The number of points at which the function f( | vedclass

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(C) The function is given by $f(x) = |x - 0.5| + |x - 1| + 	an x$.$1$. The absolute value function $|x - a|$ is not differentiable at $x = a$.Therefo

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$3$

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(C) The function is given by $f(x) = |x - 0.5| + |x - 1| + 	an x$.$1$. The absolute value function $|x - a|$ is not differentiable at $x = a$.Therefore,$|x - 0.5|$ is not differentiable at $x = 0.5$,and $|x - 1|$ is not differentiable at $x = 1$.$2$. The function $	an x$ is not defined (and thus not differentiable) at $x = \frac{\pi}{2}$.$3$. We check if these points lie in the interval $(0, 2)$:- $0.5 \in (0, 2)$- $1 \in (0, 2)$- $\frac{\pi}{2} \approx 1.57 \in (0, 2)$Since all three points $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$ lie within the interval $(0, 2)$,the function $f(x)$ is not differentiable at these $3$ points.Thus,the total number of points is $3$.

The number of points at which the function $f(x) = |x - 0.5| + |x - 1| + \tan x$ does not have a derivative in the interval $(0, 2)$ is:

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