`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt`
`f(x)` is not invertible for

Options

A. `x epsilon [-(pi)/2-tan^(-1)2,(pi)/2-tan^(-1)2]`
B. `x epsilon[tan^(-1)1/2,pi+"tan"^(-1)1/2]`
C. `x epsilon[pi+cot^(-1)2, 2pi+cot^(-1)2]`
D. none of these

Explanation

To solve the problem, we need to analyze the function given by:

\[ f(x) = \sin x + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin x + t \cos x) f(t) \, dt \]

### Step 1: Simplify the Integral

We can break the integral into two parts:

\[ f(x) = \sin x + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin x f(t) \, dt + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \cos x f(t) \, dt \]

### Step 2: Factor Out Constants

The first integral can be factored since \(\sin x\) is constant with respect to \(t\):

\[ f(x) = \sin x \left( 1 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t) \, dt \right) + \cos x \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt \]

Let:

\[ A = 1 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t) \, dt \]
\[ B = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t f(t) \, dt \]

Thus, we can rewrite \(f(x)\) as:

\[ f(x) = A \sin x + B \cos x \]

### Step 3: Find the Constants \(A\) and \(B\)

1. **Finding \(A\)**:

Substituting \(f(t) = A \sin t + B \cos t\) into the integral for \(A\):

\[ A = 1 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (A \sin t + B \cos t) \, dt \]

The integral of \(\sin t\) over the symmetric interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is zero, and the integral of \(\cos t\) is:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos t \, dt = 2 \]

Thus,

\[ A = 1 + 0 + 2B \implies A = 1 + 2B \]

2. **Finding \(B\)**:

Substituting \(f(t)\) into the integral for \(B\):

\[ B = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t (A \sin t + B \cos t) \, dt \]

Again, the integral of \(t \sin t\) over this symmetric interval is zero, and for \(t \cos t\):

\[ B = 0 + B \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \cos t \, dt \]

The integral of \(t \cos t\) can be computed using integration by parts, which results in a non-zero constant. For simplicity, let’s denote this integral as \(C\):

\[ B = B \cdot C \]

### Step 4: Solve the System of Equations

Now we have two equations:

1. \(A = 1 + 2B\)
2. \(B = B \cdot C\)

From the second equation, if \(B \neq 0\), we can divide by \(B\):

\[ 1 = C \implies B = 0 \]

Substituting \(B = 0\) into the first equation gives:

\[ A = 1 \]

### Step 5: Final Form of \(f(x)\)

Thus, we have:

\[ f(x) = \sin x \]

### Step 6: Determine Non-Invertibility

The function \(f(x) = \sin x\) is not invertible in the intervals where it is not one-to-one, specifically:

\[ f(x) \text{ is not invertible for } x \in [\frac{\pi}{2} + 2k\pi, \frac{3\pi}{2} + 2k\pi], \text{ where } k \in \mathbb{Z} \]

### Conclusion

The function \(f(x)\) is not invertible for the specified intervals.

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