Question

$(x^2-2x\cos \frac{\pi }{n}+1\left)\right(x^2-2x\cos \frac{2\pi }{n}+1)...(x^2-2x\cos \frac{n-1}{n}\pi +1)$ equals

• A. $1+x+...+x^{n-1}$
• B. $1-x+x^2-...+(-1{)}^{n-1}{x}^{n-1}$
• C. $1+x+...+x^{2n-2}$
• D. $0$

Answer 1

The correct option is C $1+x+...+x^{2n-2}$

${\alpha }_1,{\alpha }_2,.....,{\alpha }_m$ are the roots of polynomial equation,

We know,

$\Rightarrow f\left(x\right)=a_0{x}^m+a_1{x}^{m-1}+....+a_{m-1}{x}^m+a_m=0$

then, $x^m=(x-{\alpha }_1)$

$\Rightarrow x^{m-1}=(x-{\alpha }_1-1)$

then, $x^{\prime }=(x-{\alpha }_m)$

$\therefore \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=\frac{1}{(x-{\alpha }_1)}+....+\frac{1}{(x-{\alpha }_m)}$

$\Rightarrow (x^2-2x\cos \frac{\pi }{n}+1)(x^2-2x\cos \frac{2\pi }{n}+1).....(x^2-2x\cos \frac{n-1}{n}\pi +1)$

$=x^2[1-\frac{2}{x}\cos \frac{\pi }{n}+\frac{1}{x^2}](1-\frac{2}{x}\cos \frac{2\pi }{n}+\frac{1}{x^2})...$

$=1-\frac{2}{x}\cos \frac{\pi }{n}+\frac{1}{x^2}$

$\Rightarrow \frac{2}{x}\cos \frac{\pi }{n}=1-\frac{1}{x^2}$

$1+x+........x^{2n-2}$ By the given condition.

Hence, the answer is $1+x+........x^{2n-2}.$