A) \[f+\frac{1}{f}\]
B) \[f-\frac{1}{f}\]
C) \[\frac{1}{f}-f\]
D) None
Correct Answer: C
Solution :
[c] Let \[G={{(\sqrt{2}-1)}^{2n+1}}\] ??....(i) 0<G<1 \[R-G={{\left( \sqrt{2}+1 \right)}^{2n+1}}-{{\left( \sqrt{2}-1 \right)}^{2n+1}}\] \[f+\left[ R \right]-G=\]An even Integer \[f-G=0\Rightarrow f=G(0<f<1)\] \[\frac{1}{f}=\frac{1}{G}=\frac{1}{{{(\sqrt{2}-1)}^{2n+1}}}={{(\sqrt{2}+1)}^{2n+1}}=R\] \[\frac{1}{f}=R\Rightarrow \frac{1}{f}=[R]+f\] \[[R]=\frac{1}{f}-f\]
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