Question

Suppose cubic equation $x^3-px+q=0$ has three distinct & real roots $(p,q>0)$ then which of the following is true for $f\left(x\right)=x^3-px+q=0$

• A. maximum at $-\sqrt{p/3}$ & minimum at $\sqrt{p/3}$
• B. maximum at both $\sqrt{p/3}$ and $-\sqrt{p/3}$
• C. minimum at both $\sqrt{p/3}$ and $-\sqrt{p/3}$
• D. None of the above

Answer 1

The correct option is A maximum at $-\sqrt{p/3}$ & minimum at $\sqrt{p/3}$

$f\left(x\right)=x^3-px+q=0$ $(p,q>0)$

$y^{\prime }=f^{\prime }\left(x\right)=3x^2-p$, $f^{\prime \prime }\left(x\right)=3\left(2x\right)$

for maximum & minimum values $f^{\prime }\left(x\right)=0$

$\Rightarrow$ $x=\sqrt{\frac{p}{3}}$, $x=-\sqrt{\frac{p}{3}}$

Now ${(\frac{d^{2}y}{d{x}^2}=f^{\prime \prime }\left(x\right)=y^{\prime \prime })}_{x=\sqrt{\frac{p}{3}}}=6\sqrt{\frac{p}{3}}>0$ as $p>0$

$\therefore$ $f\left(x\right)$ is minimum at $x=\sqrt{\frac{p}{3}}$

Similarly $f^{\prime \prime }(-\sqrt{\frac{p}{3}})=6(-\sqrt{\frac{p}{3}})<0$ as $p>0$

$\therefore f\left(x\right)$ is maximum at $x=-\sqrt{\frac{p}{3}}$

Ans: A