karthikpandian19 wrote:Let [x] represent the average of the greatest integer less than or equal to x and the least integer greater than or equal to x. Is 0 <= |a| <= 1?
1. [x] - x = a
2. 0 < [a] < 1
We are given that [x] is the average of two numbers, lets call them x1 and x2.
x1 is greatest integer less than equal to x.
Examples:
For x=1.00, x1 = 1
For x=1.01, x1 = 1
For x=1.99, x1 = 1
x2 is the least integer greater than or equal to x.
For x=1.00, x1 = 1
For x=1.01, x1 = 2
For x=1.99, x1 = 2
So for average of x1 and x2, we can see that:
For integer value of x = 1, average = x
For non-integer values of x (1.01) or 1.99 or 1.453, [x] = (x1+x2)/2 = 1.5
I wrote this to justify that:
a) For integer values of x, [x] = x
b) For non-integer values of x, [x]=Integer part of x + 0.5
We need to determine whether 0<=|a|<=1.
With that in mind, let's look at the options:
1. [x] - x = a
For ease of explanation, I am going to use integer part of x=1, these explanations work for any number x.
Here we already know that for integer values, [x]-x = 0
Also, we know that:
max value of [x]-x occurs when x=1.0000000000...1 and
min. value of [x]-x occurs when x is 1.99999999... .
In each of those cases [x] = 1.5 Hence [x]-x lies between -0.5 and +0.5.
so -0.5<a<0.5
=> |a| < 0.5 . Hence 0<=|a| <=1. Sufficient.
2. 0 < [a] < 1
If [a] is between 0 and 1, it can have only 1 value, which is 0.5. This means that "a" itself is between 0 and 1(See logic used in explaining 1). So 0<=|a|<=1 condition is satisfied. Sufficient again.
Hence
D is the final answer.
