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If \mathrm{f(x)=x^3+a x^2+b x+c} attains it's local minima at certain negative real number then

Option: 1

\mathrm{a^2-3 b>0, a<0, b<0}


Option: 2

\mathrm{a^2-3 b>0, a<0, b>0}


Option: 3

\mathrm{a^2-3 b>0, a>0, b<0}


Option: 4

\mathrm{a^2-3 b>0, a>0, b>0}


Answers (1)

best_answer

\mathrm{f^{\prime}(x)=3 x^2+2 a x+b=3\left(x-x_1\right)\left(x-x_2\right)} where \mathrm{x_1<x_2}.

Clearly  \mathrm{f^{\prime}(x)<0 \forall x \in\left(x_1, x_2\right)\, and \, f^{\prime}(x)>0 \forall x \in\left(-\infty, x_1\right) \cup\left(x_2, \infty\right)}.

Thus \mathrm{x=x_1} is the point of local maxima and \mathrm{x=x_2}  is the point of local minima.
Thus bigger root of \mathrm{f^{\prime}(x)=0} must be negative.

Hence \mathrm{a^2-3 b>0, a>0, b>0}.

Posted by

Irshad Anwar

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