Q53 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
If $\alpha_1, \alpha_2, \ldots \ldots, \alpha_n$ are the roots of $x^n+p x+q=0$, then $\left(\alpha_n-\alpha_1\right)\left(\alpha_n-\alpha_2\right) \ldots . .\left(\alpha_n-\alpha_{n-1}\right)=$
- A$n \alpha_n^{n-1}+q$
- B$\alpha_1^2+\alpha_2^2+\ldots+\alpha_{n-1}^2$
- C$\alpha_n^{n-1}+p$
- Correct answerD$n \alpha_n^{n-1}+p$
Solution
We have,
$$
x^n+p x+q=0
$$
If $\alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_n$ are roots of given equation
$$
\begin{array}{r}
\text { So, } x^n+p x+q=\left(x-\alpha_1\right)\left(x-\alpha_2\right)\left(x-\alpha_3\right) \ldots\left(x-\alpha_n\right) \\
\Rightarrow \frac{x^n+p x+q}{x-\alpha_n}=\left(x-\alpha_1\right)\left(x-\alpha_2\right)\left(x-\alpha_3\right) \\
\ldots\left(x-\alpha_{n-1}\right)
\end{array}
$$
$$
\begin{array}{r}
\therefore \quad \lim _{x \rightarrow \alpha_n} \frac{x^n+p x+q}{x-\alpha_n}=\lim _{x \rightarrow \alpha_n}\left(x-\alpha_1\right)\left(x-\alpha_2\right)\left(x-\alpha_3\right) \\
\quad \lim _{x \rightarrow \alpha_n} \frac{n x^{n-1}+P}{1}=\left(\alpha_n-\alpha_1\right)\left(\alpha_n-\alpha_2\right) \ldots\left(\alpha_n-\alpha_{n-1}\right) \\
\Rightarrow \quad n \alpha_n{ }^{n-1}+p=\left(\alpha_n-\alpha_1\right)\left(\alpha_n-\alpha_2\right) \ldots\left(\alpha_n-\alpha_{n-1}\right)
\end{array}
$$
Hence, the correct answer is option: 'D'.