The total number of points of non-differentia | vedclass

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(D) To find the points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in the interval $(0, 2\pi)$,we analyze the inte

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$12$

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(D) To find the points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in the interval $(0, 2\pi)$,we analyze the intersections of the graphs of $y = |\sin x|$,$y = |\cos x|$,and $y = \frac{1}{4}$.
$1$. The function $f(x)$ is the lower envelope of these three curves.
$2$. The points of non-differentiability occur at the intersection points of these curves where the slope changes abruptly or where the function is not smooth.
$3$. By plotting the graphs of $y = |\sin x|$,$y = |\cos x|$,and $y = \frac{1}{4}$ on the interval $(0, 2\pi)$,we identify the points where the minimum switches from one function to another.
$4$. The intersection points are:
- $|\sin x| = |\cos x|$ at $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
- $|\sin x| = \frac{1}{4}$ at $x = \arcsin(\frac{1}{4}), \pi - \arcsin(\frac{1}{4}), \pi + \arcsin(\frac{1}{4}), 2\pi - \arcsin(\frac{1}{4})$.
- $|\cos x| = \frac{1}{4}$ at $x = \arccos(\frac{1}{4}), \pi - \arccos(\frac{1}{4}), \pi + \arccos(\frac{1}{4}), 2\pi - \arccos(\frac{1}{4})$.
$5$. By examining the graph,the function $f(x)$ consists of segments of these curves. The points where the minimum switches from one function to another are the points of non-differentiability. Counting these points in $(0, 2\pi)$,we find there are $12$ such points where the curves intersect and the minimum changes. Thus,the correct answer is $12$.

The total number of points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in $(0, 2\pi)$ is

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