Suppose the cubic $x^{3} - p x + q = 0$ has three distinct real roots where $p > 0$ and $q > 0$ . Then which one of the following holds?

Moderate

A

The cubic has minima at $- \sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$
B

The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $- \sqrt{\frac{p}{3}}$
C

The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $- \sqrt{\frac{p}{3}}$
D

The cubic has minima at $\sqrt{\frac{p}{3}}$ and maxima at $- \sqrt{\frac{p}{3}}$

Solution

The given equation is $f (x) = x^{3} - p x + q$
$\Rightarrow$ $f' (x) = 3 x^{2} - p$

$\Rightarrow$ $f'' (x) = 6 x$

for max. or min. $f' (x) = 0$

$\Rightarrow 3 x^{2} - p = 0$ $\Rightarrow x = \pm \sqrt{\frac{p}{3}}$

Then $f'' (\sqrt{\frac{p}{3}}) = 6 \sqrt{(\frac{p}{3})} > 0$

and $f'' (- \sqrt{\frac{p}{3}}) = - 6 \sqrt{\frac{p}{3}} < 0$

hence, given cubic minima at $x = \sqrt{\frac{P}{3}}$ and maxima at $x = - \sqrt{\frac{P}{3}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME