Question

If $x_1,x_2,x_3\cdots x_n$ are roots of $x^n+ax+b=0$, then the value of $(x_1-x_2)(x_1-x_3)(x_1-x_4)\cdots (x_1-x_n)$ =

• A. $(n{x}_1{)}^{n-1}+b$
• B. $n{x}_1+b$
• C. $n(x_1{)}^{n-1}$
• D. $n(x_1{)}^{n-1}+a$

Answer 1

The correct option is D

$n(x_1{)}^{n-1}+a$

$x^n+ax+b=(x-x_1)(x-x_2)(x-x_3)\cdots (x-x_n)\Rightarrow \frac{x^n+ax+b}{x-x_1}=(x-x_2)(x-x_3)(x-x_4)\cdots (x-x_n)\text{Now}\underset{x\to x_1}{lim}\frac{x^n+ax+b}{x-x_1}=\underset{x\to x_1}{lim}(x-x_2)(x-x_3)(x-x_4)\cdots (x-x_4)\therefore n{x}_1+a=(x_1-x_2)(x_1-x_3)(x_1-x_4)\cdots (x_1-x_n)$