Consider the function f(x) = [x] + |1 - x| fo | vedclass

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(A) Given $f(x) = [x] + |1 - x|$ for $x \in [-1, 3]$.For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = -1 + (1 - x) = -x$.For $x \in [0, 1)$,$[x] = 0$,so $f(x

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Statement $1$ is true; Statement $2$ is false.

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(A) Given $f(x) = [x] + |1 - x|$ for $x \in [-1, 3]$.For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = -1 + (1 - x) = -x$.For $x \in [0, 1)$,$[x] = 0$,so $f(x) = 0 + (1 - x) = 1 - x$.For $x \in [1, 2)$,$[x] = 1$,so $f(x) = 1 + (x - 1) = x$.For $x \in [2, 3)$,$[x] = 2$,so $f(x) = 2 + (x - 1) = x + 1$.At $x = 3$,$f(3) = [3] + |1 - 3| = 3 + 2 = 5$.Checking continuity:At $x=0$: $LHL = \lim_{x 	o 0^-} (-x) = 0$,$RHL = \lim_{x 	o 0^+} (1 - x) = 1$. Since $LHL 
eq RHL$,$f$ is discontinuous at $x=0$.At $x=1$: $LHL = \lim_{x 	o 1^-} (1 - x) = 0$,$RHL = \lim_{x 	o 1^+} (x) = 1$. Since $LHL 
eq RHL$,$f$ is discontinuous at $x=1$.At $x=2$: $LHL = \lim_{x 	o 2^-} (x) = 2$,$RHL = \lim_{x 	o 2^+} (x + 1) = 3$. Since $LHL 
eq RHL$,$f$ is discontinuous at $x=2$.At $x=3$: $LHL = \lim_{x 	o 3^-} (x + 1) = 4$,$f(3) = 5$. Since $LHL 
eq f(3)$,$f$ is discontinuous at $x=3$.Thus,Statement $1$ is true. Statement $2$ provides an incorrect definition of $f(x)$,so Statement $2$ is false.

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