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Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be
$\left(4000,4^{1000}\right)$
$\left(4000, \frac{1}{4^{1000}}\right)$
$\left(4^{1000}, \frac{1}{4^{1000}}\right)$
$\left(4000, \frac{1}{4000}\right)$
Solution
(b)
We have,
$2 x^2+2 x+1=0$
$\Rightarrow x-2 \pm \sqrt{4-8}$
$x=-\frac{2 \pm 2 i}{4}=\frac{-1 \pm i}{2}$
$x$ satisfies the equation $(x+1)^n-r=0$
$\therefore \quad\left(\frac{-1 \pm i}{2}+1\right)^n-r=0 \quad[r \in R]$
$\Rightarrow \quad\left(\frac{-1 \pm i+2}{2}\right)^n=r$
$\Rightarrow \quad\left(\frac{1 \pm i}{2}\right)^n=r$
$\Rightarrow \quad \frac{1}{2^n}\left[(1 \pm i)^2\right]^{n / 2}=r$
$\Rightarrow \quad \frac{1}{2^n}(\pm 2 i)^{n / 2}=r$
$\Rightarrow \quad \frac{1}{2^{n / 2}}(\pm i)^{n / 2}=r$
$r$ is real
$\therefore \quad n \in 4\,m$
$\therefore \quad n=4000$
and $r=\frac{1}{2^{\frac{4000}{2}}}=\frac{1}{4^{1000}}$
$2$