If f(x)={sin⁡x,x≤0tan⁡x,0<x<2πcos⁡x,2π≤x<3π3π,x=3πf(x) = \begin{cases} \sin x, & x \le 0 \\ \tan x, & 0 < x < 2\pi \\ \cos x, & 2\pi \le x < 3\pi \\ 3\pi, & x = 3\pi \end{cases}…

Mathematics

Problem

If f(x)={sin⁡x,x≤0tan⁡x,0<x<2πcos⁡x,2π≤x<3π3π,x=3πf(x) = \begin{cases} \sin x, & x \le 0 \\ \tan x, & 0 < x < 2\pi \\ \cos x, & 2\pi \le x < 3\pi \\ 3\pi, & x = 3\pi \end{cases}, then f(x)f(x) is discontinuous at a. π2,3π2,2π\frac{\pi}{2}, \frac{3\pi}{2}, 2\pi b. 0,π2,π,3π2,3π0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 3\pi c. π2,2π\frac{\pi}{2}, 2\pi d. None of these

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