a-5 = 99\\ & 2a = 104\\ & a = 52 \\\\ \text{(c)} & g(x) = \displaystyle\frac{x+5}{2}\\ & g(3) = \displaystyle\frac{3+5}{2} = 4\\ & g(0) = \displaystyle\frac{0+5}{2} = \displaystyle\frac{5}{2}\\ & g(-3) = \displaystyle\frac{-3+5}{2} = 1\\ & g(x)=0\\ & \displaystyle\frac{x+5}{2} = 0\\ & x=-5\\\\ \text{(d)} & g(x) = 3x-1\\ & g(x) = 20\\ & 3x-1 = 20\\ & x = 7 \end{array}$

8.           A function $f$ from $A$ to $A$, where $A$ is the set of positive integers, is given by $f(x)=$ the sum of all possible divisors of $x$.

For example $f(6)=1+2+3+6=12$

              (a)     Find the values of $f(2), f(5), f(13), f(18)$.

              (b)     Show that $f(14)=f(15)$ and $f(3) \cdot f(5)=f(15)$.

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$\begin{array}{l}f(x)=\text{the sum of all possible divisors of }x\\\\f(2)=1+2=3\\\\f(5)=1+5=6\\\\f(13)=1+13=14\\\\f(15)=1+3+5+15=24\\\\f(14)=1+2+7+14=24\\\\\therefore \ f(15)=f(14)\\\\f(3)=1+3=4\\\\f(3)\cdot f(5)=4\times 6=24\\\\\therefore \ f(3)\cdot f(5)=f(15)\end{array}$

9.           Let $A=$ the set of positive integers greater than 3 and $B=$ the set of all positive integers. Let $d: A \rightarrow B$ be a function given by $d(n)=\frac{1}{2} n(n-3)$ the number of diagonals of a polygon of $n$ sides.

              (a)     Find $d(6), d(8), d(10), d(12)$.

              (b)     How many diagonals will a polygon of 20 sides have?

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$ \begin{array}{l}\text{(a)}\ \ \ \ A=\text{ the set of positive integers greater than 3}\\\\\ \ \ \ \ \ \ A=\{4,\ 5,\ 6,\ ...\}\\\\\ \ \ \ \ \ \ B=\text{ the set of positive all integers }\\\\\ \ \ \ \ \ \ B=\{1,\ 2,\ 3,\ ...\}\\\\\ \ \ \ \ \ \ d:A\to B\ \\\\\ \ \ \ \ \ d(n)=\displaystyle \frac{1}{2}n(n-3)\\\\\ \ \ \ \ \ d(6)=\displaystyle \frac{1}{2}\cdot 6\cdot (6-3)=9\\\\\ \ \ \ \ \ d(8)=\displaystyle \frac{1}{2}\cdot 8\cdot (8-3)=20\\\\\ \ \ \ \ \ d(10)=\displaystyle \frac{1}{2}\cdot 10\cdot (10-3)=35\\\\\ \ \ \ \ \ d(12)=\displaystyle \frac{1}{2}\cdot 12\cdot (12-3)=54\\\\\text{(b)}\ \ \ d(20)=\displaystyle \frac{1}{2}\cdot 20\cdot (20-3)=170\end{array}$

10.           Determine whether $f$ and $g$ are equal functions or not. Give reason:

              (a)     $f(x)=x^{2}+2,\ g(x)=(x+2)^{2}$.

              (b)     $f(x)=x^{2},\ g(x)=|x|^{2}$.

              (c)     $f(x)=\displaystyle\frac{x^{2}-1}{x+1},\ g(x)=x-1$.

              (d)     $f(x)=\displaystyle\frac{x+2}{x^{2}-4},\ g(x)=\frac{1}{x-2}$.

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$\begin{array}{l} \text{(a)}\ \ f(x)={{x}^{2}}+2,\;g(x)={{(x+2)}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ f(1)={{1}^{2}}+2=3\\\ \ \ \ \ g(1)={{(1+2)}^{2}}=9\\ \ \ \ \ \ \therefore \ \operatorname{ran}(f)\ne \operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(b)}\ \ f(x)={{x}^{2}},\;g(x)=|x{{|}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ \operatorname{ran}(f)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \,\,\,\ \ \ \operatorname{ran}(g)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)=\operatorname{dom}(g)\ \ \text{and }\operatorname{ran}(f)=\operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f=g\\\\ \text{(c)}\ \ f(x)=\displaystyle \frac{{{{x}^{2}}-1}}{{x+1}},\;g(x)=x-1\\ \ \ \ \ f(x)\ \text{exists only when }x+1\ne 0,\ \text{i}\text{.e,}\ x\ne -1\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{-1\},\ \operatorname{dom}(g)=\mathbb{R}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(d)}\ \ f(x)=\displaystyle \frac{{x+2}}{{{{x}^{2}}-4}},\;g(x)=\displaystyle \frac{1}{{x-2}}\\ \ \ \ \ \ f(x)\ \text{exists only when }\\ \,\ \ \ \ {{x}^{2}}-4\ne 0\\ \ \ \ \ \ {{x}^{2}}\ne 4\\ \ \ \ \ \ \therefore x\ne \pm \ 2\\ \ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{\pm \ 2\}\\ \ \ \ \ \ g(x)\ \text{exists only when }\\ \,\ \ \ \ x-2\ne 0\\\ \ \ \ \ x\ne 2\\ \ \ \ \ \ \therefore \ \operatorname{dom}(g)=\mathbb{R}\smallsetminus \{2\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\end{array}$

11.           State the domain of the following functions.

              (a)     $f(x)=\sqrt{x-2}$.

