Animated Solution for Mathematics - Differentiation: 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?

Imagine you are standing before a vast, rolling landscape. This is the graph of our cubic function, $f(x)=x^3-px+q$.

In the world of JEE Advanced, we don't just solve equations; we visualize the geometry of the functions we manipulate. A cubic function with three distinct real roots is not just a collection of terms; it is a beautiful, oscillating wave that pierces the x-axis three separate times.

To achieve this, the function must rise to a peak (a local maximum), dip down into a valley (a local minimum), and then rise again. Our mission is to pinpoint exactly where these peaks and valleys occur.

Finding the Turning Points

To find these turning points, we need to identify where the function stops rising and starts falling, or vice versa. In the language of calculus, this is where the slope of the tangent line is perfectly horizontal—that is, where the slope is zero.

We invoke the power of the first derivative, $f^{\prime }(x)$. Taking the derivative of our function $f(x)=x^3-px+q$ with respect to $x$, we apply the power rule. The term $x^3$ becomes $3x^2$, the term $-px$ becomes $-p$, and the constant $q$ vanishes into zero.

We are left with:

This elegant expression is our 'slope detector.' To find the critical points, we set $f^{\prime }(x)=0$:

Solving for $x$, we find:

We have found two critical points: $x=-\sqrt{\frac{p}{3}}$ and $x=\sqrt{\frac{p}{3}}$. But which is the peak, and which is the valley? This is where the mystery deepens, and we need a more powerful tool.

The Second Derivative

To distinguish between the maximum and the minimum, we look at the concavity of the curve. Does the curve 'smile' (concave up) or 'frown' (concave down)?

We calculate the second derivative, $f^{\prime \prime }(x)$, by differentiating $f^{\prime }(x)$ once more:

This is our 'concavity detector.' If $f^{\prime \prime }(x)>0$, the curve is concave up, indicating a local minimum. If $f^{\prime \prime }(x)<0$, the curve is concave down, indicating a local maximum.

Let us test our critical points:

1. For $x=-\sqrt{\frac{p}{3}}$:

Since $p>0$, this value is clearly negative. A negative second derivative means the curve is concave down—a peak! Thus, we have a local maximum at $x=-\sqrt{\frac{p}{3}}$.

2. For $x=\sqrt{\frac{p}{3}}$:

This value is positive. A positive second derivative means the curve is concave up—a valley! Thus, we have a local minimum at $x=\sqrt{\frac{p}{3}}$.

We have successfully navigated the landscape of this cubic function. By using the first derivative to find the critical points and the second derivative to classify them, we have proven that the cubic has a local maximum at $x=-\sqrt{\frac{p}{3}}$ and a local minimum at $x=\sqrt{\frac{p}{3}}$.

This aligns perfectly with the first option provided. Remember, in JEE Advanced, the math is the map, but your intuition is the compass. Keep practicing, keep visualizing, and the curves will always reveal their secrets to you.