Solved Examples on Greatest integer function-I

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1. Domain of  f(x) = \frac{1}{{\sqrt {[x] - x} }} where [ ] denotes greatest integer function.

 We Know that ,Domain is the Possible point of x for which we can get finite and real value of y, to know more [click here]
Consider,

 f(x) = \frac{1}{{\sqrt {[x] - x} }}

 for finite and real value,
{\rm{denominator}} \ne {\rm{0}} , {[x] - x} should not be negative,

 therefore,[x] - x > 0 ( denominator cant be equal to zero, so it is strictly greater than zero)

 or, [x] > x

 but we know that , [x] \le x ( to know more about Greatest Integer Function click here )

 => for no value of x, f(x) will be real and finite.

 so domain of f(x) = { } = \phi

we can also see this from graph,


Here, we can see that for a point on x - axis or for a value of x , graph of y = x is above the graph of y = [x] and both the graph touch only on integer points.
           You can check this by taking a point on x- axis and draw a vertical line through this in upward direction, you can see that the line first cut the graph of y = [x] and then cut the graph of y = x ,
this shows for any value of x , x \ge [x]
2.Domain of f(x) = \frac{1}{{[|x - 2|] + [|x - 10|] - 8 \ne 0}}

 Since, {\rm{denominator}} \ne {\rm{0}}

 {[|x - 2|] + [|x - 10|] - 8 \ne 0}, 

 Therefore, domain of the f(x) would not have the values of x for which {[|x - 2|] + [|x - 10|] - 8 = 0}

 We have to solve this and subtract for Real numbers and get the domain

 {[|x - 2|] + [|x - 10|] - 8 = 0} 

 We know how to solve the modulus equality, to know how to solve [ click here ]

 two behaviour changing points 2 and 10,
therefore,
(1) x \le 2

 2) 2 \le x \le 10

 3) x \ge 10

 1)When x \le 2

 \begin{array}{l} \Rightarrow [2 - x] + [10 - x] - 8 = 0\\ 2 + [ - x] + 10 + [ - x] - 8 = 0\\ 2[ - x] = - 4\\ [ - x] = - 2 \end{array} (you can put out the integer, to see the properties of [x] click )

 Recall , if [x] = I , x belongs to [I,I+1) 

 or,  - x \in [ - 2, - 1)

 or, x \in (1,2] , since it satisfies x \le 2, therefore x \in (1,2] is a solution of the equation......(i)

 2) When, 2 \le x \le 10

 \begin{array}{l} \Rightarrow [x - 2] + [10 - x] - 8 = 0\\ [x] - 2 + 10 + [ - x] - 8 = 0 \end{array}

 or, [x] + [ - x] = 0

 We know from the properties of the greatest integer function that [ - x] = - [x] when x \in Integer 
to see the properties of [x] click here

 Therefore, [x] + [ - x] = 0
=> x \in Integer

 but we take 2 \le x \le 10

 therefore solution in the interval 2 \le x \le 10 is 
x \in \{ 2,3,4,5,6,7,8,9,10\}  ....................................................(ii)

 3) When x \ge 10

 \begin{array}{l} \Rightarrow [x - 2] + [x - 10] - 8 = 0\\ [x] - 2 + [x] - 10 - 8 = 0\\ 2[x] = 20\\ [x] = 10 \end{array}

 \Rightarrow x \in [10,11) and which is in interval x \ge 10 
therefore, x \in [10,11) is a solution of the equation .............(iii)

 combining all i.e. (i), (ii) and (iii) , we get the solution of the equation as

 x \in (1,2]\bigcup {\{ 2,3,4,5,6,7,8,9,10} \} \bigcup {[10,11)}

or, x \in (1,2)\bigcup {\{ 2,3,4,5,6,7,8,9,10} \} \bigcup {(10,11)}  ( since 2 and 10 are already included in the middle set)
Therefore domain of f(x) is

 x \in R - (1,2)\bigcup {\{ 2,3,4,5,6,7,8,9,10} \} \bigcup {(10,11)}








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