              (b)     $f(x)=\displaystyle\frac{1}{2 x-1}$.

              (c)     $f(x)=\displaystyle\frac{4}{x-3}$.

              (d)     $f(x)=\displaystyle\frac{2}{x^{2}-1}$.

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$\begin{array}{l}\text{(a)}\ \ f(x)=\sqrt{{x-2}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-2\ge 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ge 2\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ge 2,\ x\in \mathbb{R}\}\\\\\text{(b)}\ \ f(x)=\displaystyle \frac{1}{{2x-1}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ 2x-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne \displaystyle \frac{1}{2}\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \displaystyle \frac{1}{2},\ x\in \mathbb{R}\}\\\\\text{(c)}\ \ f(x)=\displaystyle \frac{4}{{x-3}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-3\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne 3\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne 3,\ x\in \mathbb{R}\}\\\\\text{(d)}\ \ f(x)=\displaystyle \frac{2}{{{{x}^{2}}-1}}\\\ \ \ \ \ f(x)\ \text{exists only when }\\\,\ \ \ \ {{x}^{2}}-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ {{x}^{2}}\ne 1\\\ \ \ \ \ \therefore x\ne \pm \ 1\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \pm \ 1,\ x\in \mathbb{R}\}\end{array}$

Quadratic_Function

Binary Operation : Problems and Solutions

1.        State whether the operation $ \displaystyle x\odot y$ = the greatest integer less than $ \displaystyle \frac{x}{y}+ \frac{y}{x}$ on the set of positive integers is a binary operation. If it is a binary operation, find $ \displaystyle (3\odot2)$ and $ \displaystyle 2 \odot (3\odot4)$.

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        $ \displaystyle \ \ \ \ \ x\odot y=\text{the greatest integer less than}\ \frac{x}{y}+\frac{y}{x}$

        $ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since any image is the greatest integer,}\\\ \ \ \ \ \text{closure property is satisfied}\text{.}\\\\\therefore \ \ \ \odot \text{is a binary operation}\text{.}\end{array}$

        $ \displaystyle \ \ \ \ \ \text{3}\odot 2=\text{the greatest integer less than}\ \left( {\frac{3}{2}+\frac{2}{3}} \right)$

        $ \displaystyle \begin{array}{l}\therefore \ \ \ \text{3}\odot 2=\text{the greatest integer less than}\ \left( {2.167} \right)\\\\\therefore \ \ \ \text{3}\odot 2=2\end{array}$

        $ \displaystyle \ \ \ \ \ \text{3}\odot 4=\text{the greatest integer less than}\ \left( {\frac{3}{4}+\frac{4}{3}} \right)$

        $ \displaystyle \begin{array}{l}\therefore \ \ \ \text{3}\odot 4=\text{the greatest integer less than}\ \left( {2.083} \right)\\\\\therefore \ \ \ \text{3}\odot 4=2\ \ \end{array}$

        $ \displaystyle \ \ \ \ \ 2\odot \left( {\text{3}\odot 4} \right)=2\odot 2=\ \text{the greatest integer less than}\ \left( {\frac{2}{2}+\frac{2}{2}} \right)$

        $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \text{the greatest integer less than}\ 2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1\ \ \ \ \end{array}$

2.        Let $ \displaystyle g:{{Z}^{+}}\times {{Z}^{+}}\to N$ be defined by $ \displaystyle (x, y)\mapsto g (x , y)$ = the remainder when $ \displaystyle x^y$ is divided by $ \displaystyle 3$. Find $ \displaystyle g(g(2, 3),\ 4)$. ($ \displaystyle {Z}^{+}$ = the set of positive integers and $ \displaystyle N$ = the set of natural numbers)

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        $ \displaystyle \begin{array}{l}\ \ \ \ g:{{Z}^{+}}\times {{Z}^{+}}\to N\\\\\ \ \ \ g(x,y)=\text{the remainder when }{{x}^{y}}\text{ is divided by 3}\\\\\therefore \ \ g(2,3)=\text{the remainder when }{{2}^{3}}\ \text{is divided by 3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\\\\\therefore \ \ g(g(2,3),4)=g(\text{2},4)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\text{the remainder when }{{2}^{4}}\text{ is divided by 3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1\end{array}$

3.        Let $ \displaystyle f : R\to R$ be defined by $ \displaystyle f(x) = x^3$ and $ \displaystyle g : N\to N$ be defined by $ \displaystyle g(y) = y^2 +y$. Define $ \displaystyle x\odot y = f(x) g(y)$. $ \displaystyle \odot$ is to be a binary operation, find possible domain and codomain. If $ \displaystyle \odot$ is a binary operation, prove it. Determine that $ \displaystyle \odot$ iscommutative or not.

