The period of the function f(x) = frac{x}{3} | vedclass

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(B) We know that the fractional part function is defined as $\{x\} = x - [x]$.Given the function $f(x) = \left( \frac{x}{3} - \left[ \frac{x}{3} - 5 \

Vedclass · Accepted Answer

$60$

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(B) We know that the fractional part function is defined as $\{x\} = x - [x]$.Given the function $f(x) = \left( \frac{x}{3} - \left[ \frac{x}{3} - 5 \right] \right) + \left( \frac{x}{4} - \left[ \frac{x}{4} - 5 \right] \right) + \left( \frac{x}{5} - \left[ \frac{x}{5} - 5 \right] \right)$.Since $[x - n] = [x] - n$ for any integer $n$,we have $\left[ \frac{x}{k} - 5 \right] = \left[ \frac{x}{k} \right] - 5$.Substituting this into the function:$f(x) = \left( \frac{x}{3} - \left[ \frac{x}{3} \right] + 5 \right) + \left( \frac{x}{4} - \left[ \frac{x}{4} \right] + 5 \right) + \left( \frac{x}{5} - \left[ \frac{x}{5} \right] + 5 \right)$.$f(x) = \left\{ \frac{x}{3} \right\} + \left\{ \frac{x}{4} \right\} + \left\{ \frac{x}{5} \right\} + 15$.The period of $\{ax\}$ is $\frac{1}{|a|}$.So,the periods of the individual terms are $T_1 = \frac{1}{1/3} = 3$,$T_2 = \frac{1}{1/4} = 4$,and $T_3 = \frac{1}{1/5} = 5$.The period of the sum of these functions is the $L.C.M.$ of their individual periods.$T = L.C.M.(3, 4, 5) = 60$.

The period of the function $f(x) = \frac{x}{3} - \left[ \frac{x}{3} - 5 \right] + \frac{x}{4} - \left[ \frac{x}{4} - 5 \right] + \frac{x}{5} - \left[ \frac{x}{5} - 5 \right]$ is (where $[.]$ is the $G.I.F.$).

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