Let the functions f:(-1,1) rightarrow R and g | vedclass

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(B) Given $f(x) = |2x-1| + |2x+1|$ and $g(x) = x - [x] = \{x\}$.For $x \in (-1, 1)$,the fractional part function $g(x) = \{x\}$ is defined as:$g(x) =

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$4$

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(B) Given $f(x) = |2x-1| + |2x+1|$ and $g(x) = x - [x] = \{x\}$.For $x \in (-1, 1)$,the fractional part function $g(x) = \{x\}$ is defined as:$g(x) = x+1$ for $x \in (-1, 0)$ and $g(x) = x$ for $x \in [0, 1)$.Thus,the composite function $h(x) = f(g(x)) = |2\{x\}-1| + |2\{x\}+1|$.Since $\{x\} \in [0, 1)$,we have $2\{x\}+1 \geq 1$,so $|2\{x\}+1| = 2\{x\}+1$.Then $h(x) = |2\{x\}-1| + 2\{x\} + 1$.If $0 \leq \{x\} \leq 1/2$,then $h(x) = -(2\{x\}-1) + 2\{x\} + 1 = 2$.If $1/2 Analyzing the interval $(-1, 1)$:For $x \in (-1, 0)$,$\{x\} = x+1$. So $h(x) = 2$ if $x+1 \leq 1/2$ (i.e.,$x \leq -1/2$) and $h(x) = 4(x+1)$ if $x > -1/2$.For $x \in [0, 1)$,$\{x\} = x$. So $h(x) = 2$ if $x \leq 1/2$ and $h(x) = 4x$ if $x > 1/2$.Discontinuity: The function $h(x)$ is discontinuous at $x=0$ because $\lim_{x 	o 0^-} h(x) = 4(0+1) = 4$ and $h(0) = 2$. Thus,$c=1$.Non-differentiability: The function is non-differentiable at $x = -1/2$ (corner point),$x = 0$ (discontinuity),and $x = 1/2$ (corner point). Thus,$d=3$.Therefore,$c+d = 1+3 = 4$.

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