A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
| \[{{x}^{2}}-3x+1=0\] |
| \[\therefore \] \[x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\] \[\left[ \begin{align} & \text{If}\,\,\text{a}{{x}^{2}}+bx+2c=0,\,\,\text{then} \\ & x=\frac{-\,\,b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ \end{align} \right]\] |
| Let \[x=\frac{3+\sqrt{5}}{2}\] |
| \[\therefore \]\[\frac{1}{x}=\frac{2}{3+\sqrt{5}}=\frac{2\,\,(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{2\,\,(3-\sqrt{5})}{9-5}\] \[=\frac{3-\sqrt{5}}{2}\] |
| \[\therefore \]\[x+\frac{1}{x}=\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}=\frac{3+\sqrt{5}+3-\sqrt{5}}{2}\] \[=\frac{6}{2}=3\] |
| When, \[x=\frac{3-\sqrt{5}}{2},\frac{1}{x}=\frac{3+\sqrt{5}}{2}\] |
| \[\therefore \] \[x+\frac{1}{x}=\frac{3-\sqrt{5}}{2}+\frac{3+\sqrt{5}}{2}=3\] |
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