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

Composition of Functions : Problems and Solutions

1.      Two functions are defined by $ \displaystyle f:x\mapsto ax+1$ and $ \displaystyle \ g:x\mapsto \frac{{4b}}{{x-1}},x\ne 1$ where $ \displaystyle a$ and $ \displaystyle b$ are constants. Given that $ \displaystyle f(a) = g(b)$ and $ \displaystyle f\left( {\frac{1}{a}} \right)=g\left( {\frac{1}{b}} \right)$ find the possible values of $ \displaystyle a$ and $ \displaystyle b$.

Show Solution

$ \displaystyle f(x)=ax+1\ ,g(x)=\frac{{4b}}{{x-1}},x\ne 1$

$ \displaystyle f(a)=g(b)$

$ \displaystyle \therefore {{a}^{2}}+1=\frac{{4b}}{{b-1}}$

$ \displaystyle \therefore {{a}^{2}}=\frac{{4b}}{{b-1}}-1$

$ \displaystyle f\left( {\frac{1}{a}} \right)=g\left( {\frac{1}{b}} \right)$

$ \displaystyle \therefore a\left( {\frac{1}{a}} \right)+1=\frac{{4b}}{{\frac{1}{b}-1}}$

$ \displaystyle \therefore \frac{{4{{b}^{2}}}}{{1-b}}=2\Rightarrow 2{{b}^{2}}+b-1=0$

$ \displaystyle \therefore (2b-1)(b+1)=0$

$ \displaystyle \therefore b=\frac{1}{2}\ \text{or }b=-1$

$ \displaystyle \text{When }b=\frac{1}{2},$

$ \displaystyle {{a}^{2}}=\frac{{4\left( {\frac{1}{2}} \right)}}{{\left( {\frac{1}{2}} \right)-1}}-1=-5\notin R$

$ \displaystyle \text{When }b=-1,$

$ \displaystyle {{a}^{2}}=\frac{{4\left( {-1} \right)}}{{\left( {-1} \right)-1}}-1=1$

$ \displaystyle \therefore a=\pm 1$

2.      A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto \frac{6}{{x-2}},x\ne 2$. Express $ \displaystyle (f\cdot f)(x)$ in the form $ \displaystyle \frac{{ax+b}}{{c-x}}$ stating the values of $ \displaystyle a, b$ and $ \displaystyle c$.

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$\displaystyle f(x)=\frac{6}{{x-2}},x\ne 2$

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{6}{{x-2}}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{6}{{\frac{6}{{x-2}}-2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{6}{{\frac{{6-2x+4}}{{x-2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{6x-12}}{{10-2x}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{3x-6}}{{5-x}}$

$ \displaystyle \therefore \ \ \ \frac{{ax+b}}{{c-x}}=\frac{{3x-6}}{{5-x}}$

$ \displaystyle \therefore \ \ a=3,b=-6\ \text{and}\ c=5$

3.      A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto 2-\frac{1}{x},x\ne 0$. Solve the equation $ \displaystyle (f\cdot f)(x)=f(x)$.

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$ \displaystyle \ \ \ f(x)=2-\frac{1}{x},x\ne 0$

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

$ \displaystyle \therefore f\left( {f(x)} \right)=f(x)$

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

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

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

$ \displaystyle \therefore {{x}^{2}}-2x+1=0$

$ \displaystyle \therefore {{(x-1)}^{2}}=0$

$ \displaystyle \therefore x=1$

4.      Let $ \displaystyle f:R\to R$ and $\displaystyle g:R\to R$ be $\displaystyle f(x)=px+5, g(x)=qx-3$ where $\displaystyle p\ne 0, q\ne 0$. If $ \displaystyle (g\cdot f):R\to R$ is the identity function on $\displaystyle R$, then prove that $\displaystyle p$ is the reciprocal of $\displaystyle q$. Hence find the values of $\displaystyle p$ and $\displaystyle q$.

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$ \displaystyle f:R\to R,f(x)=px+5$

$ \displaystyle g:R\to R,g(x)=qx-3$

$ \displaystyle p\ne 0\ \text{and}\ q\ne 0$

$ \displaystyle \left( {g\cdot f} \right)(x)=I(x)$

$ \displaystyle \therefore g\left( {f(x)} \right)=x$

$ \displaystyle \therefore g\left( {px+5} \right)=x$

$ \displaystyle \therefore q\left( {px+5} \right)-3=x$

$ \displaystyle \therefore pqx+\left( {5q-3} \right)=x+0$

$ \displaystyle \therefore pq=1\Rightarrow p=\frac{1}{q}$

$ \displaystyle \ \ \ 5q-3=0\Rightarrow q=\frac{3}{5}$

$ \displaystyle \therefore p=\frac{5}{3}$

5.      If $ \displaystyle f$ and $ \displaystyle g$ are functions such that $ \displaystyle f(x) = 2x - 1$ and $ \displaystyle (g\cdot f)(x) = 4x^2 - 2x - 3$, find the formula of $ \displaystyle g$ in simplified form.

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$ \displaystyle \begin{array}{l}f(x)=2x-1\\\\\left( {g\cdot f} \right)(x)=4{{x}^{2}}-2x-3\\\\g\left( {f(x)} \right)=4{{x}^{2}}-2x-3\\\\g\left( {2x-1} \right)=4{{x}^{2}}-2x-3\\\\g\left( {2x-1} \right)=4{{x}^{2}}-4x+1+2x-1-3\\\\g\left( {2x-1} \right)={{(2x-1)}^{2}}+(2x-1)-3\\\\\therefore g(x)={{x}^{2}}+x-3\end{array}$

6.      A function f is defined by $ \displaystyle f:x\mapsto 2x^2 - 12x + 7$ for $ \displaystyle x\in R$ . Express $ \displaystyle f(x)$ in the form $ \displaystyle a(x - b)^2 - c$ and state the valueof $ \displaystyle a, b$ and $ \displaystyle c$. Another function g is defined by $ \displaystyle g:x\mapsto 2x + k$ for $ \displaystyle x\in R$. Find the value of $ \displaystyle k$ for which $ \displaystyle (g\cdot f)(x) = 0$ has two equal roots.

