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If $x=(2+\sqrt{3})^{n}$ , then the value of $x-x^{2}+x[x]$ , where [.] denotes the greatest integer function.
Option: 1
2

Option: 2
3

Option: 3
0

Option: 4
1

Answers (1)

best_answer

$x-x^{2}+x[x]=x-x(x-[x])=x-x\{x\}$

$=x(1-\{x\})$

$x=(2+\sqrt{3})^{n}={ }^{n} C_{0} 2^{n}+{ }^{n} C_{1} 2^{n-1} \sqrt{3}+{ }^{n} C_{2} 2^{n-2}(\sqrt{3})^{2}+\ldots \ldots.$

Let $x_{1}=(2-\sqrt{3})^{n}$

$={ }^{\mathrm{n}} \mathrm{C}_{0} 2^{\mathrm{n}}-{ }^{\mathrm{n}} \mathrm{C}_{1} 2^{\mathrm{n}-1} \sqrt{3}+{ }^{\mathrm{n}} \mathrm{C}_{2} 2^{\mathrm{n}-2}(\sqrt{3})^{2}+\ldots \ldots$ $x+x_{1}=2\left({ }^{n} C_{0} 2^{n}+{ }^{n} C_{2} 2^{n-2}(\sqrt{3})^{2}+\ldots \ldots\right)$ = even integer

So $0<x_{1}<1$ for all $n \in N$

$[x]+\{x\}+x_{1}=$ even integer

$\Rightarrow\{x\}+x_{1}=$ integer

$0<\{x\}+x_{1}<2$

Hence $\{x\}+x_{1}=1$

$x_{1}=1-\{x\}$

Hence $x(1-\{x\})=x \cdot x_{1}=(2+\sqrt{3})^{n}(2-\sqrt{3})^{n}=1$ .

Posted by

Sanket Gandhi

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