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Question: Let\[x = {\left( {5\sqrt 3 + 8} \right)^{2n + 1}},n \in N\], where \[\left[ x \right]\] is greatest ...

Letx=(53+8)2n+1,nNx = {\left( {5\sqrt 3 + 8} \right)^{2n + 1}},n \in N, where [x]\left[ x \right] is greatest integer less than or equal to xx and \left\\{ x \right\\} = x - \left[ x \right], then
A) [x]\left[ x \right] is even
B) [x]\left[ x \right] is odd
C) x\left\\{ x \right\\} = {\left( {11} \right)^{2n + 1}}
D) x\left\\{ x \right\\} = {\left( {13} \right)^{2n + 1}}

Explanation

Solution

Here we are given an equation and we are required to find if the greatest integer value of xx is even or odd and we also have to find the value of product of xx and factorial part of xx. We solve it by subtracting the argument of xx fromxx. 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

(x+y)n=nC1xn1y1+nC2xn2y2+....+nCnx0yn (x+y)n=nC1xn1y1+nC2xn2y2....+(1)nnCnx0yn  {\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
(a2b2)=(a+b)(ab)\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 xx as,
\left\\{ x \right\\} = x - \left[ x \right]
From it we obtain xx as,
x = \left\\{ x \right\\} + \left[ x \right]
Now we see that

538=5×1.738 538=0.65 538<1  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 (538)2n+1{\left( {5\sqrt 3 - 8} \right)^{2n + 1}} as \left\\{ {{x'}} \right\\}
So, we subtract \left\\{ {{x'}} \right\\} from xx 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.