The minimum value of f(x) = |x - 1| + |2x - 1 | vedclass

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(A) The function is given by $f(x) = \sum_{k=1}^{119} |kx - 1|$.We can rewrite this as $f(x) = \sum_{k=1}^{119} k |x - \frac{1}{k}|$.This is a sum of

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$\frac{1}{84}$

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(A) The function is given by $f(x) = \sum_{k=1}^{119} |kx - 1|$.We can rewrite this as $f(x) = \sum_{k=1}^{119} k |x - \frac{1}{k}|$.This is a sum of the form $\sum_{k=1}^{n} c_k |x - a_k|$,where $c_k = k$ and $a_k = \frac{1}{k}$.The function $f(x)$ attains its minimum at the weighted median of the values $a_k = \frac{1}{k}$ with weights $c_k = k$.The values $a_k$ are $\frac{1}{1}, \frac{1}{2}, \dots, \frac{1}{119}$,which are in decreasing order: $1 > \frac{1}{2} > \dots > \frac{1}{119}$.The weights are $c_1=1, c_2=2, \dots, c_{119}=119$.The total weight is $S = \sum_{k=1}^{119} k = \frac{119 	imes 120}{2} = 119 	imes 60 = 7140$.The median occurs when the cumulative weight reaches $\frac{S}{2} = \frac{7140}{2} = 3570$.We look for $m$ such that $\sum_{k=1}^{m} k \ge 3570$.Using the formula $\frac{m(m+1)}{2} \ge 3570$,we get $m(m+1) \ge 7140$.Since $84 	imes 85 = 7140$,the cumulative weight reaches $3570$ exactly at $m = 84$.Thus,the minimum occurs at $x = a_{84} = \frac{1}{84}$.

The minimum value of $f(x) = |x - 1| + |2x - 1| + |3x - 1| + \dots + |119x - 1|$ occurs at $x$. Then $x$ is equal to

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