If the equation `cosx + 3cos(2Kx) = 4` has exactly one solution, then

Options

A. (a) K is rational number of the form `P/(P+1), P ne -1`
B. (b)K is irrational number whose rational approximation does not exceed 2.
C. (c)K is irrational number
D. (d)K is rational number of the form `P/(P-1), P ne 1`.

Answer

To solve the problem, we need to analyze the equation given:

**Equation:**
\[ \cos x + 3 \cos(2Kx) = 4 \]

### Step 1: Analyze the equation
The maximum value of \(\cos x\) is 1. Therefore, for the equation to hold true, we need:
\[ \cos x + 3 \cos(2Kx) = 4 \]
This implies:
\[ \cos x \leq 1 \]
\[ 3 \cos(2Kx) \leq 3 \]
Thus, the only way for the left-hand side to equal 4 is if:
\[ \cos x = 1 \quad \text{and} \quad \cos(2Kx) = 1 \]

### Step 2: Set conditions for \(\cos x\) and \(\cos(2Kx)\)
From \(\cos x = 1\), we have:
\[ x = 2n\pi \quad (n \in \mathbb{Z}) \]
From \(\cos(2Kx) = 1\), we have:
\[ 2Kx = 2m\pi \quad (m \in \mathbb{Z}) \]
This simplifies to:
\[ Kx = m\pi \]

### Step 3: Substitute \(x\) into the equation
Substituting \(x = 2n\pi\) into \(Kx = m\pi\):
\[ K(2n\pi) = m\pi \]
This gives:
\[ K = \frac{m}{2n} \]

### Step 4: Condition for exactly one solution
For the equation to have exactly one solution, the values of \(K\) must be such that the ratio \(\frac{m}{2n}\) leads to a unique value of \(K\). This occurs when \(n\) and \(m\) are such that they do not allow for multiple integer solutions.

### Step 5: Determine the form of \(K\)
The problem states that \(K\) must be a rational number of the form \(\frac{p}{p+1}\) where \(p \neq 1\).

### Step 6: Check the options
1. **Option A:** \(K = \frac{p}{p+1}\) where \(p \neq 1\)
2. **Option B:** \(K\) is an irrational number whose rational approximation does not exceed 2.
3. **Option C:** (not specified)
4. **Option D:** (not specified)

From the analysis, we can conclude that for the equation to have exactly one solution, \(K\) must be in the form where it does not allow for multiple integer solutions.

### Conclusion
The only suitable option that fits the requirement of having exactly one solution is:
**Option B:** \(K\) is an irrational number whose rational approximation does not exceed 2.