Unlocking Greatest Integer Function ( [X] ) - Gaurav Sharma

• e.g. [9.638] = 9,  = 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 √(x2−10x+23)
a) 19819
b) 19820
c) 19818
d) NOT

f(x)=√(x2 − 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) 1/2

y = (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 [log71] + [log72] + ... + [log7N] = N , then find the value of N ?
a) 52
b) 53
c) 54
d) 55

If [loga1] + [loga2] + ... + [logaN] = N

Then, N = a2 + a − 2

Thus, N = 72 + 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) NOT

We 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) 3

We 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) CBD

Let the sides of triangle be a,ar,ar2

So , a + ar > ar2

r2 − 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 , [ 5001/5] + [4991/5]+ ... +[1001/5] will be equal to

We know that , 35 = 243 and 45 = 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

1

1

1

1

1

27

14

1