Unlocking Greatest Integer Function ( [X] )  Gaurav Sharma

e.g. [9.638] = 9, [7] = 7, [0.286] = 0, [3.432] = 4, [0.432] = 1
In general, n ≤ x < n+1 , where n is an integer, [x] = n
Now, we can say that for any real number x, x = I + f
where,
I = Integral Part denoted by [x]
f = Fractional Part denoted by {x}
Thus , {x} = x – [x]If n is any integer and x is any real number between n and n+1 , then the greatest integer function has the following properties.
Find the value of [3/4] + [3/4 + 1/100] + [3/4+ 2/100] +…+ [3/4 + 99/100].
[3/4] + [3/4+ 1/100] + [3/4+ 2/100] +… [3/4+25/100]+[3/4+26/100]+ ⋯ + [3/4+ 99/100]
First 25 terms are 0 and the next 75 terms are 1
So, sum of all the terms = (25 × 0) + (75 × 1) = 75.
Let [x] denote the greatest integer less than or equal to . If x=19824 , compute √(x^{2}−10x+23)
a) 19819
b) 19820
c) 19818
d) NOTf(x)=√(x^{2 }− 10x + 23)
=√((x−5)^{2} − 2) < x−5
Thus, f(x) will be less than 19824−5 = 19819
∴ c is the answer
Let [x] denotes the greatest integer function and let x be any number selected randomly such that [√x] =10, find the probability that [√(100 x) ]=100.
[√x]=10
⟹ 100 ≤ x < 121
⟹1000 ≤ 100x ≤ 12100
Now, [ √(100x) ]=100
But (101)^{2}=10201
∴ 10000 ≤ 100 x < 10201
Thus , Probability = (10201−10000) / (12100−10000) = 67/700
Find the domain of f(x)=1 / √( [x]−x)
Here , f(x) exists if [x]−x > 0 i.e. [x] > x
But , we know that [x] ≤ x
Thus, it is not possible
If y = (x − [x]) / (1 − [x] + x) , then which of the following value of y is not possible
a) 0
b) 1/4
c) 1/3
d) 1/2y = (x−[x]) / (1 − [x] + x) = {x} / (1 + {x})
y+ y {x} = {x}
{x} = y/(1−y)
Also , 0 ≤ {x} < 1
0 ≤ y/(1−y) < 1
0 ≤ y < 1/2
Thus, y can take all values from 0 to less than 1/2 . Hence 1/2 is not possible .
If [√x] = 6 and [√y] = 7, where x and y are natural numbers , what can be the greatest possible value of x + y ?
[ √x ] = 6 then x = 48 (maximum)
[ √y ] = 7 then y = 63 (maximum)
Thus, maximum value of x + y = 48 + 63 = 111
If [log_{7}1] + [log_{7}2] + ... + [log_{7}N] = N , then find the value of N ?
a) 52
b) 53
c) 54
d) 55If [log_{a}1] + [log_{a}2] + ... + [log_{a}N] = N
Then, N = a^{2} + a − 2
Thus, N = 7^{2} + 7 − 2
N = 54
Find the value of [1/3] + [1/3+1/100] + [1/3+2/100] + ... + [1/3+99/100]
[1/3]+[1/3+1/100]+[1/3+2/100]+⋯+[1/3+66/100]+[1/3+67/100]+ ... +[1/3+99/100]
Value of first 66 terms is 0 each and Value of last 33 terms is 33 each
Sum of first 66 terms is 0 and Sum of last 33 terms is 33 .
Value of the given series is 33
If {x} and [x] represent fractional and integral part of x, then find the value of f(x)=[x]+ Σ{x+r}/2000 ( from r=1 to 2000)
a) x
b) x/2
c) 2x
d) NOTWe know that , {x+r}={x} , r = > integer
f(x) = [x] + [{x} / 2000 + {x}/2000 + ... 2000 times]
= [x] + 2000{x}/2000 = x
Solve 4 {x} = x + [x]
4 {x} = [x] + {x} + [x]
3 {x} = 2 [x]
{x} = 2/3[x]
Also , 0 ≤ {x} < 1
0≤ 2/3[x] < 1
0 ≤ [x] < 3/2
⟹ [x] = 0 or 1
⟹ If [x] = 1, {x}=2/3 ⟹ x=5/3
⟹ If [x] = 0, {x}=0 ⟹ x=0
Number if integral solutions of {x+1}+2x = 4[x+1] − 6 is
a) 0
b) 1
c) 2
d) 3We know that, {x}=x−[x]
{x+1}+2x=4[x+1]−6
x+1−[x+1]+2x=4[x+1]−6
3x+1=5[x+1]−6
3x+1=5 ( [x]+1 )−6
3x=5[x]−2
3 ([x]+{x}) = 5[x]−2
3 {x} = 2[x] − 2
Also, 0 ≤ {x} < 1
0 ≤ 3{x} < 3
0 ≤ 2[x]−2 < 3
2 ≤ 2 [x] < 5
1 ≤ [x] < 5/2
[x]=1 ,2
[x] = 1 ⟹ x = 1
[x] = 2 ⟹ x = 8/3, which is not an integer
Thus, x = 1 is only integral solution
If three successive terms of a GP with common ratio r > 1 form the sides of ABC Triangle and [r] denotes greatest integer function find [r]+[−r] .
a) 0
b) 1
c) 1
d) CBDLet the sides of triangle be a,ar,ar^{2}
So , a + ar > ar^{2}
r^{2 }− r − 1 < 0
(1−√5)/2 < −r < (1+√5)/2
So , 1 < r < (1+√5)/2
[r]=1
Also , −1 > −r > (−1+√5)/2
[−r]= −2
[r] + [−r] = 1 − 2 = −1
If [x] is the greatest integer function less than or equal to x then , [ 500^{1/5}] + [499^{1/5}]+ ... +[100^{1/5}] will be equal to
We know that , 3^{5} = 243 and 4^{5} = 1024
Terms from 100 to 242 i.e. 143 terms will be equal to 2 and terms from 243 to 500 i.e 258 terms will be equal to 3 .
258×3+143×2 = 1060