Tiêu đề GREATEST INTEGER FUNCTION AND fractional part function.pdf ✅

What you already know • Domain and range of a function • Types of functions • Value of a function • Functional equations What you will learn • Greatest integer function • Fractional part function NOTES RELATIONS AND FUNCTIONS MATHEMATICS GREATEST INTEGER FUNCTION AND FRACTIONAL PART FUNCTION Solution Step 1: ( ) ( ) ( ) − − − x 1 x x 1 x xx x x xx x x x 4 4 f x + f 1 x = + 4 + 2 4 + 2 4 4 4 4 4 4 + 2 = + = + = =1 4 + 2 4 4 + 2 2 2 + 4 4 + 2 + 2 4 Step 2: r =          ∑                          − −               2001 1 r 1 2 1001 2000 2001 f = f + f + ...+ f + ...+ f + f 2002 2002 2002 2002 2002 2002 1 2 1001 2 1 = f + f + ...+ f + ...+ f 1 + f 1 2002 2002 2002 2002 20                   02 1001 = 1 + 1 + 1 + ... + f 2002 1 = 1000 + f 2 If ( ) then, find r =   ∑     x 2001 x 1 4r f x = f . 4 + 2 2002 © 2021, BYJU'S. All rights reserved
Step 3:       1 2 1 2 1 4 2 1000 + f = 1000 + = 1000 + = 1000.5 2 2 + 2 4 +2 Note Greatest Integer Function (G.I.F) The expression y = [x] is the greatest integer function, which is less than or equal to x. This is also referred to as a step function. Domain: R Range: y ∈ Z Examples: [2.3]= 2 (The greatest integer less than or equal to 2.3 is 2) [7.99]=7 (The greatest integer less than or equal to 7.99 is 7) [–3.7]= –4 (The greatest integer less than or equal to –3.7 is –4) Properties • x – 1 < [x] ≤ x Example: Let x = 2.3 2.3 –1 < [2.3] ≤ 2.3 • [x+m] = [x] + m, for m ∈ Z Example: [2.3+1] = [2.3] + 1= 2+1 = 3 • [ ] [ ]  ∈  − ∉ 0, x x + -x = 1, x   In case of ( ) , is half of the upper l imit, i.e.,   ∑     x 2001 x r=1 a r 2001 f x = f a + a 2002 2 – 1 1 1 2 2 3 3 – 3 – 2 – 1 – 2 – 3 X Y – 1 1 1 2 2 – 2 – 1 3 – 2 X Y y = x y = x – 1 © 2021, BYJU'S. All rights reserved 02
Example: [7] + [–7] = 7 – 7 = 0 [7.1] + [–7.1] = 7 – 8 = –1 • [x] = m, ⇒ x ∈ [m, m + 1) Example: [x] = –1 ⇒ x ∈ [–1, 0) • [ ] [ ]      −           1 2 n 1 x + x + + x + + ... + x + = nx nn n • [ ]          x x + 1 + = x 2 2 • If [x] ≤ n, then x ∈ (–∞, n + 1) Example: 1. If [x] ≤ 2, then find x. x is an integer, which is less than or equal to 2. [x] ≤ 2 ⇒ x ∈ (– ∞, 3) 2. If –2 ≤ [x] ≤ 6, find x. x is an integer lying between –2 and 6. –2 ≤ [x] ≤ 6 ⇒ x ∈ [–2, 7) 3. If [x] > 2, find x. x is an integer, which is greater than 2. [x] > 2 ⇒ x ≥ 3 4. If [x] < 3, find x. x is an integer, which is less than 3. [x] < 3 ⇒ x < 3 Solution                      Find value of the following : 1 1 1 1 2 1 99 + + + + + ... + + 4 4 100 4 100 4 100 [ ] [ ] [ ] By comparing with we get and ,                                           1 2 n-1 x + x + + x + + ... + x + = nx nn n 1 x = n = 100 4 1 1 1 1 2 1 99 1 + + + + + ... + + = 100 × = 25 = 25 4 4 100 4 100 4 100 4 3 4 5 6 7 –3 –2 –1 0 1 2 3 –2 –1 0 1 2 3 –2 –1 0 1 2 3 4 5 6 7 © 2021, BYJU'S. All rights reserved 03
Solution                Find value of the following : 5 1 5 2 5 199 + + + + ... + + 3 200 3 200 3 200 Solution ,                            If find x+8 x+7 x+6 x+5 x+4 x+3 x+2 x+1 + + + + + + + = 24 x. 22222222 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ) ⇒ ⇒ ⇒⇒ Step 2 : x +7 + x +5 + x +3 + x +1 = 24 x +7+ x +5+ x +3+ x +1 = 24 4 x = 8 x = 2 x = 2, 3 [ ] By adding and subtracting ,               −                          − −                 5 we get 3 5 5 1 5 2 5 199 5 + + + + + ... + + 3 3 200 3 200 3 200 3 5 5 1 2 n 1 = × 200 x + x + + x + + ... + x + 3 3 nn n  [ ] [ ] [ ]            − −−       = nx 1000 5 = = 333.33 1.66 = 333 1 = 332 3 3 [ ] ( ) ( ) ( ) ( ) [ ] [ ] [ ] [ ] We know that                   ⇒                               Step 1 : x x + 1 + = x 2 2 x+7 +1 x+7 x+5 +1 x+5 x+3 +1 x+3 + + ++ + 222222 x+1 +1 x+1 + + 2 2 = x +7 + x +5 + x +3 + x +1 © 2021, BYJU'S. All rights reserved 04
Solution [ ] [ [ [ ]]] [ ] − then find 4 x + 2 1 = ; x . x + x + x 12 3 [ ] [ [ [ ]]] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ) −   − − −   ⇒ − ⇒ − ⇒ − ⇒ − ⇒ ∈− − 4 x + 2 4 x + 8 4 x + 8 4 x + 8 === x + x + x 12 x + x + x 12 x + x + x 12 3 x 12 4 x + 8 1 = 3 x 12 3 12 x + 24 = 3 x 12 9 x = 36 x = 4 x 4, 3 Solution                   If and The sum of possible values of is , then find x x x 4x 0 < x < 300 + + = . x m 2 5 10 5 m . 1000 [ ] ( ( ) ) We know that L.C.M is only possible, when all and are integers. Now, they will be int                            Step 1 : I = I x x x 5x + 2x + x 4x + + = = , 2, 5, 10 = 10 2 5 10 10 5 xx x , 2 5 10  ( ) egers for all multiples of ∴ 10. x = 10, 20, 30, ..., 290 0 < x < 300  ( ) ( ) Now [ ]          ∑ Step 2 : 10×29 x =10+ 20+ ...+ 290 = 10 1+ 2+... + 29 = 1+ 29 = 4350 2 m 4350 = = 4.35 = 4 1000 1000 © 2021, BYJU'S. All rights reserved 05