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

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(C) The function is given by $f(x) = |x - 0.5| + |x - 1| + 	an x$. We need to check the points where the function is not differentiable in the interv

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$3$

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(C) The function is given by $f(x) = |x - 0.5| + |x - 1| + 	an x$. We need to check the points where the function is not differentiable in the interval $(0, 2)$. The absolute value functions $|x - 0.5|$ and $|x - 1|$ are not differentiable at $x = 0.5$ and $x = 1$ respectively,as these are the points where the expression inside the modulus becomes zero. Both $x = 0.5$ and $x = 1$ lie within the interval $(0, 2)$. The function $	an x$ is not defined (and thus not differentiable) at $x = \frac{\pi}{2} \approx 1.57$. Since $x = \frac{\pi}{2}$ also lies within the interval $(0, 2)$,the function $f(x)$ is not differentiable at this point as well. Therefore,the points where $f(x)$ is not differentiable are $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$. There are $3$ such points in the interval $(0, 2)$.

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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