Question
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?
A.
The cubic has minima at $$\sqrt {\frac{p}{3}} $$ and maxima at $$ - \sqrt {\frac{p}{3}} $$
B.
The cubic has minima at $$ - \sqrt {\frac{p}{3}} $$ and maxima at $$\sqrt {\frac{p}{3}} $$
C.
The cubic has minima at both $$\sqrt {\frac{p}{3}} $$ and $$ - \sqrt {\frac{p}{3}} $$
D.
The cubic has maxima at both $$\sqrt {\frac{p}{3}} $$ and $$ - \sqrt {\frac{p}{3}} $$
Answer :
The cubic has minima at $$\sqrt {\frac{p}{3}} $$ and maxima at $$ - \sqrt {\frac{p}{3}} $$
Solution :
$$\eqalign{
& {\text{Let }}y = {x^3} - px + q \Rightarrow \frac{{dy}}{{dx}} = 3{x^2} - p \cr
& {\text{For }}\frac{{dy}}{{dx}} = 0 \Rightarrow 3{x^2} - p = 0 \Rightarrow x = \pm \sqrt {\frac{p}{3}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 6x \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{x = \sqrt {\frac{p}{3}} }} = + ve{\text{ and }}{\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{x = - \sqrt {\frac{p}{3}} }} = - ve \cr
& \therefore y{\text{ has minima at }}x = \sqrt {\frac{p}{3}} {\text{ and maxima at }}x = - \sqrt {\frac{p}{3}} \cr} $$