A) 196
B) 197
C) 198
D) 199
Correct Answer: B
Solution :
[b] Let \[{{(\sqrt{2}+1)}^{6}}=k+f,\] where k is integral part and f the fraction \[(0\le f<1)\] Let \[{{(\sqrt{2}-1)}^{6}}=f',(0<f'<1),\] Since \[0<(\sqrt{2}-1)<1\] Now, \[k+f+f'={{(\sqrt{2}+1)}^{6}}+{{(\sqrt{2}-1)}^{6}}\] \[\therefore \,\,\,\,f+f'=198-k\] = an integer But \[0\le f<1\] and \[0<f'<1\Rightarrow 0<(f+f')<2\] \[\Rightarrow f+f'=1,(\because f+f'\,\,is\,\,an\,\,\operatorname{int}eger)\] \[\therefore \,\,\,\,By(i),I=198-(f+f')=198-1=197\]
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