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        $ \displaystyle \begin{array}{l}\ \ \ \ f:R\to R,\ f(x)={{x}^{3}}\\\\\ \ \ \ g:R\to R,\ g(y)={{y}^{2}}+y\\\\\ \ \ \ x\odot y=f(x)g(y)\\\\\therefore \ \ x\odot y={{x}^{3}}\left( {{{y}^{2}}+y} \right)\\\\\ \ \ \ \text{Since}\ {{x}^{3}}\in R\ \text{for every }x\in R\ \text{and}\\\\\ \ \ \ {{y}^{2}}+y\in R\ \text{for every }y\in R,\\\\\therefore \ \ \text{Domain of }x\odot y=R\times R\\\\\ \ \ \ \text{Codomain of }x\odot y=R\\\\\therefore \ \ \odot :R\times R\to R\\\\\therefore \ \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \ \odot \text{ is a binary operation}\text{.}\\\\\ \ \ \ 1\odot 2={{1}^{3}}\left( {{{2}^{2}}+2} \right)=6\\\\\ \ \ \ 2\odot 1={{2}^{3}}\left( {{{1}^{2}}+1} \right)=16\\\\\therefore \ \ \odot \text{ is not commutative}\text{.}\end{array}$

4.        A binary operation $ \displaystyle \odot$ on the set $ \displaystyle R$ of real numbers is defined by $ \displaystyle a\odot b = a + b + ab$. Show that the binary operation $ \displaystyle \odot$ is (i) commutative (ii) associative.

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        $ \displaystyle \begin{array}{l}\ \ \ \ a\odot b=a+b+ab\ \text{where}\ a\in R,b\in R\\\\\ \ \ \ b\odot a=b+a+ba=a+b+ab\\\\\therefore \ \ a\odot b=b\odot a\\\\\therefore \ \ \odot \text{ is commutative}\text{.}\\\\\therefore \ \ \ \left( {a\odot b} \right)\odot c=(a+b+ab)\odot c\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(a+b+ab)+c+(a+b+ab)c\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+ab+c+ac+bc+abc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+ab+ac+bc+abc\\\\\therefore \ \ \ a\odot \left( {b\odot c} \right)=a\odot (b+c+bc)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+bc+a(b+c+bc)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+bc+ab+ac+abc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+ab+ac+bc+abc\\\\\therefore \ \ \ \left( {a\odot b} \right)\odot c=\ a\odot \left( {b\odot c} \right)\\\\\therefore \ \ \odot \text{ is associative}\text{.}\end{array}$

5.        Let $ \displaystyle Z$ be the set of all integers and $ \displaystyle A = \left\{ { 0 , 1 } \right\}.$ Let $ \displaystyle {{\odot }_{1}}:A\times A\to Z$ be defined by $ \displaystyle(x , y )\mapsto \odot_1 (x , y) = x^2 + y$. Let $ \displaystyle {{\odot }_{2}}:A\times A\to Z$ be defined by $ \displaystyle (x , y )\mapsto \odot_2 (x , y) = x^2 y$. Is $ \displaystyle \odot_1$ a binary operation on $ \displaystyle A$? Why? Is $ \displaystyle \odot_2$ a binary operation on $ \displaystyle A$? Why ?

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        $ \displaystyle \begin{array}{l}\ \ \ \ A=\left\{ {0,1} \right\}\\\\\ \ \ \ {{\odot }_{1}}:A\times A\to Z,\\\\\ \ \ \ (x,y)\mapsto {{\odot }_{1}}(x,y)={{x}^{2}}+y\\\\\ \ \ \ {{\odot }_{2}}:A\times A\to Z\\\\\ \ \ \ (x,y)\mapsto {{\odot }_{2}}(x,y)={{x}^{2}}y\\\\\ \ \ \ A\times A=\left\{ {(0,0),(0,1),(1,0),(1,1)} \right\}\\\\\ \ \ \ \ x{{\odot }_{1}}\ y={{x}^{2}}+y\\\\\therefore \ \ \ 0{{\odot }_{1}}0={{0}^{2}}+0=0\\\\\ \ \ \ \ 0{{\odot }_{1}}1={{0}^{2}}+1=1\\\\\ \ \ \ \ 1{{\odot }_{1}}0={{1}^{2}}+0=1\\\\\ \ \ \ \ 1{{\odot }_{1}}1={{1}^{2}}+1=2\\\\\therefore \ \ \ \text{Range of }{{\odot }_{1}}=\{0,1,2\}\\\\\therefore \ \ \ \text{Range of }{{\odot }_{1}}\not\subset A\\\\\therefore \ \ \ \text{Closure property is not satisfied}\text{.}\\\\\therefore \ \ \ {{\odot }_{1}}\ \text{is not a binary operation}\text{.}\\\\\ \ \ \ \ x{{\odot }_{2}}\ y={{x}^{2}}y\\\\\therefore \ \ \ 0{{\odot }_{2}}0={{0}^{2}}(0)=0\\\\\ \ \ \ \ 0{{\odot }_{2}}1={{0}^{2}}(1)=0\\\\\ \ \ \ \ 1{{\odot }_{2}}0={{1}^{2}}(0)=0\\\\\ \ \ \ \ 1{{\odot }_{2}}1={{1}^{2}}(1)=1\\\\\therefore \ \ \ \text{Range of }v=\{0,1\}\\\\\therefore \ \ \ \text{Range of }{{\odot }_{2}}\subset A\\\\\therefore \ \ \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \ \ {{\odot }_{2}}\ \text{is a binary operation}\text{.}\end{array}$

6.        The operation $ \displaystyle \odot$ on $ \displaystyle R$ is defined by $ \displaystyle x\odot y=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy$ for all real numbers $ \displaystyle x$ and $ \displaystyle y$. Prove that $ \displaystyle \odot$ is commutative but not associative.

