The set of points where the function `f ` given by `f(x) - |2x-1|` sinx is differentiable is

Options

A. `R`
B. `R - {1/2}`
C. `(0,oo)`
D. None of these

Explanation

To determine the set of points where the function \( f(x) = |2x - 1| \sin x \) is differentiable, we need to analyze the function carefully.

### Step 1: Define the function based on the absolute value
The absolute value function \( |2x - 1| \) can be expressed in piecewise form:
- For \( x < \frac{1}{2} \), \( |2x - 1| = -(2x - 1) = 1 - 2x \)
- For \( x \geq \frac{1}{2} \), \( |2x - 1| = 2x - 1 \)

Thus, we can write:
\[ f(x) = \begin{cases} (1 - 2x) \sin x & \text{if } x < \frac{1}{2} \\ (2x - 1) \sin x & \text{if } x \geq \frac{1}{2} \end{cases} \]

### Step 2: Differentiate the function in each piece
Now, we differentiate \( f(x) \) in each of the two cases.

**Case 1:** For \( x < \frac{1}{2} \):
\[ f(x) = (1 - 2x) \sin x \]
Using the product rule:
\[ f'(x) = (1 - 2x) \cos x + (-2) \sin x = (1 - 2x) \cos x - 2 \sin x \]

**Case 2:** For \( x \geq \frac{1}{2} \):
\[ f(x) = (2x - 1) \sin x \]
Using the product rule:
\[ f'(x) = (2) \sin x + (2x - 1) \cos x = 2 \sin x + (2x - 1) \cos x \]

### Step 3: Check differentiability at the point \( x = \frac{1}{2} \)
To check if \( f(x) \) is differentiable at \( x = \frac{1}{2} \), we need to see if the left-hand derivative equals the right-hand derivative at this point.

**Left-hand derivative at \( x = \frac{1}{2} \):**
\[ f'(\frac{1}{2}^-) = (1 - 2 \cdot \frac{1}{2}) \cos(\frac{1}{2}) - 2 \sin(\frac{1}{2}) = 0 \cdot \cos(\frac{1}{2}) - 2 \sin(\frac{1}{2}) = -2 \sin(\frac{1}{2}) \]

**Right-hand derivative at \( x = \frac{1}{2} \):**
\[ f'(\frac{1}{2}^+) = 2 \sin(\frac{1}{2}) + (2 \cdot \frac{1}{2} - 1) \cos(\frac{1}{2}) = 2 \sin(\frac{1}{2}) + 0 \cdot \cos(\frac{1}{2}) = 2 \sin(\frac{1}{2}) \]

### Step 4: Compare the derivatives
We find:
- \( f'(\frac{1}{2}^-) = -2 \sin(\frac{1}{2}) \)
- \( f'(\frac{1}{2}^+) = 2 \sin(\frac{1}{2}) \)

Since \( -2 \sin(\frac{1}{2}) \neq 2 \sin(\frac{1}{2}) \), the left-hand and right-hand derivatives are not equal.

### Conclusion
Thus, the function \( f(x) \) is not differentiable at \( x = \frac{1}{2} \). Therefore, the set of points where \( f \) is differentiable is:
\[ \text{The function } f \text{ is differentiable for } x \in \mathbb{R} \setminus \left\{ \frac{1}{2} \right\} \]