JEE MAIN - Mathematics (2008 - No. 12)
Suppose the cubic $${x^3} - px + q$$ has three distinct real roots
where $$p>0$$ and $$q>0$$. Then which one of the following holds?
where $$p>0$$ and $$q>0$$. Then which one of the following holds?
The cubic has minima at $$\sqrt {{p \over 3}} $$ and maxima at $$-\sqrt {{p \over 3}} $$
The cubic has minima at $$-\sqrt {{p \over 3}} $$ and maxima at $$\sqrt {{p \over 3}} $$
The cubic has minima at both $$\sqrt {{p \over 3}} $$ and $$-\sqrt {{p \over 3}} $$
The cubic has maxima at both $$\sqrt {{p \over 3}} $$ and $$-\sqrt {{p \over 3}} $$
Explanation
Let $$y = {x^3} - px + q$$
$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$
For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$
$$ \Rightarrow x = \pm \sqrt {{p \over 3}} $$
$${{{d^2}y} \over {d{x^2}}} = 6x$$
$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$$ and
$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$$
$$\therefore$$ $$y$$ has ninima at $$x = \sqrt {{p \over 3}} $$
and maxima at $$x = - \sqrt {{p \over 3}} $$
$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$
For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$
$$ \Rightarrow x = \pm \sqrt {{p \over 3}} $$
$${{{d^2}y} \over {d{x^2}}} = 6x$$
$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$$ and
$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$$
$$\therefore$$ $$y$$ has ninima at $$x = \sqrt {{p \over 3}} $$
and maxima at $$x = - \sqrt {{p \over 3}} $$
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