A) 1
B) 2
C) \[{{2}^{2n}}\]
D) \[{{2}^{n}}\]
Correct Answer: A
Solution :
| [a] \[x={{\left( 2+\sqrt{3} \right)}^{n}}\] |
| \[={{\,}^{n}}{{C}_{0}}{{2}^{n}}+{{\,}^{n}}{{C}_{1}}{{2}^{n-1}}\sqrt{3}+{{\,}^{n}}{{C}_{2}}{{2}^{n-1}}{{\left( \sqrt{3} \right)}^{2}}+...\] |
| Let \[{{x}_{1}}={{\left( 2-\sqrt{3} \right)}^{n}}\] |
| \[={{\,}^{n}}{{C}_{0}}{{2}^{n}}-{{\,}^{n}}{{C}_{1}}{{2}^{n-1}}\sqrt{3}+{{\,}^{n}}{{C}_{2}}{{2}^{n-2}}{{\left( \sqrt{3} \right)}^{2}}+...\] |
| \[x+{{x}_{1}}=2\left( ^{n}{{C}_{0}}{{2}^{n}}+{{\,}^{n}}{{C}_{2}}{{2}^{n-2}}{{\left( \sqrt{3} \right)}^{2}}+... \right)\] |
| = Even integer. |
| Clearly, \[x,\in (0,1)\forall n\in N\] |
| \[\Rightarrow [x]+\{x\}+{{x}_{1}}=\] Even integer |
| \[\Rightarrow \{x\}+{{x}_{1}}=\,\,Integer\,\,\{x\}\in (0,\,\,1),{{x}_{1}}\in (0,\,\,1)\] |
| \[\Rightarrow \{x\}+{{x}_{1}}\in (0,2)\] |
| \[\Rightarrow \{x\}+{{x}_{1}}=1\Rightarrow {{x}_{1}}=1-\{x\}\] |
| \[\Rightarrow x(1-\{x\})=x.{{x}_{1}}={{\left( 2+\sqrt{3} \right)}^{n}}{{\left( 2-\sqrt{3} \right)}^{n}}=1\] |
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