In maths , there exists the following property

\(\left\lfloor \left\lfloor \frac { x }{ 2 } \right\rfloor +\left\lfloor { x }^{ 2 } \right\rfloor \right\rfloor =\left\lfloor \frac { x }{ 2 } \right\rfloor +\left\lfloor { x }^{ 2 } \right\rfloor \)

but if you do

\(\left\{ \left\{ \frac { x }{ 2 } \right\} +\left\{ { x }^{ 2 } \right\} \right\} =\left\{ \frac { x }{ 2 } \right\} +\left\{ { x }^{ 2 } \right\} \)

then this is a big **Mistake!!!**

but the above expression is true for some set k where k contains all possible values of x from -2 <x<2 and sum of end points of all intervals of k can be expressed as **s**

**Find** \(\left\lfloor 100\left\{ s \right\} \right\rfloor \)

**Details and assumptions**

1.{x} is fractional part of x, which we will define as \( x - \lfloor x \rfloor \), even for negative numbers.

2.\(\left\lfloor \quad \right\rfloor \) is greatest integer less than or equal to x.

3.For example if your answer is

k\(= \left[ -\frac { 3 }{ 2 } ,-1 \right) \cup \left( \sqrt { 2 } ,2 \right) \)

then s = \(-\frac { 3 }{ 2 } -1+\sqrt { 2 } +2\)

4.Consider both cases whethere end points are open ( ) or closed [ ]

5.This is my second problem ever

6.This problem is inspired from the set Mistakes Give Rise to Problems

×

Problem Loading...

Note Loading...

Set Loading...