# If both roots of the equation x^2-2ax+a^2-1=0 lie between -3 and 4 , then [a] cannot be ( where [.] denote the greatest integer function)

Solution:  The equation is  $x^{2}-2ax+a^{2}-1=0$

$\\ \\ \Rightarrow \hspace{1cm} (x-a)^{2}=1\\ \\ \Rightarrow \hspace{1cm}x=a\pm1\Rightarrow a-1< a+1\\ \\ \Rightarrow \hspace{1cm}a-1> -3 \hspace{0.5cm}and \hspace{0.5cm}a+1< 4\\ \\ \Rightarrow \hspace{1cm}a> -2 \hspace{0.5cm},a< 3\\ \\ \Rightarrow \hspace{1cm}-2< a< 3\therefore [a]=-2,-1,0,1,2$

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