The number of critical points of `f (x)= (int _(0) ^(x) (cos ^(2) t- ""^(3) sqrtt)dt)+3/4 x ^(4//3)-(x+1)/(2)` in ` (0, 6pi]` is:

Options

A. 10
B. 8
C. 6
D. 12

Answer

To find the number of critical points of the function

\[ f(x) = \int_{0}^{x} \left( \cos^2 t - t^{\frac{1}{3}} \right) dt + \frac{3}{4} x^{\frac{4}{3}} - \frac{x + 1}{2} \]

in the interval \( (0, 6\pi] \), we follow these steps:

### Step 1: Differentiate the Function
To find the critical points, we first need to differentiate the function \( f(x) \). According to the Fundamental Theorem of Calculus, the derivative of an integral can be expressed as:

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x} g(t) dt \right) = g(x) \]

where \( g(t) = \cos^2 t - t^{\frac{1}{3}} \). Thus, we have:

\[ f'(x) = \cos^2 x - x^{\frac{1}{3}} + \frac{3}{4} \cdot \frac{4}{3} x^{\frac{1}{3}} - \frac{1}{2} \]

### Step 2: Simplify the Derivative
Now, simplify \( f'(x) \):

\[ f'(x) = \cos^2 x - x^{\frac{1}{3}} + x^{\frac{1}{3}} - \frac{1}{2} \]

The \( -x^{\frac{1}{3}} \) and \( +x^{\frac{1}{3}} \) cancel each other out:

\[ f'(x) = \cos^2 x - \frac{1}{2} \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set \( f'(x) = 0 \):

\[ \cos^2 x - \frac{1}{2} = 0 \]

This simplifies to:

\[ \cos^2 x = \frac{1}{2} \]

### Step 4: Solve for \( x \)
Taking the square root gives:

\[ \cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

The solutions for \( \cos x = \frac{\sqrt{2}}{2} \) are:

\[ x = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{4} + 2k\pi \]

The solutions for \( \cos x = -\frac{\sqrt{2}}{2} \) are:

\[ x = \frac{3\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{4} + 2k\pi \]

### Step 5: Find Critical Points in the Interval \( (0, 6\pi] \)
Now we need to find all the values of \( x \) in the interval \( (0, 6\pi] \):

1. For \( k = 0 \):
- \( x = \frac{\pi}{4} \)
- \( x = \frac{7\pi}{4} \)
- \( x = \frac{3\pi}{4} \)
- \( x = \frac{5\pi}{4} \)

2. For \( k = 1 \):
- \( x = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \)
- \( x = \frac{7\pi}{4} + 2\pi = \frac{15\pi}{4} \)
- \( x = \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4} \)
- \( x = \frac{5\pi}{4} + 2\pi = \frac{13\pi}{4} \)

3. For \( k = 2 \):
- \( x = \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} \) (not in the interval)
- \( x = \frac{7\pi}{4} + 4\pi = \frac{23\pi}{4} \) (not in the interval)
- \( x = \frac{3\pi}{4} + 4\pi = \frac{19\pi}{4} \) (not in the interval)
- \( x = \frac{5\pi}{4} + 4\pi = \frac{21\pi}{4} \) (not in the interval)

The valid critical points in \( (0, 6\pi] \) are:

- \( \frac{\pi}{4} \)
- \( \frac{3\pi}{4} \)
- \( \frac{5\pi}{4} \)
- \( \frac{7\pi}{4} \)
- \( \frac{9\pi}{4} \)
- \( \frac{11\pi}{4} \)
- \( \frac{13\pi}{4} \)
- \( \frac{15\pi}{4} \)

### Total Number of Critical Points
Counting these gives us a total of **12 critical points** in the interval \( (0, 6\pi] \).

### Final Answer
The number of critical points of \( f(x) \) in \( (0, 6\pi] \) is **12**.

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