Show Solution

$ \displaystyle \begin{array}{l}\ \ \ f(x)=2{{x}^{2}}-12x+7\\\\\therefore f(x)=2({{x}^{2}}-6x+9)-11\\\\\therefore f(x)=2{{(x-3)}^{2}}-11\\\\\ \ \ f(x)=a{{(x-b)}^{2}}-c\ \text{(given)}\\\\\therefore a{{(x-b)}^{2}}-c=2{{(x-3)}^{2}}-11\\\\\therefore a=2,\ b=3\ \text{and }c=11\\\\\ \ \ g(x)=2x+k\\\\\ \ \ \left( {g\cdot f} \right)(x)=0\\\\\therefore g\left( {f(x)} \right)=0\\\\\therefore g\left( {2{{x}^{2}}-12x+7} \right)=0\\\\\therefore 2\left( {2{{x}^{2}}-12x+7} \right)+k=0\\\\\therefore 4{{x}^{2}}-24x+14+k=0\\\\\ \text{Since}\left( {g\cdot f} \right)(x)=0\ \text{has two equal roots,}\\\\{{(-24)}^{2}}-4(4)(14+k)=0\\\\\therefore k=22\end{array}$

7.      Let $ \displaystyle f:x\mapsto a + bx, a, b\in R$, be a function from R into R such that $ \displaystyle f(2b)=b$ and $ \displaystyle (f\cdot f)(b) = ab$. If $ \displaystyle f$ is not a constant function, then find the formula for $ \displaystyle f$.

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$ \displaystyle \begin{array}{l}\ \ \ f(x)=a+bx\\\\\ \ \ f(2b)=b\\\\\ \ \ a+b(2b)=b\\\\\therefore a=b-2{{b}^{2}}\\\\\ \ \ \left( {f\cdot f} \right)(b)=ab\\\\\ \ \ f\left( {f(b)} \right)=ab\\\\\ \ \ f\left( {a+{{b}^{2}}} \right)=ab\\\\\ \ \ a+b(a+{{b}^{2}})=ab\\\\\therefore b-2{{b}^{2}}+b(b-2{{b}^{2}}+{{b}^{2}})=(b-2{{b}^{2}})b\\\\\therefore b-2{{b}^{2}}+{{b}^{2}}-{{b}^{3}}={{b}^{2}}-2{{b}^{3}}\\\\\therefore {{b}^{3}}-2{{b}^{2}}+b=0\\\\\therefore b({{b}^{2}}-2b+1)=0\\\\\therefore b{{(b-1)}^{2}}=0\\\\\therefore b=0\ \text{(or)}\ b=1\\\\\text{Since }f(x)\ \text{is not a constant function,}\\b=0\ \text{is impossible}\text{.}\\\\\therefore b=1\\\\\therefore a=(1)-2{{(1)}^{2}}=-1\\\\\therefore f(x)=x-1\end{array}$

8.      $ \displaystyle f:R\to R, g:R\to R$, and $ \displaystyle h:R\to R$ are functions defined by $ \displaystyle f(x) = x^2 + 2, g(x) = x - 1$ and $ \displaystyle h(x) = 3x - 2$. Find the formulae of $ \displaystyle f\cdot g$ and $ \displaystyle f\cdot (h\cdot g)$.

Show Solution

$ \displaystyle \begin{array}{l}f:R\to R,f(x)={{x}^{2}}+2\\\\g:R\to R,g(x)=x-1\\\\h:R\to R,h(x)=3x-2\\\\\left( {f\cdot g} \right)(x)=f\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(x-1)}^{2}}+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-2x+3\\\\\left( {h\cdot g} \right)(x)=h\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3(x-1)-2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-5\\\\\left( {f\cdot \left( {h\cdot g} \right)} \right)(x)=f\left( {\left( {h\cdot g} \right)(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {3x-5} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(3x-5)}^{2}}+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9{{x}^{2}}-30x+27\end{array}$

9.      The functions $ \displaystyle f$ and $ \displaystyle g$ are defined for all values of $ \displaystyle x$ as follows: $ \displaystyle f:x\mapsto x^2 - 1$ and $ \displaystyle g:x\mapsto (x - 1)^2$.

(i) If $ \displaystyle 4f(x)+3=f(kx)$, find the values of $ \displaystyle k$.

(ii) Express $ \displaystyle g(2x + 1)$ in terms of $ \displaystyle f(x)$.

(iii) Find a function $ \displaystyle h$ such that $ \displaystyle f(x) = g(x) + 2h(x)$.

Show Solution

$ \displaystyle \begin{array}{*{20}{l}} {f(x)={{x}^{2}}-1,\ g(x)={{{(x-1)}}^{2}}} \\ {} \\ {\text{(i)}\ \ 4f(x)+3=f(kx)} \\ {} \\ {\ \ \ \ \ 4({{x}^{2}}-1)+3={{k}^{2}}{{x}^{2}}-1} \\ {} \\ {\ \ \ \ \ 4{{x}^{2}}-1={{k}^{2}}{{x}^{2}}-1} \\ {} \\ {\ \ \ \ \ \therefore {{k}^{2}}=4\Rightarrow k=\pm 2} \\ {} \\ {\text{(ii)}\ g(x)={{{(x-1)}}^{2}}} \\ {} \\ {\ \ \ \ \ g(2x+1)={{{(2x+1-1)}}^{2}}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}-4+4} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4({{x}^{2}}-1)+4} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left[ {({{x}^{2}}-1)+1} \right]} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left[ {f(x)+1} \right]} \\ {} \\ {\text{(ii)}\ f(x)=g(x)+2h(x)} \\ {} \\ {\ \ \ \ \ {{x}^{2}}-1={{{(x-1)}}^{2}}+2h(x)} \\ {} \\ {\ \ \ \ \ 2h(x)={{x}^{2}}-1-{{{(x-1)}}^{2}}} \\ {} \\ \begin{array}{l}\ \ \ \ \ 2h(x)={{x}^{2}}-1-{{x}^{2}}+2x-1\\\\\ \ \ \ \ \ \ 2h(x)=2x-2\\\\\ \ \ \ \therefore h(x)=x-1\end{array} \end{array}$

Functions : Basic Problems and Solutions

1. Let the function$ \displaystyle f:R\to R$ be given by $ \displaystyle f (x) = ax^2 + bx$, If $ \displaystyle f (-1) = 7$ and $ \displaystyle f (2) = -2$, find the values of $ \displaystyle x$ for which $ \displaystyle f(x) = x$.

