Question

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:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

To ascertain the count of local maxima for the function \(f(x) = 4 \cos^3 x + 3\sqrt{3} \cos^2 x - 10\) within the interval \((0, 2\pi)\), critical points must be identified and their nature determined.

1. Compute the derivative \(f'(x)\) using the chain and power rules:

\(f(x) = 4 (\cos x)^3 + 3\sqrt{3} (\cos x)^2 - 10\)
Let \(u = \cos x\). Then \(f(u) = 4u^3 + 3\sqrt{3}u^2 - 10\).
The derivative of \(\cos x\) with respect to x is \(\frac{d}{dx} (\cos x) = -\sin x\).

1. Differentiate the function:

\(\frac{d}{dx} f(x) = \frac{d}{dx}(4u^3 + 3\sqrt{3}u^2 - 10)\)
\(= (12u^2 + 6\sqrt{3}u) \frac{du}{dx}\)
\(= (12u^2 + 6\sqrt{3}u)(- \sin x)\)
Substitute \(u = \cos x\) back:  \(f'(x) = - (12 \cos^2 x + 6\sqrt{3} \cos x) \sin x\)

1. Set \(f'(x) = 0\) to find critical points:

\(- (12 \cos^2 x + 6\sqrt{3} \cos x) \sin x = 0\)
This equation implies either \(\sin x = 0\) or \([12 \cos^2 x + 6\sqrt{3} \cos x = 0]\).

• The condition \(\sin x = 0\) yields \(\pi\) and \(2\pi\) as solutions.
• The equation \(12 \cos^2 x + 6\sqrt{3} \cos x = 0\) can be factored as:

\(6 \cos x(2 \cos x + \sqrt{3}) = 0\)
This implies \(\cos x = 0\) or \(\cos x = -\frac{\sqrt{3}}{2}\).

1. Identify solutions within the interval \((0, 2\pi)\):
   • For \(\cos x = 0\), the solutions are \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\).
   • For \(\cos x = -\frac{\sqrt{3}}{2}\), the solutions are \(\frac{5\pi}{6}\) and \(\frac{7\pi}{6}\).
2. Determine the nature of these critical points using the second derivative test or by analyzing sign changes in \(f'(x)\).
3. Following the analysis, there are 2 local maxima at \(x = \frac{\pi}{2}\) and \(x = \frac{7\pi}{6}\) within the interval \((0, 2\pi)\).

Consequently, the function \(f\) exhibits 2 points of local maxima in the interval \((0, 2\pi)\).