The domain of the function `f(x)=sqrt(((sinx-e^(x))(cosx-3)("In"x+2))/((x^(4)-x+1/2)(2x-7)^(30)))` belongs to

Options

A. `[1/(e^(2)),7/2)cup(7/2,oo)`
B. `(-1/2,0)`
C. `(0,1/(e^(5)))`
D. `phi`

Answer

To find the domain of the function

\[ f(x) = \sqrt{\frac{(\sin x - e^x)(\cos x - 3)(\ln x + 2)}{(x^4 - x + \frac{1}{2})(2x - 7)^{30}}} \]

we need to ensure that the expression inside the square root is non-negative, meaning:

\[ (\sin x - e^x)(\cos x - 3)(\ln x + 2) \geq 0 \]

and

\[ (x^4 - x + \frac{1}{2})(2x - 7)^{30} > 0. \]

### Step 1: Analyze \( (2x - 7)^{30} \)

The term \( (2x - 7)^{30} \) must be positive. This occurs when:

\[ 2x - 7 \neq 0 \]
\[ x \neq \frac{7}{2}. \]

Thus, \( x \) must be greater than \( \frac{7}{2} \) or less than \( \frac{7}{2} \).

### Step 2: Analyze \( \ln x + 2 \)

The term \( \ln x + 2 \) must be non-negative:

\[ \ln x + 2 \geq 0 \]
\[ \ln x \geq -2 \]
\[ x \geq e^{-2} = \frac{1}{e^2}. \]

### Step 3: Analyze \( \sin x - e^x \)

For \( \sin x - e^x \), we need to determine when this expression is non-negative. Since \( e^x \) grows faster than \( \sin x \), we can analyze specific values:

- For \( x = 0 \): \( \sin(0) - e^0 = 0 - 1 < 0 \).
- For \( x = 1 \): \( \sin(1) - e^1 \approx 0.84 - 2.71 < 0 \).
- For larger \( x \), \( e^x \) will dominate \( \sin x \).

Thus, \( \sin x - e^x \) is negative for \( x > 0 \).

### Step 4: Analyze \( \cos x - 3 \)

The term \( \cos x - 3 \) is always negative since \( \cos x \) ranges from -1 to 1. Therefore, \( \cos x - 3 < 0 \) for all \( x \).

### Step 5: Combine the conditions

From the analysis, we have:

1. \( \ln x + 2 \geq 0 \) gives \( x \geq \frac{1}{e^2} \).
2. \( 2x - 7 \neq 0 \) gives \( x \neq \frac{7}{2} \).
3. \( \sin x - e^x < 0 \) for \( x > 0 \).
4. \( \cos x - 3 < 0 \) for all \( x \).

### Final Domain

Since \( \sin x - e^x \) and \( \cos x - 3 \) are both negative, the product \( (\sin x - e^x)(\cos x - 3) \) is positive. Thus, the only condition we need to satisfy is:

\[ \ln x + 2 \geq 0 \text{ and } x \neq \frac{7}{2}. \]

This leads us to the final domain:

\[ x \in \left[\frac{1}{e^2}, \frac{7}{2}\right) \cup \left(\frac{7}{2}, \infty\right). \]