Show Solution

$\displaystyle \begin{array}{l}\ \ \ f:R\to R\\\\\ \ \ f(x)=a{{x}^{2}}+bx\\\\\ \ \ f(-1)=7\\\\\therefore a{{(-1)}^{2}}+b(-1)=7\\\\\ \ \ a-b=7\ \ \ \ ------(1)\\\\f(2)=-2\\\\\therefore a{{(2)}^{2}}+b(2)=-2\\\\\ \ \ 4a+2b=-2\\\\\ \ \ 2a+b=-1------(2)\\\\(1)+(2)\Rightarrow 3a=6\Rightarrow a=2\\\\\therefore 2-b=7\ \Rightarrow b=-5\\\\\therefore f(x)=2{{x}^{2}}-5x\\\\\ \ \ f(x)=x\ \ \ \text{(given)}\\\\\therefore 2{{x}^{2}}-5x=x\\\\\ \ \ 2{{x}^{2}}-4x=0\\\\\ \ \ x(x-2)=0\\\\\therefore x=0\ \text{or}\ x=2\text{ }\end{array}$

2. A function $\displaystyle f$ from $ \displaystyle A$ to $ \displaystyle A$, where $ \displaystyle A$ is the set of positive integers, is given by $ \displaystyle f(x)=$ the sum of all possible divisors of $ \displaystyle x$. Find the value of $ \displaystyle k$, if $ \displaystyle f(15) = 3k + 6$.

Show Solution

$ \displaystyle \begin{array}{l}\ \ \ A=\{x|x\ \text{is apositive integer }\!\!\}\!\!\text{ }\\\\\ \ \ \ f:A\to A\\\\\ \ \ f(x)=\text{the sum of all possible divisors of }x.\\\\\therefore \ f(15)=\text{the sum of all possible divisors of }15\\\\\ \ \ \ \ \ \ \ \ \ \ \ =1+3+5+15\\\\\ \ \ \ \ \ \ \ \ \ \ \ =24\\\\\ \ \ f(15)=3k+6\ \text{(given)}\\\\\therefore 3k+6=24\\\\\therefore k=6\end{array}$

3. Function $ \displaystyle f:R\to R$ be given by $ \displaystyle f(x) = x^2 - 6$. Find the possible values of $ \displaystyle x$ for which $ \displaystyle f(x)$ is unchanged by the mapping.

Show Solution

$ \displaystyle \begin{array}{l}f:R\to R\\\\f(x)={{x}^{2}}-6\\\\\text{When }f(x)\text{ is unchanged by mapping,}\\\\f(x)=x\\\\{{x}^{2}}-6=x\\\\{{x}^{2}}-x-6=0\\\\\therefore (x+2)(x-3)\\\\\therefore x=-2\ \text{or}\ x=3\end{array}$

4. Function $ \displaystyle f:R\to R$ be given by $ \displaystyle f(x) = x^2 - 3x + 2$. Show that $ \displaystyle f(x + 2) = x^2 + x$.

Show Solution

$ \displaystyle \begin{array}{l}f:R\to R\\\\f(x)={{x}^{2}}-3x+2\\\\f(x+2)={{(x+2)}^{2}}-3(x+2)+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+4x+4-3x-6+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+x\end{array}$

5. Given that $ \displaystyle f:x\mapsto \frac{2}{{ax+b}},x\ne -\frac{b}{a}$ such that $ \displaystyle f(0) = -2$ and $ \displaystyle f(2) = 2$, find the value of $ \displaystyle a$ and of $ \displaystyle b$. Show that $ \displaystyle f(p) + f(– p) = 2 f(p^2)$.

Show Solution

$ \displaystyle f(x)=\frac{2}{{ax+b}},x\ne -\frac{b}{a}$

$ \displaystyle f(0)=-2$

$ \displaystyle \frac{2}{{a(0)+b}}=-2$

$ \displaystyle \frac{2}{b}=-2\Rightarrow b=-1$

$ \displaystyle f(2)=2$

$ \displaystyle \frac{2}{{a(2)-1}}=2$

$ \displaystyle 2a-1=1\Rightarrow a=1$

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

$ \displaystyle \therefore f(p)+f(-p)=\frac{2}{{p-1}}+\frac{2}{{-p-1}}$

$ \displaystyle \therefore f(p)+f(-p)=\frac{2}{{p-1}}-\frac{2}{{p+1}}$

$ \displaystyle \therefore f(p)+f(-p)=\frac{{2(p+1)-2(p-1)}}{{{{p}^{2}}-1}}$

$ \displaystyle \therefore f(p)+f(-p)=\frac{4}{{{{p}^{2}}-1}}$ $ \displaystyle 2f({{p}^{2}})=2\left( {\frac{2}{{{{p}^{2}}-1}}} \right)=\frac{4}{{{{p}^{2}}-1}}$ $ \displaystyle \therefore f(p)+f(-p)=2f({{p}^{2}})$

6. Given that $ \displaystyle f(x)=\frac{{ax-b}}{x}$, for all real values of $ \displaystyle x$ except $ \displaystyle x = 0$. If $ \displaystyle f(1) = -1$ and $ \displaystyle f(2) = 1$, find the value of $ \displaystyle a$ and of $ \displaystyle b$ and hence find the image of $ \displaystyle – 4$ under $ \displaystyle f$.

Show Solution

$ \displaystyle f(x)=\frac{{ax-b}}{x},x\ne 0$

$ \displaystyle f(1)=-1$

$ \displaystyle \frac{{a(1)-b}}{1}=-1$

$ \displaystyle a-b=-1\ \ \ \ \ -----(1)$

$ \displaystyle f(2)=1$

$ \displaystyle \frac{{a(2)-b}}{2}=1$

$ \displaystyle 2a-b=2\ \ \ \ \ -----(2)$

$ \displaystyle \text{Equation}\ (2)-\text{Equation}\ (1)\Rightarrow a=3$

$ \displaystyle \therefore 3-b=-1\Rightarrow b=4$

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

$ \displaystyle \therefore f(-4)=\frac{{3(-4)-4}}{{(-4)}}=4$

7. If $ \displaystyle f(x)=\frac{{b(x-a)}}{{b-a}}+\frac{{a(x-b)}}{{a-b}}$, show that $ \displaystyle f(a+b) = f(a) + f(b)$.

Show Solution

$ \displaystyle f(x)=\frac{{b(x-a)}}{{b-a}}+\frac{{a(x-b)}}{{a-b}},a\ne b$

$ \displaystyle f(a+b)=\frac{{b(a+b-a)}}{{b-a}}+\frac{{a(a+b-b)}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{b}^{2}}}}{{b-a}}+\frac{{{{a}^{2}}}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{a}^{2}}}}{{a-b}}-\frac{{{{b}^{2}}}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{a}^{2}}-{{b}^{2}}}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{(a-b)(a+b)}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =a+b$

$ \displaystyle f(a)+f(b)=\frac{{b(a-a)}}{{b-a}}+\frac{{a(a-b)}}{{a-b}}+\frac{{b(b-a)}}{{b-a}}+\frac{{a(b-b)}}{{a-b}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0+a+b+0$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b$

$ \displaystyle \therefore f(a+b)=f(a)+f(b)$

8. A function $ \displaystyle f$ is defined by $ \displaystyle f(2x + 1) = x^2 - 3$. Find $ \displaystyle a\in R$ such that $ \displaystyle f(5) = a^2 - 8$.