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     $ \displaystyle \ \ \ \ x\odot y=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy,\ x,y\in R$

     $ \displaystyle \ \ \ \ y\odot x=\frac{{{{y}^{2}}+{{x}^{2}}}}{2}-yx$

     $ \displaystyle \therefore \ \ y\odot x=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy$

     $ \displaystyle \begin{array}{l}\therefore \ \ x\odot y=y\odot x\\\\\therefore \ \ \odot \ \text{is commutative}.\end{array}$

     $ \displaystyle \ \ \ \ 1\odot 2=\frac{{{{1}^{2}}+{{2}^{2}}}}{2}-1(2)=\frac{1}{2}$

     $ \displaystyle \ \ \ \ \left( {1\odot 2} \right)\odot 3=\frac{1}{2}\odot 3$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\left(      {\frac{1}{2}} \right)}}^{2}}+{{3}^{2}}}}{2}-\frac{1}{2}(3)$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{25}}{8}$

     $ \displaystyle \ \ \ \ 2\odot 3=\frac{{{{2}^{2}}+{{3}^{2}}}}{2}-2(3)=\frac{1}{2}$

     $ \displaystyle \ \ \ \ 1\odot \left( {2\odot 3} \right)=1\odot \frac{1}{2}$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{1}^{2}}+{{{\left( {\frac{1}{2}} \right)}}^{2}}}}{2}-1\left( {\frac{1}{2}} \right)$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{8}$

     $ \displaystyle \begin{array}{l}\therefore \ \ \left( {1\odot 2} \right)\odot 3\ne 1\odot \left( {2\odot 3} \right)\\\\\therefore \ \ \odot \ \text{is not associative}.\end{array}$

7.        The operation $ \displaystyle \odot$ is defined by $ \displaystyle (2a + b) \odot(a + 2b) = a^2 + ab + b^2$. Find $ \displaystyle 6\odot 9$.

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     $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ (2a+b)\odot (a+2b)={{a}^{2}}+ab+{{b}^{2}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ 2a+b=6\ \ \ \ \ \ \ \ \ \ \ \ -----(1)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a+2b=9\ \ \ \ \ \ \ \ \ \ \ -----(2)\\\\\ \ \ \ \ \ \ \ \text{By Equation(1)+Equation(2),}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3a+3b=15\ \ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a+b=5\ \ \ \ \ \ \ \ \ \ \ \ \ -----(3)\\\\\ \ \ \ \ \ \ \ \text{By Equation(1)}-\text{Equation(2),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a-b=-3\ \ \ \ \ \ \ \ \ \ \ -----(4)\\\\\ \ \ \ \ \ \ \ \text{By Equation(3)}+\text{Equation(4),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2a=2\ \Rightarrow a=1\\\\\ \ \ \ \ \ \ \ \text{By Equation(3)}-\text{Equation(4),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2b=8\ \Rightarrow b=4\\\\\ \ \ \ \ \ \ \ \therefore \ 6\odot 9=\left[ {2(1)+4} \right]\odot \left[ {1+2(4)} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(1)}^{2}}+(1)(4)+{{(4)}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =21\end{array}$

8.        An operation $ \displaystyle \odot$ is defined by $ \displaystyle a\odot b = a^2 + ab + b^2,\ a,\ b\in R$. Solve the equation $ \displaystyle (6\odot k) - (k\odot 2) = 8 - 8k$.

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     $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ a\odot b={{a}^{2}}+ab+{{b}^{2}}\\\\\ \ \ \ \ \ \ \ 6\odot k={{6}^{2}}+6k+{{k}^{2}}\\\\\ \ \ \ \ \ \ \ k\odot 2={{k}^{2}}+2k+{{2}^{2}}\\\\\ \ \ \ \therefore \ \ (6\odot k)-(k\odot 2)=8-8k\\\\\ \ \ \ \therefore \ \ ({{6}^{2}}+6k+{{k}^{2}})-({{k}^{2}}+2k+{{2}^{2}})=8-8k\\\\\ \ \ \ \therefore \ \ 4k-32=8-8k\\\\\ \ \ \ \therefore \ \ 12k=40\end{array}$

     $\displaystyle \ \ \ \ \therefore \ \ k=\frac{{10}}{3}$

9.        A binary operation $ \displaystyle \odot$ on $ \displaystyle R$ is defined by $ \displaystyle x\odot y = (2x - 3y)^2 - 5y^2$. Show that the binary operation is commutative. Find the values of $ \displaystyle k$ for which $ \displaystyle(-2)\odot k = 80$.

