Answer:
Option D
Explanation:
We have,
$x^{n}+px+q=0$,
If $\alpha_{1},\alpha_{2},.......,\alpha_{n}$ are roots of given equation so,
$x^{n}+px+q=(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n})$
$\Rightarrow$ $\frac{x^{n}+px+q}{x-\alpha_{n}}=$ $(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n-1})$
$\therefore$ $\lim_{x \rightarrow \alpha_{n}}\frac{x^{n}+px+q}{x-\alpha_{n}}=\lim_{x \rightarrow \alpha_{n}}$ $(x-\alpha_{1})(x-\alpha_{2}) (x-\alpha_{3}).......(x-\alpha_{n-1})$
$\lim_{x \rightarrow \alpha_{n}}\frac{nx^{n-1}+P}{1}=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$
$\Rightarrow$ $n \alpha_{n}^{n-1}+p=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$
