Q. 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 maxima at both $\frac{p}{3}\ and - \frac{p}{3}$
B) The cubic has minima at $\frac{p}{3}\ and\ maxima\ at - \frac{p}{3}$
C) The cubic has minima at$- \frac{p}{3}\ and\ maxima\ at\frac{p}{3}$
D) The cubic has minima at both $\frac{p}{3}\ and - \frac{p}{3}$
Explanation
(b)
Let $f(x) = x^{3} - px + q$
Then $f^{'}(x) = 3x^{2} - p$
Put $f^{'}(x) = 0\ \ \ \Rightarrow \ \ \ \ \ \ x = \sqrt{\frac{p}{3}},\ \ - \sqrt{\frac{p}{3}}$
Now, $f^{''}(x) = 6x$
$\therefore$ At $x = \sqrt{\frac{p}{3},\ \ \ f^{''}(x) = 6\sqrt{\frac{p}{3} > 0},\ }$ minima
And at $x = - \sqrt{\frac{p}{3},\ \ \ f^{''}(x)}$ $< 0$, maxima
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.