| The mean of x and 1/x is N. Then, the mean of \[{{x}^{2}}\]and \[1/{{x}^{2}}\] is [SSC (CGL) Pre 2015] |
A) \[{{N}^{2}}-2\]
B) \[2\,{{N}^{2}}-2\]
C) \[4\,{{N}^{2}}-2\]
D) \[{{N}^{2}}\]
Correct Answer: B
Solution :
| According to the question, |
| \[\frac{1}{2}\left( x+\frac{1}{x} \right)=N\]\[\Rightarrow \]\[\left( x+\frac{1}{x} \right)=2N\] |
| \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=4{{N}^{2}}\] [squaring both sides] |
| \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=4{{N}^{2}}-2=2\,(2{{N}^{2}}-1)\] |
| \[\Rightarrow \] \[\frac{1}{2}\,\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=(2{{N}^{2}}-1)\] |
| [divide both sides by 2] |
| Hence, mean of \[{{x}^{2}}\] and \[\frac{1}{{{x}^{2}}}\] is \[(2{{N}^{2}}-1).\] |
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