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
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