Find all the values of the parameter a for which the point of minimum of the function f(x)=1+a²x-x³ satisfies the inequality ((x²+x+2)/(x²+5x+6))>0

Dear harshit

((x²+x+2)/(x²+5x+6))>0

x²+x+2 is always positive

so x²+5x+6 should be >0

 which give

  x<-3  and x>-2

Now

f(x)=1+a²x-x³

  for minimum value differentiate

f'(x)=a²-3x²

  f''(x)=-6x

equate f'(x) =0

  fives x=+-a/√3

case 1

a>0

x=-a/√3 gives minimum value

so    -a/√3 <-3    and   -a/√3 >-2

           a>3√3        and     a<2√3

 so    0<a<2√3   and    a>3√3

case 2 

a<0

x=a/√3 gives minimum value

so    a/√3 <-3    and   a/√3 >-2

           a<-3√3        and     a>-2√3

 so    a<-3√3   and    -2√3<a<0