The values of parameter `a` for which the point of minimum of the function `f(x)=1+a^2x-x^3` satisfies the inequality `(x^2+x+2)/(x^2+5x+6)<0a r e` `(2sqrt(3),3sqrt(3))` (b) `-3sqrt(3),-2sqrt(3))` `(-2sqrt(3),3sqrt(3))` (d) `(-2sqrt(2),2sqrt(3))`

Options

A. `2sqrt(3),3sqrt(3)`
B. `-3sqrt(3),-2sqrt(3)`
C. `-2sqrt(3),3sqrt(3)`
D. `-3sqrt(2),2sqrt(3)`

Answer

1,2

Explanation

Given that `(x^(2)+x+2)/(x^(2)+5x+6)lt0 or x in (-3.-2)`
We have ot find the extreama ofr the functin
f(x) =`1+a^(2)x-x^(3)`
For maximum or minimum f(X)=0
or `a^(2)-3x^(2)=0 or x = pm(a)/(sqrt(3))`
and f(X) =-6x is +ve when x is negative
If a is positive then the point of minima is `(a)/(sqrt(3))` i.e
`-3lt-(a)/(sqrt(3))lt-2 or 2sqrt(3)ltalt3sqrt(3)`
If a is negative then the point of minima is `(a)/(sqrt(3))` i.e
`-3lt-(a)/(sqrt(3))lt-2 or 2sqrt(3)ltalt3sqrt(3)`
Then a in `(3sqrt)(3),2sqrt(3)cup(2sqrt(3),3sqrt(3))`