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`        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`
8 years ago

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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8 years ago
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