Question

If $x={(7+4\sqrt{3})}^{2n}=\left[x\right]+f$, then $x(1-f)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

We have
$(7-4\sqrt{3})=\frac{1}{7+4\sqrt{3}}$
$\therefore 0<7-4\sqrt{3}<1\Rightarrow 0<{(7-4\sqrt{3})}^{2n}<1$
Let $F={(7-4\sqrt{3})}^{2n},0<F<1$. Then
$x+F={(7+4\sqrt{3})}^{2n}+{(7-4\sqrt{3})}^{2n}$
$=2[{{}^{2n}C}_0{7}^{2n}+{{}^{2n}C}_2{7}^{2n-2}{\left(4\sqrt{3}\right)}^2+{{}^{2n}C}_4{7}^{2n-4}{\left(4\sqrt{3}\right)}^4+......+{{}^{2n}C}_{2n}{\left(4\sqrt{3}\right)}^4]$
$=2m$, where $m$ is some positive integer.

So, $x=2m-F$

$\therefore \left[x\right]=[2m-F]=2m-1$, since $2m$ is an integer and $0<F<1$.
$\Rightarrow \left[x\right]+f+F=2m$ $\Rightarrow f+F=2m-\left[x\right]=2m-(2m-1)=1$.

Thus,
$x(1-f)=xF={(7+4\sqrt{3})}^{2n}{(7-4\sqrt{3})}^{2n}$
$={(49-48)}^{2n}=1^{2n}=1$