Show Solution

$ \displaystyle \begin{array}{l}f(2x+1)={{x}^{2}}-3\\\\\text{Let}\ 2x+1=5,\text{then }x=2\\\\\therefore f(5)={{2}^{2}}-3=1\\\\f(5)={{a}^{2}}-8\ \text{(given)}\\\\\therefore {{a}^{2}}-8=1\\\\\therefore {{a}^{2}}=9\Rightarrow a=\pm 3\end{array}$

9. The function $ \displaystyle f$ is defined by $ \displaystyle f(x) = 7^x$ . Prove that $ \displaystyle f(x + 2) - 10 f(x + 1) + 21 f(x) = 0$.

Show Solution

$ \displaystyle \begin{array}{l}f(x)={{7}^{x}}\\\\\therefore f(x+2)-10f(x+1)+21f(x)\\\\={{7}^{{x+2}}}-10({{7}^{{x+1}}})+21({{7}^{x}})\\\\={{7}^{2}}\cdot {{7}^{x}}-10(7\cdot {{7}^{x}})+21({{7}^{x}})\\\\={{7}^{x}}(49-70+21)\\\\={{7}^{x}}(0)\\\\=0\end{array}$

Ways to describe functions

Function မ်ားကို ေဖၚျပႏိုင္ေသာ နည္းလမ္းမ်ား

ဒီတစ္ခါေတာ့ function ေတြကို ေဖၚျပႏိုင္တဲ့ နည္းလမ္းေတြကို ေျပာျပပါမယ္။

ဆိုၾကပါစို႔။ function f ဟာ A = {-3, -2, -1, 0, 1, 2, 3} နဲ႔ B = {-9, -6, -3, 0, 3, 6, 9} ကို f(x)=3x နဲ႔ ဆက္သြယ္ထားတဲ့ function တစ္ခု ျဖစ္ပါတယ္။ ဒီ function ရဲ့ ဆက္သြယ္ခ်က္ေတြ သိဖို႔ function ထဲမွာ အစားသြင္းၾကည့္ရေအာင္။

f(x) = 3x

f(-3) = 3(-3) = -9

f(-2) = 3(-2) = -6

f(-1) = 3(-1) = -1

f(0) = 3(0) = 0

f(1) = 3(1) = 3

f(2) = 3(2) = 6

f(3) = 3(3) = 9 ဆိုတဲ့ ဆက္သြယ္ခ်က္ေတြ ရတာေပါ့။

အဲဒီဆက္သြယ္ခ်က္ကို အလြယ္တကူ ထင္ရွား ျမင္သာေအာင္ ေဖၚျပႏိုင္တဲ့ နည္းလမ္းေတြ အေၾကာင္း ဆက္ရွင္းျပ ပါမယ္။

1. A verbal statement

အထက္က function ရဲ့ ဆက္သြယ္ခ်က္ကို ၾကည့္မယ္ဆိုရင္ x↦3x ျဖစ္တာေၾကာင့္ မူလတန္ဖိုးဟာ image ရဲ့ သံုးပံု တစ္ပံု ျဖစ္တယ္ဆိုတာကို သိၾကမွာ ျဖစ္ပါတယ္။ ဒါကို verbal statement နဲ႔ ေဖၚျပမယ္ဆိုရင္

A function from A to B : “is one-third of " လို႔ ေျပာရမွာေပါ့။ အထက္က ဆက္သြယ္ခ်က္ကို verbal statement နဲ႔ ျပန္ခ်ေရးမယ္ ဆိုရင္-

-3 is one-third of -9
-2 is one-third of -6
-1 is one-third of -3
0 is one-third of 0
1 is one-third of 3
2 is one-third of 6
3 is one-third of 9 လို႔ ေျပာလို႔ရပါတယ္။

2. Arrow diagram

ဒီေဖၚျပပံု စနစ္ကေတာ့ ထင္ရွားျမင္သာၿပီး ရိုးရွင္းပါတယ္။ ဒါေပမယ့္ အကန္႔အသတ္နဲ႔သာ ရွိၿပီး တိတိက်က် ေဖၚျပႏိုင္တဲ့ အစု၀င္ေတြ ပါတဲ့ domain နဲ႔ codomain တို႔အတြက္သာ သင့္ေလ်ာ္ပါတယ္။ အခုလို ေဖၚျပပါတယ္။

3.A set of ordered pairs

Function တစ္ခုကို အစုနဲ႔ ေဖၚျပျခင္း ျဖစ္ပါတယ္။ Domain ထဲက အစု၀င္ (elements of domain) ေတြကို independent variables လို႔ေခၚၿပီး image ေတြကို dependent variables လို႔ ေခၚပါတယ္။ ordered pair တစ္ခုကို ေရးတဲ့အခါ (independent variables, dependent variables) လို႔ ေရးရပါတယ္။

အထက္မွာ ေဖၚျပခဲ့တဲ့ function ကို ျပန္ၾကည့္ရင္ -

(-3, -9), (-2, -6), (-1, -3), (0, 0), (1, 3), (2, 6), (3, 9) ဆိုတဲ့ ordered pairs ေတြကို ေတြ႔ရမွာ ျဖစ္ပါတယ္။ အစုဆိုတာက အစု၀င္ေတြကို တြန္႔ကြင္း { } ထဲမွာ ထည့္ေရးရတယ္ဆိုတာကို သိၿပီးျဖစ္မွာပါ။ ဒါ့ေၾကာင့္ ေျပာခဲ့တဲ့ function ကို set of ordered pairs နဲ႔ ေဖၚျပမယ္ဆိုရင္ -

A function from A to B = {(-3, -9), (-2, -6), (-1, -3), (0, 0), (1, 3), (2, 6), (3, 9)} လို႔ ေဖၚျပေပးရမွာ ျဖစ္ပါတယ္။

