Q1-If-f-R-R-is-defined-by-f-x-x-x-1-2-x-2-3-3x-5-where-x-is-the-integral-part-of-x-then-a-period-of-f-is-A-1-B-2-3-C-1-2-D-

Question Number 119634 by Ar Brandon last updated on 25/Oct/20

Q 1

If f : R \to R is defined by

f (x) = [x] + [x + \frac{1}{2}] + [x + \frac{2}{3}] - 3 x + 5

where [x] is the integral part of x, then a period of f is

(A) 1 (B) 2 / 3 (C) 1 / 2 (D) 1 / 3

Q 2

Let a < c < b such that c - a = b - c . If f : R \to R is a

function satisfying the relation

f (x + a) + f (x + b) = f (x + c) for all x \in R

then a period of f is

(A) (b - a) (B) 2 (b - a)

(C) 3 (b - a) (D) 4 (b - a)

Answered by Olaf last updated on 26/Oct/20

Q 2 .

c - a = b - c

c = \frac{a + b}{2}

f (x + c) = f (x + a) + f (x + b)

f (x) = f (x + a - c) + f (x + b - c)

f (x) = f (x + a - \frac{a + b}{2}) + f (x + b - \frac{a + b}{2})

f (x) = f (x - \frac{1}{2} (b - a)) + f (x + \frac{1}{2} (b - a))

f (x - \frac{1}{2} (b - a)) = f (x - (b - a)) + f (x)

f (x) - f (x + \frac{1}{2} (b - a)) = f (x - (b - a)) + f (x)

f (x + \frac{1}{2} (b - a)) = - f (x - (b - a))

f (x + \frac{3}{2} (b - a)) = - f (x)

f (x + 3 (b - a)) = - f (x + \frac{3}{2} (b - a))

f (x + 3 (b - a)) = - [- f (x)] = f (x)

Period is 3 (b - a)

Answered by Olaf last updated on 26/Oct/20

Q 1 .

f (x) = [x] + [x + \frac{1}{2}] + [x + \frac{2}{3}] - 3 x + 5

f (x + 1) = [x + 1] + [(x + \frac{1}{2}) + 1]

+ [(x + \frac{2}{3}) + 1] - 3 (x + 1) + 5

f (x + 1) = [x] + 1 + [x + \frac{1}{2}] + 1

+ [x + \frac{2}{3}] + 1 - 3 (x + 1) + 5

f (x + 1) = [x] + [x + \frac{1}{2}] + [x + \frac{2}{3}] - 3 x + 5

f (x + 1) = f (x)

f is clearly at least 1 - periodic .

f (0) = [0] + [\frac{1}{2}] + [\frac{2}{3}] - 3 (0) + 5 = 5

f (\frac{1}{3}) = [\frac{1}{3}] + [\frac{5}{6}] + [1] - 3 (\frac{1}{3}) + 5 = 5

f (\frac{1}{3}) = f (0)

By extension, in the general case :

f (x + \frac{1}{3}) = f (x)

f is \frac{1}{3} - periodic .

Commented by Ar Brandon last updated on 26/Oct/20

Thank you so very much Sir ��