A) \[(-1,0)\cup (0,1)\]
B) \[(-2,-1)\]
C) \[(-\infty ,-2)\cup (2,\infty )\]
Correct Answer: A
Solution :
\[-3{{\left( x-[x] \right)}^{2}}+2\left( x-[x] \right)+{{a}^{2}}=0\] \[-3{{\{x\}}^{2}}+2\{x\}+{{a}^{2}}=0\] ?.(1) \[\therefore \]\[\{x\}=\frac{-2\pm \sqrt{4+12{{a}^{2}}}}{-6}=\frac{1\overline{+}\sqrt{1+3{{a}^{2}}}}{3}\] As\[0\le \{x\}<1\] \[0\le \frac{1+\sqrt{1+3{{a}^{2}}}}{3}<1\] \[-1\le \sqrt{1+3{{a}^{2}}}<2\] \[1+3{{a}^{2}}<4\] \[3({{a}^{2}}-1)<0\] \[a\in (-1,1)\] but it a = 0then equation (1)has integral solution So, \[a\ne 0\] \[a\in (-1,0)\cup (0,1)\] (b)\[(1,2)\]
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