First slide

Algebra of real valued functions

Question

If [x] stands for the greatest integer function, then
$[\frac{1}{2} + \frac{1}{1000}] + [\frac{1}{2} + \frac{2}{1000}] + ..... + [\frac{1}{2} + \frac{999}{1000}] =______$

Easy

A

250
B

500
C

750
D

600

Solution

if $1 \leq x < 500 = 1 if 500 \leq x \leq 999$
G.E = 0 + 0 + …. + 1 + 1 + 1 + ….. + 1 (500 times) = 500

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