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     $\displaystyle \begin{array}{l}\ \ \ \ x\odot y={{(2x-3y)}^{2}}-5{{y}^{2}}\\\\\ \ \ \ x\odot y=4{{x}^{2}}-12xy+9{{y}^{2}}-5{{y}^{2}}\\\\\therefore \ \ x\odot y=4{{x}^{2}}-12xy+4{{y}^{2}}\\\\\therefore \ \ x\odot y=4\left( {{{x}^{2}}-3xy+{{y}^{2}}} \right)\\\\\therefore \ \ y\odot x=4\left( {{{y}^{2}}-3yx+{{x}^{2}}} \right)\\\\\therefore \ \ y\odot x=4\left( {{{x}^{2}}-3xy+{{y}^{2}}} \right)\\\\\therefore \ \ x\odot y=y\odot x\\\\\therefore \ \ \odot \ \text{is commutative}.\\\\\ \ \ \ (-2)\odot k=80\ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 4\left[ {{{{(-2)}}^{2}}-3(-2)k+{{k}^{2}}} \right]=80\\\\\therefore \ \ {{k}^{2}}+6k-16=0\\\\\therefore \ \ (k+8)(k-2)=0\\\\\therefore \ \ k=-8\ \text{or}\ k=2\end{array}$

10.     Let $ \displaystyle A=\{0,\ 1,\ 2,\ 3,\ 4\}$. The binary operation $ \displaystyle {{\oplus }_{5}}$ on the set $ \displaystyle A$ is defined by $ \displaystyle x\ {{\oplus }_{5}}\ y$ = the remainder when $ \displaystyle x + 2y$ is divided by 5. Make a Cayley table.

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     $ \displaystyle \begin{array}{l}\ \ A=\{0,\ 1,\ 2,\ 3,\ 4\}\\\\\ \ x\ {{\oplus }_{5}}\ y=\text{the remainder when}\ x+2y\ \text{is divided by 5}\text{.}\\\\\ \ 0\ {{\oplus }_{5}}\ 0=0\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 0=1\ \ \ \ \ 2\ {{\oplus }_{5}}\ 0=2\ \ \ \ \ 3\ {{\oplus }_{5}}\ 0=3\ \ \ \ \ \ 4\ {{\oplus }_{5}}\ 0=4\ \\\\\ \ 0\ {{\oplus }_{5}}\ 1=2\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 1=3\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 1=4\ \ \ \ \ 3\ {{\oplus }_{5}}\ 1=0\ \ \ \ \ \ 4\ {{\oplus }_{5}}\ 1=1\ \\\\\ \ 0\ {{\oplus }_{5}}\ 2=4\ \ \ \ \ 1\ {{\oplus }_{5}}\ 2=0\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 2=1\ \ \ \ \ 3\ {{\oplus }_{5}}\ 2=2\ \ \ \ \ 4\ {{\oplus }_{5}}\ 2=3\ \\\\\ \ 0\ {{\oplus }_{5}}\ 3=1\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 3=2\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 3=3\ \ \ \ \ 3\ {{\oplus }_{5}}\ 3=4\ \ \ \ \ 4\ {{\oplus }_{5}}\ 3=0\ \\\\\ \ 0\ {{\oplus }_{5}}\ 4=3\ \ \ \ \ 1\ {{\oplus }_{5}}\ 4=4\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 4=0\ \ \ \ \ 3\ {{\oplus }_{5}}\ 4=1\ \ \ \ \ 4\ {{\oplus }_{5}}\ 4=2\ \end{array}$

     $ \displaystyle \text {Cayley Table}$

⊕5	0	1	2	3	4
0	0 ⊕50	0 ⊕51	0 ⊕52	0 ⊕53	0 ⊕54
1	1 ⊕50	1 ⊕51	1 ⊕52	1 ⊕53	1 ⊕54
2	2 ⊕50	2 ⊕51	2 ⊕52	2 ⊕53	2 ⊕54
3	3 ⊕50	3 ⊕51	3 ⊕52	3 ⊕53	3 ⊕54
4	4 ⊕50	4 ⊕51	4 ⊕52	4 ⊕53	4 ⊕54

⊕5	0	1	2	3	4
0	0	2	4	1	3
1	1	3	0	2	4
2	2	4	1	3	0
3	3	0	2	4	1
4	4	1	3	0	2

11.     The binary operation $ \displaystyle \odot_1$ and $ \displaystyle \odot_2$ on R defined by $ \displaystyle x\odot_1 y = x^2 - y^2$ and $ \displaystyle x\odot_2 y = 7x + 4y$. Find $ \displaystyle(2 \odot_2 1)\odot_1 4$. Find also $ \displaystyle x$ if $ \displaystyle (- 3\odot_1 2) \odot_2 (1\odot_1 x) = 3$.

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.     $ \displaystyle \begin{array}{l}\ \ \ \ x{{\odot }_{1}}y={{x}^{2}}-{{y}^{2}}\\\\\ \ \ \ x{{\odot }_{2}}y=7x+4y\\\\\therefore \ \ \ 2{{\odot }_{2}}1=7(2)+4(1)=18\\\\\therefore \ \ \ \left( {2{{\odot }_{2}}1} \right){{\odot }_{1}}4=18{{\odot }_{1}}4\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{18}^{2}}-{{4}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =306\\\\\ \ \ \ \ -3{{\odot }_{1}}2={{(-3)}^{2}}-{{2}^{2}}=5\\\\\ \ \ \ \ 1{{\odot }_{1}}x={{1}^{2}}-{{x}^{2}}=1-{{x}^{2}}\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ (-3{{\odot }_{1}}2){{\odot }_{2}}(1{{\odot }_{1}}x)=3\\\\\therefore \ \ \ 5{{\odot }_{2}}(1-{{x}^{2}})=3\\\\\ \ \ \ 7(5)+4(1-{{x}^{2}})=3\\\\\therefore \ \ \ 39-{{x}^{2}}=3\Rightarrow {{x}^{2}}=36\Rightarrow x=\pm 6\end{array}$

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Inverse Function : Problems and Solutions

1.       Given that $ \displaystyle f(x)={{e}^{{x+3}}}$ where $ \displaystyle x\in R$, find $ \displaystyle {f}^{-1}(x)$ and state the domain of $ \displaystyle {f}^{-1}$. Hence solve the equation $ \displaystyle {f}^{-1}(x)= \ln \left( {\frac{1}{x}} \right)$.

