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harshit agarwal Grade: 12
        

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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


 





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Badiuddin

7 years ago
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