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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{1}{84}$

Vedclass · Answer

(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

Similar Questions

The domain of the function $f(x) = ^{16 - x}C_{2x - 1} + ^{20 - 3x}P_{4x - 5}$,where the symbols have their usual meanings,is the set

Consider the following two statements:
$I.$ If $n$ is a composite number,then $n$ divides $(n-1)!$.
$II.$ There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$.

The number of times the digit $3$ will be written when listing the integers from $1$ to $1000$ is

If $4$-digit numbers greater than $5,000$ are randomly formed from the digits $0, 1, 3, 5,$ and $7$,what is the probability of forming a number divisible by $5$ when the repetition of digits is not allowed?

$(2n + 1) (2n + 3) (2n + 5) \dots (4n - 1)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module