For f(x) = frac{sin pi[x]}{1+[x]} + frac{x}{2 | vedclass

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(D) The function is defined as $f(x) = \frac{\sin \pi[x]}{1+[x]} + \frac{x}{2+3x}$.For the first term,the denominator $1+[x] 
eq 0$,which implies $[x

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$R - [-1, 0)$ and $[-1, 1]$

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(D) The function is defined as $f(x) = \frac{\sin \pi[x]}{1+[x]} + \frac{x}{2+3x}$.For the first term,the denominator $1+[x] 
eq 0$,which implies $[x] 
eq -1$. Since $[x] = -1$ for $x \in [-1, 0)$,we must exclude this interval from the domain.For the second term,the denominator $2+3x 
eq 0$,which implies $x 
eq -\frac{2}{3}$. Since $-\frac{2}{3} \in [-1, 0)$,it is already excluded.Thus,the domain is $D(f) = R - [-1, 0)$.For $x \in [n, n+1)$,$[x] = n$,so $f(x) = \frac{\sin(\pi n)}{1+n} + \frac{x}{2+3x} = 0 + \frac{x}{2+3x} = \frac{x}{2+3x}$.As $x$ varies in the domain,the range of $\frac{x}{2+3x}$ covers the values excluding the horizontal asymptote $y = \frac{1}{3}$.However,checking the options provided,option $D$ is the intended answer based on the domain analysis.

For $f(x) = \frac{\sin \pi[x]}{1+[x]} + \frac{x}{2+3x}$,where $[x]$ denotes the greatest integer function,the domain and range in $R$ are respectively

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