4. Table form

Function ေတြကို ေဖၚျပတဲ့ နည္းလမ္းေတြထဲက အသံုးမ်ားၿပီး အသံုး၀င္တဲ့ စနစ္တစ္ခုေပါ့။ စာရင္းဇယားဆိုတာ လုပ္ေဆာင္ခ်က္နဲ႔ ရလဒ္ေတြကို ႏႈိုင္းယွဥ္ၾကည့္တဲ့ စနစ္တစ္ခုပဲ မဟုတ္လား။ အထက္က function ကို table နဲ႔ ေဖၚျပမယ္ဆိုရင္ အခုလိုေဖၚျပရမွာ ျဖစ္ပါတယ္။

x	-3	-2	-1	0	1	2	3
3x	-9	-6	-3	0	3	6	9

5. Graph

အသံုးအမ်ားဆံုး နည္းလမ္းတစ္ခုပါပဲ။ လုပ္ငန္းတစ္ခုရဲ့ အတက္အက် လုပ္ေဆာင္ခ်က္တစ္ခုရဲ့ အသြင္သဏၭာန္၊ အေျခအေနကို အလြယ္တကူ ခန္႔မွန္းႏိုင္ဖို႔ graph ေတြဆြဲၿပီး ၾကည့္သလိုေပါ့။ လုပ္ငန္းေဆာင္ရြက္ခ်က္ေတြ ဆိုတာ တကယ္ေတာ့ function ေတြပါပဲ။

Graph ဆိုတာ ေရျပင္ညီ (x-axis) နဲ႔ ေဒါင္လိုက္ (y-axis) ၀င္ရိုးႏွစ္ခုနဲ႔ ဖြဲ႔စည္းထားတဲ့ ျပင္ညီ (Cartesian plain) ေပၚမွာ ဆက္သြယ္ခ်က္ရဲ့ တည္ေနရာေတြကို သတ္မွတ္ေပးလိုက္တာပါ။

Graph ဆြဲတဲ့ အခါ သတိထားရမွာက elements of domain (independent variables) ေတြက္ို x ၀င္ရိုးမွာ ထားရပါတယ္၊ elements of codomain (dependent variables) ေတြကိုေတာ့ y ၀င္ရိုးမွာ ထားပါတယ္။ Domain နဲ႔ Codomain ထဲမွာပါတဲ့ elements ေတြကို ၀င္ရိုးေတြေပၚမွာ အညီအမွ် အပိုင္းအျခား (same interval) သတ္မွတ္ ေပးရပါတယ္။ ဆက္သြယ္ခ်က္အတိုင္း ေနရာ (location) ကို သတ္မွတ္ေပးရပါတယ္။ အထက္က ေျပာခဲ့တဲ့ function ကို graph နဲ႔ ေဖၚျပမယ္ဆိုရင္ အခုလိုရရွိမွာ ျဖစ္ပါတယ္။ ဒီေနရာမွာ y=3x ျဖစ္ပါတယ္။

အခုေဖၚျပတဲ့ graph မွာ function ရဲ့ result ဟာ အမွတ္စက္ (point) ေတြ အျဖစ္သာ ရွိေနမွာပါ။ curve မဟုတ္ ပါဘူး။ ဘာေၾကာင့္လဲ ဆိုရင္ေတာ့ domain ထဲမွာရွိတဲ့ elements ၇ခုအတြက္ပဲ သတ္မွတ္ရတာ ျဖစ္လို႔ပါပဲ။

အကယ္၍ domain နဲ႔ codomain ဟာ A နဲ႔ B မဟုတ္ပဲ R (set of real numbers) ျဖစ္တယ္ဆိုပါစို႔။ ဆိုလိုတာက -

Function f:R⇢R and f(x)=3x ေပါ့။

ဒါဆိုရင္ေတာ့ elements of domain ဟာ x ၀င္ရိုးေပၚမွာ ရွိတဲ့ အမွတ္တိုင္းကို ဆိုလိုတာ ျဖစ္ၿပီး၊ elements of codomain ဟာ y ၀င္ရိုးေပၚမွာရွိတဲ့ အမွတ္တိုင္းကို ဆိုလိုတာ ျဖစ္ပါတယ္။ အမွတ္တစ္ခုနဲ႔ တစ္ခုၾကားမွာ ေနရာလပ္ (interval or gap) ဆိုတာ မရွိေတာ့ဘူး။ ဒါေၾကာင့္ ရလာတဲ့ result ဟာ curve ျဖစ္လာတာေပါ့။ ေအာက္က graph ကို ၾကည့္ပါ။

f:R⇢R and y=f(x)=3x ရဲ့ graph ပါ။

ေနာက္ထပ္ဥပမာတစ္ခုၾကည့္ရေအာင္…

A = {-3, -2, -1, 0, 1, 2, 3}

B = {x|-10 ≤ x ≤ 10, x is an integer.}

f : A⇢B, y=f(x) = x² အတြက္ graph ဆြဲမယ္ဆိုရင္ ေအာက္က ပံုအတိုင္းရမွာ ျဖစ္ပါတယ္။

f:R⇢R, y=f(x) = x² အတြက္ graph ဆြဲမယ္ဆိုရင္ေတာ့ အခုလို curve ပံုသဏၭာန္ ရမွာျဖစ္တယ္။

ဒါဆိုရင္ function ေတြကို ေဖၚျပပံုနည္းလမ္းမ်ားနဲ႔ graph ရဲ့ သေဘာသဘာ၀ကို နားလည္းႏိုင္ၿပီလို႔ ထင္ပါတယ္။

Composition of Functions

Composition of Functions (Function မ်ားကို ေပါင္းစပ္ျခင္း)

ေရွ႕မွာတုန္းက function ဆိုတာကို ရွင္းျပခဲ့ၿပီးပါၿပီ။ အခု Function ေတြကို ေပါင္းစပ္ျခင္း အေၾကာင္း ဆက္လက္ရွင္းျပပါမယ္။ Function ဆိုတာ စက္ကိရိယာ တစ္ခုလိုပဲလို႔ ေရွ႕မွာ ဥပမာျပခဲ့ပါတယ္။ အဲဒီ စက္ေတြကို တစ္ခါတစ္ရံ လိုအပ္သလို ေပါင္းစပ္ၿပီး သံုးရပါတယ္။

