Gujarati
4-2.Quadratic Equations and Inequations
normal

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

A

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

B

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

C

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

D

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

(KVPY-2010)

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

$2$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.