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**Periodic Function**

A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such that f^{ }(x^{ }+^{ }T) = f(x), for all values of x and x + T within the domain of f(x). The least positive period is called the principal or fundamental period of f.**Example:** The function sin x & cos x both are periodic over 2π & tan x is periodic over π

**Remark:****(a)** A constant function is always periodic, with no fundamental period.**(b)** If f(x) has a period p, then 1/f(x) and √f(x) also has a period p.**(c)** If f(x) has a period T then f(ax + b) has a period T/a (a > 0).**(d)** If f(x) has a period T_{1} & g(x) also has a period T_{2} then period of f(x) ± g(x) or f(x)/g(x) s L.C.M of T_{1} & T_{2 }provided their L.C.M. exists. However that L.C.M. (if exists) need not to be fundamental period. If L.C.M. does not exists then f(x) ± g(x) or f(x). g(x) or f(x)/g(x) is non-periodic e.g. |sin x| has the period p, |cos x| also has the period π |sin x| + |cos x| also has a period p. But the fundamental period of |sin x| + |cos x| is π/2.**(e)** If g is a function such that gof is defined on the domain of f and f is periodic with T, then gof is also periodic with T as one of its periods. Further if # g is one-one, then T is the period of gof # g is also periodic with T' as the period and the range of f is a subset of [0, T'], then T is the period of gof**(f)** Inverse of a periodic function does not exist.**Example 1. Find period of the following functions****(ii) f(x) = {x} + sin x****(iii) f(x) = cos x . cos 3x**** ****Solution.****(i)** Period of sin x/2 is 4p while period of cos x/3 is 6p. Hence period of sin x/2 + cos x/3 is 12 π {L.C.M. or 4 & 6 is 12}**(ii)** Period of sin x = 2π ; Period of {x} = 1; but L.C.M. of 2π & 1 is not possible ∴ it is a periodic**(iii)** f(x) = cos x . cos 3x ; Period of f(x) is L.C.M. of (2π, 2π/3) = 2π but 2π may or may not be the fundamental period. The fundamental period can be 2π/n where n ∈ Np may or may not be the fundamental period. The fundamental period can be f(π + x) = (–cos x) (– cos 3x) = f(x)**Example 2. If f(x) = sin x + cos ax is a periodic function, show that a is a rational number.****Solution.** Given f(x) = sin x + cos ax

Period of sin x = 2π/1 and period of cos ax 2π/a

Hence period of f(x) = L.C.M.

or

where k = H.C.F. of 1 and a

1/k = integer = q (say), (≠0) and k a/k = integer = p (say)

a = p/q

a = rational number**Example 3. Given below is a partial graph of an even periodic function f whose period is 8. If [*] denotes greatest integer function then find the value of the expression.**

f (-3) + 2 | f (-1) | + [f(7/8)] + f (0) + arc cos (f(-2)) + f (-7) + f (20)

f (-3) = f (3) = 2 [f(x) is an even function, ∴ f(-x) = f (x)]

again f (-1) = f (1) = - 3

∴ 2 | f (-1) | = 2 | f (1) | = 2 | -3 | = 6

from the graph, -3 < f(7/8) < -3 ∴ [f(7/8)] = -3

f (0) = 0 (obviously from the graph)

cos -1 (f(-2)) = cos

f (-7) = f (-7 + 8) = f (1) = - 3 [f (x) has period 8]

f (20) = f (4 + 16) = f (4) = 3 [f (nT + X) = f (x)]

sum = 2 + 6 - 3 + 0 + 0 - 3 + 3 = 5

We have f(x + 1) + f(x – 1) = √3 f(x) ∀ x ∈ R

Replacing x by x – 1 and x + 1 in (1) then f(x) + f(x – 2) = √3 f(x – 1) ...(2)

Adding (2) and (3), we get 2f(x) + f(x – 2) + f(x + 2) = √3 (f (x – 1) + f(x + 1)) ...(3)

2f (x) + f(x – 2) + f(x + 2) = √3 . √3 f(x) [From (1)]

f(x + 2) + f(x – 2) = f(x) ...(4)

Replacing x by x + 2 in equation (4) then f (x + 4) + f (x) = f (x + 2) ...(5)

Adding equations (4) and (5), we get f(x + 4) + f (x - 2) = 0 ...(6)

Again replacing x by x + 6 in (6) then f (x + 10) + f (x + 4) = 0 ...(7)

Subtracting (6) from (7), we get f (x + 10) - f (x - 2) = 0 ...(8)

Replacing x by x + 2 in (8) then f (x + 12) - f(x) = 0 or f (x + 12) = f(x)

Hence f(x) is periodic function with period 12.

Let f : A → B be a one-one & onto function, then their exists a unique function

g : B → A such that f(x) = y ⇔ g(y) = x, x ∈ A & y ∈ B. Then g is said to be inverse of f.

Thus g = f

Given f(x) = ln (x

which is a strictly increasing function. Thus f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Now let y = f(x) = ln (x^{2} + 3x + 1) then x = f^{–1} (y) ...(1)

and y = ln (x^{2} + 3x + 1) ⇒ e^{y} = x^{ 2} + 3x + 1 ⇒ x^{2} + 3x + 1 – e^{y} = 0

From (1) and (2), we get

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