Question

Let $r$ be a real number and $n\in N$ be such that the polynom ial $2x^2+2x+1$ divides the polynomial ${\left(x+1\right)}^n-r$ Then, $(n,r)$ can be

• A. $(4000,4^{1000})$
• B. $\left(4000,\frac{1}{4^{1000}}\right)$
• C. $\left(4^{1000},\frac{1}{4^{1000}}\right)$
• D. $\left(4000,\frac{1}{4000}\right)$

Answer 1

Answer: (B)

We have,
$2x^2+2x+1=0$
$\Rightarrow x=-\frac{2\pm \sqrt{4-8}}{4}$
$x=-\frac{2\pm 2i}{4}=\frac{-1\pm i}{2}$
x satisfies the equation ${\left(x+1\right)}^n-r=0$
$\therefore {\left(\frac{-1\pm i}{2}+1\right)}^n-r=0\left[r\in R\right]$
$\Rightarrow {\left(\frac{-1\pm i+2}{2}\right)}^n=r$
$\Rightarrow {\left(\frac{1\pm i}{2}\right)}^n=r$
$\Rightarrow \frac{1}{2^n}{\left[{\left(1\pm i\right)}^2\right]}^{n2}=r$
$\Rightarrow \frac{1}{2^n}{\left(\pm 2i\right)}^{n2}=r$
$\Rightarrow \frac{1}{2^{n2}}{\left(\pm i\right)}^{n2}=r$
r is real
$\therefore n\in 4m$
$\therefore n=4000$
and $r=\frac{1}{2\frac{4000}{2}}=\frac{1}{4^{1000}}$