Question

If $(2+\sqrt{3}{)}^n=I+f,n\in N$, where $I$ is integral part and $f$ is fractional part, then $(I+f)(1-f)$ is

• A. $0$
• B. $1$
• C. $n$
• D. $-1$

Answer 1

The correct option is B $1$

Given
$(2+\sqrt{3}{)}^n=I+f$
So,
$0<f<1$
Assuming
$f^{\prime }=(2-\sqrt{3}{)}^n$
Clearly,
$0<f^{\prime }<1\Rightarrow 0<f+f^{\prime }<2$
Now,
$I+f+f^{\prime }=(2+\sqrt{3}{)}^n+(2-\sqrt{3}{)}^n=2k,k\in N\Rightarrow f+f^{\prime }=2k-I=\text{integer}$
So, $f+f^{\prime }=1(\because 0<f+f^{\prime }<2)$
Now,
$(I+f)(1-f)=(I+f)\left(f^{\prime }\right)=(2+\sqrt{3}{)}^n\times (2-\sqrt{3}{)}^n=(4-3{)}^n=1$