Answer

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**Hint:**Here we are given an equation and we are required to find if the greatest integer value of \[x\] is even or odd and we also have to find the value of product of \[x\] and factorial part of \[x\]. We solve it by subtracting the argument of \[x\] from\[x\]. We do this as the argument is less than one. We solve further to get the desired results.

Formula used: We have used the following to solve this question

\[

{\left( {x + y} \right)^n} = {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + .... + {}^n{C_n}{x^0}{y^n} \\

{\left( {x + y} \right)^n} = - {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} - .... + {\left( { - 1} \right)^n}{}^n{C_n}{x^0}{y^n} \\

\]

And

\[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\]

**Complete step-by-step solution:**

We are given the fractional part of \[x\] as,

\[\left\{ x \right\} = x - \left[ x \right]\]

From it we obtain \[x\] as,

\[x = \left\{ x \right\} + \left[ x \right] \]

Now we see that

\[

5\sqrt 3 - 8 = 5 \times 1.73 - 8 \\

\Rightarrow 5\sqrt 3 - 8 = 0.65 \\

\Rightarrow 5\sqrt 3 - 8 < 1 \\

\]

So we consider \[{\left( {5\sqrt 3 - 8} \right)^{2n + 1}}\] as \[\left\{ {{x'}} \right\}\]

So, we subtract \[\left\{ {{x'}} \right\}\] from \[x\] as,

\[\left\{ x \right\} + \left[ x \right] - \left\{ {{x'}} \right\} = {\left( {5\sqrt 3 + 8} \right)^{2n + 1}} - {\left( {5\sqrt 3 - 8} \right)^{2n + 1}}\]

We know that,

\[

{\left( {x + y} \right)^n} = {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + .... + {}^n{C_n}{x^0}{y^n} \\

{\left( {x - y} \right)^n} = - {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} - .... + {\left( { - 1} \right)^n}{}^n{C_n}{x^0}{y^n} \]

Using these formulas we get,

\[ \Rightarrow \left\{ x \right\} + \left[ x \right] - \left\{ {{x'}} \right\} =

\left[ {{}^{2n + 1}{C_1}5{{\sqrt 3 }^{2n}}{8^1} + {}^{2n + 1}{C_2}5{{\sqrt 3 }^{2n - 1}}{8^2} + {}^{2n + 1}{C_3}5{{\sqrt 3 }^{2n - 2}}{8^3} + .....} \right] - \left[ { - {}^{2n + 1}{C_1}5{{\sqrt 3 }^{2n}}{8^1} + {}^{2n + 1}{C_2}5{{\sqrt 3 }^{2n - 1}}{8^2} - {}^{2n + 1}{C_3}5{{\sqrt 3 }^{2n - 2}}{8^3} + .....} \right]\]

On solving it further,

\[ \Rightarrow \left\{ x \right\} + \left[ x \right] - \left\{ {{x'}} \right\} = 2\left[ {{}^{2n + 1}{C_1}5{{\sqrt 3 }^{2n}}{8^1} + {}^{2n + 1}{C_3}5{{\sqrt 3 }^{2n - 2}}{8^3} + ...} \right]\]

This shows that \[\left\{ x \right\} + \left[ x \right] - \left\{ {{x'}} \right\}\] is an even integer.

This means that \[\left\{ x \right\} - \left\{ {{x'}} \right\}\] must also be an integer,

As \[0 < \left\{ x \right\} < 1\] and \[0 < \left\{ {{x'}} \right\} < 1\], this means that \[\left\{ x \right\} - \left\{ {{x'}} \right\} = 0\].

Which means that \[\left[ x \right]\] is an even integer

Now since \[\left\{ x \right\} - \left\{ {{x'}} \right\} = 0\], this means that

\[\left\{ x \right\} = \left\{ {{x'}} \right\}\]

\[

\Rightarrow x\left\{ x \right\} = x\left\{ {{x'}} \right\} \\

\Rightarrow x\left\{ x \right\} = {\left( {5\sqrt 3 + 8} \right)^{2n + 1}}^{} \cdot {\left( {5\sqrt 3 - 8} \right)^{2n + 1}} \\

\Rightarrow x\left\{ x \right\} = {\left[ {\left( {5\sqrt 3 + 8} \right)\left( {5\sqrt 3 - 8} \right)} \right]^{2n + 1}} \\

\Rightarrow x\left\{ x \right\} = {\left[ {{{(5\sqrt 3 )}^2} - {8^2}} \right]^{2n + 1}} \\

\Rightarrow x\left\{ x \right\} = {\left[ {25 \times 3 - 64} \right]^{2n + 1}} \\

\Rightarrow x\left\{ x \right\} = {\left[ {75 - 64} \right]^{2n + 1}} \\

\Rightarrow x\left\{ x \right\} = {\left[ {11} \right]^{2n + 1}} \]

**Thus the answer of this question comes out to be options A) and C).**

**Note:**This is to note that we have used the argument of the given function here as the base of the argument is also less than one. We should know that any number is formed from two parts, first part is the fraction part and second part is the integer part.

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