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          $ \displaystyle \begin{array}{l}f(x)={{e}^{{x+3}}},\ x\in R\\\\{{f}^{{-1}}}(y)=x\Leftrightarrow f(x)=y\\\\\therefore {{e}^{{x+3}}}=y\\\\\therefore x+3={{\log }_{e}}y\\\\\therefore x+3=\ln y\\\\\therefore x=\ln y-3\\\\\therefore {{f}^{{-1}}}(y)=\ln y-3\\\\\therefore {{f}^{{-1}}}(x)=\ln x-3\\\\\text{Domain of }{{f}^{{-1}}}=\{x|x>0,\ x\in R\}\\\\{{f}^{{-1}}}(x)=\ln \left( {\frac{1}{x}} \right)\\\\\ln x-3=\ln \left( {\frac{1}{x}} \right)\\\\\ln x-\ln \left( {\frac{1}{x}} \right)=3\\\\\ln {{x}^{2}}=3\\\\2\ln x=3\\\\\ln x=\frac{3}{2}\\\\x={{e}^{{\frac{3}{2}}}}\end{array}$

2.        A function f is defined by $ \displaystyle f(3x-2) = 5+6x$. Find the value of $ \displaystyle {f}^{-1}(29)$.

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          $ \displaystyle \begin{array}{l}f(3x-2)=5+6x\\\\\therefore {{f}^{{-1}}}(5+6x)=3x-2\\\\\text{Let}\ 5+6x=29,\text{then}\\\\6x=24\Rightarrow x=4\\\\\therefore {{f}^{{-1}}}(29)=3(4)-2=10\end{array}$

3.        A function f is defined by $ \displaystyle f(x)=\frac{{x-3}}{{2x-5}}$.

(i) State the value of $ \displaystyle x$ for which $ \displaystyle f$ is not defined.

(ii) Find the value of $ \displaystyle x$ for which $ \displaystyle f(x) = 0$.

(iii) Find the inverse function$ \displaystyle {f}^{-1}$ and state the domain of $ \displaystyle {f}^{-1}$.

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         $ \displaystyle \ \ \ \ \ \ \ \ f(x)=\frac{{x-3}}{{2x-5}}$

         $ \displaystyle \text{(i)}\ \ \ \ f\ \text{is not defined when}$

         $ \displaystyle \ \ \ \ \ \ \ 2x-5=0\Rightarrow x=\frac{5}{2}$

         $ \displaystyle \text{(ii)}\ \ \ f(x)=0$

         $ \displaystyle \ \ \ \ \ \ \ \frac{{x-3}}{{2x-5}}=0$

         $ \displaystyle \ \ \ \ \ \ \ \text{Since }2x-5\ne 0,$

         $ \displaystyle \ \ \ \ \ \ \ x-3=0\Rightarrow x=3$

         $ \displaystyle \text{(iii)}\ \ \text{Let }{{f}^{{-1}}}(x)=y,\ \text{then}$

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

         $ \displaystyle \ \ \ \ \ \ \ \frac{{y-3}}{{2y-5}}=x$

         $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ y-3=2xy-5x\\\\\ \ \ \ \ \ \ y-2xy=3-5x\\\\\ \ \ \ \ \ \ y(1-2x)=3-5x\end{array}$

         $ \displaystyle \ \ \ \ \ \ \ y=\frac{{3-5x}}{{1-2x}},\ x\ne \frac{1}{2}$

         $ \displaystyle \ \ \ \ \ \ \ \text{Domain of }{{f}^{{-1}}}=\{x|x\in R,x\ne \frac{1}{2}\}$

4.       A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto \frac{a}{x}+1,\ x\ne 0$ where $ \displaystyle a$ is a constant. Given that $ \displaystyle 6( f \cdot f )(-1) +{f}^{-1}(2) = 0$, find the possible values of $ \displaystyle a$.

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         $ \displaystyle f(x)=\frac{a}{x}+1,\ x\ne 0$

         $ \displaystyle (f\cdot f)(-1)=f\left( {f(-1)} \right)$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{a}{{-1}}+1} \right)$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {1-a} \right)$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a}{{1-a}}+1$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{{1-a}}$

         $ \displaystyle {{f}^{{-1}}}(2)=k\Leftrightarrow f(k)=2$

         $ \displaystyle \therefore \frac{a}{k}+1=2$

         $ \displaystyle \ \ \ \frac{a}{k}=1\Rightarrow k=a$

         $ \displaystyle \therefore {{f}^{{-1}}}(2)=a$

         $ \displaystyle 6(f\cdot f)(-1)+{{f}^{{-1}}}(2)=0$

         $ \displaystyle \therefore \frac{6}{{1-a}}+a=0$

         $ \displaystyle \ \ \ 6+a-{{a}^{2}}=0$

         $ \displaystyle \therefore (3-a)(2+a)=0$

         $ \displaystyle \therefore a=3\ \text{or}\ a=-2$

5.       A function $ \displaystyle g$ is defined by $ \displaystyle g:x\mapsto \frac{{x+1}}{{x-2}},\ x\ne 2,x\ne 5$ and $ \displaystyle h$ is defined by is defined by $ \displaystyle h:x\mapsto \frac{{ax+3}}{{x}},\ x\ne 0$. Given that $ \displaystyle (h\cdot {g}^{–1})(4) = 6$, calculate the value of $ \displaystyle a$.

