Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$x={(7+4\sqrt{3})}^{2n}$

$x=3+f$

$f$ is fractional part of $x$

$\Rightarrow 0<f<1$

Let $f^{\prime }={(7-4\sqrt{3})}^{2n}$

$0<7-4\sqrt{3}<1$

$0<{(7-4\sqrt{3})}^n<1$

$\Rightarrow 0<f^{\prime }<1$

$I+f+f^{\prime }=2[{}^{2n}{C}_0{\left(7\right)}^{2n}{+}^{2n}{C}_2{\left(7\right)}^{2n-2}+....]$

$=$ even

$0+f+f^{\prime }<2$ is integer.

${}^{\prime }{I}^{\prime }$ is integer and $f+f^{\prime }$ is also integer.

$\Rightarrow f+f^{\prime }=1$

$f^{\prime }=1-f$

$x(1-f)=x{f}^{\prime }={(7+4\sqrt{3})}^{2n}{(7-4\sqrt{3})}^{2n}$

$=1$