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 and that the denominator is not zero.

### Step 1: Identify the conditions for the denominator

The denominator is given by

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

1. **Condition 1:** The term \( (2x - 7)^{30} \) must not be zero. Thus, we have:

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

2. **Condition 2:** The term \( x^4 - x + \frac{1}{2} \) must not be zero. To analyze this, we note that:

- The polynomial \( x^4 - x + \frac{1}{2} \) is a continuous function.
- Since \( x^4 \) grows faster than \( x \) for large \( x \), and the constant term is positive, this polynomial does not cross zero for any real \( x \). Thus, it is always positive.

### Step 2: Identify the conditions for the numerator

The numerator is given by

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

1. **Condition 3:** The term \( \ln x + 2 \) must be non-negative:

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

2. **Condition 4:** The term \( \cos x - 3 \) is always negative since \( \cos x \) ranges from -1 to 1. Thus, \( \cos x - 3 < 0 \) for all \( x \).

3. **Condition 5:** The term \( \sin x - e^x \) must be non-negative. Since \( e^x \) grows exponentially while \( \sin x \) oscillates between -1 and 1, for \( x > 0 \), \( e^x \) will always be greater than \( \sin x \). Therefore, \( \sin x - e^x < 0 \) for \( x > 0 \).

### Step 3: Combine conditions

From the conditions derived:

- \( x \geq \frac{1}{e^2} \) (from \( \ln x + 2 \geq 0 \))
- \( x \neq \frac{7}{2} \) (from the denominator)

Thus, the domain of \( f(x) \) can be expressed as:

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

### Final Answer

The domain of the function \( f(x) \) is:

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