(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 $r=\frac{1}{2^{\frac{4000}{2}}}=\frac{1}{4^{1000}}$

$2$