Consider the equation `x^(4) + 2ax^(3) + x^(2)+2ax + 1 =0`, where `a in R`.
If exactly two roots are positive and two roots are negative, then the number of intergal values of a is

Options

A. `(3//4,oo)`
B. `(-5//4,oo)`
C. `(-oo, 1//4)`
D. none of these

Answer

1

Explanation

Given equations is
`x^(4) + 2ax^(3) + x^(2) + 2ax + 1 =0`
or `(x + (1)/(x^(2))) + 2a(x+(1)/(x))+ 1 =0`
or `(x +(1)/(x))^(2) + 2a(x+(1)/(x)) - 1=0`
`rArr t^(2) + 2at -1=0`
Where `t = x+ (1//x)`. Now,
`(x+(1)/(x))^(2) ge 2 or (x+(1)/(x)) le-2`
`therefore tge 2 or t le-2`
Now, Eq.(1) will have at least two roots negative, when at least one roots of Eq.(2) will be less than `- 2`. If `alpha, beta ge -2`,then
`f(-2) ge 0 and -(B)/(2A) gt -2`
`therefore 4-4a-1 ge 0 and -(2a)/(2) gt -2`
`therefore a le(3)/(4) and lt 2`
`therefore a le (3)/(4)`
Combining (3) and (5), at lest one roots will be less than `-2` for Eq.(2) if
`a gt (3)/(4)`
`therfore a in ((3)/(4), oo)`