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

We have, $f(x) = |2x - 1|\sin x$

At $x = \frac{1}{2},$ $f(x)$ is not differentiable.

Hence, $f(x)$ is differentiable in $R - \left( {\frac{1}{2}} \right)$.

$Rf'\left( {\frac{1}{2}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {\frac{1}{2} + h} \right) - f\left( {\frac{1}{2}} \right)}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{\left| {2\left( {\frac{1}{2} + h} \right) - 1} \right|\sin \left( {\frac{1}{2} + h} \right) - 0}}{h}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{|2h| \cdot \sin \left( {\frac{{1 + 2h}}{2}} \right)}}{h} = 2 \cdot \sin \frac{1}{2}$
and $Lf'\left( {\frac{1}{2}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {\frac{1}{2} - h} \right) - f\left( {\frac{1}{2}} \right)}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( {\frac{1}{2} - h} \right)}^{ - 1}}\mid - \sin \left( {\frac{1}{2} - h} \right) - 0}}{{ - h}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{|0 - 2h| - \sin \left( {\frac{1}{2} - h} \right)}}{{ - h}} = - 2\sin \left( {\frac{1}{2}} \right)$

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