The set of points where the function f given by f (x) =| 2x – 1| sin x is differentiable is

The correct option is R \ {1/2} (represented in the options as R = {1/2}).

To determine the set of points where the function $f\left(x\right)=|2x-1|\sin x$ is differentiable, we examine the differentiability of its individual components.

The function is a product of two functions:
1. $g\left(x\right)=|2x-1|$
2. $h\left(x\right)=\sin x$

The trigonometric function $h\left(x\right)=\sin x$ is differentiable everywhere on the set of real numbers $\mathbb{R}$.

The absolute value function $g\left(x\right)=|2x-1|$ is differentiable everywhere except at the point where the expression inside the absolute value becomes zero. Let's find this critical point:

$2x-1=0⟹x=\frac{1}{2}$

Since both components are differentiable for all $x\ne \frac{1}{2}$, their product $f\left(x\right)$ is also differentiable for all $x\ne \frac{1}{2}$. We now need to check the differentiability of $f\left(x\right)$ specifically at the point $x=\frac{1}{2}$ using the definition of derivatives.

Let's find the Left-Hand Derivative (LHD) at $x=\frac{1}{2}$:

$\text{LHD}=\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$

Since $f\left(\frac{1}{2}\right)=\left|2\right(\frac{1}{2})-1|\sin \left(\frac{1}{2}\right)=0$, and for $h<0$ we have $|2\left(\frac{1}{2}+h\right)-1|=\left|2h\right|=-2h$:

$\text{LHD}=\underset{h\to 0^{-}}{\lim }\frac{-2h\sin \left(\frac{1}{2}+h\right)}{h}=-2\sin \left(\frac{1}{2}\right)$

Now, let's find the Right-Hand Derivative (RHD) at $x=\frac{1}{2}$:

$\text{RHD}=\underset{h\to 0^{+}}{\lim }\frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$

For $h>0$, we have $|2\left(\frac{1}{2}+h\right)-1|=\left|2h\right|=2h$:

$\text{RHD}=\underset{h\to 0^{+}}{\lim }\frac{2h\sin \left(\frac{1}{2}+h\right)}{h}=2\sin \left(\frac{1}{2}\right)$

Since $\sin \left(\frac{1}{2}\right)\ne 0$, we have:

$\text{LHD}\ne \text{RHD}$

Because the left-hand derivative and right-hand derivative are not equal at $x=\frac{1}{2}$, the function $f\left(x\right)$ is not differentiable at this point.

Thus, the function is differentiable at all points in $\mathbb{R}$ except at $x=\frac{1}{2}$. The set of points of differentiability is $\mathbb{R}\setminus \left\{\frac{1}{2}\right\}$ (given in the option format as $\mathbb{R}=\left\{\frac{1}{2}\right\}$ due to character rendering of the set subtraction symbol).