The set of values of `lambda` for which the function `f(x) = (4lambda - 3) (x + log 5) + 2(lambda-7).cot"" (x)/2 sin^2 "" (x)/2` does not possess critical point is :

Options

A. `[1, oo]`
B. `(2, oo)`
C. `(-oo,-4//3)`
D. `(-oo,-1)`

Answer

To solve the problem, we need to find the set of values of \( \lambda \) for which the function

\[ f(x) = (4\lambda - 3)(x + \log 5) + 2(\lambda - 7) \cdot \frac{\cot\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \]

does not possess critical points. A critical point occurs where the derivative of the function is zero or undefined.

### Step-by-Step Solution:

1. **Simplify the Function**:
We start by rewriting the function \( f(x) \):
\[ f(x) = (4\lambda - 3)(x + \log 5) + 2(\lambda - 7) \cdot \frac{\cot\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} \]

We can express \( \cot\left(\frac{x}{2}\right) \) in terms of sine and cosine:
\[ \cot\left(\frac{x}{2}\right) = \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \]

Thus, we have:
\[ f(x) = (4\lambda - 3)(x + \log 5) + 2(\lambda - 7) \cdot \frac{\cos\left(\frac{x}{2}\right)}{\sin^3\left(\frac{x}{2}\right)} \]

2. **Find the Derivative**:
We differentiate \( f(x) \):
\[ f'(x) = (4\lambda - 3) + 2(\lambda - 7) \cdot \frac{d}{dx}\left(\frac{\cot\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}\right) \]

The derivative of \( \cot\left(\frac{x}{2}\right) \) is:
\[ -\frac{1}{2}\csc^2\left(\frac{x}{2}\right) \]

Thus, we have:
\[ f'(x) = 4\lambda - 3 + (\lambda - 7) \cdot \cos\left(\frac{x}{2}\right) \cdot \left(-\frac{1}{2}\csc^2\left(\frac{x}{2}\right)\right) \]

3. **Set Derivative to Zero**:
For critical points, we set \( f'(x) = 0 \):
\[ 4\lambda - 3 + (\lambda - 7) \cdot \cos\left(\frac{x}{2}\right) = 0 \]

Rearranging gives:
\[ \cos\left(\frac{x}{2}\right) = \frac{3 - 4\lambda}{\lambda - 7} \]

4. **Analyze the Range of Cosine**:
Since \( \cos\left(\frac{x}{2}\right) \) must be in the range \([-1, 1]\), we need:
\[ -1 \leq \frac{3 - 4\lambda}{\lambda - 7} \leq 1 \]

This gives us two inequalities to solve:
- \( \frac{3 - 4\lambda}{\lambda - 7} \geq -1 \)
- \( \frac{3 - 4\lambda}{\lambda - 7} \leq 1 \)

5. **Solve the Inequalities**:
- For \( \frac{3 - 4\lambda}{\lambda - 7} \geq -1 \):
\[ 3 - 4\lambda + \lambda - 7 \geq 0 \implies -3\lambda - 4 \geq 0 \implies \lambda \leq -\frac{4}{3} \]

- For \( \frac{3 - 4\lambda}{\lambda - 7} \leq 1 \):
\[ 3 - 4\lambda \leq \lambda - 7 \implies 10 \leq 5\lambda \implies \lambda \geq 2 \]

6. **Combine the Results**:
From the inequalities, we have:
- \( \lambda \leq -\frac{4}{3} \)
- \( \lambda \geq 2 \)

Thus, the set of values of \( \lambda \) for which the function does not possess critical points is:
\[ (-\infty, -\frac{4}{3}] \cup [2, 7) \cup (7, \infty) \]

### Final Answer:
The set of values of \( \lambda \) for which the function does not possess critical points is:
\[ (-\infty, -\frac{4}{3}] \cup [2, 7) \cup (7, \infty) \]