The number of positive solutions of the equat | vedclass

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(B) Let $f(x) = \int_{0}^{x} (t - \{t\})^2 dt$. Since $t - \{t\} = \lfloor t \rfloor$,the integral becomes $\int_{0}^{x} \lfloor t \rfloor^2 dt$.For $

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$2$

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(B) Let $f(x) = \int_{0}^{x} (t - \{t\})^2 dt$. Since $t - \{t\} = \lfloor t \rfloor$,the integral becomes $\int_{0}^{x} \lfloor t \rfloor^2 dt$.For $n \le x Case $1$: $0 \le x Case $2$: $1 \le x Case $3$: $2 \le x Since $x = 2.5$ is in $[2, 3)$,it is a valid solution.For $x > 3$,the function $\int_{0}^{x} \lfloor t \rfloor^2 dt$ grows much faster than $2x - 2$,so no further solutions exist.The solutions are $x = 1$ and $x = 2.5$. Thus,there are $2$ solutions.

The number of positive solutions of the equation $\int_{0}^{x} (t - \{t\})^2 dt = 2(x - 1)$,where $\{ \}$ denotes the fractional part function,is:

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