GAZETA MATEMATICĂ Seria B N0 2/1981

           Faceti click pe imagine pentru descarcare.

GAZETA MATEMATICĂ Seria B N0 5/2022

           Faceti click pe imagine pentru descarcare.

SUPLIMENTUL cu EXERCIȚII al GMB N0 5/2022

           Faceti click pe imagine pentru descarcare.

Infimum and supremum of $\;\;\frac{a+b}{\sqrt{a^2+b^2}}$

         Following Infimum and supremum we will prove

$$1=\underset{a>0,b>0}{inf} \;\frac{a+b}{\sqrt{a^2+b^2}}<\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \underset{a>0,b>0}{sup }\;\frac{a+b}{\sqrt{a^2+b^2}}=\frac{a+b}{\sqrt{a^2+b^2}}\mid \underset{(a,b)=(\delta,\delta){\mid}\delta>0}{}=\sqrt{2}$$

           For the inequality on the right we have, for $a>0,\;b>0$

$(a-b)^2 \geqslant 0\;\Rightarrow\;2ab \leqslant a^2+b^2\;\Rightarrow \;a^2+b^2+2ab\leqslant 2(a^2+b^2)$

$\;\Rightarrow\; (a+b)^2\leqslant 2\cdot (a^2+b^2)\;\Rightarrow\;a+b \leqslant \sqrt{2}\cdot \sqrt{a^2+b^2}\;\Rightarrow\;\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \sqrt{2};$

the $=$ sign occurs when $a=b$.

            For the inequality on the left we have, for $a>0,b>0$

$0<2ab\;\Rightarrow \;a^2+b^2<a^2+b^2+2ab\;\Rightarrow a^2+b^2<(a+b)^2\;\Rightarrow\;$  

$\Rightarrow\;\sqrt{a^2+b^2}<a+b\;\Rightarrow\;1<\frac{a+b}{\sqrt{a^2+b^2}}.$ The value $1$ of the fraction $\frac{a+b}{\sqrt{a^2+b^2}}$ is never reached, but whatever $\varepsilon >0$ is, the fraction takes for certain $a_{\varepsilon}$ and $b_{\varepsilon}$ the value $1+\varepsilon$.

     Indeed, let $a_{\varepsilon}=1+(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2}$ and $b=2\varepsilon+\varepsilon^2$. We have 

$a_{\varepsilon}+b_{\varepsilon}=1+2\varepsilon+\varepsilon^2+(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2})=(1+\varepsilon)(1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2})$

and

$a_{\varepsilon}^2+b_{\varepsilon}^2=1+(1+\varepsilon)^2(1-2\varepsilon-\varepsilon^2)+2(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2}+(2\varepsilon+\varepsilon^2)^2=$

$\underset{\alpha=2\varepsilon+\varepsilon^2}{=}1+(1+\alpha)(1-\alpha)+2(1+\varepsilon)\sqrt{1-\alpha}+\alpha^2=2+2(1+\varepsilon)(\sqrt{1-\alpha}=$

$=1+\alpha+2(1+\varepsilon)\sqrt{1-\alpha}+1-\alpha \;\;\underset{1+\alpha=(1+\varepsilon)^2}{=}\;(1+\varepsilon)^2+2(1+\varepsilon)\sqrt{1-\alpha}+(1-\alpha)=$

$=(1+\varepsilon+\sqrt{1-\alpha})^2.$

 So, $\sqrt{a_{\varepsilon}^2+b_{\varepsilon}^2}=1+\varepsilon+\sqrt{1-\alpha}=1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}.$

     We finally get $\frac{a_{\varepsilon}+b_{\varepsilon}}{\sqrt{a_{\varepsilon}^2+b_{\varepsilon}^2}}=\frac{(1+\varepsilon)(1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}}{1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}}=1+\varepsilon$.

$\blacksquare \blacksquare$

Три математичка задатка из Додатка вежби Математичког гласника серије Б

           In the magazine here, on page 9, problems proposed for the 8th grade

           In translation, thanks to DeepL :

             

             PROBLEM S:E22.152

          " Determine the functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that check the relation: for any real numbers $x$ and $y$, $f(x+2y-1)+f(x-y)=2x+y.$"

          ANSWER CiP 

$$f(x)=x+\frac{1}{2}.$$

          Solution CiP

          Let's note with

$$u=x+2y-1,\;\;\;v=x-y. \tag{1}$$

Adding the two equations (1) we get $2x+y-1=u+v\;\;\Rightarrow \;2x+y=u+v+1.$

     The given relationship is written

$$f(u)+f(v)=u+v+1\;(\forall)u,v \in \mathbb{R}\;\Leftrightarrow$$

$$\Leftrightarrow\;f(u)-u=v+1-f(v)\;(\forall)u,v \in \mathbb{R}. \tag{2}$$

In the equation (2), the left member has only the variable $u$ and the right member only the variable $v$. This is only possible if both members are constant. So there is $k \in \mathbb{R}$, so that

$$f(u)-u=k=v+1-f(v)\;(\forall) u,v \in \mathbb{R}.$$

          Hence $f(u)=u+k,\;f(v)=v+1-k.$ But we still have $f(v)=v+k$(if we transcribe the first relation with the letter $v$ instead of $u$), so we must have the equation

$$v+k=v+1-k,\;\;(\forall) v \in \mathbb{R}.$$

 It follows from the above $k=1-k\;\Leftrightarrow \;k+k=1\;\Leftrightarrow\;k=\frac{1}{2}.$

          We get the answer, because the function $f(x)=x+\frac{1}{2}$ check the given relationship.

$\blacksquare$

 

               PROBLEM S:E22.153

              "Show that $\frac{a+b}{\sqrt{a^2+b^2}}\in  (\frac{\sqrt{2}-1}{\sqrt{2}}, \sqrt{2}\;] $, for any positive real numbers $a$ and $b$."

           ANSWER CiP

           More precise inequality occurs:

$$1-\frac{1}{\sqrt{2}}<1<\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \sqrt{2}\;\;\;\;(\forall) a,b >0.$$

          Solution CiP

           See here for inequality $1<\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \sqrt{2}.$ Because $\frac{\sqrt{2}-1}{\sqrt{2}}=1-\frac{1}{\sqrt{2}}<1$, then we have the inclusion of intervals

$$( 1, \sqrt{2}] \subset (\frac{\sqrt{2}-1}{\sqrt{2}}, \sqrt{2}\;]$$

and the problem is solved.

$\blacksquare\;\blacksquare$

 

           PROBLEM S:E22.154

        "Give an example of a non-constant function $f:\mathbb{R}\rightarrow \mathbb{R}$, satisfying the relation $f(x+y)+f(x-y)=2f(f(x)+f(y))+4$, for any real numbers $x$ and $y$.

 

           ANSWER CiP

 

           Solution CiP

 

 

$\blacksquare\;\blacksquare\;\blacksquare$

SUPLIMENTUL cu EXERCIȚII al GMB N0 4/2022

  Faceti click pe imagine pentru descarcare.

Vezi ERATA

 

GAZETA MATEMATICĂ Seria B N0 4/2022

 Faceti click pe imagine pentru descarcare.

 Vezi ERATA

Vezi ERATA