JEE Main · 2008 · 27 Apr 2008 · Offline · Application of Derivatives

1. (A)The cubic has minima at $\sqrt{\frac{p}{3}}$ and maxima at $-\sqrt{\frac{p}{3}}$✓ Correct
2. (B)The cubic has minima at $-\sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$
3. (C)The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$
4. (D)The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

Answer: (A)

Solution

The critical points of $f(x)=x^3-px+q$ come from $f^{\prime }(x)=3x^2-p=0$, giving $x=\pm \sqrt{p/3}$. The second derivative $f^{\prime \prime }(x)=6x$ is positive at $x=+\sqrt{p/3}$, marking a minimum there, and negative at $x=-\sqrt{p/3}$, marking a maximum. So the cubic has its minimum at $\sqrt{p/3}$ and its maximum at $-\sqrt{p/3}$.

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option