The set of number where the function f given by `f(x)=|2x-1|` cos x is differentiable is

Options

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

Answer

To determine the set of numbers where the function \( f(x) = |2x - 1| \cos x \) is differentiable, we will follow these steps:

### Step 1: Identify the points of non-differentiability
The function \( f(x) = |2x - 1| \cos x \) contains an absolute value, which can lead to non-differentiability at points where the expression inside the absolute value is zero.

Set the expression inside the absolute value to zero:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]
Thus, we need to check the differentiability at \( x = \frac{1}{2} \).

### Step 2: Break the function into cases
We can express \( f(x) \) in piecewise form based on the value of \( x \):
\[ f(x) = \begin{cases} -(2x - 1) \cos x & \text{if } x < \frac{1}{2} \\ (2x - 1) \cos x & \text{if } x \geq \frac{1}{2} \end{cases} \]

### Step 3: Check continuity at \( x = \frac{1}{2} \)
To check if \( f(x) \) is differentiable at \( x = \frac{1}{2} \), we first need to ensure it is continuous at that point. We will calculate the left-hand limit (LHL), right-hand limit (RHL), and the function value at \( x = \frac{1}{2} \).

**Left-hand limit:**
\[ \text{LHL} = \lim_{x \to \frac{1}{2}^-} f(x) = \lim_{x \to \frac{1}{2}^-} -(2x - 1) \cos x = -0 \cdot \cos\left(\frac{1}{2}\right) = 0 \]

**Right-hand limit:**
\[ \text{RHL} = \lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}^+} (2x - 1) \cos x = 0 \cdot \cos\left(\frac{1}{2}\right) = 0 \]

**Function value:**
\[ f\left(\frac{1}{2}\right) = |2 \cdot \frac{1}{2} - 1| \cos\left(\frac{1}{2}\right) = 0 \cdot \cos\left(\frac{1}{2}\right) = 0 \]

Since LHL = RHL = \( f\left(\frac{1}{2}\right) = 0 \), the function is continuous at \( x = \frac{1}{2} \).

### Step 4: Check differentiability at \( x = \frac{1}{2} \)
Next, we need to find the left-hand derivative (LHD) and right-hand derivative (RHD) at \( x = \frac{1}{2} \).

**Left-hand derivative:**
\[ f'(x) = \frac{d}{dx}[-(2x - 1) \cos x] = -[2 \cos x - (2x - 1)(\sin x)] \]
Evaluating at \( x = \frac{1}{2} \):
\[ \text{LHD} = -\left[2 \cos\left(\frac{1}{2}\right) - 0 \cdot \sin\left(\frac{1}{2}\right)\right] = -2 \cos\left(\frac{1}{2}\right) \]

**Right-hand derivative:**
\[ f'(x) = \frac{d}{dx}[(2x - 1) \cos x] = [2 \cos x - (2x - 1)(\sin x)] \]
Evaluating at \( x = \frac{1}{2} \):
\[ \text{RHD} = 2 \cos\left(\frac{1}{2}\right) - 0 \cdot \sin\left(\frac{1}{2}\right) = 2 \cos\left(\frac{1}{2}\right) \]

### Step 5: Compare LHD and RHD
Since:
\[ \text{LHD} = -2 \cos\left(\frac{1}{2}\right) \quad \text{and} \quad \text{RHD} = 2 \cos\left(\frac{1}{2}\right) \]
These two values are not equal, hence the function is not differentiable at \( x = \frac{1}{2} \).

### Conclusion
The function \( f(x) \) is differentiable everywhere except at \( x = \frac{1}{2} \). Therefore, the set of numbers where the function is differentiable is:
\[ \mathbb{R} \setminus \left\{\frac{1}{2}\right\} \]