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


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


    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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