If one the roots fo the equation `x^(2) +x f(a) + a=0` is the cube of the othere for all `x in R`, then f(x)=

Options

A. `x ^(1//4) + x ^(3//4)`
B. ` - (x^(1//4) + x ^(3//4))`
C. ` x + x ^(3)`
D. none of these

Answer

B

Explanation

Let `alpha ` and `alpha^(3)` be the roots of the equation `x^(2) + x f(a) + a=0`. Then,
`alpha + alpha^(3) = - f (a) and alpha^(4) = a`
`rArr f (a) = -alpha a alpha^(3) = - (a_^(1//4) - a^(3//4)`
`rArr f(a) = - (a^(1//4) + a^(3//4))`, where `a = alpha^(4) gt 0`
`rArr f(x) = - (x^(1//4) + x^(3//4)), x gt 0`