The set of points of discontinuity of the fun | vedclass

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(D) function $f(x)$ is discontinuous where it is undefined. $1$. The denominator $x(x-3)(x-5)$ is zero at $x = 0, 3, 5$. Thus,the function is undefine

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$\{0, 1, 3, 5\} \cup \{(2n+1)\frac{\pi}{2}, n \in Z\}$

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(D) function $f(x)$ is discontinuous where it is undefined. $1$. The denominator $x(x-3)(x-5)$ is zero at $x = 0, 3, 5$. Thus,the function is undefined at these points. $2$. The term $	an x$ is undefined when $x = (2n+1)\frac{\pi}{2}$ for $n \in Z$. $3$. The term $	an^{-1}\left(\frac{1}{x-1}ight)$ is defined for all $x 
eq 1$. At $x = 1$,the limit $\lim_{x 	o 1^+} 	an^{-1}\left(\frac{1}{x-1}ight) = \frac{\pi}{2}$ and $\lim_{x 	o 1^-} 	an^{-1}\left(\frac{1}{x-1}ight) = -\frac{\pi}{2}$. Since the left-hand limit and right-hand limit are not equal,the function is discontinuous at $x = 1$. Combining all these,the set of points of discontinuity is $\{0, 1, 3, 5\} \cup \{(2n+1)\frac{\pi}{2}, n \in Z\}$.

The set of points of discontinuity of the function $f(x) = \frac{\tan x \cdot \tan^{-1}\left(\frac{1}{x-1}\right)}{x(x-3)(x-5)}$ is

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