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 determine the number of points of local maxima for the function \(f(x) = 4 \cos^3 x + 3\sqrt{3} \cos^2 x - 10\) in the interval \((0, 2\pi)\), we need to find the critical points and evaluate the nature of these points.

1. First, find the derivative \(f'(x)\). We use the chain rule and power rule for differentiation:

\(f(x) = 4 (\cos x)^3 + 3\sqrt{3} (\cos x)^2 - 10\)
Let \(u = \cos x\), then \(f(u) = 4u^3 + 3\sqrt{3}u^2 - 10\).
So, \(\frac{d}{dx} (\cos x) = -\sin x\).

1. Differentiate:

\(f'(x) = \frac{d}{dx}(4u^3 + 3\sqrt{3}u^2 - 10)\)
\(= (12u^2 + 6\sqrt{3}u)(- \sin x)\)
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 becomes: \([\sin x = 0] \; \text{or} \; [12 \cos^2 x + 6\sqrt{3} \cos x = 0]\)

• \(\sin x = 0\) gives \(x = \pi, 2\pi\).
• \(12 \cos^2 x + 6\sqrt{3} \cos x = 0\) becomes:

Factor out the expression: \(6 \cos x(2 \cos x + \sqrt{3}) = 0\)
Hence, \(\cos x = 0\) or \(\cos x = -\frac{\sqrt{3}}{2}\).

1. Finding solutions in the interval \((0, 2\pi)\):
   • \(\cos x = 0\) gives \(x = \frac{\pi}{2}, \frac{3\pi}{2}\).
   • \(\cos x = -\frac{\sqrt{3}}{2}\) gives \(x = \frac{5\pi}{6}, \frac{7\pi}{6}\).
2. Evaluate the nature of these critical points using the second derivative test or analyzing the sign change of \(f'(x)\).
3. Using the second derivative test or analyzing, we find that there are 2 points of local maxima at \(x = \frac{\pi}{2}\) and \(x = \frac{7\pi}{6}\) in the interval \((0, 2\pi)\).

Therefore, the number of points of local maxima of \(f\) in the interval \((0, 2\pi)\) is 2.

Answer 2

The given function is:

\[ f(x) = 4 \cos^3(x) + 3\sqrt{3}\cos^2(x) - 10, \quad x \in (0, 2\pi). \]

Step 1: Taking the derivative:

\[ f'(x) = 12 \cos^2(x)(- \sin(x)) + 3\sqrt{3}[2\cos(x)(- \sin(x))], \] \[ f'(x) = -6\sin(x)\cos(x)[2\cos(x) + \sqrt{3}]. \]

Step 2: Critical points occur when:

\[ \sin(x) = 0 \quad \text{or} \quad 2\cos(x) + \sqrt{3} = 0. \]

Step 3: Solving these equations:

\[ \sin(x) = 0 \implies x = 0, \pi, 2\pi, \] \[ \cos(x) = -\frac{\sqrt{3}}{2} \implies x = \frac{5\pi}{6}, \frac{7\pi}{6}. \]

Step 4: Checking the interval \((0, 2\pi)\):

The local maxima occur at: \[ x = \frac{5\pi}{6}, \frac{7\pi}{6}. \]

Final Answer:

\[ \text{2.} \]