If f(x) = xsqrt{1 - [x]^2},then (where [.] de | vedclass

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(A) $1$. For $x \in (0, 1)$,$[x] = 0 \Rightarrow f(x) = x\sqrt{1 - 0} = x$. Since $f'(x) = 1 > 0$,$f(x)$ is increasing in $(0, 1)$.$2$. At $x = 1$,$f(

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$f(x)$ is increasing in $x \in (0, 1)$

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(A) $1$. For $x \in (0, 1)$,$[x] = 0 \Rightarrow f(x) = x\sqrt{1 - 0} = x$. Since $f'(x) = 1 > 0$,$f(x)$ is increasing in $(0, 1)$.$2$. At $x = 1$,$f(1) = 1\sqrt{1 - 1^2} = 0$. For $x \in (0, 1)$,$f(x) = x$. As $x 	o 1^-$,$f(x) 	o 1$. Since $f(1) = 0$ and $f(x) > 0$ for $x$ near $1$ from the left,$x = 1$ is not a point of local maxima.$3$. The domain is determined by $1 - [x]^2 \ge 0 \Rightarrow [x]^2 \le 1 \Rightarrow -1 \le [x] \le 1$. This implies $x \in [-1, 2)$.$4$. The function is defined as:$f(x) = \begin{cases} 0, & -1 \le x Since $f(x) \ge 0$ for all $x$ in the domain,it is not a negative function.$5$. Since $f(x)$ is discontinuous at $x = 1$,Rolle's theorem is not applicable on $[0, 1]$.

If $f(x) = x\sqrt{1 - [x]^2}$,then (where $[.]$ denotes the greatest integer function):

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