`f(x)` satisfies the relation `f(x)-lambdaint_0^(pi//2)sinx*costf(t)dt=sinx` If `lambda > 2` then `f(x)` decreases in

Options

A. `(0,pi)`
B. `(pi/2,3pi//2)`
C. `(-pi//2,pi//2)`
D. none of these

Explanation

To solve the problem, we start with the given relation for the function \( f(x) \):

\[ f(x) - \lambda \int_0^{\frac{\pi}{2}} \sin x \cos t f(t) dt = \sin x \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express \( f(x) \):

\[ f(x) = \sin x + \lambda \int_0^{\frac{\pi}{2}} \sin x \cos t f(t) dt \]

### Step 2: Define the Integral
Let us define the integral:

\[ A = \int_0^{\frac{\pi}{2}} \cos t f(t) dt \]

Then we can rewrite \( f(x) \):

\[ f(x) = \sin x (1 + \lambda A) \]

### Step 3: Substitute for \( A \)
Now, if we substitute \( f(t) \) back into the integral \( A \):

\[ A = \int_0^{\frac{\pi}{2}} \cos t \left( \sin t (1 + \lambda A) \right) dt \]

This simplifies to:

\[ A = (1 + \lambda A) \int_0^{\frac{\pi}{2}} \sin t \cos t dt \]

### Step 4: Evaluate the Integral
The integral \( \int_0^{\frac{\pi}{2}} \sin t \cos t dt \) can be evaluated as:

\[ \int_0^{\frac{\pi}{2}} \sin t \cos t dt = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2t) dt = \frac{1}{2} \left[-\frac{1}{2} \cos(2t)\right]_0^{\frac{\pi}{2}} = \frac{1}{2} \]

Thus, we have:

\[ A = (1 + \lambda A) \cdot \frac{1}{2} \]

### Step 5: Solve for \( A \)
Rearranging gives:

\[ 2A = 1 + \lambda A \]

\[ 2A - \lambda A = 1 \]

\[ A(2 - \lambda) = 1 \]

\[ A = \frac{1}{2 - \lambda} \]

### Step 6: Substitute Back to Find \( f(x) \)
Now substituting \( A \) back into the expression for \( f(x) \):

\[ f(x) = \sin x \left(1 + \lambda \cdot \frac{1}{2 - \lambda}\right) \]

This simplifies to:

\[ f(x) = \sin x \left(\frac{2 - \lambda + \lambda}{2 - \lambda}\right) = \sin x \left(\frac{2}{2 - \lambda}\right) \]

### Step 7: Determine the Decreasing Condition
To determine when \( f(x) \) is decreasing, we differentiate \( f(x) \):

\[ f'(x) = \frac{2}{2 - \lambda} \cos x \]

For \( f(x) \) to be decreasing, we need:

\[ f'(x) < 0 \]

This implies:

\[ \frac{2}{2 - \lambda} \cos x < 0 \]

Since \( \frac{2}{2 - \lambda} > 0 \) when \( \lambda < 2 \), we need \( \cos x < 0 \). This occurs in the intervals:

\[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]

### Conclusion
Thus, for \( \lambda > 2 \), the function \( f(x) \) is decreasing in the interval:

\[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]