Question

If $x={(5+2\sqrt{6})}^n,n\in N$, $x-x^2+x\left[x\right]=$_____ where $[.]$ denotes the greatest integer function.

• A. $0$
• B. $-1$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$1$

Solution:

Since, $x=(5+2\sqrt{6}{)}^n=[x]+f$ $................\left(i\right)$

where $\left[x\right]\le x$ and $0\le f\le 1$

Now let $f^{\prime }=(5-2\sqrt{6}{)}^n,$ where $0\le f^{\prime }\le 1$

Also, $\left[x\right]f+f^{\prime }=(5+2\sqrt{6}{)}^n+(5-2\sqrt{6}{)}^n$

$=2(5^n+{}^n{C}_2(5{)}^{n-2}(2\sqrt{6}{)}^2+{}^n{C}_4(5{)}^{n-4}(2\sqrt{6}{)}^4+.............)$

$=2$ (Integer)

$\therefore \left[x\right]+f+f^{\prime }=$ Even integer $=2k$ (say) $.............\left(ii\right)$

$\therefore$ L.H.S is also an integer.

$\therefore f+f^{\prime }$ is an integer.

therefore, $f+f^{\prime }=1$ $(\because 0<f+f^{\prime }<2)$

Hence $\left[x\right]+1=2k$ (From (ii))

$\left[x\right]2k-1=$ odd integer.

Now, $x-x^2+x\left[x\right]=$

$=x-x(x-[x\left]\right)$

$=x-xf$

$=x(1-f)$

$=x{f}^{\prime }$ $(\because f+f^{\prime }=1)$

$=(5+2\sqrt{6}{)}^n(5-2\sqrt{6}{)}^n$

$=(25-24{)}^n$

$=1^n$

$=1$

Hence, C is the correct option.