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,
$2x^{2}+2x+1=0$
$\Rightarrow x=-\frac{2\pm\sqrt{4-8}}{4}$
$x=-\frac{2 \pm2i}{4}=\frac{-1\pm i}{2}$
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]^{n 2} =r $
$\Rightarrow \frac{1}{2^{n}}\left(\pm\,2i\right)^{n 2}=r $
$\Rightarrow \frac{1}{2^{n 2}}\left(\pm i\right)^{n 2}=r $
r is real
$\therefore n \in\,4 m $
$\therefore n=4000$
and $r=\frac{1}{2\frac{4000}{2}}=\frac{1}{4^{1000}}$