Question

If the cube equation $x^3-px+q$ has three distinct real roots, where $p>0$ and $q>0$.

Then, which one of the following is correct?

• A. The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$
• B. The cubic has minima at $\sqrt{\frac{p}{3}}$ and maxima at $-\sqrt{\frac{p}{3}}$
• C. The cubic has minima at $-\sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$
• D. The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

Answer 1

The correct option is A The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

$f\left(x\right)=x^3-px+q{f}^{\prime }\left(x\right)=3x^2-p\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow 3x^2-p=0\Rightarrow p=3x^2\Rightarrow x=\pm \sqrt{\frac{p}{3}}$

The critical points are: $-\sqrt{\frac{p}{3}}$ & $\sqrt{\frac{p}{3}}$

$\therefore f(-\sqrt{\frac{p}{3}})={(-\sqrt{\frac{p}{3}})}^3+p\sqrt{\frac{p}{3}}+q=-\frac{p}{3}\sqrt{\frac{p}{3}}+p\sqrt{\frac{p}{3}}+q=\sqrt{\frac{p}{3}}(p-\frac{p}{3})+q=\frac{2p}{3}\sqrt{\frac{p}{3}}+qandf\left(\sqrt{\frac{p}{3}}\right)={\left(\sqrt{\frac{p}{3}}\right)}^3-p\sqrt{\frac{p}{3}}+q=\frac{p}{3}\sqrt{\frac{p}{3}}-p\sqrt{\frac{p}{3}}+q=\frac{-2p}{3}\sqrt{\frac{p}{3}}+q$

Since $p>0$ & $q=0$ and $f^{\prime \prime }\left(x\right)=6x\Rightarrow f^{\prime \prime }\left(x\right)>0forx>0$

$<0forx<0$

Since $f^{\prime \prime }\left(x\right)>0$ for $x>0,f\left(x\right)$ will have local minimum at critical point which is greater than zero.

$\therefore$ minima at $x=\sqrt{\frac{p}{3}}$

maxima at $x=-\sqrt{\frac{p}{3}}$