Consider an unknow polynomial which divided by `(x - 3)` and `(x-4)` leaves remainder 2 and 1, respectively. Let R(x) be the remainder when this polynomial is divided by `(x-3)(x-4)`.
If equations `R(x) = x^(2)+ ax +1` has two distint real roots, then exhaustive values of a are.

Options

A. `(-2,2)`
B. `(-oo,-2) uu(2,oo)`
C. `(-2,oo)`
D. all real numbers

Answer

4

Explanation

Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then
`P(x) = (x-3)(x-4) Q(x) + R(x)`
Then we have
`R(x) = ax + b`
` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b`
Given that `P(3) =2 ` and `P(4)=1` . Hence,
`3a + b = 2 and 4a + b = 1`
`rArr a = -1 and b= 5`
`rArr R(x) = 5 -x`
`5-x = x^(2) + ax + 1 `
`rArr x^(2) + (a + 1) x - 4 =0`
Given that roots and reall distinct. Therefore,
`D lt 0 rArr (a+1)^(2) + 16 lt 0`
Which is true for all real a.