If a in R and the equation -3(x - [x])^2 + 2( | vedclass

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(A) Let $t = x - [x] = \{x\}$. Since $0 \leq \{x\} < 1$,we have $0 \leq t < 1$.The equation becomes $-3t^2 + 2t + a^2 = 0$,or $3t^2 - 2t - a^2 = 0$.So

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$(-1, 0) \cup (0, 1)$

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(A) Let $t = x - [x] = \{x\}$. Since $0 \leq \{x\} The equation becomes $-3t^2 + 2t + a^2 = 0$,or $3t^2 - 2t - a^2 = 0$.Solving for $t$ using the quadratic formula: $t = \frac{2 \pm \sqrt{4 - 4(3)(-a^2)}}{2(3)} = \frac{2 \pm \sqrt{4 + 12a^2}}{6} = \frac{1 \pm \sqrt{1 + 3a^2}}{3}$.Since $t \geq 0$,we must take the positive root: $t = \frac{1 + \sqrt{1 + 3a^2}}{3}$.For the equation to have no integral solution,$t$ must not be $0$ (because if $t=0$,then $x$ is an integer,which would be a solution). Thus,$t > 0$.Also,we require $t $\frac{1 + \sqrt{1 + 3a^2}}{3} Squaring both sides: $1 + 3a^2 This implies $-1  0$,we check $a=0$: if $a=0$,$t = 1/3$,which is valid. However,if $a=0$,the equation is $-3t^2 + 2t = 0 \Rightarrow t(2-3t)=0$,so $t=0$ or $t=2/3$. $t=0$ gives an integral solution. Thus $a 
eq 0$.Therefore,$a \in (-1, 0) \cup (0, 1)$.

If $a \in R$ and the equation $-3(x - [x])^2 + 2(x - [x]) + a^2 = 0$ (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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