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


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    Questions from topic functions are always tricky and require an extra effort over other topics especially in case of students without a mathematical background. Questions related to greatest integer function have been seen very frequent in past few years in various exams and mocks. So let us put an effort to unlock the mystery of Greatest Integer Function. Greatest Integer Function, [x] indicated the integral part of x which is nearest and smallest integer to x. It is also known as floor of x.

    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 √(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
     


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