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
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
.\\
, assume
, and choose
such that
. The given condition implies that
Next, if
This, coupled with the previous result, implies that
, which is a contradiction. Hence we have
. Proceeding similarly, it follows that
.