If `int_(e)^(x) t f(t)dt=sin x-x cos x-(x^(2))/(2)` for all `x in R-{0}`, then the value of `f((pi)/(6))` will be equal to

Options

A. 0
B. 1
C. `-(1)/(2)`
D. none of these

Explanation

To solve the problem, we start with the given equation:

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

for all \( x \in \mathbb{R} \setminus \{0\} \). We want to find the value of \( f\left(\frac{\pi}{6}\right) \).

### Step 1: Differentiate both sides with respect to \( x \)

Using Leibniz's rule for differentiation under the integral sign, we differentiate the left-hand side:

\[ \frac{d}{dx} \left( \int_{e}^{x} t f(t) \, dt \right) = x f(x) \]

Now, differentiate the right-hand side:

\[ \frac{d}{dx} \left( \sin x - x \cos x - \frac{x^2}{2} \right) = \cos x - \left( \cos x - x \sin x \right) - x = x \sin x - x \]

### Step 2: Set the derivatives equal to each other

Now we equate the derivatives from both sides:

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

### Step 3: Simplify the equation

We can factor out \( x \) from the right-hand side:

\[ x f(x) = x (\sin x - 1) \]

Assuming \( x \neq 0 \), we can divide both sides by \( x \):

\[ f(x) = \sin x - 1 \]

### Step 4: Find \( f\left(\frac{\pi}{6}\right) \)

Now we substitute \( x = \frac{\pi}{6} \) into the function:

\[ f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) - 1 \]

We know that:

\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

So we have:

\[ f\left(\frac{\pi}{6}\right) = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Final Answer

Thus, the value of \( f\left(\frac{\pi}{6}\right) \) is:

\[ \boxed{-\frac{1}{2}} \]