The set of points where the function

Solution:

Let $g(x)=|2 x-1|$ and $h(x)=\sin x$

We know that, the trigonometric functions are differentiable in their respective domain.

So, $h(x)=\sin x$ is differentiable for all $x \in \mathrm{R}$.

Now,

$g(x)=|2 x-1|= \begin{cases}2 x-1, & x \geq \frac{1}{2} \\ -(2 x-1), & x<\frac{1}{2}\end{cases}$

$(2 x-1)$ and $-(2 x-1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} . \mathrm{So}, f(x)$ is differentiable for all $x<\frac{1}{2}$ and for all $x>\frac{1}{2}$.

So, we need to check the differentiability of $g(x)$ at $x=\frac{1}{2}$.

We have

$L g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{g\left(\frac{1}{2}-h\right)-g\left(\frac{1}{2}\right)}{-h}$

$\Rightarrow L g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{-\left[2\left(\frac{1}{2}-h\right)-1\right]-\left(2 \times \frac{1}{2}-1\right)}{-h}$

$\Rightarrow L g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{2 h}{-h}$

$\Rightarrow L g^{\prime}\left(\frac{1}{2}\right)=-2$

And

$R g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{g\left(\frac{1}{2}+h\right)-g\left(\frac{1}{2}\right)}{h}$

$\Rightarrow R g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{\left[2\left(\frac{1}{2}+h\right)-1\right]-\left(2 \times \frac{1}{2}-1\right)}{h}$

$\Rightarrow R g^{\prime}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0} \frac{2 h}{h}$

$\Rightarrow R g^{\prime}\left(\frac{1}{2}\right)=2$

$\therefore L g^{\prime}\left(\frac{1}{2}\right) \neq R g^{\prime}\left(\frac{1}{2}\right)$

So, $g(x)=|2 x-1|$ is not differentiable at $x=\frac{1}{2}$.

The function $g(x)=|2 x-1|$ is differentiable for all $x \in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

We know that, the product of two differentiable functions is differentiable.

$\therefore f(x)=g(x) \times h(x)=|2 x-1| \sin x$ is differentiable for all $x \in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

Thus, the set of points where the function $f(x)=|2 x-1| \sin x$ is differentiable is $\mathrm{R}-\left\{\frac{1}{2}\right\}$.

Hence, the correct answer is option (b).