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# ISI MStat PSB 2014 Problem 2 | Properties of a Function

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 !!

## Problem- ISI MStat PSB 2014 Problem 2

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 $1 \leq j \leq n$

### Prerequisites

Differentiability

Mod function

continuity

## Solution :

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 $1 \leq j \leq n$ .

## Food For Thought

$a<b \in \mathbb{R} .$ 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 $f(x)=x$

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### One comment on “ISI MStat PSB 2014 Problem 2 | Properties of a Function”

1. abhishek sinha says:

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 $a_i=b_i, \forall 1\leq i \leq n$.

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