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(B) Given $f(x) = \begin{cases} |x| + [x], & -1 \le x < 1 \ x + |x|, & 1 \le x < 2 \ x + [x], & 2 \le x \le 3 \end{cases}$.For $-1 \le x < 0$,$f(x)

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(B) Given $f(x) = \begin{cases} |x| + [x], & -1 \le x For $-1 \le x For $0 \le x For $1 \le x For $2 \le x For $x = 3$,$f(3) = 3 + [3] = 3 + 3 = 6$.Checking continuity:At $x = 0$: $LHL = \lim_{x 	o 0^-} -(x+1) = -1$,$RHL = \lim_{x 	o 0^+} x = 0$. Since $LHL 
eq RHL$,$f$ is discontinuous at $x = 0$.At $x = 1$: $LHL = \lim_{x 	o 1^-} x = 1$,$RHL = \lim_{x 	o 1^+} 2x = 2$. Since $LHL 
eq RHL$,$f$ is discontinuous at $x = 1$.At $x = 2$: $LHL = \lim_{x 	o 2^-} 2x = 4$,$RHL = \lim_{x 	o 2^+} (x+2) = 4$. Since $LHL = RHL = f(2) = 4$,$f$ is continuous at $x = 2$.At $x = 3$: $LHL = \lim_{x 	o 3^-} (x+2) = 5$,$f(3) = 6$. Since $LHL 
eq f(3)$,$f$ is discontinuous at $x = 3$.Thus,$f$ is discontinuous at $x = 0, 1, 3$ (three points).

Let $f:[-1, 3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \le x < 1 \\ x + |x|, & 1 \le x < 2 \\ x + [x], & 2 \le x \le 3 \end{cases}$ where $[t]$ denotes the greatest integer function. Then $f$ is discontinuous at:

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