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

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(A) Given the equation $-3(x - [x])^2 + 2(x - [x]) + a^2 = 0$.Let $t = x - [x] = \{x\}$,where $0 \leq t < 1$.The equation becomes $-3t^2 + 2t + a^2 =

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

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(A) Given the equation $-3(x - [x])^2 + 2(x - [x]) + a^2 = 0$.Let $t = x - [x] = \{x\}$,where $0 \leq t 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 $t = \{x\}$,we must have $0 \leq t Case $1$: $t = \frac{1 + \sqrt{1 + 3a^2}}{3}$. For $t Case $2$: $t = \frac{1 - \sqrt{1 + 3a^2}}{3}$. Since $\sqrt{1 + 3a^2} \geq 1$,$t \leq 0$. For $t$ to be a valid fractional part,$t$ must be $0$ (which corresponds to $x$ being an integer). If $t=0$,then $a^2=0$,so $a=0$. If $a=0$,$t=0$ or $t=2/3$. $t=0$ gives integral solutions. To have no integral solutions,we exclude $a=0$.Thus,$a \in (-1, 0) \cup (0, 1)$.

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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