Question

If $x={(8+3\sqrt{7})}^n$, where $n\in N$,then $x-x^2+x\left[x\right]=$_____, where [.] denotes the greatest integer function.

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

$Sincex={(8+3\sqrt{7})}^n=\left[x\right]+f........\left(i\right)where\left[x\right]\le xand0\le f<1Now,let{f}^{\prime }={(8-3\sqrt{7})}^n,where0<f^{\prime }<1Also,\left[x\right]+f+f^{\prime }=(8+3\sqrt{7})+{(8-3\sqrt{7})}^n=2\{8^n+{{}^{n}C}_2{\left(8\right)}^{n-2}{\left(3\sqrt{7}\right)}^2+{{}^{n}C}_4{\left(8\right)}^{n-4}{\left(3\sqrt{7}\right)}^4+...\}=2\left(Integer\right)\left[x\right]+f+f^{\prime }=Eveninteger=2k\left(say\right)........\left(ii\right)\therefore L.H.Sisalsoaninteger\therefore f+f^{\prime }isaninteger\Rightarrow f+f^{\prime }=1(\because 0<f+f^{\prime }<2)Hence\left[x\right]+1=2k\left(from\right(ii\left)\right)\left[x\right]=2k-1=oddintegerN$