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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 and be real numbers such

that

Show that and for

Differentiability

Mod function

continuity

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 .

Let be a continuous and differentiable on (a,b) . Suppose that for all for some Then prove that there exists unique such that

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 and be real numbers such

that

Show that and for

Differentiability

Mod function

continuity

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 .

Let be a continuous and differentiable on (a,b) . Suppose that for all for some Then prove that there exists unique such that

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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 .