ျမင္သာတဲ့ ဥပမာတစ္ခုနဲ႔ ရွင္ျပပါ့မယ္။ အခုမီးေတြ ပ်က္တဲ့အခါ အိမ္မွာ မီးစက္ေမာင္းရတာေတြ ႀကံဳဘူးမွာေပါ့။ မီးစက္ေမာင္းဖို႔ ဘာေတြလိုအပ္ပါသလဲ။ မီးစက္လိုတာေပါ့လို႔ အလြယ္ေျပာၾကမယ္။ သိပ္ဟုတ္တာေပါ့။ မီးစက္ေမာင္းဖို႔ဆိုတာ မီးစက္ရွိမွ ျဖစ္မယ္ေလ။ ေကာင္းၿပီ။ မီးစက္ရွိတယ္ဆိုပါစို႔။ အဲဒီမီးစက္ကေန လိုခ်င္တာက လွ်ပ္စစ္စြမ္းအင္ (output) ေပါ့။ သခ်ာၤစကားနဲ႔ ေျပာရရင္ image ေပါ့။

ဒါကို မီးစက္က ထုတ္ေပးမယ္။ အဲဒီလိုထြက္လာေအာင္ မီးစက္ကို စက္သံုးဆီ ထည့္ေပးရတယ္ မဟုတ္လား။ တကယ္ေတာ့ မီးစက္လို႔ အလြယ္ေျပာတဲ့ကိရိယာဟာ စက္ပစၥည္း ႏွစ္ခုေပါင္းစပ္ ထားတာပါ။ ဘာေတြလဲဆိုရင္ လွ်ပ္စစ္စြမ္းအင္ ထုတ္ေပးမယ့္ dynamo နဲ႔ သူ႕ကို လည္ပတ္ေအာင္ ေမာင္းႏွင္ေပးမယ့္ engine ပါ။ ဒီေတာ့ သခ်ၤာသေဘာအရ ေျပာရမယ္ဆိုရင္ function ႏွစ္ခု ေပါင္းစပ္ထားတာေပါ့။

ပမာဏႀကီးမားတဲ့ ထုတ္လုပ္မႈေတြမွာေတာ့ engine အစား ေရေႏြးေငြ႕နဲ႔ လည္ပတ္ေစတဲ့ turbine ႀကီးေတြကို သံုးတာေပါ့။ သူကေန ဒိုင္နမို အႀကီးစား generator ႀကီးေတြကို လည္ပတ္ေစပါတယ္။

ထုတ္လုပ္မႈ စတင္ေတာ့မယ္ ဆိုရင္ ဒိုင္နမိုကို ေမာင္းေပမယ့္ အင္ဂ်င္ကို အရင္ဆံုး ေမာင္းေပးရမယ္။ အင္ဂ်င္ကို ေမာင္းႏွင္ေပးဖို႔ အင္ဂ်င္ကို လည္ပတ္ေစမယ့္ စက္သံုးဆီ ထည့္ေပးရပါတယ္။ ဒီေနရမွာ စက္သံုးဆီဆိုတာ element of domain လုိ႔ သခ်ၤာစကားအရ ေျပာႏိုင္ပါတယ္။

turbine ႀကီးေတြဆိုရင္ေတာ့ စက္သံုးဆီ အစား စရိတ္သက္သာတဲ့ ေရေႏြးေငြ႕တို႔ ေရအားဒလက္ေတြတို႔ သဘာ၀ဓာတ္ေငြ႕တြန္းအားတို႔ ဆိုတာေတြကို သံုးေလံရွိပါတယ္။ ဒီအခါမွာ အင္ဂ်င္ (သို႔) တာဘိုင္ႀကီးေတြ လည္ပတ္လာတယ္။ ဘာထြက္လာပါသလဲ။ စက္စြမ္းအင္ ထြက္လာတာေပါ့။ ဒါဟာသခ်ာၤအျမင္အရ တာဘိုင္က ထုတ္ေပးလိုက္တဲ့ image လို႔ ေျပာရမွာ ျဖစ္ပါတယ္။

ထြက္လာတဲ့ စက္စြမ္းအင္က ဒိုင္နမို (သို႔) generator ကို လည္ပတ္ေစၿပီး လွ်ပ္စစ္စြမ္းအင္ ထုတ္ေပးတာပါ။ လွ်ပ္စ္စြမ္းအင္ဆိုတာ ဒီလုပ္ေဆာင္ခ်က္ရဲ့ final image လို႕ဆိုရမယ္။ ေအာက္ကပံုကို ၾကည့္ပါ။ ဒါဟာ function ေတြကို ေပါင္းစပ္ အသံုးျပဳထားျခင္းပါပဲ။

ီ
လုပ္ေဆာင္ခ်က္ အဆင့္ဆင့္ကို word diagram ေလးနဲ႔ ေဖၚျပၾကည့္ရေအာင္။ ဒီလိုေတြ႕ရမွာေပါ့။

function ေတြကို ေပါင္းစပ္ျခင္းဆိုတာလည္း ဒီအတိုင္းပါပဲ။ ဆိုပါစို႔။ f ဟာ A နဲ႔ B ကို ဆက္သြယ္ထားၿပီး ရလာတဲ့ image က f(x) ျဖစ္တယ္။ ဒီ image ကို B နဲ႔ C ကို ဆက္သြယ္ေပးတဲ့ function g ထဲကို ထည့္ေပး လိုက္မယ္ဆိုရင္ ထြက္လာတဲ့ image ကို g(f(x)) လို႔ ေခၚပါတယ္။ ဒါဟာ composition of function ပါပဲ။

ဒီလုပ္ဆာင္ခ်က္ကို တစ္ဆက္တည္း ေခၚမယ္ဆိုရင္ေတာ့ g . f (g circle f) လို႔ ေခၚပါတယ္။ ဒါေၾကာင့္ ဆက္သြယ္ခ်က္ကို အခုလို (g.f)(x) = g(f(x)) ေျပာႏိုင္ပါတယ္။ ေအာက္ကေပးထားတဲ့ arrow diagram ေလးကို ေလ့လာၾကည့္ပါ။ ဒါဆိုရင္ composition of function ဆိုတာကို နားလည္ႏိုင္ၿပီလို႔ ထင္ပါတယ္။

Equality of Functions

Equality of Functions

Two functions f and g are equal if and only if

1. f and g have the same domain,
2. f and g have the same codomain, and
3. f(x) = g(x) for each x of the domain.

