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Q.
If the point of minima of the function, $f ( x )=1+ a ^2 x - x ^3$ satisfy the inequality $\frac{x^2+x+2}{x^2+5 x+6}<0$, then 'a' must lie in the interval:
Application of Derivatives
Solution:
$f ^{\prime}( x )=0 \Rightarrow x =\frac{ a }{\sqrt{3}}$
or $-\frac{ a }{\sqrt{3}} f ^{\prime \prime}( x )=-6 x$
Case I if $a>0 \Rightarrow x=-\frac{a}{\sqrt{3}}$ is minima
Case II if $a <0 \Rightarrow x =\frac{ a }{\sqrt{3}}$ is minima
put $x=\frac{a}{\sqrt{3}}$ and then $x=-\frac{a}{\sqrt{3}}$ in the given inequality to get the result