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  • If , then the value of <img alt="x-x^{2}+x[x]" src="https://entra

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 \ldotsx+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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