Animated Solution for Mathematics - Differentiation: Let $f(x)=4{\cos }^{3}x+3\sqrt{3}{\cos }^{2}x-10$. The number of points of local maxima of $f$ in interval $(0,2\pi )$ is

My dear student, welcome to the arena. Today, we are not just solving a calculus problem; we are embarking on a journey to map the topography of a trigonometric landscape.

We are given the function $f(x)=4{\cos }^{3}x+3\sqrt{3}{\cos }^{2}x-10$, and our mission is to find the number of local maxima within the open interval $(0,2\pi )$. This is a classic JEE Advanced challenge—it tests not just your ability to differentiate, but your patience, your precision, and your ability to visualize the behavior of functions.

To understand where a function reaches its peak, we must first understand its rate of change. We need the derivative, $f^{\prime }(x)$.

Let us apply the chain rule with care. For the first term, $4{\cos }^{3}x$, the derivative is $12{\cos }^{2}x\cdot (-\sin x)$. For the second term, $3\sqrt{3}{\cos }^{2}x$, the derivative is $6\sqrt{3}\cos x\cdot (-\sin x)$. The constant $-10$ vanishes into the void, as all constants do when we seek change.

So, we have:

Now, do not rush to calculate values yet. Look at the expression. It is begging to be factored. Let us pull out the common terms: $-6\sin x\cos x$.

Inside the brackets, we are left with $(2\cos x+\sqrt{3})$. And here is the elegance of trigonometry: we recognize that $2\sin x\cos x$ is the double angle identity for $\sin 2x$. Thus, our derivative simplifies beautifully to:

This compact form is our compass. It tells us exactly where the function stops moving—the critical points.

To find the critical points, we set $f^{\prime }(x)=0$. This gives us two distinct paths to follow:

1. $\sin 2x=0$ 2. $2\cos x+\sqrt{3}=0⟹\cos x=-\frac{\sqrt{3}}{2}$

Let us solve these within our domain $(0,2\pi )$. For $\sin 2x=0$, we know that $2x$ must be a multiple of $\pi$. Thus, $2x=\pi ,2\pi ,3\pi$, which gives us $x=\frac{\pi }{2},\pi ,\frac{3\pi }{2}$.

For $\cos x=-\frac{\sqrt{3}}{2}$, we look to the second and third quadrants. The reference angle is $\frac{\pi }{6}$. In the second quadrant, $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$. In the third quadrant, $x=\pi +\frac{\pi }{6}=\frac{7\pi }{6}$.

We have five critical points: $\frac{\pi }{2},\frac{5\pi }{6},\pi ,\frac{7\pi }{6},\frac{3\pi }{2}$. These points divide our domain into six distinct intervals. This is where the detective work begins.

We must now test the sign of $f^{\prime }(x)=-3\sin 2x(2\cos x+\sqrt{3})$ in each interval. Remember, a local maximum occurs when the derivative changes from positive to negative.

In $(0,\frac{\pi }{2})$: $\sin 2x>0$ and $(2\cos x+\sqrt{3})>0$. With the $-3$ multiplier, $f^{\prime }(x)<0$. The function is decreasing.

In $(\frac{\pi }{2},\frac{5\pi }{6})$: $\sin 2x<0$ and $(2\cos x+\sqrt{3})>0$. The product of two negatives (the $-3$ and the $\sin 2x$) makes $f^{\prime }(x)>0$. The function is increasing. Since we went from decreasing to increasing, $x=\frac{\pi }{2}$ is a local minimum.

In $(\frac{5\pi }{6},\pi )$: $\sin 2x<0$ and $(2\cos x+\sqrt{3})<0$. We have three negative factors ($-3$, $\sin 2x$, and the cosine term). The result is $f^{\prime }(x)<0$. The function is decreasing. Since we went from increasing to decreasing at $x=\frac{5\pi }{6}$, we have found our first local maximum!

In $(\pi ,\frac{7\pi }{6})$: $\sin 2x>0$ and $(2\cos x+\sqrt{3})<0$. The derivative $f^{\prime }(x)>0$. The function is increasing. At $x=\pi$, we went from decreasing to increasing—another local minimum.

In $(\frac{7\pi }{6},\frac{3\pi }{2})$: $\sin 2x>0$ and $(2\cos x+\sqrt{3})>0$. The derivative $f^{\prime }(x)<0$. The function is decreasing. At $x=\frac{7\pi }{6}$, we went from increasing to decreasing—our second local maximum!

In $(\frac{3\pi }{2},2\pi )$: $\sin 2x<0$ and $(2\cos x+\sqrt{3})>0$. The derivative $f^{\prime }(x)>0$. The function is increasing. At $x=\frac{3\pi }{2}$, we went from decreasing to increasing—a local minimum.

After carefully traversing the intervals, we see that the function peaks exactly twice: at $x=\frac{5\pi }{6}$ and $x=\frac{7\pi }{6}$.

The total number of points of local maxima is 2.