If in a $\Delta$ le $ABC,A=\large\frac{\pi}{4}$ and $\tan B\tan C=\alpha$ then the value of $\alpha$ lies in

$\begin{array}{1 1}(a)\;(-\infty,3-2\sqrt 2)\\(b)\;(3-2\sqrt 2,3+2\sqrt 2)\\(c)\;(-\infty,3-2\sqrt 2)\cup (3+2\sqrt 2,\infty)\\(d)\;None\;of\;the\;above\end{array}$

As $\angle A=\large\frac{\pi}{4}$
$B+C=\large\frac{3\pi}{4}$
$\tan(B+C)=\tan\large\frac{3\pi}{4}$
$\Rightarrow \large\frac{\tan B+\tan C}{1-\tan B\tan C}$$=-1 \Rightarrow \tan B+\large\frac{\alpha}{\tan B}$$=-1(1-\alpha)$
$\Rightarrow \tan^2B+(1-\alpha)\tan B+\alpha=0$
Since $B$ is real.
$\therefore (1-\alpha)^2-4\alpha\geq 0$
$\Rightarrow \alpha^2-6\alpha+1\geq 0$
$\Rightarrow \{\alpha-(3-2\sqrt 2)\}\{\alpha-(3+2\sqrt 2)\}\geq 0$
$\Rightarrow \alpha\in (-\infty,3-2\sqrt 2)\cup [3+2\sqrt 2,\infty]$
Hence (c) is the correct answer.