`f(x)` satisfies the relation `f(x)-lamda int_(0)^(pi//2)sinxcostf(t)dt=sinx`
If `f(x)=2` has the least one real root, then

Options

A. `lamda epsilon[1,4]`
B. `lamda epsilon[-1,2]`
C. `lamda epsilon[0,1]`
D. `lamda epsilon [1,3]`

Explanation

To solve the problem step by step, we will analyze the given equation and derive the conditions for the function \( f(x) \).

### Step 1: Write down the given relation
We start with the relation given in the problem:
\[ f(x) - \lambda \int_{0}^{\frac{\pi}{2}} \sin x \cos t f(t) dt = \sin x \]
Rearranging this, we get:
\[ f(x) = \sin x + \lambda \int_{0}^{\frac{\pi}{2}} \sin x \cos t f(t) dt \]

### Step 2: Define \( a \)
Let:
\[ a = \lambda \int_{0}^{\frac{\pi}{2}} \cos t f(t) dt \]
Then we can rewrite \( f(x) \) as:
\[ f(x) = a + \sin x \]

### Step 3: Substitute \( f(t) \)
Now, substituting \( f(t) = a + \sin t \) into the integral:
\[ a = \lambda \int_{0}^{\frac{\pi}{2}} \cos t (a + \sin t) dt \]
This expands to:
\[ a = \lambda \left( a \int_{0}^{\frac{\pi}{2}} \cos t dt + \int_{0}^{\frac{\pi}{2}} \cos t \sin t dt \right) \]

### Step 4: Calculate the integrals
Calculating the integrals:
1. \(\int_{0}^{\frac{\pi}{2}} \cos t dt = 1\)
2. \(\int_{0}^{\frac{\pi}{2}} \cos t \sin t dt = \frac{1}{2}\)

Substituting these values, we have:
\[ a = \lambda \left( a \cdot 1 + \frac{1}{2} \right) \]
This simplifies to:
\[ a = \lambda a + \frac{\lambda}{2} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ a - \lambda a = \frac{\lambda}{2} \]
Factoring out \( a \):
\[ a(1 - \lambda) = \frac{\lambda}{2} \]
Thus,
\[ a = \frac{\lambda}{2(1 - \lambda)} \quad \text{(if \( \lambda \neq 1 \))} \]

### Step 6: Substitute \( a \) back into \( f(x) \)
Now substituting \( a \) back into the expression for \( f(x) \):
\[ f(x) = \frac{\lambda}{2(1 - \lambda)} + \sin x \]

### Step 7: Finding the condition for \( f(x) = 2 \)
We need to find the condition under which \( f(x) = 2 \) has at least one real root:
\[ \frac{\lambda}{2(1 - \lambda)} + \sin x = 2 \]
Rearranging gives:
\[ \sin x = 2 - \frac{\lambda}{2(1 - \lambda)} \]

### Step 8: Determine the range of \( \sin x \)
Since \( \sin x \) must be between -1 and 1, we set up the inequalities:
\[ -1 \leq 2 - \frac{\lambda}{2(1 - \lambda)} \leq 1 \]

### Step 9: Solve the inequalities
1. From \( 2 - \frac{\lambda}{2(1 - \lambda)} \geq -1 \):
\[ \frac{\lambda}{2(1 - \lambda)} \leq 3 \implies \lambda \leq 6(1 - \lambda) \implies 7\lambda \leq 6 \implies \lambda \leq \frac{6}{7} \]

2. From \( 2 - \frac{\lambda}{2(1 - \lambda)} \leq 1 \):
\[ \frac{\lambda}{2(1 - \lambda)} \geq 1 \implies \lambda \geq 2(1 - \lambda) \implies 3\lambda \geq 2 \implies \lambda \geq \frac{2}{3} \]

### Step 10: Combine the results
Combining the results gives:
\[ \frac{2}{3} \leq \lambda \leq \frac{6}{7} \]

### Final Result
Thus, the values of \( \lambda \) for which \( f(x) = 2 \) has at least one real root are:
\[ \lambda \in \left[\frac{2}{3}, \frac{6}{7}\right] \]