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          $ \displaystyle \ \ \ \ \ \ g(x)=\frac{{x+1}}{{x-2}},\ x\ne 2,x\ne 5$

          $ \displaystyle \ \ \ \ \ \ h(x)=\frac{{ax+3}}{x},\ x\ne 0$

          $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \left( {h\cdot {{g}^{{-1}}}} \right)(4)=6\\\\\ \ \ \ \ \ {{g}^{{-1}}}(4)=p\Leftrightarrow g(p)=4\end{array}$

          $ \displaystyle \therefore \ \ \ \ \frac{{p+1}}{{p-2}}=4$

          $ \displaystyle \begin{array}{l}\ \ \ \ \ \ p+1=4p-8\\\\\ \ \ \ \ \ 3p=9\Rightarrow p=3\\\\\therefore \ \ \ \ {{g}^{{-1}}}(4)=3\\\\\ \ \ \ \ \ \left( {h\cdot {{g}^{{-1}}}} \right)(4)=6\\\\\ \ \ \ \ \ h\left( {{{g}^{{-1}}}(4)} \right)=6\\\\\ \ \ \ \ \ h(3)=6\end{array}$

          $ \displaystyle \ \ \ \ \ \ \frac{{a(3)+3}}{3}=6$

          $ \displaystyle \therefore \ \ \ \ a+1=6\Rightarrow a=5$

6.       Let $ \displaystyle f:R\to R$ and $ \displaystyle g:R\to R$ be defined by $ \displaystyle f(x) = 3x - 1$ and $ \displaystyle g(x) = x + 7$. Find $ \displaystyle ({f}^{-1}\cdot g)(x)$ and what is the value of $ \displaystyle a\in R$ for which $ \displaystyle ({f}^{-1}\cdot g)(a)=3$.

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         $ \displaystyle \begin{array}{l}f:R\to R,\ f(x)=3x-1\\\\g:R\to R,\ g(x)=x+7\\\\({{f}^{{-1}}}\cdot g)(x)={{f}^{{-1}}}\left( {g(x)} \right)\\\\\text{Let }{{f}^{{-1}}}\left( {g(x)} \right)=y\ \text{then }g(x)=f(y).\\\\\therefore x+7=3y-1\end{array}$

         $ \displaystyle \ \ \ y=\frac{{x+8}}{3}$

         $ \displaystyle \therefore ({{f}^{{-1}}}\cdot g)(x)=\frac{{x+8}}{3}$

         $ \displaystyle \ \ \ ({{f}^{{-1}}}\cdot g)(a)=3$

         $ \displaystyle \ \ \ \frac{{a+8}}{3}=3\Rightarrow a=1$

7.       For the function $ \displaystyle f(x)=\frac{{2x}}{{3x+1}},\ x\ne -\frac{1}{3}$ find $ \displaystyle {f}^{-1}$ and verify that $ \displaystyle (f\cdot {f}^{-1})$ and $ \displaystyle ({f}^{-1}\cdot f)$ both equal $ \displaystyle I$.

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          $ \displaystyle \ \ \ \ f(x)=\frac{{2x}}{{3x+1}},\ x\ne -\frac{1}{3}$

          $ \displaystyle \ \ \ \ {{f}^{{-1}}}(x)=y\Leftrightarrow f(y)=x$ $ \displaystyle \ \ \ \ \frac{{2y}}{{3y+1}}=x$

          $ \displaystyle \begin{array}{l}\ \ \ \ 2y=3xy+x\\\\\ \ \ \ y(2-3x)=x\end{array}$

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

          $ \displaystyle \therefore \ \ {{f}^{{-1}}}(x)=\frac{x}{{2-3x}},\ x\ne \frac{2}{3}$

          $ \displaystyle \therefore \ \ (f\cdot {{f}^{{-1}}})(x)=f\left( {{{f}^{{-1}}}(x)} \right)$

          $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{x}{{2-3x}}} \right)$

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

          $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2x}}{{2-3x}}\times \frac{{2-3x}}{{3x+2-3x}}$

          $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\\\\\therefore \ \ ({{f}^{{-1}}}\cdot f)(x)={{f}^{{-1}}}\left( {(x)} \right)\end{array}$

          $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {\frac{{2x}}{{3x+1}}} \right)$

          $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{2x}}{{3x+1}}}}{{2-\frac{{6x}}{{3x+1}}}}$

          $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2x}}{{3x+1}}\times \frac{{3x+1}}{{6x+2-6x}}$

          $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\\\\\therefore \ \ (f\cdot {{f}^{{-1}}})(x)=({{f}^{{-1}}}\cdot f)(x)=I(x)\end{array}$

8.       Functions $ \displaystyle f$ and $ \displaystyle g$ are defined, for $ \displaystyle x\in R$, by $ \displaystyle f : x\mapsto 5x - 2, g:x\mapsto \frac{1}{2x-1},\ x\ne \frac{1}{2}$. Find the value of $ \displaystyle x$ for which

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         $ \displaystyle \text{(i)}\ f(x)={f}^{-1}(x)$.