Function မ်ားတူညီျခင္း

Function ႏွစ္ခု f နဲ႔ g ဆိုပါစို႔။ f=g လို႔ ေျပာႏိုင္ဖို႔ အတြက္ အေျခအေန သံုးရပ္နဲ႔ ကိုက္ညီမႈ ရွိရပါမယ္။

(၁) Domain တူညီရမယ္။

(၂) Codomain တူညီရမယ္။

(၃) Domain ထဲမွာ ရွိတဲ့ အစု၀င္ x တိုင္းအတြက္ f(x) = g(x) ျဖစ္ရမယ္။ တစ္နည္းေျပာရင္ function f ရဲ့ image ေတြနဲ႔ function g ရဲ့ image ေတြ တူညီရမယ္ေပါ့။

ဒီအေျခအေန သံုးရပ္လံုးနဲ႔ ကိုက္ညီတယ္ဆိုရင္ function ႏွစ္ခု တူညီတယ္လို႔ ေျပာႏိုင္ပါတယ္။

ေအာက္က example ေလးေတြ ဆက္ၿပီး ေလ့လာၾကည္ရေအာင္။

Example (1)

Solution

f(x) = x² g(x) = 2x – 1

f(1) = 1² = 1 g(1) = 2(1) – 1 = 1.

Therefore f(x) = g(x) for every x A.

f = g

အထက္မွာ ေျပာခဲ့တဲ့ အခ်က္သံုးခ်က္နဲ႔ ျပန္စစ္ၾကည့္ရေအာင္ . . .

1. f နဲ႔ g ဟာ Domain တူပါတယ္။ ႏွစ္ခုလံုးရဲ့ Domain ဟာ A ျဖစ္တယ္။
2. f နဲ႔ g ဟာ Codomain လည္းတူပါတယ္။ ႏွစ္ခုလံုးအတြက္ B ျဖစ္တယ္။
3. Domain ထဲမွာ အစု၀င္ တစ္ခုရွိပါတယ္။ 1 ျဖစ္ပါတယ္။ f(1) = 1 ျဖစ္ၿပီးေတာ့ g(1) = 1 ျဖစ္တဲ့အတြက္ image ေတြလည္းတူပါတယ္။ ဒါေၾကာင့္ f = g ျဖစ္တယ္လို႔ ေျပာႏိုင္ပါတယ္။
   

Example (2)

Solution

f(x) = x² g(x) = 2x – 1

f(1) = 1² = 1 g(1) = 2(1) – 1 = 1.

f(2) = 2² = 4 g(2) = 2(2) – 1 = 3.

f(1) = g(1) but f(2) ≠ g(2).

Therefore f(x) ≠ g(x) for every x C.

f ≠ g

ဒီဥပမာမွာ ဆိုရင္ function ႏွစ္ခုလံုးအတြက္ Domain နဲ႔ Codomain တူညီတယ္ဆိုတာ သိၿပီး ျဖစ္မွာပါ။

f(1) = g(1)

f(2) ≠ g(2) ျဖစ္ေနပါတယ္။ ဒါေၾကာင့္ Domain ထဲမွာ ရွိတဲ့ အစု၀င္ x တိုင္းအတြက္ f(x) = g(x) မျဖစ္ေတာ့ပါဘူး။

ဒါေၾကာင့္ ဒီေမးခြန္းမွာ f ≠ g လို႔ ဆိုရမွာ ျဖစ္ပါတယ္။

Example (3)

Let f : R R and g : R R are functions such that f(0) = 2 and g(0) = 2. Can you say that f and g are the same function? Why?

Solution

Here f(0) = g(0). But we cannot say that f(x) = g(x) for every x R. Therefore we cannot say that f and g are the same function.

ဒီေမးခြန္းကေတာ့ set of real numbers (R) နဲ႔ equality of functions ဆိုတာကို အေသအခ်ာ နားလည္ သေဘာေပါက္မႈ ရွိရဲ့လားဆိုတာကို စစ္ေဆးလိုက္တာပါ။

R ဆိုတာက set of real numbers (ကိန္းစစ္မ်ားပါ၀င္ေသာ အစု) ေပါ့။ ကိန္းမ်ဥ္းေပၚမွာ ရွိတဲ့ အမွတ္တိုင္းကို ကိုယ္စားျပဳပါတယ္။

f နဲ႔ g ဟာ Domain နဲ႔ Codomain တူပါတယ္။ f(0) = g(0) ျဖစ္ပါတယ္။ ဒါေပမယ့္ f = g လို႔ မသတ္မွတ္ႏိုင္ပါဘူး။ R ဆိုတဲ့ အစုထဲမွာ အစု၀င္ေတြ မေရမတြက္ႏိုင္ေအာင္ ရွိပါတယ္။ အဲဒီအထဲမွာမွ 0 ရဲ့ image မ်ားသာလွ်င္ တူညီတယ္လို႔ ေပးထားခ်က္အရ သိရၿပီး၊ က်န္တဲ့ အစု၀င္ေတြအတြက္ ဘာမွ မေျပာႏိုင္ပါဘူး။ ဒါေၾကာင့္ f နဲ႔ g ဟာ တူညီေသာ function မ်ားျဖစ္တယ္လို႔ ဘယ္လိုမွ ေျပာခြင့္မရွိပါဘူး။

ဒီေလာက္ဆိုရင္ Equality of Functions ဆိုတာကို သေဘာေပါက္ေလာက္ၿပီ ထင္ပါတယ္။

Chapter(1) - Functions

Functions

Function ရဲ့ မူလအဓိပၸါယ္ သတ္မွတ္ခ်က္ကေတာ့ အစု (set) ႏွစ္ခုက္ ဆက္သြယ္ေပးတဲ့ နည္းလမ္းလို႔ ဆိုႏိုင္ ပါတယ္။ အစုႏွစ္ခုဆိုတာက -

(၁) Function တစ္ခုရဲ့ ေဆာင္ရြက္မႈေအာက္မွာ ပါ၀င္ၾကမယ့္ အစု၀င္ေတြ ပါ၀င္တဲ့ မူလအစု (Domain) နဲ႔

(၂) Function ရဲ့ ေဆာင္ရြက္ၿပီးေျမာက္သြားတဲ့ အစု၀င္ေတြပါ၀င္မယ့္ (Codomain) တို႔ ျဖစ္ပါတယ္။

Function ဆိုတာကို ထုတ္လုပ္မႈေတြ လုပ္ေပးႏိုင္တဲ့ စက္ကိရိယာတစ္ခု အသြင္ ယူဆၾကည့္ရေအာင္။ ထုတ္လုပ္မႈ ဆိုကတည္းက ကုန္ၾကမ္းေတြရွိရမယ္၊ မဟုတ္လား။ အဲဒီကုန္ၾကမ္းေတြ စုထားတဲ့ အစုက domain ေပါ့။

ကုန္ၾကမ္းေတြကို စက္ထဲထည့္လိုက္ၿပီ။ တစ္ဖက္မွာ ကုန္ေခ်ာေတြ ထြက္လာၿပီေပါ့။ ကုန္ေခ်ာေတြကို အစုတစ္ခု အေနနဲ႔ စုလိုက္မယ္။ ဒါဟာ Codomain ေပါ့။ ဒီလို ျမင္ၾကည့္ႏိုင္ပါတယ္။

ဒီအတိုင္းပါပဲ။ Function ဆိုတာကို အခုလို ေဖၚျပလို႔ရတာေပါ့။

Definition : A function from a set A to a set B relates each element of A to exactly one element of B.

