Matematické Fórum

stuart clark · 05. 06. 2014 05:59

The positive values of $a$ for which the equation $\lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution is,

where $\lfloor x \rfloor =$ floor function of $x$

Options

$(a)\;\;: (0,1)$

$(b)\;\;: (1,2)$

$(c)\;\;: (0,2)$

$(d)$ None of these

sugyman · 15. 06. 2014 12:57

For $x=-1, a=\frac{1}{2}$ we get $\lfloor x+a \rfloor =\lfloor\frac{-1}{2} \rfloor=-1$ and $\sin \left(\frac{\pi x}{2}\right)=\sin \left(\frac{-\pi}{2}\right)=-1$ . So it's not $(a)$ or $(c)$ .
Let's make assumption now that $1<a<2$ and multiply it $-1>a>-2$ . Since $\lfloor x+a \rfloor$ is a whole number, then $\sin \left(\frac{\pi x}{2}\right)$ is either $1,-1$ or $0$ .
If $\sin \left(\frac{\pi x}{2}\right)=1 \Rightarrow x=4k+1, k\in \mathbb{Z}$ . Also $1\leq x+a<2 \Rightarrow 0 \leq 4k+a<1$ . Using our assumption $-2<-a<-1$ and adding it to $0 \leq 4k+a<1$ , we get:
$-2<4k<0$ , which is contradiction because there is no whole $k$ satisfying this inequality.
Similary with other cases we get condtradictions.
The answer is $(b)$ .

Matematické Fórum

#1 05. 06. 2014 05:59

floor function equation

#2 15. 06. 2014 12:57

Re: floor function equation

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