Let r be a real number and n in N be such tha | vedclass

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Question

(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 $(

Vedclass · Accepted Answer

$\left(4000, \frac{1}{4^{1000}}\right)$

Vedclass · Answer

(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 $(x+1)^n-r=0$

$	herefore \quad\left(\frac{-1 \pm i}{2}+1ight)^n-r=0 \quad[r \in R]$

$\Rightarrow \quad\left(\frac{-1 \pm i+2}{2}ight)^n=r$

$\Rightarrow \quad\left(\frac{1 \pm i}{2}ight)^n=r$

$\Rightarrow \quad \frac{1}{2^n}\left[(1 \pm i)^2ight]^{n / 2}=r$

$\Rightarrow \quad \frac{1}{2^n}(\pm 2 i)^{n / 2}=r$

$\Rightarrow \quad \frac{1}{2^{n / 2}}(\pm i)^{n / 2}=r$

$r$ is real

$	herefore \quad n \in 4\,m$

$	herefore \quad n=4000$

and $r=\frac{1}{2^{\frac{4000}{2}}}=\frac{1}{4^{1000}}$

$2$

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

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