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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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The correct answer is:
3
Given, $f(x)=|x-0.5|+|x-1|+\tan x$
Clearly $f(x)$ is not differentiable at $x=0.5,1$ and $\frac{\pi}{2}$ for $x \in(0,2)$.
$\therefore$ Number of Non differentiable points of $f(x)$ are 3 .
$\therefore$ Hence, option (c) is correct.
Clearly $f(x)$ is not differentiable at $x=0.5,1$ and $\frac{\pi}{2}$ for $x \in(0,2)$.
$\therefore$ Number of Non differentiable points of $f(x)$ are 3 .
$\therefore$ Hence, option (c) is correct.
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