If a in R and the equation - 3{left( {x - le | vedclass

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Question

Here, $a \in R$ and equation is

$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$

$	ext { Let } \quad t=x-[x]$

$	herefore-3 t^{2}+2 t+a^{2}=0$

$\Righ

Vedclass · Accepted Answer

$\left( { - 1,0} \right) \cup \left( {0,1} \right)$

Vedclass · Answer

Here, $a \in R$ and equation is

$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$

$	ext { Let } \quad t=x-[x]$

$	herefore-3 t^{2}+2 t+a^{2}=0$

$\Rightarrow t=\frac{1 \pm \sqrt{1+3 a^{2}}}{3}$

$\because \quad t=x-[x]=\{x\} \quad[	ext { fractional part }]$

$	herefore \quad 0 \leq t \leq 1$

$\Rightarrow 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3} \leq 1$

Taking positive sign

$\because \quad[\{x\}>0]$

$	herefore 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3}<1$

$\Rightarrow \sqrt{1+3 a^{2}}<2$

$\Rightarrow \quad 1+3 a^{2}<4$

$\Rightarrow a^{2}-1<0$

$\Rightarrow \quad(a+1)(a-1)<0$

For no integral solution of a we consider the interval $(-1,0) \cup(0,1)$

If $a \in R$ and the equation $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ (where $[x]$ denotes the greatest integer $\leq\,x$)has no integral solution ,then all possible values of $a$ lie in the interval

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