If `x_(1),x_(2),x_(3),…,x_(n)` are the roots of the equation `x^(n)+ax+b=0`, the value of
`(x_(1)-x_(2))(x_(1)-x_(3))(x_(1)-x_(4))…….(x_(1)-x_(n))` is

Options

A. `nx_(1)+b`
B. `n(x_(1))^(n-1)`
C. `n(x_(1)^(n-1)+a`
D. `n(x_(1))^(n-1)+b`

Explanation

`:'x^(n)+ax+b=(x-x_(1))L(x-x_(2))(x-x_(3))…..(x-x_(n))`
`implies(x-x_(2))(x-x_(3))…….(x-x_(n))=(x^(n)+ax+b)/(x-x_(1))`
On taking `lim_(xtox_(1))` both sides we get
`(x_(1)-x_(2))(x_(1)-x_(3))……………(x_(1)-x_(n))=lim_(xtox_(1))(x^(n)+ax+b)/(x-x_(1))[ 0/0 "from"]`
`=lim_(xtox_(1))(nx^(n-1)+a)/1=n(x_(1))^(n-1)+a`