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(C) The function $f(x)$ is defined as:$f(x) = \begin{cases} x-1, & x \leq 1 \ 2-x^2, & 1 < x \leq 3 \ x-10, & 3 < x < 5 \ 2x, & x \geq 5 \end{cases

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$\{1,5\}$

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(C) The function $f(x)$ is defined as:$f(x) = \begin{cases} x-1, & x \leq 1 \ 2-x^2, & 1 $f(x)$ is continuous in the intervals $(-\infty, 1), (1, 3), (3, 5),$ and $(5, \infty)$. We check the continuity at the transition points $x=1, 3,$ and $5$.$(i)$ At $x=1$:$\lim_{x ightarrow 1^-} f(x) = 1-1 = 0$$\lim_{x ightarrow 1^+} f(x) = 2-(1)^2 = 1$Since $\lim_{x ightarrow 1^-} f(x) 
eq \lim_{x ightarrow 1^+} f(x)$,$f$ is discontinuous at $x=1$.(ii) At $x=3$:$\lim_{x ightarrow 3^-} f(x) = 2-(3)^2 = 2-9 = -7$$f(3) = 2-(3)^2 = -7$$\lim_{x ightarrow 3^+} f(x) = 3-10 = -7$Since $\lim_{x ightarrow 3^-} f(x) = f(3) = \lim_{x ightarrow 3^+} f(x)$,$f$ is continuous at $x=3$.(iii) At $x=5$:$\lim_{x ightarrow 5^-} f(x) = 5-10 = -5$$\lim_{x ightarrow 5^+} f(x) = 2(5) = 10$Since $\lim_{x ightarrow 5^-} f(x) 
eq \lim_{x ightarrow 5^+} f(x)$,$f$ is discontinuous at $x=5$.Thus,the set of points of discontinuity is $\{1, 5\}$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x-1, & \text{for } x \leq 1 \\ 2-x^2, & \text{for } 1 < x \leq 3 \\ x-10, & \text{for } 3 < x < 5 \\ 2x, & \text{for } x \geq 5 \end{cases}$,then the set of points of discontinuity of $f$ is

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