Question

If ${\alpha }_1,{\alpha }_2,{\alpha }_3\dots ..{\alpha }_n$ are the roots of $x^n+ax+b=0$, then $({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)\dots ..({\alpha }_1-{\alpha }_n)=$

• A. $n{\alpha }_1^{n-1}$
• B. $a$
• C. $n{\alpha }_1^{n-1}+a$
• D. $n{\alpha }_1^{n-1}-a$

Answer 1

Answer: (C)

$n{\alpha }_1^{n-1}+a$

Since,${\alpha }_1,{\alpha }_2,{\alpha }_3\dots ..{\alpha }_n$ are the roots of $x^n+ax+b=0$
$\Rightarrow (x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3).....(x-{\alpha }_n)=x^n+ax+b$
$\Rightarrow (x-{\alpha }_2)(x-{\alpha }_3).....(x-{\alpha }_n)=\frac{x^n+ax+b}{x-{\alpha }_1}$
$\Rightarrow \underset{x\to {\alpha }_1}{lim}(x-{\alpha }_2)(x-{\alpha }_3).....(x-{\alpha }_n)=\underset{x\to {\alpha }_1}{lim}\frac{x^n+ax+b}{x-{\alpha }_1}$
applying L'Hospital's rule in RHS as it is of the form $\frac{0}{0}$
$\Rightarrow ({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)\dots ..({\alpha }_1-{\alpha }_n)=\underset{x\to {\alpha }_1}{lim}\frac{n{x}^{n-1}+a}{1}$
$\Rightarrow ({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)\dots ..({\alpha }_1-{\alpha }_n)=n{{\alpha }_1}^{n-1}+a$