If alpha_1, alpha_2, ldots, alpha_n are the r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the polynomial equation $x^n+px+q=0$ with roots $\alpha_1, \alpha_2, \ldots, \alpha_n$,we can write it as: $x^n+px+q = (x-\alpha_1)(x-\alpha

Vedclass · Accepted Answer

$n \alpha_n^{n-1}+p$

Vedclass · Answer

(D) Given the polynomial equation $x^n+px+q=0$ with roots $\alpha_1, \alpha_2, \ldots, \alpha_n$,we can write it as: $x^n+px+q = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$. Dividing both sides by $(x-\alpha_n)$,we get: $\frac{x^n+px+q}{x-\alpha_n} = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1})$. Taking the limit as $x 	o \alpha_n$ on both sides: $\lim_{x 	o \alpha_n} \frac{x^n+px+q}{x-\alpha_n} = (\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1})$. Using $L$'$H$ôpital's Rule on the left side: $\lim_{x 	o \alpha_n} \frac{\frac{d}{dx}(x^n+px+q)}{\frac{d}{dx}(x-\alpha_n)} = \lim_{x 	o \alpha_n} (nx^{n-1}+p) = n\alpha_n^{n-1}+p$. Thus,$(\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1}) = n\alpha_n^{n-1}+p$.

If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are the roots of $x^n+px+q=0$,then $(\alpha_n-\alpha_1)(\alpha_n-\alpha_2) \ldots (\alpha_n-\alpha_{n-1})=$

Similar Questions

The value of $\lim _{x \rightarrow 2} \frac{1}{x-2} \int_{2}^{x} 3 t^{2} dt$ is

If $f(1) = 1$ and $f'(1) = 2$,then $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

Evaluate the limit: $\mathop {\text{Limit}}\limits_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$,where $0 < \alpha < \frac{\pi}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module