Domain of function f(x) = log|5{x} - 2x| is x | vedclass

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(B) The function $f(x) = \log|5\{x\} - 2x|$ is defined when $5\{x\} - 2x 
eq 0$.We need to find the values of $x$ for which $5\{x\} - 2x = 0$.Since $

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$2$

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(B) The function $f(x) = \log|5\{x\} - 2x|$ is defined when $5\{x\} - 2x 
eq 0$.We need to find the values of $x$ for which $5\{x\} - 2x = 0$.Since $x = [x] + \{x\}$,we have $\{x\} = x - [x]$.Substituting this into the equation: $5(x - [x]) - 2x = 0$.$5x - 5[x] - 2x = 0 \Rightarrow 3x = 5[x] \Rightarrow [x] = \frac{3x}{5}$.Since $[x]$ is an integer,$\frac{3x}{5}$ must be an integer. Let $\frac{3x}{5} = k$,where $k \in Z$. Then $x = \frac{5k}{3}$.Also,we know that $k \le x Substituting $x = \frac{5k}{3}$: $k \le \frac{5k}{3} $3k \le 5k From $3k \le 5k$,we get $2k \ge 0 \Rightarrow k \ge 0$.From $5k Since $k$ is an integer,$k$ can be $0$ or $1$.If $k = 0$,$x = \frac{5(0)}{3} = 0$.If $k = 1$,$x = \frac{5(1)}{3} = \frac{5}{3}$.Thus,the set $A = \{0, \frac{5}{3}\}$.The number of elements $n(A) = 2$.

Domain of function $f(x) = \log|5\{x\} - 2x|$ is $x \in R - A$,then $n(A)$ is (where $\{.\}$ denotes fractional part function)

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