A) \[{{M}^{2}}\]
B) \[{{M}^{2}}\text{/}4\]
C) \[2{{M}^{2}}-1\]
D) \[2{{M}^{2}}+1\]
Correct Answer: C
Solution :
Mean of\[x\]and\[\frac{1}{x},\]\[M=\frac{x+\frac{1}{x}}{2}\] \[\Rightarrow x+\frac{1}{x}=2M\] We have \[\Rightarrow {{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2x\left( \frac{1}{x} \right)\] \[={{(2M)}^{2}}-2=4{{M}^{2}}-2=2(2{{M}^{2}}-1)\] Now, the mean of\[{{x}^{2}}\]and \[\frac{1}{{{x}^{2}}}=\frac{{{x}^{2}}+\frac{1}{{{x}^{2}}}}{2}\] \[=\frac{2(2{{M}^{2}}-1)}{2}=2{{M}^{2}}-1\]
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