A) \[{{x}^{2}}+x-1=0\]
B) \[{{x}^{2}}+x+1=0\]
C) \[{{x}^{2}}-x-1=0\]
D) None of these
Correct Answer: C
Solution :
\[\alpha ,\beta \] are the roots of the equation \[{{x}^{2}}-3x+1=0\] \[\therefore \alpha +\beta =3\] and \[\alpha \beta =1\] \[S=\frac{1}{\alpha -2}+\frac{1}{\beta -2}=\frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4}\] \[=\frac{3-4}{1-2.3+4}=1\] and \[P=\frac{1}{(\alpha -2)(\beta -2)}=\frac{1}{\alpha \beta -2(\alpha +\beta )+4}=-1\] Hence the equation whose roots are \[\frac{1}{\alpha -2}\] and \[\frac{1}{\beta -2}\] are \[{{x}^{2}}-Sx+P=0\Rightarrow {{x}^{2}}-x-1=0\].
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