The number of points in the interval (0,2) at | vedclass

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(C) Given the function $f(x) = |x - 0.5| + |x - 1| + 	an x$. We know that the modulus function $|x - a|$ is not differentiable at $x = a$. Therefore,

Vedclass · Accepted Answer

$3$

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(C) Given the function $f(x) = |x - 0.5| + |x - 1| + 	an x$. We know that the modulus 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 $x = 1$. Both $x = 0.5$ and $x = 1$ lie within the interval $(0, 2)$. Additionally,the function $	an x$ is not defined (and hence not differentiable) at $x = \frac{\pi}{2}$. Since $\pi \approx 3.14$,we have $\frac{\pi}{2} \approx 1.57$,which also lies within the interval $(0, 2)$. Thus,the function $f(x)$ is not differentiable at $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$. There are $3$ such points in the interval $(0, 2)$. Therefore,option $C$ is correct.

The number of points in the interval $(0,2)$ at which $f(x)=|x-0.5|+|x-1|+\tan x$ is not differentiable is

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