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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 and be real numbers such
that Show that and for ### Prerequisites

Differentiability

Mod function

continuity

## Solution :

Let , Then , is not differentiable at ---(1)

As we know the function is not differentiable at .

Again we have , it also not differentiable at ----(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 and .

So , we have for .

## Food For Thought Let be a continuous and differentiable on (a,b) . Suppose that for all for some Then prove that there exists unique such that ## Subscribe to Cheenta at Youtube

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 and be real numbers such
that Show that and for ### Prerequisites

Differentiability

Mod function

continuity

## Solution :

Let , Then , is not differentiable at ---(1)

As we know the function is not differentiable at .

Again we have , it also not differentiable at ----(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 and .

So , we have for .

## Food For Thought Let be a continuous and differentiable on (a,b) . Suppose that for all for some Then prove that there exists unique such that ## Subscribe to Cheenta at Youtube

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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 , the given condition implies that: i.e., If , this linear equation has only one solution. But the above equation should be valid for any of the infinitely many . Hence we must have and .\\
Next, if , assume , and choose such that . The given condition implies that This, coupled with the previous result, implies that , which is a contradiction. Hence we have . Proceeding similarly, it follows that .

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