The domain of differentiability of the function `f(x)=|2x+1|sin3x` is

Options

A. R
B. `R-{-(1)/(2)}`
C. `(-(1)/(2),oo)`
D. `(0,oo)`

Answer

To find the domain of differentiability of the function \( f(x) = |2x + 1| \sin(3x) \), we will analyze the differentiability of each component of the function.

### Step 1: Identify the components of the function
The function can be expressed as the product of two functions:
- \( g(x) = |2x + 1| \)
- \( t(x) = \sin(3x) \)

### Step 2: Analyze the function \( g(x) = |2x + 1| \)
The function \( g(x) \) is the absolute value function, which can be expressed as:
\[ g(x) = \begin{cases} 2x + 1 & \text{if } 2x + 1 \geq 0 \\ -(2x + 1) & \text{if } 2x + 1 < 0 \end{cases} \]
To find where \( g(x) \) is not differentiable, we need to find the point where the expression inside the absolute value equals zero:
\[ 2x + 1 = 0 \implies x = -\frac{1}{2} \]
At \( x = -\frac{1}{2} \), \( g(x) \) has a sharp corner, which means it is not differentiable at this point.

### Step 3: Analyze the function \( t(x) = \sin(3x) \)
The function \( t(x) = \sin(3x) \) is a smooth trigonometric function. It is differentiable everywhere on the real line \( \mathbb{R} \).

### Step 4: Determine the differentiability of \( f(x) = g(x) t(x) \)
The product of two functions is differentiable at a point if both functions are differentiable at that point. Since \( t(x) \) is differentiable everywhere, the differentiability of \( f(x) \) depends solely on \( g(x) \).

### Conclusion
The function \( f(x) = |2x + 1| \sin(3x) \) is not differentiable at \( x = -\frac{1}{2} \). Therefore, the domain of differentiability of \( f(x) \) is:
\[ \text{Domain of differentiability} = \mathbb{R} \setminus \left\{ -\frac{1}{2} \right\} \]