         $ \displaystyle \text{(ii)}\ (f\cdot g)(x)+3g(x)=0$.

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ f(x)=5x-2,$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ g(x)=\frac{1}{{2x-1}},\ x\ne \frac{1}{2}$

         $ \displaystyle \begin{array}{l}\text{(i)}\ \ \ \ \ \ \ f(x)={{f}^{{-1}}}(x)\\\\\ \ \ \ \ \therefore \ \ \ f\left( {f(x)} \right)=x\\\\\ \ \ \ \ \therefore \ \ \ f\left( {5x-2} \right)=x\\\\\ \ \ \ \ \therefore \ \ \ 5(5x-2)-2=x\\\\\ \ \ \ \ \therefore \ \ \ 24x=12\end{array}$

         $ \displaystyle \ \ \ \ \ \therefore \ \ \ x=\frac{1}{2}$

         $ \displaystyle \begin{array}{l}\text{(ii)}\ \ \ \ \ \ (f\cdot g)(x)+3g(x)=0\\\\\ \ \ \ \ \ \ \ \ \ f\left( {g(x)} \right)+3g(x)=0\end{array}$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ f\left( {\frac{1}{{2x-1}}} \right)+\frac{3}{{2x-1}}=0$

         $ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{5}{{2x-1}}-2+\frac{3}{{2x-1}}=0$

         $ \displaystyle \ \ \ \ \ \therefore \ \ \ \frac{8}{{2x-1}}=2$

         $ \displaystyle \ \ \ \ \ \therefore \ \ \ 2x-1=4$

        $ \displaystyle \ \ \ \ \ \therefore \ \ \ x=\frac{5}{2}$

မဂၤလာပါ

Trigonometric Function and Solving Trigonometric Equation

Let $ \displaystyle f(x)=\frac{{\cos x}}{{1+\sin x}}+\frac{{1+\sin x}}{{\cos x}}.$

(a) Prove that $ \displaystyle f(x)=2\sec x.$

(b) Hence solve the equation $ \displaystyle f(x)=4$ where 0 < x < 2π.

 Solution 

(a)  $ \displaystyle f(x)=\frac{{\cos x}}{{1+\sin x}}+\frac{{1+\sin x}}{{\cos x}}$

     $ \displaystyle \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}x+{{{(1+\sin x)}}^{2}}}}{{(1+\sin x)\cos x}}$

     $ \displaystyle \ \ \ \ \ \ \ =\frac{{{{{\cos }}^{2}}x+1+2\sin x+\sin^{2}{{x}}}}{{(1+\sin x)\cos x}}$

    $ \displaystyle \ \ \ \ \ \ \ =\frac{{\sin {{x}^{2}}+{{{\cos }}^{2}}x+1+2\sin x}}{{(1+\sin x)\cos x}}$

    $ \displaystyle \ \ \ \ \ \ \ =\frac{{1+1+2\sin x}}{{(1+\sin x)\cos x}}$

    $ \displaystyle \ \ \ \ \ \ \ =\frac{{2(1+\sin x)}}{{(1+\sin x)\cos x}}$

    $ \displaystyle \ \ \ \ \ \ \ =\frac{2}{{\cos x}}$

    $ \displaystyle \ \ \ \ \ \ \ =2\sec x$

(b)    $\displaystyle \  f(x)=4$

     $ \displaystyle \therefore 2\sec x=4$

     $ \displaystyle \ \ \ \sec x=2$

    $ \displaystyle \ \ \ x=\frac{\pi }{3}$ or $ \displaystyle \ \ \ x=\frac{5\pi }{3}$

Problems Study

 

1. Functions f and g are such that  g^-1 (x) = x  -${\displaystyle \frac{1}{3 }}$ and (f ∘ g ) (x) = 3x - 1.
    Find (g ^-1 ∘ f ) (x) where x ∈ R.

   Solution 

   Let g ^-1(x ) = y then g (y) = x.
   Hence x - ${\displaystyle \frac{1}{3}}$ = y
            x = y + ${\displaystyle \frac{1}{3}}$
            g (y) = y + ${\displaystyle \frac{1}{3}}$
            g (x) = x + ${\displaystyle \frac{1}{3}}$
           (f ∘ g) (x) = 3x - 1
           f (g ) (x)   = 3x - 1
           f (x + ${\displaystyle \frac{1}{3}}$)  = 3x - 1
                          = 3 (x + ${\displaystyle \frac{1}{3}}$) - 2
   Hence f (x ) = 3x - 2
           (g ^-1 ∘ f ) (x) = g ^-1 ( f   (x ) )
                             = g ^-1 ( 3x - 2)
                             = 3x - ${\displaystyle \frac{7}{3}}$

2. If x, y and z are any three consecutive even numbers, Show that x² + y² + z² = 3y² + 8.

   Solution 

   Let x = 2a where a is an integer.

   Since x, y and z are any three consecutive even numbers,

   y = 2a + 2 and z = 2a + 4

   x² + y² + z² = (2a)² + (2a + 2)² + (2a + 4)² 

                     = 4a² + 4a² + 8a + 4 + 4a² + 16a + 16

                     = 12a² + 24a + 20

                     = 12a² + 24a + 12 + 8

                     = 3 (4a² + 8a + 4) + 8 

                     = 3 (2a + 2)² + 8 

                     = 3 y² + 8