If `a_0, a_1, a_2, a_3` are all the positive, then `4a_0x^3+3a_1x^2+2a_2x+a_3=0` has least one root in `(-1,0)` if `a_0+a_2=a_1+a_3a n d4a_0+2a_2>3a_1+a_3` `4a_0+2a_2<3a_1+a_3` `4a_0+2a_2=3a_1+a_0a n d4a_0+a_2

Options

A. `a_(0) + a_(2) = a_(1) + a_(3) and 4_(a_(0)) + 2a_(2) gt 3a_(1) + a_(3)`
B. `4a_(0) + 2a_(2) lt 3a_(1) + a _(3)`
C. `4a_(0) + 2a_(2) = 3a_(1) + a_(3) and a_(0) + a_(2) lt a_(1) + a_(3)`
D. none of these

Answer

1

Explanation

`P(x) = 4a_(0) x^(3) a_(1) x^(2) + 2a_(1) x + a_(3)` is a polynomial, so it is
continouns for all x .
` P (x) = 0 ` has a root in (-1, 0)
` P (-1) P (0) lt 0 `
Now ` P (0) = a_(3) gt 0 `,
`rArr P (-1) = - 4a_(0) + 3a_(1) - 2a_(2) + a_(3) lt 0 `
`rArr 4a_(0) + 2a_(2) gt 3a_(1) + a _(3)`
Also using Rolle's theorem,
`Q(x) int p (x) dx = a_(0) x^(4) + a_(1) x^(3) + a_(2) x^(2) + a_(3)x + d`
For ` Q' () = p (x) = 0`
`rArr a_(0) - a_(1) + a_(2) - a_(3) + d = d `
or ` a_(0) + a_(2) = a_(1) + a_(3)` .