A) \[2,\,\,2\pm \sqrt{2}\]
B) \[2\pm \sqrt{3}\,,\,\,3\]
C) \[3\pm \sqrt{2}\,,\,\,2\]
D) None of these
Correct Answer: A
Solution :
Since \[(7+4\sqrt{3})(7-4\sqrt{3})=\,\,1\] \[\therefore \] The given equation becomes \[\,y+\frac{1}{y}=14\] where \[y={{(7-4\sqrt{3})}^{{{x}^{2}}-4x+3}}\] \[\Rightarrow \,\,\,{{y}^{2}}-14y+1=0\Rightarrow \,\,y=7\pm 4\sqrt{3}\] Now \[y=\,\,7+\,\,4\sqrt{3}\,\,\Rightarrow {{x}^{2}}-4x+3=-1\,\,\Rightarrow \,\,x=2\,\,,\,\,2\] Also \[y=\,\,7-\,4\sqrt{3}\,\,\Rightarrow {{x}^{2}}-4x+3=1\,\,\Rightarrow \,\,x=2\pm \,\,,\,\,\sqrt{2}\]
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