A) \[-2+\sqrt{2}\]
B) \[4-3\sqrt{2}\]
C) \[4-2\sqrt{3}\]
D) \[2-\sqrt{3}\]
Correct Answer: B
Solution :
Here, \[3{{m}^{2}}{{x}^{2}}+m(m-4)x+2=0\] Let \[\alpha ,\beta \] be the roots of the given equation \[\therefore \]\[\lambda =\frac{\alpha }{\beta },\alpha +\beta =\frac{m(4-m)}{3{{m}^{2}}}=\frac{4-m}{3m},\alpha \beta =\frac{2}{3{{m}^{2}}}\] Given,\[\lambda +\frac{1}{\lambda }=1\] \[\Rightarrow \]\[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=1\Rightarrow \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=1\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=\alpha \beta \] Now, \[{{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =3\alpha \beta \] \[\Rightarrow \]\[\frac{{{(4-m)}^{2}}}{9{{m}^{2}}}=\frac{2}{{{m}^{2}}}\Rightarrow {{(4-m)}^{2}}=18\] \[\Rightarrow \]\[4-m=\pm 3\sqrt{2}\Rightarrow m=4\pm 3\sqrt{2}\] \[\therefore \]Least value of\[m=4-3\sqrt{2}\]
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