Question

The equation ${(x-n)}^m+{(x-n^2)}^m+{(x-n^3)}^m+...+{(x-n^m)}^m=0$ $(m$ is odd positive integer$),$ has

• A. all real roots
• B. one real and $(n-1)$ imaginary roots
• C. one real and $(m-1)$ imaginary roots
• D. No real roots

Answer 1

The correct option is C one real and $(m-1)$ imaginary roots

Let $f\left(x\right)={(x-n)}^m+{(x-n^2)}^m+{(x-n^3)}^m+...+{(x-n^m)}^m$

$\Rightarrow f^{\prime }\left(x\right)=m{(x-n)}^{m-1}+m{(x-n^2)}^{m-1}+m{(x-n^3)}^{m-1}+...+m{(x-n^m)}^{m-1}$

Since $m$ is odd, $(m-1)$ is even.

Therefore, $f^{\prime }\left(x\right)=0$ has no real root.

$\Rightarrow f\left(x\right)=0$ has one real and $(m-1)$ imaginary roots.