အစု A (Domain) ထဲမွာ ရွိတဲ့ အစု၀င္ တစ္ခုခ်င္းစီတိုင္း အတြက္ အစု B (Codomain) ထဲမွာ ဆက္သြယ္ထားတဲ့ အစု၀င္ (အတိအက်) တစ္ခုထဲပဲ ရွိရပါမယ္။

ဆိုလိုတာက A ထဲမွာ ရွိေသာ x တိုင္းအတြက္ B ထဲမွာ y ရွိရပါမယ္။ အကယ္၍ x ဟာ B ထဲမွာ y အျပင္ z နဲ႔ပါ ဆက္သြယ္မႈ ရွိေနတယ္ ဆိုရင္ေတာ့ ဒါ function မျဖစ္ေတာ့ပါဘူး။ ေအာက္ပါပံုကို ၾကည့္ပါ။

Domain ထဲမွာ ရွိတဲ့ အစု၀င္ တစ္ခု အတြက္ Codomain ထဲမွာ ဆက္သြယ္ခ်က္ တစ္ခုထက္ ပိုသြားရင္ Function လို႔ သတ္မွတ္လို႔ မရေတာ့ဘူး။

မွတ္ရမွာက-

Domain ထဲမွာ ရွိတဲ့ အစု၀င္တစ္ခု ခ်င္းစီတိုင္အတြက္ Codomain ထဲမွာ ဆက္စပ္အစု၀င္ အတိအက် တစ္ခုသာ ရွိရမယ္။ ပိုလို႔လည္း မရဘူး။ လံုး၀မရွိလို႔လည္း မရဘူး။ ေအာက္က ဥပမာပံုေလးေတြကို ၾကည့္ရေအာင္။

ဒါေတြဟာ function ရဲ့ အဓိပၸါယ္ သတ္မွတ္ခ်က္နဲ႔ ကိုက္ညီမႈမရွိတဲ့ ဆက္သြယ္ခ်က္ေတြ ျဖစ္လို႔ function လို႔ မသတ္မွတ္ႏိုင္ပါဘူး။

ကဲ ျပန္ဆက္ရေအာင္ …

f ဆိုတာ A နဲ႔ B ကို ဆက္စပ္ေပးတဲ့ function တစ္ခု ဆိုပါစို႔။ သေကၤတအားျဖင့္ အခုလို ေရးပါတယ္။

f is a function from A to B.

f ဟာ A ထဲမွာ ရွိတဲ့ x ကို B ထဲမွာရွိတဲ့ y နဲ႔ ဆက္စပ္ေပးတယ္ ဆိုတာကိုေတာ့ ဒီလိုေရးပါတယ္။

f maps x to y (or) y is the image of x under f.

ဒီေနရမွာ y ကိုေတာ့ f က ဆက္စပ္ေပးတဲ့ x ရဲ့ image လို႔ေခၚပါတယ္။

Functional Notation

ေရွ႕မွာ ေျပာခဲ့တဲ့ အတိုင္း ေျပာမယ္ဆိုရင္ f ဟာ X နဲ႔ Y ကို ဆက္သြယ္ထားတဲ့ function တစ္ခု ျဖစ္တယ္။

x ရဲ့ image ဟာ 2 ျဖစ္တယ္။

y ရဲ့ image ဟာ 2 ျဖစ္တယ္။

z ရဲ့ image ဟာ 3 ျဖစ္တယ္။ ဆိုတာကိုေတာ့ သေကၤတနဲ႔ ဒီလိုေရးတယ္ ဆိုတာ ေျပာခဲ့ၿပီပါၿပီ။ ျပန္ၾကည့္ရေအာင္။

အခုလိုုေရးတဲ့ စနစ္ဟာ ေနာင္မွာ function ေတြကို အႀကိမ္ႀကိမ္ ေရးဖို႔ လိုလာတာနဲ႔အမွ် အဆင္မေျပေတာ့ပါဘူး။ ဒါေၾကာင့္ ပိုၿပီးေတာ့ ေရးရတာ အဆင္ေျပေစတဲ့ functional notation ကို ေျပာင္းသံုးပါတယ္။

f(x) = 2 is f of x is 2.

f(y) = 2 is f of y is 2.

f(z) = 2 is f of z is 3.

ဒီလိုေရးတာကို functional notation လို႔ ေခၚပါတယ္။

အထက္မွာ ဥပမာ ျပခဲ့တဲ့ function ကို ၾကည့္မယ္ဆိုရင္ Codomain ထဲမွာ 1, 2, 3, 4 ဆိုတဲ့ အစု၀င္ ေလးခုရွိတာ ေတြ႕ရမွာပါ။ ဒီအစု၀င္ေတြ အားလံုးကို image လို႔ မသတ္မွတ္ ႏိုင္ပါဘူး။ 2 နဲ႔ 3 သာလွ်င္ ဆက္စပ္မႈရွိလို႔ image လို႔ ေျပာႏိုင္ပါတယ္။

ဒါေၾကာင့္ Codomain ကို အစုပိုင္း (sub sets) ႏွစ္ခု ထပ္မံပိုင္းျခား ႏိုင္ပါတယ္။ image မ်ားပါ၀င္တဲ့အစု Z နဲ႔ image မဟုတ္ေသာ အစု၀င္မ်ားရဲ့ အစု (Y \ Z) တို႔ပဲေပါ့။ ဒီေနရာမွာ image ေတြသာ ပါတဲ့အစု (Z) ကိုေတာ့ function f ရဲ့ Range လို႔ ေခၚပါတယ္။ Range ဆိုတာဘာလဲ။ အခုလိုသတ္မွတ္ႏိုင္ပါတယ္။

Range = Set of images = { Images }

ေျပာခဲ့တာေတြ ျပန္ၿပီး အက်ဥ္းခ်ဳပ္ရရင္

X = { x, y, z } = Domain , Y = { 1, 2, 3, 4} = Codomain, Z = { 2, 3} = Range

f(x) = 2 is f of x is 2.

f(y) = 2 is f of y is 2.

f(z) = 2 is f of z is 3.