This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let's give it a try !!
Let \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) be real numbers such
that \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R} \)
Show that \(m=n\) and \(a_{j}=b_{j}\) for \(1 \leq j \leq n\)
Differentiability
Mod function
continuity
Let , \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R} \)
Then , \( f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right| \) is not differentiable at \( x=a_1,a_2, \cdots , a_m \) ---(1)
As we know the function \(|x-a_i|\) is not differentiable at \(x=a_i\) .
Again we have , \( f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right| \) it also not differentiable at \( x= b_1,b_2, \cdots , b_n \) ----(2)
Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .
And also the points where f is not differentiable must be same in both (1) and (2) .
As we have the restriction that \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) .
So , we have \(a_{j}=b_{j}\) for \(1 \leq j \leq n\) .
\(a<b \in \mathbb{R} .\) Let \(f:[a, b] \rightarrow[a, b]\) be a continuous and differentiable on (a,b) . Suppose that \(\left|f^{\prime}(x)\right| \leq \alpha<1\) for all \(x \in(a, b)\) for some \(\alpha .\) Then prove that there exists unique \(x \in[a, b]\) such that \(f(x)=x\)
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.
Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.
This problem has an absolutely elementary solution. For any $x < \min (a_1, b_1)$, the given condition implies that:
\[\sum_{i=1}^m a_i - mx= \sum_{j=1}^n b_j - nx.\]
i.e.,
\[(m-n)x= \sum_{i=1}^m a_i - \sum_{j=1}^n b_j.\]
If $m\neq n$, this linear equation has only one solution. But the above equation should be valid for any of the infinitely many $x<\min (a_1, b_1)$. Hence we must have $m=n$ and $\sum_{i=1}^na_i =\sum_{i=1}^n b_i$.\\
Next, if $a_1 \neq b_1$, assume $a_1 < b_1$, and choose $x$ such that $a_1<x< \min(a_2, b_1)$. The given condition implies that
\[x-a_1 + \sum_{i=2}^m a_i - (m-1)x = \sum_{i=1}^nb_i -nx.\]
This, coupled with the previous result, implies that $x=a_1$, which is a contradiction. Hence we have $a_1=b_1$. Proceeding similarly, it follows that $a_i=b_i, \forall 1\leq i \leq n$.